- (i) The signed definite integral , which is usually interpreted as the Riemann integral (or equivalently, the Darboux integral), which can be defined as the limit (if it exists) of the Riemann sums
where is some partition of , is an element of the interval , and the limit is taken as the maximum mesh size goes to zero. It is convenient to adopt the convention that for ; alternatively one can interpret as the limit of the Riemann sums (1), where now the (reversed) partition goes leftwards from to , rather than rightwards from to .

- (ii) The
*unsigned definite integral*, usually interpreted as the Lebesgue integral. The precise definition of this integral is a little complicated (see e.g. this previous post), but roughly speaking the idea is to approximate by simple functions for some coefficients and sets , and then approximate the integral by the quantities , where is the Lebesgue measure of . In contrast to the signed definite integral, no orientation is imposed or used on the underlying domain of integration, which is viewed as an “undirected” set . - (iii) The
*indefinite integral*or antiderivative , defined as any function whose derivative exists and is equal to on . Famously, the antiderivative is only defined up to the addition of an arbitrary constant , thus for instance .

There are some other variants of the above integrals (e.g. the Henstock-Kurzweil integral, discussed for instance in this previous post), which can handle slightly different classes of functions and have slightly different properties than the standard integrals listed here, but we will not need to discuss such alternative integrals in this course (with the exception of some improper and principal value integrals, which we will encounter in later notes).

The above three notions of integration are closely related to each other. For instance, if is a Riemann integrable function, then the signed definite integral and unsigned definite integral coincide (when the former is oriented correctly), thus

and

If is continuous, then by the fundamental theorem of calculus, it possesses an antiderivative , which is well defined up to an additive constant , and

for any , thus for instance and .

All three of the above integration concepts have analogues in complex analysis. By far the most important notion will be the complex analogue of the signed definite integral, namely the contour integral , in which the directed line segment from one real number to another is now replaced by a type of curve in the complex plane known as a contour. The contour integral can be viewed as the special case of the more general line integral , that is of particular relevance in complex analysis. There are also analogues of the Lebesgue integral, namely the arclength measure integrals and the area integrals , but these play only an auxiliary role in the subject. Finally, we still have the notion of an antiderivative (also known as a *primitive*) of a complex function .

As it turns out, the fundamental theorem of calculus continues to hold in the complex plane: under suitable regularity assumptions on a complex function and a primitive of that function, one has

whenever is a contour from to that lies in the domain of . In particular, functions that possess a primitive must be conservative in the sense that for any closed contour. This property of being conservative is not typical, in that “most” functions will not be conservative. However, there is a remarkable and far-reaching theorem, the Cauchy integral theorem (also known as the Cauchy-Goursat theorem), which asserts that any holomorphic function is conservative, so long as the domain is simply connected (or if one restricts attention to contractible closed contours). We will explore this theorem and several of its consequences the next set of notes.

** — 1. Integration along a contour — **

The notion of a curve is a very intuitive one. However, the precise mathematical definition of what a curve actually is depends a little bit on what type of mathematics one wishes to do. If one is mostly interested in topology, then a good notion is that of a continuous (parameterised) curve. It one wants to do analysis in somewhat irregular domains, it is convenient to restrict the notion of curve somewhat, to the rectifiable curves. If one is doing analysis in “nice” domains (such as the complex plane , a half-plane, a punctured plane, a disk, or an annulus), then it is convenient to restrict the notion further, to the *piecewise smooth curves*, also known as contours. If one wished to get to the main theorems of complex analysis as quickly as possible, then one would restrict attention only to contours and skip much of this section; however we shall take a more leisurely approach here, discussing curves and rectifiable curves as well, as these concepts are also useful outside of complex analysis.

We begin by defining the notion of a continuous curve.

Definition 1 (Continuous curves)Acontinuous parameterised curve, orcurvefor short, is a continuous map from a compact interval to the complex plane . We call the curvetrivialif , andnon-trivialotherwise. We refer to the complex numbers as theinitial pointandterminal pointof the curve respectively, and refer to these two points collectively as theendpointsof the curve. We say that the curve isclosedif . We say that the curve issimpleif one has for any distinct , with the possible exception of the endpoint cases or (thus we allow closed curves to be simple). We refer to the subset of the complex plane as theimageof the curve.

We caution that the term “closed” here does *not* refer to the topological notion of closure: for any curve (closed or otherwise), the image of the curve, being the continuous image of a compact set, is necessarily a compact subset of and is thus always topologically closed.

A basic example of a curve is the directed line segment from one complex point to another , defined by

for . (Thus, contrary to the informal English meaning of the terms, we consider line segments to be examples of curves, despite having zero curvature; in general, it is convenient in mathematics to admit such “degenerate” objects into one’s definitions, in order to obtain good closure properties for these objects, and to maximise the generality of the definition.) If , this is a simple curve, while for it is (a rather degenerate, but still non-trivial) closed curve. Another important example of a curve is the anti-clockwise circle of some radius around a complex centre , defined by

This is a simple closed non-trivial curve. If we extended the domain here from to (say) , the curve would remain closed, but would no longer be simple (every point in the image is now traversed twice by the curve).

Note that it is technically possible for two distinct curves to have the same image. For instance, the anti-clockwise circle of some radius around a complex centre defined by

traverses the same image as the previous curve (2), but is considered a distinct curve from . Nevertheless the two curves are closely related to each other, and we formalise this as follows. We say that one curve is a *continuous reparameterisation* of another , if there is a homeomorphism (that is to say, a continuous invertible map whose inverse is also continuous) which is endpoint preserving (i.e., and ) such that for all (that is to say, , or equivalently ), in which case we write . Thus for instance . The relation of being a continuous reparameterisation is an equivalence relation, so one can talk about the notion of a curve “up to continuous reparameterisation”, by which we mean an equivalence class of a curve under this relation. Thus for instance the image of a curve, as well as its initial point and end point, are well defined up to continuous reparameterisation, since if then and have the same image, the same initial point, and the same terminal point. It is common to depict an equivalence class of a curve graphically, by drawing its image together with an arrow depicting the direction of motion from the initial point to its endpoint. (In the case of a non-simple curve, one may need multiple arrows in order to clarify the direction of motion, and also the possible multiplicity of the curve.)

Exercise 2Let be a continuous invertible map.

- (i) Show that is continuous, so that is a homeomorphism. (
Hint:use the fact that a continuous image of a compact set is compact, and that a subset of an interval is topologically closed if and only if it is compact.)- (ii) If , show that and that is monotone increasing. (
Hint:use the intermediate value theorem.)- (iii) Conversely, if is a continuous monotone increasing map with and , show that is a homeomorphism.

It will be important for us that we do not allow reparameterisations to reverse the endpoints. For instance, if are distinct points in the complex plane, the directed line segment is *not* a reparameterisation of the directed line segment since they do not have the same initial point (or same terminal point); the map is a homeomorphism from to but it does not preserve the initial point or the terminal point. In general, given a curve , we define its *reversal* to be the curve , thus for instance is (up to reparameterisation) the reversal of , thus

Another basic operation on curves is that of concatenation. Suppose we have two curves and with the property that the terminal point of equals the initial point of . We can reparameterise by translation to , defined by . We then define the *concatenation* or *sum* by setting

for and

for (note that these two definitions agree on their common domain point by the hypothesis . It is easy to see that this concatenation is still a continuous curve. The reader at this point is encouraged to draw a picture to understand what the concatenation operation is doing; it is much simpler to grasp it visually than the above lengthy definition may suggest. If the terminal point of does not equal the initial point of , we leave the sum undefined. (One can define more general spaces than the space of curves in which such an addition can make sense, such as the space of -currents if one assumes some rectifiability on the curves, but we will not need such general spaces here.)

Concatenation is well behaved with respect to equivalence and reversal:

Exercise 3Let be continuous curves. Suppose that the terminal point of equals the initial point of , and the terminal point of equals the initial point of .

- (i) (Concatenation well defined up to equivalence) If and , show that .
- (ii) (Concatenation associative) Show that . In particular, we certainly have
- (iii) (Concatenation and reversal) Show that .
- (iv) (Non-commutativity) Give an example in which and are both well-defined, but not equivalent to each other.
- (v) (Identity) If and denote the initial and terminal points of respectively, and is the trivial curve defined for any , show that and .
- (vi) (Non-invertibility) Give an example in which is not equivalent to a trivial curve.

Remark 4The above exercise allows one to view the space of curves up to equivalence as a category, with the points in the complex plane being the objects of the category, and each equivalence class of curves being a single morphism from the initial point to the terminal point (and with the equivalence class of trivial curves being the identity morphisms). This point of view can be useful in topology, particularly when relating to concepts such as the fundamental group (and fundamental groupoid), monodromy, and holonomy. However, we will not need to use any advanced category-theoretic concepts in this course.

Exercise 5Let and . For any integer , let denote the curve(thus for instance ).

- (i) Show that for any integer , we have .
- (ii) Show that for any non-negative integers , we have . What happens for other values of ?
- (ii) If are distinct integers, show that .

Given a sequence of complex numbers , we define the *polygonal path* traversing these numbers in order to be the curve

This is well-defined thanks to Exercise 3(ii) (actually all we really need in applications is being well-defined up to equivalence). Thus for instance would traverse a closed triangular path connecting , , and (this path may end up being non-simple if the points are collinear).

In order to do analysis, we need to restrict our attention to those curves which are rectifiable:

Definition 6Let be a curve. The arc length of the curve is defined to be the supremum of the quantitieswhere ranges over the natural numbers and ranges over the partitions of . We say that the curve is

rectifiableif its arc length is finite.

The concept is best understood visually: a curve is rectifiable if there is some finite bound on the length of polygonal paths one can form while traversing the curve in order. From Exercise 2 we see that equivalent curves have the same arclength, so the concepts of arclength and rectifiability are well defined for curves that are only given up to continuous reparameterisation.

Exercise 7Let be curves, with the terminal point of equal to the initial point of . Show thatIn particular, is rectifiable if and only if are both individually rectifiable.

It is not immediately obvious that any reasonable curve (e.g. the line segments or the circles ) are rectifiable. To verify this, we need two preliminary results.

Lemma 8 (Triangle inequality)Let be a continuous function. ThenHere we interpret as the Riemann integral (or equivalently, ).

*Proof:* We first attempt to prove this inequality by considering the real and imaginary parts separately. From the real-valued triangle inequality (and basic properties of the Riemann integral) we have

and similarly

but these two bounds only yield the weaker estimate

To eliminate this loss we can amplify the above argument by exploiting phase rotation. For any real , we can repeat the above arguments (using the complex linearity of the Riemann integral, which is easily verified) to give

But we have for any complex number , so taking the supremum of both sides in we obtain the claim.

Exercise 9Let be real numbers. Show that the interval is topologically connected, that is to say the only two subsets of that are both open and closed relative to are the empty set and all of . (Hint:if is a non-empty set that is both open and closed in and contains , consider the supremum of all such that .)

Next, we say that a non-trivial curve is *continuously differentiable* if the derivative

exists and is continuous for all (note that we are only taking right-derivatives at and left-derivatives at ).

Proposition 10 (Arclength formula)If is a continuously differentiable curve, then it is rectifiable, and

*Proof:* We first prove the upper bound

which in particular implies the rectifiability of since the right-hand side of (3) is finite. Let be any partition of . By the fundamental theorem of calculus (applied to the real and imaginary parts of ) we have

for any , and hence by Lemma 8 we have

Summing in we obtain

and taking suprema over all partitions we obtain (3).

Now we need to show the matching lower bound. Let be a small quantity, and for any , let denote the restriction of to . We will show the bound

for all ; specialising to and then sending will give the claim.

It remains to prove (4) for a given choice of . We will use a continuous version of induction known as the continuity method{continuity method}, which exploits Exercise 9.

Let denote the set of such that (4) holds for all . It is clear (using Exercise 7) that this set is topologically closed, and also contains the left endpoint of . If and , then from the differentiability of at , we have some interval such that

for all . Rearranging this using the triangle inequality, we have

Also, from the continuity of we have

for all , if is small enough. We conclude that

and hence

where is the restriction of to . Adding this to the case of (4) using (7) we conclude that (4) also holds for all . From this we see that is (relatively) open in ; from the connectedness of we conclude that , and we are done.

It is now easy to verify that the line segment is rectifiable with arclength , and that the circle is rectifiable with arclength , exactly as one would expect from elementary geometry. Finally, from Lemma 7, a polygonal path will be rectifiable with arclength , again exactly as one would expect.

Exercise 11Show that the curve defined by setting for and is continuous but not rectifiable. (Hint:it is not necessary to compute the arclength precisely; a lower bound that goes to infinity will suffice. Graph the curve to discover some convenient partitions with which to generate such lower bounds. Alternatively, one can apply the arclength formula to some subcurves of .)

Exercise 12(This exercise presumes familiarity with Lebesgue measure.) Show that the image of a rectifable curve is necessarily of measure zero in the complex plane. (In particular, space-filling curves such as the Peano curve or the Hilbert curve cannot be rectifiable.)

Remark 13As the above exercise suggests, many fractal curves will fail to be rectifiable; for instance the Koch snowflake is a famous example of an unrectifiable curve. (The situation is clarified once one develops the theory of Hausdorff dimension, as is done for instance in this previous post: any curve of Hausdorff dimension strictly greater than one will be unrectifiable.)

Much as continuous functions on an interval may be integrated by taking limits of Riemann sums, we may also integrate continuous functions on the image of a rectifiable curve :

Proposition 14 (Integration in rectifiable curves)Let be a rectifiable curve, and let be a continuous function on the image of . Then the “Riemann sums”where ranges over the partitions of , and for each , is an element of , converge as the maximum mesh size goes to zero to some complex limit, which we will denote as . In other words, for every there exists a such that

whenever .

*Proof:* In real analysis courses, one often uses the order properties of the real line to replace the rather complicated looking Riemann sums with the simpler Darboux sums, en route to proving the real-variable analogue of the above proposition. However, in our complex setting the ordering of the real line is not available, so we will tackle the Riemann sums directly rather than try to compare them with Darboux sums.

It suffices to prove that the “Riemann sums” (5) are a Cauchy sequence, in the sense that the difference

between two sums of the form (5) is smaller than any specified if the maximum mesh sizes of the two partitions and are both small enough. From the triangle inequality, and from the fact that any two partitions have a common refinement, it suffices to prove this under the additional assumption that the second partition is a refinemnt of . This means that there is an increasing sequence of natural numbers such that for . In that case, the above difference may be rearranged as

where

By telescoping series, we may rearrange further as

As is continuous and is compact, is uniformly continuous. In particular, if the maximum mesh sizes are small enough, we have

for all and . From the triangle inequality we conclude that

and hence on summing in and using the triangle inequality, we can bound

Since is finite, and can be made arbitrarily small, we obtain the required Cauchy sequence property.

One cannot simply omit the rectifiability hypothesis from the above proposition:

Exercise 15Give an example of a curve such that the Riemann sumsfail to converge to a limit as the maximum mesh size goes to zero, so the integral does not exist even though the integrand is extremely smooth. (Of course, such a curve cannot be rectifiable, thanks to Proposition 14.)

Hint:the non-rectifiable curve in Exercise (11) is a good place to start, but it turns out that this curve does not oscillate wildly enough to make the Riemann sums here diverge, because of the decay of the function near the origin. Come up with a variant of this curve which oscillates more.)

By abuse of notation, we will refer to the quantity as the *contour integral* of along , even though is not necessarily a contour (we will define this concept shortly). We have some easy properties of this integral:

Exercise 16Let be a rectifiable curve, and be a continuous function.

