Lemma 1 (Van der Corput inequality)Let be unit vectors in a Hilbert space . Then

*Proof:* The left-hand side may be written as for some unit complex numbers . By Cauchy-Schwarz we have

As a corollary, correlation becomes transitive in a statistical sense (even though it is not transitive in an absolute sense):

Corollary 2 (Statistical transitivity of correlation)Let be unit vectors in a Hilbert space such that for all and some . Then we have for at least of the pairs .

*Proof:* From the lemma, we have

One drawback with this corollary is that it does not tell us *which* pairs correlate. In particular, if the vector also correlates with a separate collection of unit vectors, the pairs for which correlate may have no intersection whatsoever with the pairs in which correlate (except of course on the diagonal where they must correlate).

While working on an ongoing research project, I recently found that there is a very simple way to get around the latter problem by exploiting the tensor power trick:

Corollary 3 (Simultaneous statistical transitivity of correlation)Let be unit vectors in a Hilbert space for and such that for all , and some . Then there are at least pairs such that . In particular (by Cauchy-Schwarz) we have for all .

*Proof:* Apply Corollary 2 to the unit vectors and , in the tensor power Hilbert space .

It is surprisingly difficult to obtain even a qualitative version of the above conclusion (namely, if correlates with all of the , then there are many pairs for which correlates with for all simultaneously) without some version of the tensor power trick. For instance, even the powerful Szemerédi regularity lemma, when applied to the set of pairs for which one has correlation of , for a single , does not seem to be sufficient. However, there is a reformulation of the argument using the Schur product theorem as a substitute for (or really, a disguised version of) the tensor power trick. For simplicity of notation let us just work with real Hilbert spaces to illustrate the argument. We start with the identity

where is the orthogonal projection to the complement of . This implies a Gram matrix inequality for each where denotes the claim that is positive semi-definite. By the Schur product theorem, we conclude that and hence for a suitable choice of signs , One now argues as in the proof of Corollary 2.A separate application of tensor powers to amplify correlations was also noted in this previous blog post giving a cheap version of the Kabatjanskii-Levenstein bound, but this seems to not be directly related to this current application.

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Proposition 1 (Classical Möbius inversion)Let be functions from the natural numbers to an additive group . Then the following two claims are equivalent:

- (i) for all .
- (ii) for all .

There is a generalisation of this formula to (finite) posets, due to Hall, in which one sums over chains in the poset:

Proposition 2 (Poset Möbius inversion)Let be a finite poset, and let be functions from that poset to an additive group . Then the following two claims are equivalent:(Note from the finite nature of that the inner sum in (ii) is vacuous for all but finitely many .)

- (i) for all , where is understood to range in .
- (ii) for all , where in the inner sum are understood to range in with the indicated ordering.

Comparing Proposition 2 with Proposition 1, it is natural to refer to the function as the Möbius function of the poset; the condition (ii) can then be written as

In fact it is not completely necessary that the poset be finite; an inspection of the proof shows that it suffices that every element of the poset has only finitely many predecessors .

It is not difficult to see that Proposition 2 includes Proposition 1 as a special case, after verifying the combinatorial fact that the quantity

is equal to when divides , and vanishes otherwise.I recently discovered that Proposition 2 can also lead to a useful variant of the inclusion-exclusion principle. The classical version of this principle can be phrased in terms of indicator functions: if are subsets of some set , then

In particular, if there is a finite measure on for which are all measurable, we haveOne drawback of this formula is that there are exponentially many terms on the right-hand side: of them, in fact. However, in many cases of interest there are “collisions” between the intersections (for instance, perhaps many of the pairwise intersections agree), in which case there is an opportunity to collect terms and hopefully achieve some cancellation. It turns out that it is possible to use Proposition 2 to do this, in which one only needs to sum over chains in the resulting poset of intersections:

Proposition 3 (Hall-type inclusion-exclusion principle)Let be subsets of some set , and let be the finite poset formed by intersections of some of the (with the convention that is the empty intersection), ordered by set inclusion. Then for any , one has where are understood to range in . In particular (setting to be the empty intersection) if the are all proper subsets of then we have In particular, if there is a finite measure on for which are all measurable, we have