- (i) (Independence of parameterisation) If is another curve equivalent to , show that .
- (ii) (Reversal) Show that .
- (iii) (Concatenation) If for some curves , show that
- (iv) (Change of variables) If is continuously differentiable, show that
- (v) (Upper bound) If there is a pointwise bound of the form for all and some , show that
- (vi) (Linearity) If is a complex number and is a continuous function, show that
and

- (vii) (Integrating a constant) If has initial point and terminal point , show that
- (viii) (Uniform convergence) If , is a sequence of continuous functions converging uniformly to as , show that converges to as . (The requirement of uniform convergence can be relaxed substantially, thanks to tools such as the dominated convergence theorem, but this weaker convergence theorem will suffice for most of our applications.)

Exercise 17(This exercise assumes familiarity with the Riemann-Stieltjes integral.) Let be a rectifiable curve. Let denote the monotone non-decreasing functionfor , where is the restriction of to . For any continuous function , define the

arclength measure integralby the formulawhere the right-hand side is a Riemann-Stieltjes integral. Establish the triangle inequality

for any continuous . Also establish the identity

and obtain an alternate proof of Exercise 16(v).

The change of variables formula (iv) lets one compute many contour integrals using the familiar Riemann integral. For instance, if are real numbers and is continuous, then the contour integral along coincides with the Riemann integral,

and on reversal we also have

Similarly, if is continuous on the circle , we have

Remark 18We caution that if is real-valued, we cannot conclude that is also real valued, unless the contour lies in the real line. This is because the complex line element may introduce some non-trivial imaginary part, as is the case for instance in (6). For similar reasons, we haveand

in general. If one wishes to mix line integrals with real and imaginary parts, it is recommended to replace the contour integrals above with the line integrals

which are defined as in Proposition 14 but where the expression

appearing in (5) is replaced by

The contour integral corresponds to the special case (or more informally, ). Line integrals are in turn special cases of the more general concept of integration of differential forms, discussed for instance in this article of mine, and which are used extensively in differential geometry and geometric topology. However we will not use these more general line integrals or differential form integrals much in this course.

In later notes it will be convenient to restrict to a more regular class of curves than the rectifiable curves. We thus give the definitions here:

Definition 19 (Smooth curves and contours)Asmooth curveis a curve that is continuously differentiable, and such that for all . Acontouris a curve that is equivalent to the concatenation of finitely many smooth curves (that is to say, apiecewise smoothcurve).

Example 20The line segments and circles are smooth curves and hence contours. Polygonal paths are usually not smooth, but they are contours. Any sum of finitely many contours is again a contour, and the reversal of a contour is also a contour.

Note here that the term “smooth” differs somewhat here from the real-variable notion of smoothness, which is defined to be “infinitely differentiable”. Smooth curves are still only assumed to just be continuously differentiable; we do not assume that the second derivative of exists. (In particular, smooth curves may have infinite curvature at some points.) In practice, this distinction tends to be minor, though, as the smooth curves that one actually uses in complex analysis do tend to be infinitely differentiable. On the other hand, for most applications one does not need to control any derivative of a contour beyond the first.

The following examples and exercises may help explain why the non-vanishing condition imbues curves with a certain degree of “smoothness”.

Example 21 (Cuspidal curve)Consider the curve defined by . Clearly is continously differentiable (and even infinitely differentiable), but we do not view this curve as smooth, because vanishes at the origin. Indeed, the image of the curve is , which looks visibly non-smooth at the origin if one plots it, due to the presence of a cusp.

Example 22 (Absolute value function)Consider the curve defined by . This curve is certainly continuously differentiable, and in fact is four times continuously differentiable, but is not smooth because vanishes at the origin. The image of this curve is , which looks visibly non-smooth at the origin (in particular, there is no unique tangent line to this curve here).

Example 23 (Spiral)Consider the curve defined by for , and . One can check (exercise!) that is continuously differentiable, even at the origin ; but it is not a smooth curve because vanishes. The image of has some rather complicated behaviour at the origin, for instance it intersects itself multiple times (try to sketch it!).

Exercise 24 (Local behaviour of smooth curves)Let be a simple smooth curve, and let be an interior point of . Let be a phase of , thus for some . Show that for all sufficiently small , the portion of the image of near looks like a rotated graph, in the sense thatfor some interval containing the origin, and some continuously differentiable function with . Furthermore, show that as (or equivalently, that as ). (

Hint:you may find it easier to first work with the model case where and . The real-variable inverse function theorem will also be helpful.)

Exercise 25Show that a curve is a contour if and only if it is equivalent to the concatenation of finitely manysimplesmooth curves.

Exercise 26Show that the cuspidal curve and absolute value curves in Examples 21, 22 are contours, but the curve in Exercise 23 is not.

** — 2. The fundamental theorem of calculus — **

Now we establish the complex analogues of the fundamental theorem of calculus. As in the real-variable case, there are two useful formulations of this theorem. Here is the first:

Theorem 27 (First fundamental theorem of calculus)Let be an open subset of , let be a continuous function, and suppose that has an antiderivative , that is to say a holomorphic function with for all . Let be a rectifiable curve in with initial point and terminal point . Then

*Proof:* If were continuously differentiable, or at least *piecewise continuously differentiable* (the concatenation of finitely many continuously differentiable curves), we could establish this theorem by using Exercise 16 to rewrite everything in terms of real-variable Riemann integrals, at which point one can use the real-variable fundamental theorem of calculus (and the chain rule). But actually we can just give a direct proof that does not need any rectifiability hypothesis whatsoever.

We again use the continuity method. Let , and for each , let be the restriction of to . It will suffice to show that

for all , as the claim then follows by setting and sending to zero.

Let denote the set of all such that (7) holds for all . As before, is clearly closed and contains ; as is connected, the only remaining task is to show that is open in . Let be such that . As is differentiable at with derivative , and is continuous, there exists with such that

for any . On the other hand, if is small enough, we have that

for all , and hence by Exercise 16(v) we have

where is the restriction of to . Applying Exercise 16(vi), (vii) we thus have

Combining this with (8) and the triangle inequality, we conclude that

and on adding this to the case of (7) and again using the triangle inequality, we conclude that (7) holds for all . This ensures that is open, as desired.

One can use this theorem to quickly evaluate many integrals by using an antiderivative for the integrand as in the real-variable case. For instance, for any rectifiable curve with initial point and terminal point , we have

and so forth. If the curve avoids the origin, we also have

since is an antiderivative of on . If is a power series with radius of convergence , and is a curve in with initial point and terminal point , we similarly have

For the second fundamental theorem of calculus, we need a topological preliminary result.

Exercise 28Let be a non-empty open subset of . Show that the following statements are equivalent:

- (i) is topologically connected (that is to say, the only two subsets of that are both open and closed relative to are the empty set and itself).
- (ii) is path connected (that is to say, for any there exists a curve with image in whose initial point is and terminal point is )
- (iii) is polygonally path connected (the same as (ii), except that is now required to also be a polygonal path).
(

Hint:to show that (i) implies (iii), pick a base point in and consider the set of all in that can be reached from by a polygonal path.)

We remark that the relationship between path connectedness and connectedness is more delicate when one does not assume that the space is open; every path-connected space is still connected, but the converse need not be true.

Remark 29There is some debate as to whether to view the empty set as connected, disconnected, or neither. I view this as analogous to the debate as to whether the natural number should be viewed as prime, composite, or neither. In both cases I personally prefer the convention of “neither” (and like to use the term “unconnected” to describe the empty set, and “unit” to describe ), but to avoid any confusion I will restrict the discussion of connectedness to non-empty sets in this course (which is all we will need in applications).

In real analysis, the second fundamental theorem of calculus asserts that if a function is continuous, then the function is an antiderivative of . In the complex case, there is an analogous result, but one needs the additional requirement that the function is conservative:

Theorem 30 (Second fundamental theorem of calculus)Let be a non-empty open connected subset of the complex numbers. Let be a continuous function which isconservativein the sense thatwhenever is a closed polygonal path in . Fix a base point , and define the function by the formula

for all , where is any polygonal path from to in (the existence of such a path follows from Exercise 28 and the hypothesis that is connected, and the independence of the choice of path for the purposes of defining follows from Exercise 16 and the conservative hypothesis (9)). Then is holomorphic on and is an antiderivative of , thus for all .

*Proof:* We mimic the proof of the real-variable second fundamental theorem of calculus. Let be any point in . As is open, it contains some disk centred at . In particular, if lies in this disk, then the line segment will have image in . If is any polygonal path from to , then we have

and

and hence by Exercise 16

(The reader is strongly advised to draw a picture depicting the situation here.) Let , then for sufficiently close to , we have for all in the image of . Thus by Exercise 16(v) we have

for sufficiently close to , which implies that

for any , and thus is an antiderivative of as required.

The notion of a non-empty open connected subset of the complex plane comes up so frequently in complex analysis that many texts assign a special term to this notion; for instance, Stein-Shakarchi refers to such sets as *regions*, and in other texts they may be called *domains*. We will stick to just “non-empty open connected subset of ” in this course.

The requirement that be conservative is necessary, as the following exercise shows. Actually it is also necessary in the real-variable case, but it is redundant in that case due to the topological triviality of closed polygonal paths in one dimension: see Exercise 33 below.

Exercise 31Let be a non-empty connected subset of , and let be continuous. Show that the following are equivalent:

- (i) possesses at least one antiderivative .
- (ii) is conservative in the sense that (9) holds for all closed polygonal paths in .
- (iii) is conservative in the sense that (9) holds for all
simpleclosed polygonal paths in .- (iv) is conservative in the sense that (9) holds for all closed contours in .
- (v) is conservative in the sense that (9) holds for all closed rectifiable curves in .
(Hint: to show that (iii) implies (ii), induct on the number of edges in the closed polygonal path, and find a way to decompose non-simple closed polygonal paths into paths with fewer edges.) Furthermoore, show that if has two antiderivatives , then there exists a constant such that .

Exercise 32Show that the function does not have an antiderivative on . (Hint:integrate on . In later notes we will see that nevertheless does have antiderivatives on many subsets of , formed by various branches of the complex logarithm.

Exercise 33If is a continuous function on an interval, show that is conservative in the sense that (9) holds for any closed polygonal path in . What happens for closed rectifiable paths?

Exercise 34Let be an open subset of (not necessarily connected).

- (i) Show that there is a unique collection of non-empty subsets of that are open, connected, disjoint, and partition : . (The elements of are known as the connected components of .)
- (ii) Show that the number of connected components of is at most countable. (
Hint:show that each connected component contains at least one complex number with rational real and imaginary parts.)- (iii) If is a continuous conservative function on , show that has at least one antiderivative .
- (iv) If has more than one connected component, show that it is possible for a function to have two antiderivatives which do not differ by a constant (i.e. there is no complex number such that ).

Exercise 35 (Integration by parts)Let be holomorphic functions on an open set , and let be a rectifiable curve in with initial point and terminal point . Prove that

Filed under: 246A - complex analysis, math.CV, math.GT Tagged: contour integration, fundamental theorem of calculus, rectifiable curve ]]>

In this set of notes we will focus solely on the concept of complex differentiation, deferring the discussion of contour integration to the next set of notes. To begin with, the theory of complex differentiation will greatly resemble the theory of real differentiation; the definitions look almost identical, and well known laws of differential calculus such as the product rule, quotient rule, and chain rule carry over *verbatim* to the complex setting, and the theory of complex power series is similarly almost identical to the theory of real power series. However, when one compares the “one-dimensional” differentiation theory of the complex numbers with the “two-dimensional” differentiation theory of two real variables, we find that the dimensional discrepancy forces complex differentiable functions to obey a real-variable constraint, namely the Cauchy-Riemann equations. These equations make complex differentiable functions substantially more “rigid” than their real-variable counterparts; they imply for instance that the imaginary part of a complex differentiable function is essentially determined (up to constants) by the real part, and vice versa. Furthermore, even when considered separately, the real and imaginary components of complex differentiable functions are forced to obey the strong constraint of being *harmonic*. In later notes we will see these constraints manifest themselves in integral form, particularly through Cauchy’s theorem and the closely related Cauchy integral formula.

Despite all the constraints that holomorphic functions have to obey, a surprisingly large number of the functions of a complex variable that one actually encounters in applications turn out to be holomorphic. For instance, any polynomial with complex coefficients will be holomorphic, as will the complex exponential . From this and the laws of differential calculus one can then generate many further holomorphic functions. Also, as we will show presently, complex power series will automatically be holomorphic inside their disk of convergence. On the other hand, there are certainly basic complex functions of interest that are *not* holomorphic, such as the complex conjugation function , the absolute value function , or the real and imaginary part functions . We will also encounter functions that are only holomorphic at some portions of the complex plane, but not on others; for instance, rational functions will be holomorphic except at those few points where the denominator vanishes, and are prime examples of the *meromorphic* functions mentioned previously. Later on we will also consider functions such as branches of the logarithm or square root, which will be holomorphic outside of a *branch cut* corresponding to the choice of branch. It is a basic but important skill in complex analysis to be able to quickly recognise which functions are holomorphic and which ones are not, as many of useful theorems available to the former (such as Cauchy’s theorem) break down spectacularly for the latter. Indeed, in my experience, one of the most common “rookie errors” that beginning complex analysis students make is the error of attempting to apply a theorem about holomorphic functions to a function that is not at all holomorphic. This stands in contrast to the situation in real analysis, in which one can often obtain correct conclusions by formally applying the laws of differential or integral calculus to functions that might not actually be differentiable or integrable in a classical sense. (This latter phenomenon, by the way, can be largely explained using the theory of distributions, as covered for instance in this previous post, but this is beyond the scope of the current course.)

Remark 1In this set of notes it will be convenient to impose some unnecessarily generous regularity hypotheses (e.g. continuous second differentiability) on the holomorphic functions one is studying in order to make the proofs simpler. In later notes, we will discover that these hypotheses are in fact redundant, due to the phenomenon ofelliptic regularitythat ensures that holomorphic functions are automatically smooth.

** — 1. Complex differentiation and power series — **

Recall in real analysis that if is a function defined on some subset of the real line , and is an interior point of (that is to say, contains an interval of the form for some ), then we say that is *differentiable* at if the limit

exists (note we have to exclude from the possible values of to avoid division by zero. If is differentiable at , we denote the above limit as or , and refer to this as the *derivative* of at . If is open (that is to say, every element of is an interior point), and is differentiable at every point of , then we say that is differentiable on , and call the *derivative* of . (One can also define differentiability at non-interior points if they are not isolated, but for simplicity we will restrict attention to interior derivatives only.)

We can adapt this definition to the complex setting without any difficulty:

Definition 2 (Complex differentiability)Let be a subset of the complex numbers , and let be a function. If is an interior point of (that is to say, contains a disk for some ), we say that iscomplex differentiableat if the limitexists, in which case we denote this limit as or , and refer to this as the

complex derivativeof at . If is open (that is to say, every point in is an interior point), and is complex differentiable at every point at , we say that iscomplex differentiableon , orholomorphicon .

In terms of epsilons and deltas: is complex differentiable at with derivative if and only if, for every , there exists such that whenever is such that . Another way of writing this is that we have an approximate *linearisation*

as approaches , where denotes a quantity of the form for in a neighbourhood of , where goes to zero as goes to .

If is differentiable at , then from the limit laws we see that

and hence

that is to say that is continuous at . In particular, holomorphic functions are automatically continuous. (Later on we will see that they are in fact far more regular than this, being smooth and even analytic.)

It is usually quite tedious to verify complex differentiability of a function, and to compute its derivative, from first principles. We will give just one example of this:

Proposition 3Let be a non-negative integer. Then the function is holomorphic on the entire complex plane , with derivative (with the convention that is zero when ).

*Proof:* This is clear for , so suppose . We need to show that for any complex number , that

But we have the geometric series identity

which is valid (in any field) whenever , as can be seen either by induction or by multiplying both sides by and cancelling the telescoping series on the right-hand side. The claim then follows from the usual limit laws.

Fortunately, we have the familiar laws of differential calculus, that allow us to more quickly establish the differentiability of functions if they arise as various combinations of functions that are already known to be differentiable, and to compute the derivative:

Exercise 4 (Laws of differentiation)Let be an open subset of , let be a point in , and let be functions that are complex differentiable at .

- (i) (Linearity) Show that is complex differentiable at , with derivative . For any constant , show that is differentiable at , with derivative .
- (ii) (Product rule) Show that is complex differentiable at , with derivative .
- (iii) (Quotient rule) If is non-zero, show that (which is defined in a neighbourhood of , by continuity) is complex differentiable at , with derivative .
- (iv) (Chain rule) If is a neighbourhood of , and is a function that is complex differentiable at , show that the composition (which is defined in a neighbourhood of ) is complex differentiable at , with derivative
(

Hint:take your favourite proof of the real-variable version of these facts and adapt them to the complex setting.)

One could also state and prove a complex-variable form of the inverse function theorem here, but the proof of that statement is a bit more complicated than the ones in the above exercise, so we defer it until later in the course when it becomes needed.

If a function is holomorphic on the entire complex plane, we call it an entire function; clearly such functions remain holomorphic when restricted to any open subset of the complex plane. Thus for instance Proposition 3 tells us that the functions are entire, and from linearity we then see that any complex polynomial

will be an entire function, with derivative given by the familiar formula

A function of the form , where are polynomials with not identically zero, is called a rational function, being to polynomials as rational numbers are to integers. Such a rational function is well defined as long as is not zero. From the factor theorem (which works over any field, and in particular over the complex numbers) we know that the number of zeroes of is finite, being bounded by the degree of (of course we will be able to say something stronger once we have the fundamental theorem of algebra). Because of these singularities, rational functions are rarely entire; but from the quotient rule we do at least see that is complex differentiable wherever the denominator is non-zero. Such functions are prime examples of meromorphic functions, which we will discuss later in the course.

Exercise 5 (Gauss-Lucas theorem)Let be a complex polynomial that is factored asfor some non-zero constant and roots (not necessarily distinct) with .

- (i) Suppose that all lie in the upper half-plane . Show that any root of the derivative also lies in the upper half-plane. (
Hint:use the product rule to decompose thelog-derivativeinto partial fractions, and then investigate the sign of the imaginary part of this log-derivative for outside the upper half-plane.)- (ii) Show that all the roots of lie in the convex hull of the set of roots of , that is to say the smallest convex polygon that contains .

Now we discuss power series, which are infinite degree variants of polynomials, and which turn out to inherit many of the algebraic and analytic properties of such polynomials, at least if one stays within the disk of convergence.

Definition 6 (Power series)Let be a complex number. Aformal power serieswith complex coefficients around the point is a formal series of the formfor some complex numbers , with an indeterminate.

One can attempt to evaluate a formal power series at a given complex number by replacing the formal indeterminate with the complex number . This may or may not produce a convergent (or absolutely convergent) series, depending on where is; for instance, the power series is always absolutely convergent at , but the geometric power series fails to be even conditionally convergent whenever (since the summands do not go to zero). As it turns out, the region of convergence is always essentially a disk, the size of which depends on how rapidly the coefficients decay (or how slowly they grow):

Proposition 7 (Convergence of power series)Let be a formal power series, and define the radius of convergence of the series to be the quantitywith the convention that is infinite if . (Note that is allowed to be zero or infinite.) Then the formal power series is absolutely convergent for any in the disk (known as the

disk of convergence), and is divergent (i.e., not convergent) for any in the exterior region .

*Proof:* The proof is nearly identical to the analogous result for real power series. First suppose that is a complex number with (this of course implies that is finite). Then by (3), we have for infinitely many , which after some rearranging implies that for infinitely many . In particular, the sequence does not go to zero as , which implies that is divergent.

Now suppose that is a complex number with (this of course implies that is non-zero). Choose a real number with , then by (3), we have for all sufficiently large , which after some rearranging implies that

for all sufficiently large . Since the geometric series is absolutely convergent, this implies that is absolutely convergent also, as required.

Remark 8Note that this proposition does not say what happens on the boundary of this disk (assuming for sake of discussion that the radius of convergence is finite and non-zero). The behaviour of power series on and near the boundary of the disk of convergence is in fact remarkably subtle; see for instance Example 11 below.

The above proposition gives a “root test” formula for the radius of convergence. The following “ratio test” variant gives a convenient lower bound for the radius of convergence which suffices in many applications:

Exercise 9 (Ratio test)If is a formal power series with the non-zero for all sufficiently large , show that the radius of convergence of the series obeys the lower boundIn particular, if the limit exists, then it is equal to . Give example to show that strict inequality can hold in both bounds in (4).

If a formal power series has a positive radius of convergence, then it defines a function in the disk of convergence by setting

We refer to such a function as a *power series*, and refer to as the *radius of convergence* of that power series. (Strictly speaking, a formal power series and a power series are different concepts, but there is little actual harm in conflating them together in practice, because of the uniqueness property established in Exercise 17 below.)

Example 10The formal power series has a zero radius of convergence, thanks to the ratio test, and so only converges at . Conversely, the exponential formal power series has an infinite radius of convergence (thanks to the ratio test), and converges of course to when evaluated at any complex number .

Example 11 (Geometric series)The formal power series has radius of convergence . If lies in the disk of convergence , then we haveand thus after some algebra we obtain the geometric series formula

as long as is inside the disk . The function does not extend continuously to the boundary point of the disk, but does extend continuously (and even smoothly) to the rest of the boundary, and is in fact holomorphic on the remainder of the complex plane. However, the geometric series diverges at every single point of this boundary (when , the coefficients of the series do not converge to zero), and of course definitely diverge outside of the disk as well. Thus we see that the function that a power series converges to can extend well beyond the disk of convergence, which thus may only capture a portion of the domain of definition of that function. For instance, if one

formallyapplies (5) with, say, , one ends up with the apparent identityThis identity does not make sense if one interprets infinite series in the classical fashion, as the series is definitely divergent. However, by formally extending identities such as (5) beyond their disk of convergence, we can

generalisethe notion of summation of infinite series to assign meaningful values to such series even if they do not converge in the classical sense. This leads to generalised summation methods such as zeta function regularisation, which are discussed in this previous blog post. However, we will not use such generalised interpretations of summation very much in this course.

Exercise 12For any complex numbers , show that the formal power series has radius of convergence (with the convention that this is infinite for ), and is equal to the function inside the disk of convergence.

Exercise 13For any positive integer , show that the formal power serieshas radius of convergence , and converges to the function in the disk . Here of course is the usual binomial coefficient.

We have seen above that power series can be well behaved as one approaches the boundary of the disk of convergence, while being divergent at the boundary. However, the converse scenario, in which the power series converges at the boundary but does not behave well as one approaches the boundary, does not occur:

Exercise 14(i) (Summation by parts formula) Let be a finite sequence of complex numbers, and let be the partial sums for . Show that for any complex numbers , that (ii) Let be a sequence of complex numbers such that is convergent (not necessarily absolutely) to zero. Show that for any , the series is absolutely convergent, and (

Hint:use summation by parts and a limiting argument to express in terms of the partial sums .)(iii) (Abel’s theorem) Let be a power series with a finite positive radius of convergence , and let be a point on the boundary of the disk of convergence at which the series converges (not necessarily absolutely). Show that . ( Hint:use various translations and rotations to reduce to the case considered in (ii).)

As a general rule of thumb, as long as one is inside the disk of convergence, power series behave very similarly to polynomials. In particular, we can generalise the differentiation formula (2) to such power series:

Theorem 15Let be a power series with a positive radius of convergence . Then is holomorphic on the disk of convergence , and the derivative is given by the power seriesthat has the same radius of convergence as .

*Proof:* From (3), the standard limit and the usual limit laws, it is easy to see that the power series has the same radius of convergence as . To show that this series is actually the derivative of , we use first principles. If lies in the disk of convergence, we consider the Newton quotient

for . Expanding out the absolutely convergent series and , we can write

The ratio vanishes for , and for it is equal to as in the proof of Proposition 3. Thus

As approaches , each summand converges to . This *almost* proves the desired limiting formula

but we need to justify the interchange of a sum and limit. Fortunately we have a standard tool for this, namely the Weierstrass -test (which works for complex-valued functions exactly as it does for real-valued functions; one could also use the dominated convergence theorem here). It will be convenient to select two real numbers with . Clearly, for close enough to , we have . By the triangle inequality we then have

On the other hand, from (3) we know that for sufficiently large , hence for sufficiently large . From the ratio test we know that the series is absolutely convergent, hence the series is also. Thus, for sufficiently close to , the summands are uniformly dominated by an absolutely summable sequence of numbers . Applying the Weierstrass -test (or dominated convergence theorem), we obtain the claim.

Exercise 16Prove the above theorem directly using epsilon and delta type arguments, rather than invoking the -test or the dominated convergence theorem.

We remark that the above theorem is a little easier to prove once we have the complex version of the fundamental theorem of calculus, but this will have to wait until the next set of notes, where we will also prove a remarkable converse to the above theorem, in that *any* holomorphic function can be expanded as a power series around any point in its domain.

A convenient feature of power series is the ability to *equate coefficients*: if two power series around the same point agree, then their coefficients must also agree. More precisely, we have:

Exercise 17 (Taylor expansion and uniqueness of power series)Let be a power series with a positive radius of convergence. Show that , where denotes the complex derivative of . In particular, if is another power series around with a positive radius of convergence which agrees with on some neighbourhood of (thus, for all ), show that the coefficients of and are identical, that is to say that for all .

Of course, one can no longer compare coefficients so easily if the power series are based around two different points. For instance, from Exercise 11 we see that the geometric series and both converge to the same function on the unit disk , but have differing coefficients. The precise relation between the coefficients of power series of the same function is given as follows:

Exercise 18 (Changing the origin of a power series)Let be a power series with a positive radius of convergence . Let be an element of the disk of convergence . Show that the formal power series , wherehas radius of convergence at least , and converges to on the disk . Here of course is the usual binomial coefficient.

Theorem 15 gives us a rich supply of complex differentiable functions, particularly when combined with Exercise 4. For instance, the complex exponential function

has an infinite radius of convergence, and so is entire, and is its own derivative:

This makes the complex trigonometric functions

entire as well, and from the chain rule we recover the familiar formulae

Of course, one can combine these functions together in many ways to create countless other complex differentiable functions with explicitly computable derivatives, e.g. is an entire function with derivative , the tangent function is holomorphic outside of the discrete set with derivative , and so forth.

Exercise 19 (Multiplication of power series)Let and be power series that both have radius of convergence at least . Show that on the disk , we havewhere the right-hand side is another power series of radius of convergence at least , with coefficients given as the convolution

of the sequences and .

** — 2. The Cauchy-Riemann equations — **

Thus far, the theory of complex differentiation closely resembles the analogous theory of real differentiation that one sees in an introductory real analysis class. But now we take advantage of the Argand plane representation of to view a function of one complex variable as a function of two real variables. This gives rise to some further notions of differentiation. Indeed, if is a function defined on an open subset of , and is a point in , then in addition to the complex derivative

already discussed, we can also define (if they exist) the partial derivatives

and

these will be complex numbers if the limits on the right-hand side exist. There is also (if it exists) the gradient (or Fréchet derivative) , defined as the vector with the property that

where is the Euclidean norm of .

These notions of derivative are of course closely related to each other. If a function is *Fréchet differentiable* at , in the sense that the gradient exists, then on specialising the limit in (7) to vectors of the form or we see that

and

leading to the familiar formula

for the gradient of a function that is Fréchet differentiable at . We caution however that it is possible for the partial derivatives of a function to exist without the function being Fréchet differentiable, in which case the formula (8) is of course not valid. (A typical example is the function defined by setting for , with ; this function has both partial derivatives existing at , but is not differentiable here.) On the other hand, if the partial derivatives exist everywhere on and are additionally known to be continuous, then the fundamental theorem of calculus gives the identity

for and sufficiently small (with the convention that if ), and from this it is not difficult to see that is then Fréchet differentiable everywhere on .

Similarly, if is complex differentiable at , then by specialising the limit (6) to variables of the form or for some non-zero real near zero, we see that

and

leading in particular to the Cauchy-Riemann equations

that must be satisfied in order for to be complex differentiable. More generally, from (6) we see that if is complex differentiable at , then

which on comparison with (7) shows that is also Fréchet differentiable with

Finally, if is Fréchet differentiable at and one has the Cauchy-Riemann equations (9), then from (7) we have

which after making the substitution gives

which on comparison with (6) shows that is complex differentiable with

We summarise the above discussion as follows:

Proposition 20 (Differentiability and the Cauchy-Riemann equations)Let be an open subset of , let be a function, and let be an element of .

- (i) If is complex differentiable at , then it is also Fréchet differentiable at , with
In particular, the Cauchy-Riemann equations (9) hold at .

- (ii) Conversely, if is Freéchet differentiable at and obeys the Cauchy-Riemann equations at , then is complex differentiable at .

Remark 21From part (ii) of the above proposition we see that if is Fre\’chet differentiable on and obeys the Cauchy-Riemann equations on , then it is holomorphic on . One can ask whether the requirement of Freéchet differentiability can be weakened. It cannot be omitted entirely; one can show, for instance, that the function defined by for non-zero and obeys the Cauchy-Riemann equations at every point , but is not complex differentiable (or even continuous) at the origin. But there is a somewhat difficult theorem of Looman and Menchoff that asserts that if is continuous on and obeys the Cauchy-Riemann equations on , then it is holomorphic. We will not prove or use this theorem in this course; generally in modern applications, when one wants to weaken the regularity hypotheses of a theorem involving classical differentiation, the best way to do so is to replace the notion of a classical derivative with that of a weak derivative, rather than insist on computing derivatives in the classical pointwise sense. See this blog post for more discussion.Combining part (i) of the above proposition with Theorem 15, we also conclude as a corollary that any power series will be smooth inside its disk of convergence, in the sense all partial derivatives to all orders of this power series exist.

Remark 22From the geometric perspective, one can interpret complex differentiability at a point as a requirement that a map is conformal and orientation-preserving at , at least in the non-degenerate case when is non-zero. In more detail: suppose that is a map that is complex differentiable at some point with . Let be a differentiable curve with ; we view this as the trajectory of some particle which passes through at time . The derivative (defined in the usual manner by limits of Newton quotients) can then be viewed as the velocity of the particle as it passes through . The map takes this particle to a new particle parameterised by the curve ; at time , this new particle passes through , and by the chain rule we see that the velocity of the new particle at this time is given byThus, if we write in polar coordinates as , the map transforms the velocity of the particle by multiplying the speed by a factor of and rotating the direction of travel counter-clockwise by . In particular, we consider two differentiable trajectories both passing through at time (with non-zero speeds), then the map preserves the angle between the two velocity vectors , as well as their orientation (e.g. if is counterclockwise to , then is counterclockwise to . This is in contrast to, for instance, shear transformations such as , which preserve orientation but not angle, or the complex conjugation map , which preserve angle but not orientation. The same preservation of angle is present for real differentiable functions , but is much less impressive in that case since the only angles possible between two vectors on the real line are and ; it is the geometric two-dimensionality of the complex plane that makes conformality a much stronger and more “rigid” property for complex differentiable functions.

If one breaks up a complex function into real and imaginary parts for some , then on taking real and imaginary parts one can express the Cauchy-Riemann equations as a system

of two partial differential equations for two functions . This gives a quick way to test if various functions are differentiable. Consider for instance the conjugation function . In this case, and . These functions, being polynomial in , is certainly Fréchet differentiable everywhere; the equation (11) is always satisfied, but the equation (10) is never satisfied. As such, the conjugation function is never complex differentiable. Similarly for the real part function , the imaginary part function , or the absolute value function . The function has real part and imaginary part ; one easily checks that the system (10), (11) is only satisfied when , so this function is only complex differentiable at the origin. In particular, it is not holomorphic on any non-empty open set.