Using the Möbius function on the poset , one can write these formulae as

and
*Proof:* It suffices to establish (2) (to derive (3) from (2) observe that all the are contained in one of the , so the effect of may be absorbed into ). Applying Proposition 2, this is equivalent to the assertion that

Example 4If with , and are all distinct, then we have for any finite measure on that makes measurable that due to the four chains , , , of length one, and the three chains , , of length two. Note that this expansion just has six terms in it, as opposed to the given by the usual inclusion-exclusion formula, though of course one can reduce the number of terms by combining the factors. This may not seem particularly impressive, especially if one views the term as really being three terms instead of one, but if we add a fourth set with for all , the formula now becomes and we begin to see more cancellation as we now have just seven terms (or ten if we count as four terms) instead of terms.

Example 5 (Variant of Legendre sieve)If are natural numbers, and is some sequence of complex numbers with only finitely many terms non-zero, then by applying the above proposition to the sets and with equal to counting measure weighted by the we obtain a variant of the Legendre sieve where range over the set formed by taking least common multiples of the (with the understanding that the empty least common multiple is ), and denotes the assertion that divides but is strictly less than . I am curious to know of this version of the Legendre sieve already appears in the literature (and similarly for the other applications of Proposition 2 given here).

If the poset has bounded depth then the number of terms in Proposition 3 can end up being just polynomially large in rather than exponentially large. Indeed, if all chains in have length at most then the number of terms here is at most . (The examples (4), (5) are ones in which the depth is equal to two.) I hope to report in a later post on how this version of inclusion-exclusion with polynomially many terms can be useful in an application.

Actually in our application we need an abstraction of the above formula, in which the indicator functions are replaced by more abstract idempotents:

Proposition 6 (Hall-type inclusion-exclusion principle for idempotents)Let be pairwise commuting elements of some ring with identity, which are all idempotent (thus for ). Let be the finite poset formed by products of the (with the convention that is the empty product), ordered by declaring when (note that all the elements of are idempotent so this is a partial ordering). Then for any , one has where are understood to range in . In particular (setting ) if all the are not equal to then we have

Morally speaking this proposition is equivalent to the previous one after applying a “spectral theorem” to simultaneously diagonalise all of the , but it is quicker to just adapt the previous proof to establish this proposition directly. Using the Möbius function for , we can rewrite these formulae as

and
*Proof:* Again it suffices to verify (6). Using Proposition 2 as before, it suffices to show that

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The Department of Mathematics at the University of California, Los Angeles, is inviting applications for the position of an Academic Administrator who will serve as the Director of the UCLA Endowed Olga Radko Math Circle (ORMC). The Academic Administrator will have the broad responsibility for administration of the ORMC, an outreach program with weekly activities for mathematically inclined students in grades K-12. Currently, over 300 children take part in the program each weekend. Instruction is delivered by a team of over 50 docents, the majority of whom are UCLA undergraduate and graduate students.

The Academic Administrator is required to teach three mathematics courses in the undergraduate curriculum per academic year as assigned by the Department. This is also intended to help with the recruitment of UCLA students as docents and instructors for the ORMC.

As the director of ORMC, the Academic Administrator will have primary responsibility for all aspects of ORMC operations:

- Determining the structure of ORMC, including the number and levels of groups
- Recruiting, training and supervising instructors, docents, and postdoctoral fellows associated with the ORMC
- Developing curricular materials and providing leadership in development of innovative ways of explaining mathematical ideas to school children
- Working with the Mathematics Department finance office to ensure timely payment of stipends and wages to ORMC instructors and docents, as appropriate
- Maintaining ORMC budget and budgetary projections, ensuring that the funds are used appropriately and efficiently for ORMC activities, and applying for grants as appropriate to fund the operations of ORMC
- Working with the Steering Committee and UCLA Development to raise funds for ORMC, both from families whose children participate in ORMC and other sources
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- Reporting to and working with the ORMC Steering Committee throughout the year

A competitive candidate should have leadership potential and experience with developing mathematical teaching materials for the use of gifted school children, as well as experience with teaching undergraduate mathematics courses. Candidates must have a Ph.D. degree (or equivalent) or expect to complete their Ph.D. by June 30, 2021.

Applications should be received by March 15, 2021. Further details on the position and the application process can be found at the application page.