The general rule of thumb that one should take away from these examples is that complex functions that are constructed purely out of “good” functions such as polynomials, the complex exponential, complex trigonometric functions, or other convergent power series are likely to be holomorphic, whereas functions that involve “bad” functions such as complex conjugation, the real and imaginary part, or the absolute value, are unlikely to be holomorphic.

Exercise 23 (Wirtinger derivatives)Let be an open subset of , and let be a Fréchet differentiable function. Define the Wirtinger derivatives , by the formulae

- (i) Show that is holomorphic on if and only if the Wirtinger derivative vanishes identically on .
- (ii) If is given by a polynomial
in both and for some complex coefficients and some natural number , show that

and

(

Hint:first establish a Leibniz rule for Wirtinger derivatives.) Conclude in particular that is holomorphic if and only if vanishes whenever (i.e. does not contain any terms that involve ).- (iii) If is a point in , show that one has the Taylor expansion
as , where denotes a quantity of the form , where goes to zero as goes to (compare with (1)). Conversely, show that this property determines the numbers and uniquely (and thus can be used as an alternate definition of the Wirtinger derivatives).

Remark 24Any polynomialin the real and imaginary parts of can be rewritten as a polynomial in and as per (17), using the usual identities

for . Thus such a non-holomorphic polynomial of one complex variable can be viewed as the restriction of a holomorphic polynomial

of two complex variables to the anti-diagonal , and the Wirtinger derivatives can then be interpreted as genuine (complex) partial derivatives in these two complex variables. More generally, Wirtinger derivatives are convenient tools in the subject of several complex variables, which we will not cover in this course.

The Cauchy-Riemann equations couple the real and imaginary parts of a holomorphic function to each other. But it is also possible to eliminate one of these components from the equations and obtain a constraint on just the real part , or just the imaginary part . Suppose for the moment that is a holomorphic function which is twice continuously differentiable (thus the second partial derivatives , , , all exist and are continuous on ); we will show in the next set of notes that this extra hypothesis is in fact redundant. Assuming continuous second differentiability for now, we have Clairaut’s theorem

everywhere on . Similarly for the real and imaginary parts . If we then differentiate (10) in the direction, (11) in the direction, and then sum, the derivatives of cancel thanks to Clairaut’s theorem, and we obtain Laplace’s equation

which is often written more compactly as

where is the Laplacian operator

A similar argument gives ; by linearity we then also have .

Functions for which are known as harmonic functions: thus we have shown that (twice continuously differentiable) holomorphic functions are automatically harmonic, as are the real and imaginary parts of such functions. The converse is not true: not every harmonic function is holomorphic. For instance, the conjugate function is clearly harmonic on , but not holomorphic. We will return to the precise relationship between harmonic and holomorphic functions shortly.

Harmonic functions have many remarkable properties. Since the second derivative in a given direction is a local measure of “convexity” of a function, we see from (13) that any convex behaviour of a harmonic function in one direction has to be balanced by an equal and opposite amount of concave behaviour in the orthogonal direction. A good example of a harmonic function to keep in mind is the function

which exhibits convex behavior in and concave behavior in in exactly opposite amounts. This function is the real part of the holomorphic function , which is of course consistent with the previous observation that the real parts of holomorphic functions are harmonic.

We will discuss harmonic functions more in later notes. For now, we record just one important property of these functions, namely the maximum principle:

Theorem 25 (Maximum principle)Let be an open subset of , and let be a continuously twice differentiable harmonic function. Let be a compact subset of , and let be the boundary of . Then

Informally, the maximum principle asserts that the maximum of a real-valued harmonic function on a compact set is always attained on the boundary, and similarly for the minimum. In particular, any bound on the harmonic function that one can obtain on the boundary is automatically inherited by the interior. Compare this with a non-harmonic function such as , which is bounded by on the boundary of the compact unit disk , but is not bounded by on the interior of this disk.

*Proof:* We begin with an “almost proof” of this principle, and then repair this attempted proof so that it is an actual proof.

We will just prove (14), as (15) is proven similarly (or one can just observe that if is harmonic then so is ). Clearly we have

so the only scenario that needs to be excluded is when

Suppose that this is the case. As is continuous and is compact, must attain its maximum at some point in ; from (16) we see that must be an interior point. Since is a local maximum of , and is twice differentiable, we must have

and similarly

This *almost*, but does not quite, contradict the harmonicity of , since it is still possible that both of these partial derivatives vanish. To get around this problem we use the trick of creating an epsilon of room, adding a tiny bit of convexity to . Let be a small number to be chosen later, and let be the modified function

Since is compact, the function is bounded on . Thus, from (16), we see that if is small enough we have

Arguing as before, must attain its maximum at some interior point of , and so we again have

and similarly

On the other hand, since is harmonic, we have

on . These facts contradict each other, and we are done.

Exercise 26 (Maximum principle for holomorphic functions)If is a continuously twice differentiable holomorphic function on an open set , and is a compact subset of , show that(

Hint:use Theorem 25 and the fact that for any complex number .) What happens if we replace the suprema on both sides by infima?

Exercise 27Recall the Wirtinger derivatives defined in Exercise 23(i).

- (i) If is twice continuously differentiable on an open subset of , show that
Use this to give an alternate proof that holomorphic functions are harmonic.

- (ii) If is given by a polynomial
in both and for some complex coefficients and some natural number , show that is harmonic on if and only if vanishes whenever and are both positive (i.e. only contains terms or that only involve one of or ).

- (iii) If is a real polynomial
in and for some real coefficients and some natural number , show that is harmonic if and only if it is the real part of a polynomial of one complex variable .

We have seen that the real and imaginary parts of any holomorphic function are harmonic functions. Conversely, let us call a harmonic function a harmonic conjugate of another harmonic function if is holomorphic on ; in the case that are continuously twice differentiable, this is equivalent by Proposition 20 to satisfying the Cauchy-Riemann equations (10), (11). Here is a short table giving some examples of harmonic conjugates:

(for the last example one of course has to exclude the origin from the domain ).

From Exercise 27(ii) we know that every harmonic polynomial has at least one harmonic conjugate; it is natural to ask whether the same fact is true for more general harmonic functions than polynomials. In the case that the domain is the entire complex plane, the answer is affirmative:

Proposition 28Let be a continuously twice differentiable harmonic function. Then there exists a harmonic conjugate of . Furthermore, this harmonic conjugate is unique up to constants: if are two harmonic conjugates of , then is a constant function.

*Proof:* We first prove uniqueness. If is a harmonic conjugate of , then from the fundamental theorem of calculus, we have

and hence by the Cauchy-Riemann equations (10), (11) we have

Similarly for any other harmonic conjugate of . It is now clear that and differ by a constant.

Now we prove existence. Inspired by the above calculations, we define to be the define explicitly by the formula

From the fundamental theorem of calculus, we see that is differentiable in the direction with

This is one of the two Cauchy-Riemann equations needed. To obtain the other one, we differentiate (18) in the variable. The fact that is continuously twice differentiable allows one to differentiate under the integral sign (exercise!) and conclude that

As is harmonic, we have , so by the fundamental theorem of calculus we conclude that

Thus we now have both of the Cauchy-Riemann equations (10), (11) in . Differentiating these equations again, we conclude that is twice continuously differentiable, and hence by Proposition 20 we have holomorphic on , giving the claim.

The same argument would also work for some other domains than , such as rectangles . To handle the general case, though, it becomes convenient to introduce the notion of contour integration, which we will do in the next set of notes. In some cases (specifically, when the underlying domain fails to be simply connected), it will turn out that some harmonic functions do not have conjugates!

Exercise 29Show that an entire function can real-valued on only if it is constant.

Exercise 30Let be a complex number. Show that if is an entire function such that for all , then for all .

Filed under: 246A - complex analysis, math.CA, math.CV Tagged: Cauchy-Riemann equations, differentiation, harmonic functions, holomorphicity, power series ]]>

- The integers are the additive completion of the natural numbers (the minimal additive group that contains a copy of ).
- The rationals are the multiplicative completion of the integers (the minimal field that contains a copy of ).
- The reals are the metric completion of the rationals (the minimal complete metric space that contains a copy of ).
- The complex numbers are the algebraic completion of the reals (the minimal algebraically closed field that contains a copy of ).

These descriptions of the standard number systems are elegant and conceptual, but not entirely suitable for *constructing* the number systems in a non-circular manner from more primitive foundations. For instance, one cannot quite define the reals from scratch as the metric completion of the rationals , because the definition of a metric space itself requires the notion of the reals! (One can of course construct by other means, for instance by using Dedekind cuts or by using uniform spaces in place of metric spaces.) The definition of the complex numbers as the algebraic completion of the reals does not suffer from such a non-circularity issue, but a certain amount of field theory is required to work with this definition initially. For the purposes of quickly constructing the complex numbers, it is thus more traditional to first define as a quadratic extension of the reals , and more precisely as the extension formed by adjoining a square root of to the reals, that is to say a solution to the equation . It is not immediately obvious that this extension is in fact algebraically closed; this is the content of the famous fundamental theorem of algebra, which we will prove later in this course.

The two equivalent definitions of – as the algebraic closure, and as a quadratic extension, of the reals respectively – each reveal important features of the complex numbers in applications. Because is algebraically closed, all polynomials over the complex numbers split completely, which leads to a good spectral theory for both finite-dimensional matrices and infinite-dimensional operators; in particular, one expects to be able to diagonalise most matrices and operators. Applying this theory to constant coefficient ordinary differential equations leads to a unified theory of such solutions, in which real-variable ODE behaviour such as exponential growth or decay, polynomial growth, and sinusoidal oscillation all become aspects of a single object, the complex exponential (or more generally, the matrix exponential ). Applying this theory more generally to diagonalise arbitrary translation-invariant operators over some locally compact abelian group, one arrives at Fourier analysis, which is thus most naturally phrased in terms of complex-valued functions rather than real-valued ones. If one drops the assumption that the underlying group is abelian, one instead discovers the representation theory of unitary representations, which is simpler to study than the real-valued counterpart of orthogonal representations. For closely related reasons, the theory of complex Lie groups is simpler than that of real Lie groups.

Meanwhile, the fact that the complex numbers are a quadratic extension of the reals lets one view the complex numbers geometrically as a two-dimensional plane over the reals (the Argand plane). Whereas a point singularity in the real line disconnects that line, a point singularity in the Argand plane leaves the rest of the plane connected (although, importantly, the punctured plane is no longer simply connected). As we shall see, this fact causes singularities in complex analytic functions to be better behaved than singularities of real analytic functions, ultimately leading to the powerful residue calculus for computing complex integrals. Remarkably, this calculus, when combined with the quintessentially complex-variable technique of *contour shifting*, can also be used to compute some (though certainly not all) definite integrals of *real*-valued functions that would be much more difficult to compute by purely real-variable methods; this is a prime example of Hadamard’s famous dictum that “the shortest path between two truths in the real domain passes through the complex domain”.

Another important geometric feature of the Argand plane is the angle between two tangent vectors to a point in the plane. As it turns out, the operation of multiplication by a complex scalar preserves the magnitude and orientation of such angles; the same fact is true for any non-degenerate complex analytic mapping, as can be seen by performing a Taylor expansion to first order. This fact ties the study of complex mappings closely to that of the conformal geometry of the plane (and more generally, of two-dimensional surfaces and domains). In particular, one can use complex analytic maps to conformally transform one two-dimensional domain to another, leading among other things to the famous Riemann mapping theorem, and to the classification of Riemann surfaces.

If one Taylor expands complex analytic maps to second order rather than first order, one discovers a further important property of these maps, namely that they are harmonic. This fact makes the class of complex analytic maps extremely rigid and well behaved analytically; indeed, the entire theory of elliptic PDE now comes into play, giving useful properties such as elliptic regularity and the maximum principle. In fact, due to the magic of residue calculus and contour shifting, we already obtain these properties for maps that are merely complex differentiable rather than complex analytic, which leads to the striking fact that complex differentiable functions are automatically analytic (in contrast to the real-variable case, in which real differentiable functions can be very far from being analytic).

The geometric structure of the complex numbers (and more generally of complex manifolds and complex varieties), when combined with the algebraic closure of the complex numbers, leads to the beautiful subject of *complex algebraic geometry*, which motivates the much more general theory developed in modern algebraic geometry. However, we will not develop the algebraic geometry aspects of complex analysis here.

Last, but not least, because of the good behaviour of Taylor series in the complex plane, complex analysis is an excellent setting in which to manipulate various generating functions, particularly Fourier series (which can be viewed as boundary values of power (or Laurent) series ), as well as Dirichlet series . The theory of contour integration provides a very useful dictionary between the asymptotic behaviour of the sequence , and the complex analytic behaviour of the Dirichlet or Fourier series, particularly with regard to its poles and other singularities. This turns out to be a particularly handy dictionary in analytic number theory, for instance relating the distribution of the primes to the Riemann zeta function. Nowadays, many of the analytic number theory results first obtained through complex analysis (such as the prime number theorem) can also be obtained by more “real-variable” methods; however the complex-analytic viewpoint is still extremely valuable and illuminating.

We will frequently touch upon many of these connections to other fields of mathematics in these lecture notes. However, these are mostly side remarks intended to provide context, and it is certainly possible to skip most of these tangents and focus purely on the complex analysis material in these notes if desired.

Note: complex analysis is a very visual subject, and one should draw plenty of pictures while learning it. I am however not planning to put too many pictures in these notes, partly as it is somewhat inconvenient to do so on this blog from a technical perspective, but also because pictures that one draws on one’s own are likely to be far more useful to you than pictures that were supplied by someone else.

** — 1. The construction and algebra of the complex numbers — **

Note: this section will be far more algebraic in nature than the rest of the course; we are concentrating almost all of the algebraic preliminaries in this section in order to get them out of the way and focus subsequently on the analytic aspects of the complex numbers.

Thanks to the laws of high-school algebra, we know that the real numbers are a field: it is endowed with the arithmetic operations of addition, subtraction, multiplication, and division, as well as the additive identity and multiplicative identity , that obey the usual laws of algebra (i.e. the field axioms).

The algebraic structure of the reals does have one drawback though – not all (non-trivial) polynomials have roots! Most famously, the polynomial equation has no solutions over the reals, because is always non-negative, and hence is always strictly positive, whenever is a real number.

As mentioned in the introduction, one traditional way to define the complex numbers is as the smallest possible extension of the reals that fixes this one specific problem:

Definition 1 (The complex numbers)Afield of complex numbersis a field that contains the real numbers as a subfield, as well as a root of the equation . (Thus, strictly speaking, a field of complex numbers is a pair , but we will almost always abuse notation and use as a metonym for the pair .) Furthermore, is generated by and , in the sense that there is no subfield of , other than itself, that contains both and ; thus, in the language of field extensions, we have .

(We will take the existence of the real numbers as a given in this course; constructions of the real number system can of course be found in many real analysis texts, including my own.)

Definition 1 is short, but proposing it as a definition of the complex numbers raises some immediate questions:

- (Existence) Does such a field even exist?
- (Uniqueness) Is such a field unique (up to isomorphism)?
- (Non-arbitrariness) Why the square root of ? Why not adjoin instead, say, a fourth root of , or the solution to some other algebraic equation? Also, could one iterate the process, extending further by adding more roots of equations?

The third set of questions can be answered satisfactorily once we possess the fundamental theorem of algebra. For now, we focus on the first two questions.

We begin with existence. One can construct the complex numbers quite explicitly and quickly using the Argand plane construction; see Remark 7 below. However, from the perspective of higher mathematics, it is more natural to view the construction of the complex numbers as a special case of the more general algebraic construction that can extend any field by the root of an irreducible nonlinear polynomial over that field; this produces a field of complex numbers when specialising to the case where and . We will just describe this construction in that special case, leaving the general case as an exercise.

Starting with the real numbers , we can form the space of (formal) polynomials

with real co-efficients and arbitrary non-negative integer in one indeterminate variable . (A small technical point: we do not view this indeterminate as belonging to any particular domain such as , so we do not view these polynomials as functions but merely as formal expressions involving a placeholder symbol (which we have rendered in Roman type to indicate its formal character). In this particular characteristic zero setting of working over the reals, it turns out to be harmless to identify each polynomial with the corresponding function formed by interpreting the indeterminate as a real variable; but if one were to generalise this construction to positive characteristic fields, and particularly finite fields, then one can run into difficulties if polynomials are not treated formally, due to the fact that two distinct formal polynomials might agree on all inputs in a given finite field (e.g. the polynomials and agree for all in the finite field ). However, this subtlety can be ignored for the purposes of this course.) This space of polynomials has a pretty good algebraic structure, in particular the usual operations of addition, subtraction, and multiplication on polynomials, together with the zero polynomial and the unit polynomial , give the structure of a (unital) commutative ring. This commutative ring also contains as a subring (identifying each real number with the degree zero polynomial ). The ring is however not a field, because many non-zero elements of do not have multiplicative inverses. (In fact, no non-constant polynomial in has an inverse in , because the product of two non-constant polynomials has a degree that is the sum of the degrees of the factors.)

If a unital commutative ring fails to be field, then it will instead possess a number of non-trivial ideals. The only ideal we will need to consider here is the principal ideal

This is clearly an ideal of – it is closed under addition and subtraction, and the product of any element of the ideal with an element of the full ring remains in the ideal .

We now define to be the quotient space

of the commutative ring by the ideal ; this is the space of cosets of in . Because is an ideal, there is an obvious way to define addition, subtraction, and multiplication in , namely by setting

and

for all ; these operations, together with the additive identity and the multiplicative identity , can be easily verified to give the structure of a commutative ring. Also, the real line embeds into by identifying each real number with the coset ; note that this identification is injective, as no real number is a multiple of the polynomial .

If we define to be the coset

then it is clear from construction that . Thus contains both and a solution of the equation . Also, since every element of is of the form for some polynomial , we see that every element of is a polynomial combination of with real coefficients; in particular, any subring of that contains and will necessarily have to contain every element of . Thus is generated by and .

The only remaining thing to verify is that is a field and not just a commutative ring. In other words, we need to show that every non-zero element of has a multiplicative inverse. This stems from a particular property of the polynomial , namely that it is irreducible in . That is to say, we cannot factor into non-constant polynomials

with . Indeed, as has degree two, the only possible way such a factorisation could occur is if both have degree one, which would imply that the polynomial has a root in the reals , which of course it does not.

Because the polynomial is irreducible, it is also prime: if divides a product of two polynomials in , then it must also divide at least one of the factors , . Indeed, if does not divide , then by irreducibility the greatest common divisor of and is . Applying the Euclidean algorithm for polynomials, we then obtain a representation of as

for some polynomials ; multiplying both sides by , we conclude that is a multiple of .

Since is prime, the quotient space is an integral domain: there are no zero-divisors in other than zero. This brings us closer to the task of showing that is a field, but we are not quite there yet; note for instance that is an integral domain, but not a field. But one can finish up by using finite dimensionality. As is a ring containing the field , it is certainly a vector space over ; as is generated by and , and , we see that it is in fact a two-dimensional vector space over , spanned by and (which are linearly independent, as clearly cannot be real). In particular, it is finite dimensional. For any non-zero , the multiplication map is an -linear map from this finite-dimensional vector space to itself. As is an integral domain, this map is injective; by finite-dimensionality, it is therefore surjective (by the rank-nullity theorem). In particular, there exists such that , and hence is invertible and is a field. This concludes the construction of a complex field .

Remark 2One can think of the action of passing from a ring to a quotient by some ideal as the action of forcing some relations to hold between the various elements of , by requiring all the elements of the ideal (or equivalently, all the generators of ) to vanish. Thus one can think of as the ring formed by adjoining a new element to the existing ring and then demanding the constraint . With this perspective, the main issues to check in order to obtain a complex field are firstly that these relations do not collapse the ring so much that two previously distinct elements of become equal, and secondly that all the non-zero elements become invertible once the relations are imposed, so that we obtain a field rather than merely a ring or integral domain.

Remark 3It is instructive to compare the complex field , formed by adjoining the square root of to the reals, with other commutative rings such as (which adjoins an additional square root of to the reals) or (which adjoins a new root of to the reals). The latter two objects are perfectly good rings, but are not fields (they contain zero divisors, and the first ring even contains a nilpotent). This is ultimately due to the reducible nature of the polynomials and in .

Uniqueness of up to isomorphism is a straightforward exercise:

Exercise 4 (Uniqueness up to isomorphism)Suppose that one has two complex fields and . Show that there is a unique field isomorphism that maps to and is the identity on .

Now that we have existence and uniqueness up to isomorphism, it is safe to designate one of the complex fields as *the* complex field; the other complex fields out there will no longer be of much importance in this course (or indeed, in most of mathematics), with one small exception that we will get to later in this section. One can, if one wishes, use the above abstract algebraic construction as the choice for “the” complex field , but one can certainly pick other choices if desired (e.g. the Argand plane construction in Remark 7 below). But in view of Exercise 4, the precise construction of is not terribly relevant for the purposes of actually doing complex analysis, much as the question of whether to construct the real numbers using Dedekind cuts, equivalence classes of Cauchy sequences, or some other construction is not terribly relevant for the purposes of actually doing real analysis. So, from here on out, we will no longer refer to the precise construction of used; the reader may certainly substitute his or her own favourite construction of in place of if desired, with essentially no change to the rest of the lecture notes.

Exercise 5Let be an arbitrary field, let be the ring of polynomials with coefficients in , and let be an irreducible polynomial in of degree at least two. Show that is a field containing an embedded copy of , as well as a root of the equation , and that this field is generated by and . Also show that all such fields are unique up to isomorphism. (This field is an example of a field extension of , the further study of which can be found in any advanced undergraduate or early graduate text on algebra, and is the starting point in particular for the beautiful topic of Galois theory, which we will not discuss here.)

Exercise 6Let be an arbitrary field. Show that every non-constant polynomial in can be factored as the product of irreducible non-constant polynomials. Furthermore show that this factorisation is unique up to permutation of the factors , and multiplication of each of the factors by a constant (with the product of all such constants being one). In other words: the polynomiail ring is a unique factorisation domain.

Remark 7 (Real and imaginary coordinates)As a complex field is spanned (over ) by the linearly independent elements and , we can writewith each element of having a unique representation of the form , thus

for real . The addition, subtraction, and multiplication operations can then be written down explicitly in these coordinates as

and with a bit more work one can compute the division operation as

if . One could take these coordinate representations as the

definitionof the complex field and its basic arithmetic operations, and this is indeed done in many texts introducing the complex numbers. In particular, one could take the Argand plane as the choice of complex field, where we identify each point in with (so for instance becomes endowed with the multiplication operation ). This is a very concrete and direct way to construct the complex numbers; the main drawback is that it is not immediately obvious that the field axioms are all satisfied. For instance, the associativity of multiplication is rather tedious to verify in the coordinates of the Argand plane. In contrast, the more abstract algebraic construction of the complex numbers given above makes it more evident what the source of the field structure on is, namely the irreducibility of the polynomial .

Remark 8Because of the Argand plane construction, we will sometimes refer to the space of complex numbers as thecomplex plane. We should warn, though, that in some areas of mathematics, particularly in algebraic geometry, is viewed as a one-dimensional complex vector space (or a one-dimensional complex manifold or complex variety), and so is sometimes referred to in those cases as acomplex line. (Similarly, Riemann surfaces, which from a real point of view are two-dimensional surfaces, can sometimes be referred to ascomplex curvesin the literature; the modular curve is a famous instance of this.) In this current course, though, the topological notion of dimension turns out to be more important than the algebraic notions of dimension, and as such we shall generally refer to as a plane rather than a line.

Elements of of the form for real are known as *purely imaginary numbers*; the terminology is colourful, but despite the name, imaginary numbers have precisely the same first-class mathematical object status as real numbers. If is a complex number, the real components of are known as the *real part* and *imaginary part* of respectively. Complex numbers that are not real are occasionally referred to as *strictly complex numbers*. In the complex plane, the set of real numbers forms the *real axis*, and the set of imaginary numbers forms the *imaginary axis*. Traditionally, elements of are denoted with symbols such as , , or , while symbols such as are typically intended to represent real numbers instead.

Remark 9We noted earlier that the equation had no solutions in the reals because was always positive. In other words, the properties of the order relation on prevented the existence of a root for the equation . As does have a root for , this means that the complex numbers cannot be ordered in the same way that the reals are ordered (that is to say, being totally ordered, with the positive numbers closed under both addition and multiplication). Indeed, one usually refrains from putting any order structure on the complex numbers, so that statements such as for complex numbers are left undefined (unless are real, in which case one can of course use the real ordering). In particular, the complex number is considered to be neither positive nor negative, and an assertion such as is understood to implicitly carry with it the claim that are real numbers and not just complex numbers. (Of course, if one really wants to, one can find some total orderings to place on , e.g. lexicographical ordering on the real and imaginary parts. However, such orderings do not interact too well with the algebraic structure of and are rarely useful in practice.)

As with any other field, we can raise a complex number to a non-negative integer by declaring inductively and for ; in particular we adopt the usual convention that (when thinking of the base as a complex number, and the exponent as a non-negative integer). For negative integers , we define for non-zero ; we leave undefined when is zero and is negative. At the present time we do not attempt to define for any exponent other than an integer; we will return to such exponentiation operations later in the course, though we will at least define the complex exponential for any complex later in this set of notes.

By definition, a complex field is a field together with a root of the equation . But if is a root of the equation , then so is (indeed, from the factorisation we see that these are the only two roots of this quadratic equation. Thus we have another complex field which differs from only in the choice of root . By Exercise 4, there is a unique field isomorphism from to that maps to (i.e. a complex field isomorphism from to ); this operation is known as complex conjugation and is denoted . In coordinates, we have

Being a field isomorphism, we have in particular that

and

for all complex numbers . It is also clear that complex conjugation fixes the real numbers, and only the real numbers: if and only if is real. Geometrically, complex conjugation is the operation of reflection in the complex plane across the real axis. It is clearly an involution in the sense that it is its own inverse:

One can also relate the real and imaginary parts to complex conjugation via the identities

Remark 10Any field automorphism of has to map to a root of , and so the only field automorphisms of that preserve the real line are the identity map and the conjugation map; conversely, the real line is the subfield of fixed by both of these automorphisms. In the language of Galois theory, this means that is a Galois extension of , withGalois groupconsisting of two elements. There is a certain sense in which one can think of the complex numbers (or more precisely, the scheme of complex numbers) as a double cover of the real numbers (or more precisely, the scheme of real numbers), analogous to how the boundary of a Möbius strip can be viewed as a double cover of the unit circle formed by shrinking the width of the strip to zero. (In this analogy, points on the unit circle correspond to specific models of the real number system , and lying above each such point are two specific models , of the complex number system; this analogy can be made precise using Grothendieck’s “functor of points” interpretation of schemes.) The operation of complex conjugation is then analogous to the operation of monodromy caused by looping once around the base unit circle, causing the two complex fields sitting above a real field to swap places with each other. (This analogy is not quite perfect, by the way, because the boundary of a Möbius strip is not simply connected and can in turn be finitely covered by other curves, whereas the complex numbers are algebraically complete and admit no further finite extensions; one should really replace the unit circle here by something with a two-element fundamental group, such as the projective plane that is double covered by the sphere , but this is harder to visualize.) The analogy between (absolute) Galois groups and fundamental groups suggested by this picture can be made precise in scheme theory by introducing the concept of an étale fundamental group, which unifies the two concepts, but developing this further is well beyond the scope of this course; see this book of Szamuely for further discussion.

Observe that if we multiply a complex number by its complex conjugate , we obtain a quantity which is invariant with respect to conjugation (i.e. ) and is therefore real. The map produced this way is known in field theory as the norm form of over ; it is clearly multiplicative in the sense that , and is only zero when is zero. It can be used to link multiplicative inversion with complex conjugation, in that we clearly have

for any non-zero complex number . In coordinates, we have

(thus recovering, by the way, the inversion formula implicit in Remark 7). In coordinates, the multiplicativity takes the form of Lagrange’s identity

** — 2. The geometry of the complex numbers — **

The norm form of the complex numbers has the feature of being positive definite: is always non-negative (and strictly positive when is non-zero). This is a feature that is somewhat special to the complex numbers; for instance, the quadratic extension of the rationals by has the norm form , which is indefinite. One can view this positive definiteness of the norm form as the one remaining vestige in of the order structure on the reals, which as remarked previously is no longer present directly in the complex numbers. (One can also view the positive definiteness of the norm form as a consequence of the topological connectedness of the punctured complex plane : the norm form is positive at , and cannot change sign anywhere in , so is forced to be positive on the rest of this connected region.)

One consequence of positive definiteness is that the bilinear form

becomes a positive definite inner product on (viewed as a vector space over ). In particular, this turns the complex numbers into an inner product space over the reals. From the usual theory of inner product spaces, we can then construct a norm

(thus, the norm is the square root of the norm form) which obeys the triangle inequality

(which implies the usual permutations of this inequality, such as ), and from the multiplicativity of the norm form we also have

(and hence also if is non-zero) and from the involutive nature of complex conjugation we have

The norm clearly extends the absolute value operation on the real numbers, and so we also refer to the norm of a complex number as its *absolute value* or *magnitude*. In coordinates, we have

thus for instance , and from (6) we also immediately have the useful inequalities

As with any other normed vector space, the norm defines a metric on the complex numbers via the definition

Note that using the Argand plane representation of as that this metric coincides with the usual Euclidean metric on . This metric in turn defines a topology on (generated in the usual manner by the open disks ), which in turn generates all the usual topological notions such as the concept of an open set, closed set, compact set, connected set, and boundary of a set; the notion of a limit of a sequence ; the notion of a continuous map, and so forth. For instance, a sequence of complex numbers converges to a limit if as , and a map is continuous if one has whenever , or equivalently if the inverse image of any open set is open. Again, using the Argand plane representation, these notions coincide with their counterparts on the Euclidean plane .

As usual, if a sequence of complex numbers converges to a limit , we write . From the triangle inequality (3) and the multiplicativity (4) we see that the addition operation , subtraction operation , and multiplication operation , thus we have the familiar limit laws

and

whenever the limits on the right-hand side exist. Similarly, from (5) we see that complex conjugation is an isometry of the complex numbers, thus

when the limit on the right-hand side exists. As a consequence, the norm form and the absolute value are also continuous, thus

whenever the limit on the right-hand side exists. Using the formula (2) for the reciprocal of a complex number, we also see that division is a continuous operation as long as the denominator is non-zero, thus

as long as the limits on the right-hand side exist, and the limit in the denominator is non-zero.

From (7) we see that

in particular

and

whenever the limit on the right-hand side exists. One consequence of this is that is complete: every sequence of complex numbers that is a Cauchy sequence (thus as ) converges to a unique complex limit . (As such, one can view the complex numbers as a (very small) example of a Hilbert space.)

As with the reals, we have the fundamental fact that any formal series of complex numbers which is absolutely convergent, in the sense that the non-negative series is finite, is necessarily convergent to some complex number , in the sense that the partial sums converge to as . This is because the triangle inequality ensures that the partial sums are a Cauchy sequence. As usual we write to denote the assertion that is the limit of the partial sums . We will occasionally have need to deal with series that are only conditionally convergent rather than absolutely convergent, but in most of our applications the only series we will actually evaluate are the absolutely convergent ones. Many of the limit laws imply analogues for series, thus for instance

whenever the series on the right-hand side is absolutely convergent (or even just convergent). We will not write down an exhaustive list of such series laws here.