]]>One of the great classical triumphs of complex analysis was in providing the first complete proof (by Hadamard and de la Vallée Poussin in 1896) of arguably the most important theorem in analytic number theory, the prime number theorem:

Theorem 1 (Prime number theorem)Let denote the number of primes less than a given real number . Then (or in asymptotic notation, as ).

(Actually, it turns out to be slightly more natural to replace the approximation in the prime number theorem by the logarithmic integral , which turns out to be a more precise approximation, but we will not stress this point here.)

The complex-analytic proof of this theorem hinges on the study of a key meromorphic function related to the prime numbers, the Riemann zeta function . Initially, it is only defined on the half-plane :

Definition 2 (Riemann zeta function, preliminary definition)Let be such that . Then we define

Note that the series is locally uniformly convergent in the half-plane , so in particular is holomorphic on this region. In previous notes we have already evaluated some special values of this function:

However, it turns out that the zeroes (and pole) of this function are of far greater importance to analytic number theory, particularly with regards to the study of the prime numbers.The Riemann zeta function has several remarkable properties, some of which we summarise here:

Theorem 3 (Basic properties of the Riemann zeta function)

- (i) (Euler product formula) For any with , we have where the product is absolutely convergent (and locally uniform in ) and is over the prime numbers .
- (ii) (Trivial zero-free region) has no zeroes in the region .
- (iii) (Meromorphic continuation) has a unique meromorphic continuation to the complex plane (which by abuse of notation we also call ), with a simple pole at and no other poles. Furthermore, the Riemann xi function is an entire function of order (after removing all singularities). The function is an entire function of order one after removing the singularity at .
- (iv) (Functional equation) After applying the meromorphic continuation from (iii), we have for all (excluding poles). Equivalently, we have for all . (The equivalence between the (5) and (6) is a routine consequence of the Euler reflection formula and the Legendre duplication formula, see Exercises 26 and 31 of Notes 1.)

*Proof:* We just prove (i) and (ii) for now, leaving (iii) and (iv) for later sections.

The claim (i) is an encoding of the fundamental theorem of arithmetic, which asserts that every natural number is uniquely representable as a product over primes, where the are natural numbers, all but finitely many of which are zero. Writing this representation as , we see that

whenever , , and consists of all the natural numbers of the form for some . Sending and to infinity, we conclude from monotone convergence and the geometric series formula that whenever is real, and then from dominated convergence we see that the same formula holds for complex with as well. Local uniform convergence then follows from the product form of the Weierstrass -test (Exercise 19 of Notes 1).The claim (ii) is immediate from (i) since the Euler product is absolutely convergent and all terms are non-zero.

We remark that by sending to in Theorem 3(i) we conclude that

and from the divergence of the harmonic series we then conclude Euler’s theorem . This can be viewed as a weak version of the prime number theorem, and already illustrates the potential applicability of the Riemann zeta function to control the distribution of the prime numbers.The meromorphic continuation (iii) of the zeta function is initially surprising, but can be interpreted either as a manifestation of the extremely regular spacing of the natural numbers occurring in the sum (1), or as a consequence of various integral representations of (or slight modifications thereof). We will focus in this set of notes on a particular representation of as essentially the Mellin transform of the theta function that briefly appeared in previous notes, and the functional equation (iv) can then be viewed as a consequence of the modularity of that theta function. This in turn was established using the Poisson summation formula, so one can view the functional equation as ultimately being a manifestation of Poisson summation. (For a direct proof of the functional equation via Poisson summation, see these notes.)

Henceforth we work with the meromorphic continuation of . The functional equation (iv), when combined with special values of such as (2), gives some additional values of outside of its initial domain , most famously

If one
From Theorem 3 and the non-vanishing nature of , we see that has simple zeroes (known as *trivial zeroes*) at the negative even integers , and all other zeroes (the *non-trivial zeroes*) inside the *critical strip* . (The non-trivial zeroes are conjectured to all be simple, but this is hopelessly far from being proven at present.) As we shall see shortly, these latter zeroes turn out to be closely related to the distribution of the primes. The functional equation tells us that if is a non-trivial zero then so is ; also, we have the identity

Conjecture 4 (Riemann hypothesis)All the non-trivial zeroes of lie on the critical line .

This conjecture would have many implications in analytic number theory, particularly with regard to the distribution of the primes. Of course, it is far from proven at present, but the partial results we have towards this conjecture are still sufficient to establish results such as the prime number theorem.