An important role in complex analysis is played by the *unit circle*

In coordinates, this is the set of points for which , and so this indeed has the geometric structure of a unit circle. Elements of the unit circle will be referred to in these notes as *phases*. Every non-zero complex number has a unique polar decomposition as where is a positive real and lies on the unit circle . Indeed, it is easy to see that this decomposition is given by and , and that this is the only polar decopmosition of . We refer to the polar components and of a non-zero complex number as the *magnitude* and *phase* of respectively.

From (4) we see that the unit circle is a multiplicative group; it contains the multiplicative identity , and if lie in , then so do and . From (2) we see that reciprocation and complex conjugation agree on the unit circle, thus

for . It is worth emphasising that this useful identity does *not* hold as soon as one leaves the unit circle, in which case one must use the more general formula (2) instead! If are non-zero complex numbers with polar decompositions and respectively, then clearly the polar decompositions of and are given by and respectively. Thus polar coordinates are very convenient for performing complex multiplication, although they turn out to be atrocious for performing complex addition. (This can be contrasted with the usual Cartesian coordinates , which are very convenient for performing complex addition and mildly inconvenient for performing complex multiplication.) In the language of group theory, the polar decomposition splits the multiplicative complex group as the direct product of the positive reals and the unit circle : .

If is an element of the unit circle , then from (4) we see that the operation of multiplication by is an isometry of , in the sense that

for all complex numbers . This isometry also preserves the origin . As such, it is geometrically obvious (see Exercise 11 below) that the map must either be a rotation around the origin, or a reflection around a line. The former operation is orientation preserving, and the latter is orientation reversing. Since the map is clearly orientation preserving when , and the unit circle is connected, a continuity argument shows that must be orientation preserving for all , and so must be a rotation around the origin by some angle. Of course, by trigonometry, we may write

for some real number . The rotation clearly maps the number to the number , and so the rotation must be a counter-clockwise rotation by (adopting the usual convention of placing to the right of the origin and above it). In particular, when applying this rotation to another point on the unit circle, this point must get rotated to . We have thus given a geometric proof of the multiplication formula

taking real and imaginary parts, we recover the familiar trigonometric addition formulae

We can also iterate the multiplication formula to give de Moivre’s formula

for any natural number (or indeed for any integer ), which can in turn be used to recover familiar identities such as the double angle formulae

or triple angle formulae

after expanding out de Moivre’s formula for or and taking real and imaginary parts.

- (i) Let be an isometry of the Euclidean plane that fixes the origin . Show that is either a rotation around the origin by some angle , or the reflection around some line through the origin. (
Hint:try to compose with rotations or reflections to achieve some normalisation of , e.g. that fixes . Then consider what must do to other points in the plane, such as . Alternatively, one can use various formulae relating distances to angle, such as the sine rule or cosine rule, or the formula for the inner product.) For this question, you may use any result you already know from Euclidean geometry or trigonometry.- (ii) Show that all isometries of the complex numbers take the form
or

for some complex number and phase .

Every non-zero complex number can now be written in polar form as

with and ; we refer to as an argument of , and can be interpreted as an angle of counterclockwise rotation needed to rotate the positive real axis to a position that contains . The argument is not quite unique, due to the periodicity of sine and cosine: if is an argument of , then so is for any integer , and conversely these are all the possible arguments that can have. The set of all such arguments will be denoted ; it is a coset of the discrete group , and can thus be viewed as an element of the -torus .

The operation of multiplying a complex number by a given non-zero complex number now has a very appealing geometric interpretation when expressing in polar coordinates (9): this operation is the composition of the operation of dilation by around the origin, and counterclockwise rotation by around the origin. For instance, multiplication by performs a counter-clockwise rotation by around the origin, while multiplication by performs instead a clockwise rotation by . As complex multiplication is commutative and associative, it does not matter in which order one performs the dilation and rotation operations. Similarly, using Cartesian coordinates, we see that the operation of adding a complex number by a given complex number is simply a spatial *translation* by a displacement of . The multiplication operation need not be isometric (due to the presence of the dilation ), but observe that both the addition and multiplication operations are conformal (angle-preserving) and also orientation-preserving (a counterclockwise loop will transform to another counterclockwise loop, and similarly for clockwise loops). As we shall see later, these conformal and orientation-preserving properties of the addition and multiplication maps will extend to the larger class of *complex differentiable maps* (at least outside of critical points of the map), and are an important aspect of the geometry of such maps.

Remark 12One can also interpret the operations of complex arithmetic geometrically on the Argand plane as follows. As the addition law on coincides with the vector addition law on , addition and subtraction of complex numbers is given by the usual parallelogram rule for vector addition; thus, to add a complex number to another , we can translate the complex plane until the origin gets mapped to , and then gets mapped to ; conversely, subtraction by corresponds to translating back to . Similarly, to multiply a complex number with another , we can dilate and rotate the complex plane around the origin until gets mapped to , and then will be mapped to ; conversely, division by corresponds to dilating and rotating back to .

When performing computations, it is convenient to restrict the argument of a non-zero complex number to lie in a fundamental domain of the -torus , such as the half-open interval or , in order to recover a unique parameterisation (at the cost of creating a branch cut at one point of the unit circle). Traditionally, the fundamental domain that is most often used is the half-open interval . The unique argument of that lies in this interval is called the *standard argument* of and is denoted , and is called the *standard branch* of the argument function. Thus for instance , , , and . Observe that the standard branch of the argument has a discontinuity on the negative real axis , which is the *branch cut* of this branch. Changing the fundamental domain used to define a branch of the argument can move the branch cut around, but cannot eliminate it completely, due to non-trivial monodromy (if one continuously loops once counterclockwise around the origin, and varies the argument continuously as one does so, the argument will increment by , and so no branch of the argument function can be continuous at every point on the loop).

The multiplication formula (8) resembles the multiplication formula

for the real exponential function . The two formulae can be unified through the famous Euler formula involving the complex exponential . There are many ways to define the complex exponential. Perhaps the most natural is through the ordinary differential equation with boundary condition . However, as we have not yet set up a theory of complex differentiation, we will proceed (at least temporarily) through the device of Taylor series. Recalling that the real exponential function has the Taylor expansion

which is absolutely convergent for any real , one is led to define the complex exponential function by the analogous expansion

noting from (4) that the absolute convergence of the real exponential for any implies the absolute convergence of the complex exponential for any . We also frequently write for . The multiplication formula (10) for the real exponential extends to the complex exponential:

Exercise 13Use the binomial theorem and Fubini’s theorem for (complex) doubly infinite series to conclude that

If one compares the Taylor series for with the familiar Taylor expansions

and

for the (real) sine and cosine functions, one obtains Euler formula

for any real number ; in particular we have the famous identities

We now see that the multiplication formula (8) can be written as a special form

of (12); similarly, de Moivre’s formula takes the simple and intuitive form

From (12) and (13) we also see that the exponential function basically transforms Cartesian coordinates to polar coordinates:

Later on in the course we will study (the various branches of) the logarithm function that inverts the complex exponential, thus converting polar coordinates back to Cartesian ones.

From (13) and (1), together with the easily verified identity

we see that we can recover the trigonometric functions from the complex exponential by the formulae

(Indeed, if one wished, one could take these identities as the *definition* of the sine and cosine functions, giving a purely analytic way to construct these trigonometric functions.) From these identities one can derive all the usual trigonometric identities from the basic properties of the exponential (and in particular (12)). For instance, using a little bit of high school algebra we can prove the familiar identity

from (16):

Thus, in principle at least, one no longer has a need to memorize all the different trigonometric identities out there, since they can now all be unified as consequences of just a handful of basic identities for the complex exponential, such as (12), (14), and (15).

In view of (16), it is now natural to introduce the complex sine and cosine functions and by the formula

These complex trigonometric functions no longer have a direct trigonometric interpretation (as one cannot easily develop a theory of complex angles), but they still inherit almost all of the algebraic properties of their real-variable counterparts. For instance, one can repeat the above high school algebra computations *verbatim* to conclude that

for all . (We caution however that this does *not* imply that and are bounded in magnitude by – note carefully the lack of absolute value signs outside of and in the above formula! See also Exercise 16 below.) Similarly for all of the other trigonometric identities. (Later on in this series of lecture notes, we will develop the concept of analytic continuation, which can explain why so many real-variable algebraic identities naturally extend to their complex counterparts.) From (11) we see that the complex sine and cosine functions have the same Taylor series expansion as their real-variable counterparts, namely

and

The formulae (17) for the complex sine and cosine functions greatly resemble those of the hyperbolic trigonometric functions , defined by the formulae

Indeed, if we extend these functions to the complex domain by defining to be the functions

then on comparison with (17) we obtain the complex identities

for all complex . Thus we see that once we adopt the perspective of working over the complex numbers, the hyperbolic trigonometric functions are “rotations by 90 degrees” of the ordinary trigonometric functions; this is a simple example of what physicists call a Wick rotation. In particular, we see from these identities that any trigonometric identity will have a hyperbolic counterpart, though due to the presence of various factors of , the signs may change as one passes from trigonometric to hyperbolic functions or vice versa (a fact quantified by Osborne’s rule). For instance, by substituting (19) or (20) into (18) (and replacing by or as appropriate), we end up with the analogous identity

for the hyperbolic trigonometric functions. Similarly for all other trigonometric identities. Thus we see that the complex exponential single-handedly unites the trigonometry, hyperbolic trigonometry, and the real exponential function into a single coherent theory!

Exercise 14

- (i) If is a positive integer, show that the only complex number solutions to the equation are given by the complex numbers for ; these numbers are thus known as the roots of unity. Conclude the identity for any complex number .
- (ii) Show that the only compact subgroups of the multiplicative complex numbers are the unit circle and the roots of unity
for . (Hint: there are two cases, depending on whether is a limit point of or not.)

- (iii) Given an example of a non-compact subgroup of .
- (iv) (Warning: this one is tricky.) Show that the only connected closed subgroups of are the whole group , the trivial group , and the one-parameter groups of the form for some non-zero complex number .

The next exercise gives a special case of the fundamental theorem of algebra, when considering the roots of polynomials of the specific form .

Exercise 15Show that if is a non-zero complex number and is a positive integer, then there are exactly distinct solutions to the equation , and any two such solutions differ (multiplicatively) by an root of unity. In particular, a non-zero complex number has two square roots, each of which is the negative of the other. What happens when ?

Exercise 16Let be a sequence of complex numbers. Show that is bounded if and only if the imaginary part of is bounded, and similarly with replaced by .

Exercise 17(This question was drawn from a previous version of this course taught by Rowan Killip.) Let be distinct complex numbers, and let be a positive real that is not equal to .

- (i) Show that the set defines a circle in the complex plane. (Ideally, you should be able to do this without breaking everything up into real and imaginary parts.)
- (ii) Conversely, show that every circle in the complex plane arises in such a fashion (for suitable choices of , of course).
- (iii) What happens if ?
- (iv) Let be a circle that does not pass through the origin. Show that the image of under the inversion map is a circle. What happens if is a line? What happens if the passes through the origin (and one then deletes the origin from before applying the inversion map)?

Filed under: 246A - complex analysis, math.CV, math.RA Tagged: complex numbers, exponentiation, trigonometry ]]>

Filed under: 246A - complex analysis, admin Tagged: announcement ]]>

for the divisor function , in particular recovering the calculation of Ingham that obtained the asymptotic

when was fixed and non-zero and went to infinity. It is natural to consider the more general correlations

for fixed and non-zero , where

is the order divisor function. The sum (1) then corresponds to the case . For , , and a routine application of the Dirichlet hyperbola method (or Perron’s formula) gives the asymptotic

or more accurately

where is a certain explicit polynomial of degree with leading coefficient ; see e.g. Exercise 31 of this previous post for a discussion of the case (which is already typical). Similarly if . For more general , there is a conjecture of Conrey and Gonek which predicts that

for some polynomial of degree which is explicit but whose form is rather complicated (one has to compute residues of a various complicated products of zeta functions and local factors). This conjecture has been verified when or , by the work of Linnik, Motohashi, Fouvry-Tenenbaum, and others, but all the remaining cases when are currently open.

In principle, the calculations of the previous post should recover the predictions of Conrey and Gonek. In this post I would like to record this for the top order term:

Conjecture 1If and are fixed, thenas , where the product is over all primes , and the local factors are given by the formula

where is the degree polynomial

where

and one adopts the conventions that and for .

For instance, if then

and hence

and the above conjecture recovers the Ingham formula (2). For , we have

and so we predict

where

Similarly, if we have

and so we predict

where

and so forth.

As in the previous blog, the idea is to factorise

where the local factors are given by

(where means that divides precisely times), or in terms of the valuation of at ,

We then have the following exact local asymptotics:

Proposition 2 (Local correlations)Let be a profinite integer chosen uniformly at random, let be a profinite integer, and let . Then

(For profinite integers it is possible that and hence are infinite, but this is a probability zero event and so can be ignored.)

Conjecture 1 can then be heuristically justified from the local calculations (2) by various pseudorandomness heuristics, as discussed in the previous post.

I’ll give a short proof of the above proposition below, basically using the recursive methods of the previous post. This short proof actually took be quite a while to find; I spent several hours and a fair bit of scratch paper working out the cases laboriously by hand (with some assistance and cross-checking from Maple). Here is an unorganised sample of some of this scratch, just to show how the sausage is actually made:

It was only after expending all this effort that I realised that it would be much more efficient to compute the correlations for all values of simultaneously by using generating functions. After performing this computation, it then became apparent that there would be a direct combinatorial proof of (6) that was shorter than even the generating function proof. (I will not supply the full generating function calculations here, but will at least show them for the easier correlation (5).)

I am confident that Conjecture 1 is consistent with the explicit asymptotic in the Conrey-Gonek conjecture, but have not yet rigorously established that the leading order term in the latter is indeed identical to the expression provided above.

We now prove (5). Here we will use the generating function method. From the binomial formula we have the formal expansion

for any . By the geometric series formula, it thus suffices to show that

in the ring of formal power series in .

With probability , is coprime to and thus . In the remaining probability event, for some with the same distribution as the original random variable , and . This leads to the identity

and the claim then follows from a brief amount of high-school algebra.

Now we prove (6). By (3), our task is to show that

The above generating function method will establish this (and, as I said above, this was how I originally discovered the formula), but we give here a combinatorial proof. We first deal with the case when is not divisible by . In this case, at least one of and must equal ; we can phrase this fact algebraically as the identity

By linearity (and translation invariance) of expectation and (5) we thus have

which gives (7) in this case.

Now suppose that for some (profinite) integer . With probability , is coprime to , so that ; in particular (8) holds here. In the remaining probability event, we can write . From (4) and Pascal triangle identities, we have

and similarly

and hence on subtracting from both equations, multiplying, and rearranging we have a variant of (8):

Putting this together, we see that

and hence by (5) as before

The claim is now easily verified from an induction on , as well as many applications of Pascal triangle identities.

Filed under: expository, math.NT, math.PR Tagged: correlation, divisor function, generating function ]]>

where denotes the estimate as . Much better error estimates are possible here, but we will not focus on the lower order terms in this discussion. For somewhat idiosyncratic reasons I will interpret this estimate (and the other analytic number theory estimates discussed here) through the probabilistic lens. Namely, if is a random number selected uniformly between and , then the above estimate can be written as

that is to say the random variable has mean approximately . (But, somewhat paradoxically, this is not the median or mode behaviour of this random variable, which instead concentrates near , basically thanks to the Hardy-Ramanujan theorem.)

Now we turn to the pair correlations for a fixed positive integer . There is a classical computation of Ingham that shows that

The error term in (2) has been refined by many subsequent authors, as has the uniformity of the estimates in the aspect, as these topics are related to other questions in analytic number theory, such as fourth moment estimates for the Riemann zeta function; but we will not consider these more subtle features of the estimate here. However, we will look at the next term in the asymptotic expansion for (2) below the fold.

Using our probabilistic lens, the estimate (2) can be written as

From (1) (and the asymptotic negligibility of the shift by ) we see that the random variables and both have a mean of , so the additional factor of represents some arithmetic coupling between the two random variables.

Ingham’s formula can be established in a number of ways. Firstly, one can expand out and use the hyperbola method (splitting into the cases and and removing the overlap). If one does so, one soon arrives at the task of having to estimate sums of the form

for various . For much less than this can be achieved using a further application of the hyperbola method, but for comparable to things get a bit more complicated, necessitating the use of non-trivial estimates on Kloosterman sums in order to obtain satisfactory control on error terms. A more modern approach proceeds using automorphic form methods, as discussed in this previous post. A third approach, which unfortunately is only heuristic at the current level of technology, is to apply the Hardy-Littlewood circle method (discussed in this previous post) to express (2) in terms of exponential sums for various frequencies . The contribution of “major arc” can be computed after a moderately lengthy calculation which yields the right-hand side of (2) (as well as the correct lower order terms that are currently being suppressed), but there does not appear to be an easy way to show directly that the “minor arc” contributions are of lower order, although the methods discussed previously do indirectly show that this is ultimately the case.

Each of the methods outlined above requires a fair amount of calculation, and it is not obvious while performing them that the factor will emerge at the end. One can at least explain the as a normalisation constant needed to balance the factor (at a heuristic level, at least). To see this through our probabilistic lens, introduce an independent copy of , then

using symmetry to order (discarding the diagonal case ) and making the change of variables , we see that (4) is heuristically consistent with (3) as long as the asymptotic mean of in is equal to . (This argument is not rigorous because there was an implicit interchange of limits present, but still gives a good heuristic “sanity check” of Ingham’s formula.) Indeed, if denotes the asymptotic mean in , then we have (heuristically at least)

and we obtain the desired consistency after multiplying by .

This still however does not explain the presence of the factor. Intuitively it is reasonable that if has many prime factors, and has a lot of factors, then will have slightly more factors than average, because any common factor to and will automatically be acquired by . But how to quantify this effect?

One heuristic way to proceed is through analysis of local factors. Observe from the fundamental theorem of arithmetic that we can factor

where the product is over all primes , and is the local version of at (which in this case, is just one plus the –valuation of : ). Note that all but finitely many of the terms in this product will equal , so the infinite product is well-defined. In a similar fashion, we can factor

where

(or in terms of valuations, ). Heuristically, the Chinese remainder theorem suggests that the various factors behave like independent random variables, and so the correlation between and should approximately decouple into the product of correlations between the local factors and . And indeed we do have the following local version of Ingham’s asymptotics:

Proposition 1 (Local Ingham asymptotics)For fixed and integer , we haveand

From the Euler formula

we see that

and so one can “explain” the arithmetic factor in Ingham’s asymptotic as the product of the arithmetic factors in the (much easier) local Ingham asymptotics. Unfortunately we have the usual “local-global” problem in that we do not know how to rigorously derive the global asymptotic from the local ones; this problem is essentially the same issue as the problem of controlling the minor arc contributions in the circle method, but phrased in “physical space” language rather than “frequency space”.

Remark 2The relation between the local means and the global mean can also be seen heuristically through the applicationof Mertens’ theorem, where is PÃ³lya’s magic exponent, which serves as a useful heuristic limiting threshold in situations where the product of local factors is divergent.

Let us now prove this proposition. One could brute-force the computations by observing that for any fixed , the valuation is equal to with probability , and with a little more effort one can also compute the joint distribution of and , at which point the proposition reduces to the calculation of various variants of the geometric series. I however find it cleaner to proceed in a more recursive fashion (similar to how one can prove the geometric series formula by induction); this will also make visible the vague intuition mentioned previously about how common factors of and force to have a factor also.

It is first convenient to get rid of error terms by observing that in the limit , the random variable converges vaguely to a uniform random variable on the profinite integers , or more precisely that the pair converges vaguely to . Because of this (and because of the easily verified uniform integrability properties of and their powers), it suffices to establish the exact formulae

in the profinite setting (this setting will make it easier to set up the recursion).

We begin with (5). Observe that is coprime to with probability , in which case is equal to . Conditioning to the complementary probability event that is divisible by , we can factor where is also uniformly distributed over the profinite integers, in which event we have . We arrive at the identity

As and have the same distribution, the quantities and are equal, and (5) follows by a brief amount of high-school algebra.

We use a similar method to treat (6). First treat the case when is coprime to . Then we see that with probability , and are simultaneously coprime to , in which case . Furthermore, with probability , is divisible by and is not; in which case we can write as before, with and . Finally, in the remaining event with probability , is divisible by and is not; we can then write , so that and . Putting all this together, we obtain

and the claim (6) in this case follows from (5) and a brief computation (noting that in this case).

Now suppose that is divisible by , thus for some integer . Then with probability , and are simultaneously coprime to , in which case . In the remaining event, we can write , and then and . Putting all this together we have

which by (5) (and replacing by ) leads to the recursive relation

and (6) then follows by induction on the number of powers of .

The estimate (2) of Ingham was refined by Estermann, who obtained the more accurate expansion

for certain complicated but explicit coefficients . For instance, is given by the formula

where is the Euler-Mascheroni constant,

The formula for is similar but even more complicated. The error term was improved by Heath-Brown to ; it is conjectured (for instance by Conrey and Gonek) that one in fact has square root cancellation here, but this is well out of reach of current methods.

These lower order terms are traditionally computed either from a Dirichlet series approach (using Perron’s formula) or a circle method approach. It turns out that a refinement of the above heuristics can also predict these lower order terms, thus keeping the calculation purely in physical space as opposed to the “multiplicative frequency space” of the Dirichlet series approach, or the “additive frequency space” of the circle method, although the computations are arguably as messy as the latter computations for the purposes of working out the lower order terms. We illustrate this just for the term below the fold.

** â€” 1. More refined heuristics â€” **

To begin with we revisit the heuristic justification of (2) and modify it a bit, in the spirit of the “modified CramÃ©r models” from this previous blog post. Our arguments here will be even less rigorous than in the preceding section, for instance we shall be quite cavalier with the use of the symbol without precisely quantifying error terms.

For this section, it will be convenient to set not to be the uniform distribution from to as before, but rather the uniform distribution from to for some fairly small (e.g. will suffice). The reason for this is that we need to start keeping track of the magnitude of to a relative accuracy of at least , and with this new choice of the magnitude is simply (up to this accuracy). By “differentiating” asymptotics such as (7) in we see that

where we ignore terms of lower order than the term and

and so our task is now to heuristically justify the statement

Let us begin by revisiting the derivation of the simpler asymptotic

without passing immediately to local factors, but instead using a modified CramÃ©r type model that will be more amenable to keeping careful track of the magnitude . For any square-free , write

and

so that we have the factorisation . From the Chinese remainder theorem we expect and to behave approximately independently, at least when is fixed and is large.

We can statistics of to deduce statistics of . This is traditionally done using the device of MÃ¶bius inversion, but we will use a recursive probabilistic approach here similar to that used to prove Proposition 1. Let’s start with understanding for some prime . With probability about , is coprime to , so that . In the remaining event of probability about , we may write where is uniformly distributed near , and then . This leads to the approximate identity

To work with this equation we make two further heuristic assumptions, which turn out to be reasonable at the level of accuracy required for justifying (10), but not for (9). Firstly, we assume that the distribution of is largely unchanged after conditioning to the event , which is plausible from the Chinese remainder theorem. Secondly, we assume that the distribution of is largely the same as that of , and similarly for . With these assumptions, (11) simplifies to

(note that this is consistent with Proposition 1 and the presumed independence of and ). Iterating this argument gives

for any fixed squarefree . Applying (1) we conclude

Now let be a moderately large number (growing very slowly with , e.g. ), and let be the primorial of . We factor

There is arithmetic coupling between the and factors; indeed from Proposition 1 and the Chinese remainder theorem we have

Meanwhile, from (13) we have

and

In analogy with the modified CramÃ©r model for the primes, we now assume that the random variables , , and behave like independent random variables; this turns out to be sufficiently accurate for predicting (10), but not (9). This leads to the prediction

and if goes to infinity with then we recover (10).

We now repeat the above analysis but expanding to the next order in , which will require keeping better track of the parameter. The Dirichlet hyperbola method actually gives the refinement

and hence

We previously computed the expectation . Now that we want to be more accurate, it turns out that it is better to compute the slightly different *conditional expectation* . Now we compute the mean of a bit more carefully. Again we begin with for prime. From (11) and (16), and by replacing with , we have

Previously we simplified this equation by assuming that the conditioning had negligible impact on , and that had similar distribution to . As it turns out, we can no longer afford to assume these simplifying assumptions at this level of accuracy, and must argue more carefully. With probability about is coprime to , and in the remaining event of probability about , we have and . This gives the approximate identity

If we refine the approximation (12) to give the ansatz

and

for some constants to be worked out momentarily, the above two equations give after some computation (and canceling out all the terms

which can be solved as

Thus we have

It is instructive (albeit lengthy) to deduce this asymptotic rigorously for fixed square-free and large by using MÃ¶bius inversion and the hyperbola method; one can also proceed a bit more quickly by using Perron’s formula and inspecting the pole of the relevant Dirichlet series at . (Not surprisingly, similar calculations appear when computing major arc contributions to (2).)

Now we are ready to predict the value of . Our starting point is again the expansion (14), except that we now strip out the factors within that divide , in order to use (19). More precisely, we adopt the factorisation of arbitrary natural numbers by the formulae

and observe that

and similarly

We are now faced with computing the expression

Previously we had assumed that the factors , , and behaved independently. However this is not quite true at the current level of accuracy, because the magnitude of the natural numbers and are influenced by the size of and respectively, and these in turn are certainly correlated with . Because of the non-trivial dependence on on the right-hand side of (19) at this level of accuracy, the size of and influences the expected value of and , causing a coupling between the three factors of (20). However, if we first condition the magnitude of and to be fixed, then we expect the three factors to be *conditionally* independent (at this level of accuracy). Heuristically, with this conditioning, behaves like a random element of size about that is coprime to , and hence by (19)

and similarly

We can thus heuristically estimate (20) by

This can be expanded as

where

and is something complicated that we do not compute here. Using (15) we have

which recovers the Ingham estimate (10). To obtain the finer approximation (9) we need to show that

It will suffice to obtain the asymptotic

together with the analogue of (21) with replaced by . (One in fact has the approximation , which reflects the Laurent approximation near , but we will not need this.)

We will just prove (21), as the analogue of (21) for is proven similarly. As this is a local calculation we may replace by . Splitting

we can write the left-hand side of (21) as a sum over primes,

The and terms decouple, so by Proposition 1 this becomes

Extracting a common factor of , we reduce to showing that

The left-hand side is a sum over primes , so we work on making the right-hand side a sum over primes as well. This is achieved by the following two lemmas:

Lemma 3We have .

*Proof:* While this identity can be proven elementarily, it is fastest to proceed using Dirichlet series. From Taylor expansion and (8) we have

for close to . Inverting, we have

and hence

The left-hand side is , giving the claim.

Lemma 4We have .

*Proof:* Again it is fastest to proceed using Dirichlet series. Setting and , we have from the fundamental theorem of arithmetic that

Differentiating this at we obtain the claim.

We remark that is an example of a derived multiplicative function, discussed in this previous blog post.

In view of these expansions, we can reduce matters to establishing a purely local estimate at a single prime , namely that

(compare with Proposition 1).

We need two preliminary estimates to handle lower order terms:

*Proof:* Once again we use recursion. With probability , is coprime to , so vanishes. Otherwise, , , and . We conclude that

From Proposition 1 we have

and from the identity we thus have

and the claim follows.

Note that one can view Lemma 5 as the (p-adic) limiting case of Lemma 6.

*Proof:* First suppose that is coprime to . Then is equal to whenever is non-zero, so the claim follows from (23) in this case. Now suppose that for some integer .

When is coprime to the left-hand side is equal to when is coprime to and otherwise, giving the claim. Now suppose that for some integer . With probability , is coprime to , so vanishes; otherwise,, , and . We conclude that

Arguing as before we conclude that

and the claim follows by induction on the number of times divides (noting that ).

Now we can prove (22). First suppose is coprime to . In this case, the random variable is only non-zero when is divisible by , which forces to equal . By Lemma 5, the left-hand side of (22) is thus equal to , thus proving (22) in this case.

Now suppose that is divisible by . With probability , is coprime to , so vanishes. Otherwise, we write and observe that

thus the left-hand side of (22) expands as

This expands into eight terms that can be computed using Proposition 1, (23), and Lemmas 5, (6) as

The claim then follows by induction on the number of times divides , together with a tedious computation using the identities

and

Filed under: expository, math.NT, math.PR Tagged: correlation, divisor function, pseudorandomness ]]>

I will assume familiarity with the notation of my paper. In Section 10, some complicated spaces are constructed for each frequency scale , and then a further space is constructed for a given frequency envelope by the formula

where is the Littlewood-Paley projection of to frequency magnitudes . Then, given a spacetime slab , we define the restrictions

where the infimum is taken over all extensions of to the Minkowski spacetime ; similarly one defines

The gap in the paper is as follows: it was implicitly assumed that one could restrict (1) to the slab to obtain the equality

(This equality is implicitly used to establish the bound (36) in the paper.) Unfortunately, (1) only gives the lower bound, not the upper bound, and it is the upper bound which is needed here. The problem is that the extensions of that are optimal for computing are not necessarily the Littlewood-Paley projections of the extensions of that are optimal for computing .

To remedy the problem, one has to prove an upper bound of the form

for all Schwartz (actually we need affinely Schwartz , but one can easily normalise to the Schwartz case). Without loss of generality we may normalise the RHS to be . Thus

for each , and one has to find a single extension of such that

for each . Achieving a that obeys (4) is trivial (just extend by zero), but such extensions do not necessarily obey (5). On the other hand, from (3) we can find extensions of such that

the extension will then obey (5) (here we use Lemma 9 from my paper), but unfortunately is not guaranteed to obey (4) (the norm does control the norm, but a key point about frequency envelopes for the small energy regularity problem is that the coefficients , while bounded, are not necessarily summable).

This can be fixed as follows. For each we introduce a time cutoff supported on that equals on and obeys the usual derivative estimates in between (the time derivative of size for each ). Later we will prove the truncation estimate

Assuming this estimate, then if we set , then using Lemma 9 in my paper and (6), (7) (and the local stability of frequency envelopes) we have the required property (5). (There is a technical issue arising from the fact that is not necessarily Schwartz due to slow decay at temporal infinity, but by considering partial sums in the summation and taking limits we can check that is the strong limit of Schwartz functions, which suffices here; we omit the details for sake of exposition.) So the only issue is to establish (4), that is to say that

for all .

For this is immediate from (2). Now suppose that for some integer (the case when is treated similarly). Then we can split

where

The contribution of the term is acceptable by (6) and estimate (82) from my paper. The term sums to which is acceptable by (2). So it remains to control the norm of . By the triangle inequality and the fundamental theorem of calculus, we can bound

By hypothesis, . Using the first term in (79) of my paper and Bernstein’s inequality followed by (6) we have

and then we are done by summing the geometric series in .

It remains to prove the truncation estimate (7). This estimate is similar in spirit to the algebra estimates already in my paper, but unfortunately does not seem to follow immediately from these estimates as written, and so one has to repeat the somewhat lengthy decompositions and case checkings used to prove these estimates. We do this below the fold.