Return now to the original region where . To take more advantage of the Euler product formula (3), we take complex logarithms to conclude that

for suitable branches of the complex logarithm, and then on taking derivatives (using for instance the generalised Cauchy integral formula and Fubini’s theorem to justify the interchange of summation and derivative) we see that From the geometric series formula we have and so (by another application of Fubini’s theorem) we have the identity for , where the von Mangoldt function is defined to equal whenever is a power of a prime for some , and otherwise. The contribution of the higher prime powers is negligible in practice, and as a first approximation one can think of the von Mangoldt function as the indicator function of the primes, weighted by the logarithm function.The series and that show up in the above formulae are examples of Dirichlet series, which are a convenient device to transform various sequences of arithmetic interest into holomorphic or meromorphic functions. Here are some more examples:

Exercise 5 (Standard Dirichlet series)Let be a complex number with .

- (i) Show that .
- (ii) Show that , where is the divisor function of (the number of divisors of ).
- (iii) Show that , where is the Möbius function, defined to equal when is the product of distinct primes for some , and otherwise.
- (iv) Show that , where is the Liouville function, defined to equal when is the product of (not necessarily distinct) primes for some .
- (v) Show that , where is the holomorphic branch of the logarithm that is real for , and with the convention that vanishes for .
- (vi) Use the fundamental theorem of arithmetic to show that the von Mangoldt function is the unique function such that for every positive integer . Use this and (i) to provide an alternate proof of the identity (8). Thus we see that (8) is really just another encoding of the fundamental theorem of arithmetic.

Given the appearance of the von Mangoldt function , it is natural to reformulate the prime number theorem in terms of this function:

Theorem 6 (Prime number theorem, von Mangoldt form)One has (or in asymptotic notation, as ).

Let us see how Theorem 6 implies Theorem 1. Firstly, for any , we can write

The sum is non-zero for only values of , and is of size , thus Since , we conclude from Theorem 6 that as . Next, observe from the fundamental theorem of calculus that Multiplying by and summing over all primes , we conclude that From Theorem 6 we certainly have , thus By splitting the integral into the ranges and we see that the right-hand side is , and Theorem 1 follows.

Exercise 7Show that Theorem 1 conversely implies Theorem 6.

The alternate form (8) of the Euler product identity connects the primes (represented here via proxy by the von Mangoldt function) with the logarithmic derivative of the zeta function, and can be used as a starting point for describing further relationships between and the primes. Most famously, we shall see later in these notes that it leads to the remarkably precise Riemann-von Mangoldt explicit formula:

Theorem 8 (Riemann-von Mangoldt explicit formula)For any non-integer , we have where ranges over the non-trivial zeroes of with imaginary part in . Furthermore, the convergence of the limit is locally uniform in .

Actually, it turns out that this formula is in some sense *too* precise; in applications it is often more convenient to work with smoothed variants of this formula in which the sum on the left-hand side is smoothed out, but the contribution of zeroes with large imaginary part is damped; see Exercise 22. Nevertheless, this formula clearly illustrates how the non-trivial zeroes of the zeta function influence the primes. Indeed, if one formally differentiates the above formula in , one is led to the (quite nonrigorous) approximation

Comparing Theorem 8 with Theorem 6, it is natural to suspect that the key step in the proof of the latter is to establish the following slight but important extension of Theorem 3(ii), which can be viewed as a very small step towards the Riemann hypothesis:

Theorem 9 (Slight enlargement of zero-free region)There are no zeroes of on the line .

It is not quite immediate to see how Theorem 6 follows from Theorem 8 and Theorem 9, but we will demonstrate it below the fold.

Although Theorem 9 only seems like a slight improvement of Theorem 3(ii), proving it is surprisingly non-trivial. The basic idea is the following: if there was a zero at , then there would also be a different zero at (note cannot vanish due to the pole at ), and then the approximation (9) becomes

But the expression can be negative for large regions of the variable , whereas is always non-negative. This conflict eventually leads to a contradiction, but it is not immediately obvious how to make this argument rigorous. We will present here the classical approach to doing so using a trigonometric identity of Mertens.In fact, Theorem 9 is basically equivalent to the prime number theorem:

Exercise 10For the purposes of this exercise, assume Theorem 6, but do not assume Theorem 9. For any non-zero real , show that as , where denotes a quantity that goes to zero as after being multiplied by . Use this to derive Theorem 9.