** — 1. Proof of truncation estimate — **

Firstly, by rescaling (and changing as necessary) we may assume that . By the triangle inequality and time translation invariance, it suffices to show an estimate of the form

where is a smooth time cutoff that equals on and is supported in , and all norms are understood to be on . We may normalise the right-hand side to be , thus is supported in frequencies , and by equation (79) of my paper one has the estimates

for all , and our objective is to show that

The bound (11) easily follows from (8), the Leibniz rule, and using the frequency localisation of to ignore spatial derivatives. Now we turn to (12). From the definition of the norms, we have

for all integers , and we need to show that

Fix . We can use Littlewood-Paley operators to split , where is supported on time frequencies and is supported on time frequencies . For the contribution of one can replace in (15) by (say) and the claim then follows from (14), the Leibniz rule, and Hölder’s inequality (again ignoring spatial derivatives). For the contribution of , we discard and observe that has an norm of (and its time derivative has a norm of ), so this contribution is then acceptable from (8) and Hölder’s inequality.

Finally we need to show (13). Similarly to before, we split . We also split , leaving us with the task of proving the four estimates

We begin with (16). The multiplier is disposable in the sense of the paper, and similarly if one replaces by a slightly larger multiplier; this lets us bound the left-hand side of (16) by

The time cutoff commutes with the spatial Fourier projection and can then be discarded by equation (66) of my paper. This term is thus acceptable thanks to (10).

Now we turn to (17). We can freely insert a factor of in front of . Applying estimate (75) from my paper, it then suffices to show that

From the Fourier support of the expression inside the norm, the left-hand side is bounded by

discarding the cutoff and using (9) we see that this contribution is acceptable.

Next, we show (18). Here we use the energy estimate from equation (27) (and (25)) of the paper. By repeating the proof of (11) (and using Lemma 4 from my paper) we see that

so it suffices to show that

Expanding out the d’Lambertian using the Leibniz rule, we are reduced to showing the estimates

For (20) we note that has an norm of , while from (9) has an norm of , so the claim follows from Hölder’s inequality. For (21) we can similarly observe that has an norm of while from (8) we see that has an norm of , so the claim again follows from Hölder’s inequality. A similar argument gives (22) (with an additional gain of coming from the second derivative on ).

Finally, for (19), we observe from the Fourier separation between and that we may replace by (in fact one could do a much more drastic replacement if desired). The claim now follows from repeating the proof of (18).

Filed under: math.AP, update Tagged: Hao Jia, wave maps ]]>

In the previous blog post, one of us (Terry) implicitly introduced a notion of rank for tensors which is a little different from the usual notion of tensor rank, and which (following BCCGNSU) we will call “slice rank”. This notion of rank could then be used to encode the Croot-Lev-Pach-Ellenberg-Gijswijt argument that uses the polynomial method to control capsets.

Afterwards, several papers have applied the slice rank method to further problems – to control tri-colored sum-free sets in abelian groups (BCCGNSU, KSS) and from there to the triangle removal lemma in vector spaces over finite fields (FL), to control sunflowers (NS), and to bound progression-free sets in -groups (P).

In this post we investigate the notion of slice rank more systematically. In particular, we show how to give lower bounds for the slice rank. In many cases, we can show that the upper bounds on slice rank given in the aforementioned papers are sharp to within a subexponential factor. This still leaves open the possibility of getting a better bound for the original combinatorial problem using the slice rank of some other tensor, but for very long arithmetic progressions (at least eight terms), we show that the slice rank method cannot improve over the trivial bound using any tensor.

It will be convenient to work in a “basis independent” formalism, namely working in the category of abstract finite-dimensional vector spaces over a fixed field . (In the applications to the capset problem one takes to be the finite field of three elements, but most of the discussion here applies to arbitrary fields.) Given such vector spaces , we can form the tensor product , generated by the tensor products with for , subject to the constraint that the tensor product operation is multilinear. For each , we have the smaller tensor products , as well as the tensor product

defined in the obvious fashion. Elements of of the form for some and will be called *rank one functions*, and the *slice rank* (or *rank* for short) of an element of is defined to be the least nonnegative integer such that is a linear combination of rank one functions. If are finite-dimensional, then the rank is always well defined as a non-negative integer (in fact it cannot exceed . It is also clearly subadditive:

For , is when is zero, and otherwise. For , is the usual rank of the -tensor (which can for instance be identified with a linear map from to the dual space ). The usual notion of tensor rank for higher order tensors uses complete tensor products , as the rank one objects, rather than , giving a rank that is greater than or equal to the slice rank studied here.

From basic linear algebra we have the following equivalences:

Lemma 1Let be finite-dimensional vector spaces over a field , let be an element of , and let be a non-negative integer. Then the following are equivalent:

- (i) One has .
- (ii) One has a representation of the form
where are finite sets of total cardinality at most , and for each and , and .

- (iii) One has
where for each , is a subspace of of total dimension at most , and we view as a subspace of in the obvious fashion.

- (iv) (Dual formulation) There exist subspaces of the dual space for , of total dimension at least , such that is orthogonal to , in the sense that one has the vanishing
for all , where is the obvious pairing.

*Proof:* The equivalence of (i) and (ii) is clear from definition. To get from (ii) to (iii) one simply takes to be the span of the , and conversely to get from (iii) to (ii) one takes the to be a basis of the and computes by using a basis for the tensor product consisting entirely of functions of the form for various . To pass from (iii) to (iv) one takes to be the annihilator of , and conversely to pass from (iv) to (iii).

One corollary of the formulation (iv), is that the set of tensors of slice rank at most is Zariski closed (if the field is algebraically closed), and so the slice rank itself is a lower semi-continuous function. This is in contrast to the usual tensor rank, which is not necessarily semicontinuous.

Corollary 2Let be finite-dimensional vector spaces over an algebraically closed field . Let be a nonnegative integer. The set of elements of of slice rank at most is closed in the Zariski topology.

*Proof:* In view of Lemma 1(i and iv), this set is the union over tuples of integers with of the projection from of the set of tuples with orthogonal to , where is the Grassmanian parameterizing -dimensional subspaces of .

One can check directly that the set of tuples with orthogonal to is Zariski closed in using a set of equations of the form locally on . Hence because the Grassmanian is a complete variety, the projection of this set to is also Zariski closed. So the finite union over tuples of these projections is also Zariski closed.

We also have good behaviour with respect to linear transformations:

Lemma 3Let be finite-dimensional vector spaces over a field , let be an element of , and for each , let be a linear transformation, with the tensor product of these maps. Then

Furthermore, if the are all injective, then one has equality in (2).

Thus, for instance, the rank of a tensor is intrinsic in the sense that it is unaffected by any enlargements of the spaces .

*Proof:* The bound (2) is clear from the formulation (ii) of rank in Lemma 1. For equality, apply (2) to the injective , as well as to some arbitrarily chosen left inverses of the .

Computing the rank of a tensor is difficult in general; however, the problem becomes a combinatorial one if one has a suitably sparse representation of that tensor in some basis, where we will measure sparsity by the property of being an antichain.

Proposition 4Let be finite-dimensional vector spaces over a field . For each , let be a linearly independent set in indexed by some finite set . Let be a subset of .

where for each , is a coefficient in . Then one has

where the minimum ranges over all coverings of by sets , and for are the projection maps.

Now suppose that the coefficients are all non-zero, that each of the are equipped with a total ordering , and is the set of maximal elements of , thus there do not exist distinct , such that for all . Then one has

In particular, if is an antichain (i.e. every element is maximal), then equality holds in (4).

*Proof:* By Lemma 3 (or by enlarging the bases ), we may assume without loss of generality that each of the is spanned by the . By relabeling, we can also assume that each is of the form

with the usual ordering, and by Lemma 3 we may take each to be , with the standard basis.

Let denote the rank of . To show (4), it suffices to show the inequality

for any covering of by . By removing repeated elements we may assume that the are disjoint. For each , the tensor

can (after collecting terms) be written as

for some . Summing and using (1), we conclude the inequality (6).

Now assume that the are all non-zero and that is the set of maximal elements of . To conclude the proposition, it suffices to show that the reverse inequality

Â holds for some covering . By Lemma 1(iv), there exist subspaces of whose dimension sums to

Let . Using Gaussian elimination, one can find a basis of whose representation in the standard dual basis of is in row-echelon form. That is to say, there exist natural numbers

such that for all , is a linear combination of the dual vectors , with the coefficient equal to one.

We now claim that is disjoint from . Suppose for contradiction that this were not the case, thus there exists for each such that

As is the set of maximal elements of , this implies that

for any tuple other than . On the other hand, we know that is a linear combination of , with the coefficient one. We conclude that the tensor product is equal to

plus a linear combination of other tensor products with not in . Taking inner products with (3), we conclude that , contradicting the fact that is orthogonal to . Thus we have disjoint from .

For each , let denote the set of tuples in with not of the form . From the previous discussion we see that the cover , and we clearly have , and hence from (8) we have (7) as claimed.

As an instance of this proposition, we recover the computation of diagonal rank from the previous blog post:

Example 5Let be finite-dimensional vector spaces over a field for some . Let be a natural number, and for , let be a linearly independent set in . Let be non-zero coefficients in . Thenhas rank . Indeed, one applies the proposition with all equal to , with the diagonal in ; this is an antichain if we give one of the the standard ordering, and another of the the opposite ordering (and ordering the remaining arbitrarily). In this case, the are all bijective, and so it is clear that the minimum in (4) is simply .

The combinatorial minimisation problem in the above proposition can be solved asymptotically when working with tensor powers, using the notion of the Shannon entropy of a discrete random variable .

Proposition 6Let be finite-dimensional vector spaces over a field . For each , let be a linearly independent set in indexed by some finite set . Let be a non-empty subset of .Let be a tensor of the form (3) for some coefficients . For each natural number , let be the tensor power of copies of , viewed as an element of . Then

and range over the random variables taking values in .

Now suppose that the coefficients are all non-zero and that each of the are equipped with a total ordering . Let be the set of maximal elements of in the product ordering, and let where range over random variables taking values in . Then

as . In particular, if the maximizer in (10) is supported on the maximal elements of (which always holds if is an antichain in the product ordering), then equality holds in (9).

*Proof:*

as , where is the projection map. Then the same thing will apply to and . Then applying Proposition 4, using the lexicographical ordering on and noting that, if are the maximal elements of , then are the maximal elements of , we obtain both (9) and (11).

We first prove the lower bound. By compactness (and the continuity properties of entropy), we can find a random variable taking values in such that

Let be a small positive quantity that goes to zero sufficiently slowly with . Let denote the set of all tuples in that are within of being distributed according to the law of , in the sense that for all , one has

By the asymptotic equipartition property, the cardinality of can be computed to be

if goes to zero slowly enough. Similarly one has

Now let be an arbitrary covering of . By the pigeonhole principle, there exists such that

which by (13) implies that

noting that the factor can be absorbed into the error). This gives the lower bound in (12).

Now we prove the upper bound. We can cover by sets of the form for various choices of random variables taking values in . For each such random variable , we can find such that ; we then place all of in . It is then clear that the cover and that

for all , giving the required upper bound.

It is of interest to compute the quantity in (10). We have the following criterion for when a maximiser occurs:

Proposition 7Let be finite sets, and be non-empty. Let be the quantity in (10). Let be a random variable taking values in , and let denote the essential range of , that is to say the set of tuples such that is non-zero. Then the following are equivalent:

- (i) attains the maximum in (10).
- (ii) There exist weights and a finite quantity , such that whenever , and such that
for all , with equality if . (In particular, must vanish if there exists a with .)

Furthermore, when (i) and (ii) holds, one has

*Proof:* We first show that (i) implies (ii). The function is concave on . As a consequence, if we define to be the set of tuples such that there exists a random variable taking values in with , then is convex. On the other hand, by (10), is disjoint from the orthant . Thus, by the hyperplane separation theorem, we conclude that there exists a half-space

where are reals that are not all zero, and is another real, which contains on its boundary and in its interior, such that avoids the interior of the half-space. Since is also on the boundary of , we see that the are non-negative, and that whenever .

By construction, the quantity

is maximised when . At this point we could use the method of Lagrange multipliers to obtain the required constraints, but because we have some boundary conditions on the (namely, that the probability that they attain a given element of has to be non-negative) we will work things out by hand. Let be an element of , and an element of . For small enough, we can form a random variable taking values in , whose probability distribution is the same as that for except that the probability of attaining is increased by , and the probability of attaining is decreased by . If there is any for which and , then one can check that

for sufficiently small , contradicting the maximality of ; thus we have whenever . Taylor expansion then gives

for small , where

and similarly for . We conclude that for all and , thus there exists a quantity such that for all , and for all . By construction must be nonnegative. Sampling using the distribution of , one has

almost surely; taking expectations we conclude that

The inner sum is , which equals when is non-zero, giving (17).

Now we show conversely that (ii) implies (i). As noted previously, the function is concave on , with derivative . This gives the inequality

for any (note the right-hand side may be infinite when and ). Let be any random variable taking values in , then on applying the above inequality with and , multiplying by , and summing over and gives

By construction, one has

and

so to prove that (which would give (i)), it suffices to show that

or equivalently that the quantity

is maximised when . Since

it suffices to show this claim for the quantity

One can view this quantity as

By (ii), this quantity is bounded by , with equality if is equal to (and is in particular ranging in ), giving the claim.

The second half of the proof of Proposition 7 only uses the marginal distributions and the equation(16), not the actual distribution of , so it can also be used to prove an upper bound on when the exact maximizing distribution is not known, given suitable probability distributions in each variable. The logarithm of the probability distribution here plays the role that the weight functions do in BCCGNSU.

Remark 8Suppose one is in the situation of (i) and (ii) above; assume the nondegeneracy condition that is positive (or equivalently that is positive). We can assign a “degree” to each element by the formula

then every tuple in has total degree at most , and those tuples in have degree exactly . In particular, every tuple in has degree at most , and hence by (17), each such tuple has a -component of degree less than or equal to for some with . On the other hand, we can compute from (19) and the fact that for that . Thus, by asymptotic equipartition, and assuming , the number of “monomials” in of total degree at most is at most ; one can in fact use (19) and (18) to show that this is in fact an equality. This gives a direct way to cover by sets with , which is in the spirit of the Croot-Lev-Pach-Ellenberg-Gijswijt arguments from the previous post.

We can now show that the rank computation for the capset problem is sharp:

Proposition 9Let denote the space of functions from to . Then the function from to , viewed as an element of , has rank as , where is given by the formula

*Proof:* In , we have

Thus, if we let be the space of functions from to (with domain variable denoted respectively), and define the basis functions

of indexed by (with the usual ordering), respectively, and set to be the set

then is a linear combination of the with , and all coefficients non-zero. Then we have . We will show that the quantity of (10) agrees with the quantity of (20), and that the optimizing distribution is supported on , so that by Proposition 6 the rank of is .

To compute the quantity at (10), we use the criterion in Proposition 7. We take to be the random variable taking values in that attains each of the values with a probability of , and each of with a probability of ; then each of the attains the values of with probabilities respectively, so in particular is equal to the quantity in (20). If we now set and

we can verify the condition (16) with equality for all , which from (17) gives as desired.

This statement already follows from the result of Kleinberg-Sawin-Speyer, which gives a “tri-colored sum-free set” in of size , as the slice rank of this tensor is an upper bound for the size of a tri-colored sum-free set. If one were to go over the proofs more carefully to evaluate the subexponential factors, this argument would give a stronger lower bound than KSS, as it does not deal with the substantial loss that comes from Behrend’s construction. However, because it actually constructs a set, the KSS result rules out more possible approaches to give an exponential improvement of the upper bound for capsets. The lower bound on slice rank shows that the bound cannot be improved using only the slice rank of this particular tensor, whereas KSS shows that the bound cannot be improved using any method that does not take advantage of the “single-colored” nature of the problem.

We can also show that the slice rank upper bound in a result of Naslund-Sawin is similarly sharp:

Proposition 10Let