This equivalence can help explain why the prime number theorem is remarkably non-trivial to prove, and why the Riemann zeta function has to be either explicitly or implicitly involved in the proof.

This post is only intended as the briefest of introduction to complex-analytic methods in analytic number theory; also, we have not chosen the shortest route to the prime number theorem, electing instead to travel in directions that particularly showcase the complex-analytic results introduced in this course. For some further discussion see this previous set of lecture notes, particularly Notes 2 and Supplement 3 (with much of the material in this post drawn from the latter).

** — 1. Meromorphic continuation and functional equation — **

We now focus on understanding the meromorphic continuation of , as well as the functional equation that that continuation satisfies. The arguments here date back to Riemann’s original paper on the zeta function. The general strategy is to relate the zeta function for to some sort of integral involving the parameter , which is manipulated in such a way that the integral makes sense for values of outside of the halfplane , and can thus be used to define the zeta function meromorphically in such a region. Often the Gamma function is involved in the relationship between the zeta function and integral. There are many such ways to connect to an integral; we present some of the more classical ones here.

One way to motivate the meromorphic continuation is to look at the continuous analogue

of (1). This clearly extends meromorphically to the whole complex plane. So one now just has to understand the analytic continuation properties of the residual For instance, using the Riemann sum type quadrature one can write this residual as since , it is a routine application of the Fubini and Morera theorems to establish analytic continuation of the residual to the half-plane , thus giving a meromorphic extension of to the region . Among other things, this shows that (the meromorphic continuation of) has a simple pole at with residue .

Exercise 11Using the trapezoid rule, show that for any in the region with , there exists a unique complex number for which one has the asymptotic for any natural number , where . Use this to extend the Riemann zeta function meromorphically to the region . Conclude in particular that and .

Exercise 12Obtain the refinement to the trapezoid rule when are integers and is continuously three times differentiable. Then show that for any in the region with , there exists a unique complex number for which one has the asymptotic for any natural number , where . Use this to extend the Riemann zeta function meromorphically to the region . Conclude in particular that .

One can keep going in this fashion using the Euler-Maclaurin formula (see this previous blog post) to extend the range of meromorphic continuation to the rest of the complex plane. However, we will now proceed in a different fashion, using the theta function

that made an appearance in previous notes, and try to transform this function into the zeta function. We will only need this function for imaginary values of the argument in the upper half-plane (so ); from Exercise 7 of Notes 2 we have the modularity relation In particular, since decays exponentially to as , blows up like as .We will attempt to apply the Mellin transform (Exercise 11 from Notes 2) to this function; formally, we have

There is however a problem: as goes to infinity, converges to one, and the integral here is unlikely to be convergent. So we will compute the Mellin transform of : The function decays exponentially as , and blows up like as , so this integral will be absolutely integrable when . Since we can write By the Fubini–Tonelli theorem, the integrand here is absolutely integrable, and hence From the Bernoulli definition of the Gamma function (Exercise 29(ii) of Notes 1) and a change of variables we have and hence by (1) we obtain the identity whenever . Replacing by , we can rearrange this as a formula for the function (4), namely whenever .Now we exploit the modular identity (12) to improve the convergence of this formula. The convergence of is much better near than near , so we use (13) to split

and then transform the first integral using the change of variables to obtain Using (12) we can write this as Direct computation shows that and thus whenever . However, the integrand here is holomorphic in and exponentially decaying in , so from the Fubini and Morera theorems we easily see that the right-hand side is an entire function of ; also from inspection we see that it is symmetric with respect to the symmetry . Thus we can define as an entire function, and hence as a meromorphic function, and one verifies the functional equation (6).It remains to establish that is of order . From (11) we have so from the triangle inequality

From the Stirling approximation (Exercise 30(v) from Notes 1) we conclude that for (say), and hence is of order at most as required. (One can show that has order exactly one by inspecting what happens to as , using that in this regime.) This completes the proof of Theorem 3.

Exercise 13 (Alternate derivation of meromorphic continuation and functional equation)

- (i) Establish the identity whenever .
- (ii) Establish the identity whenever , is not an integer, , where is the branch of the logarithm with real part in , and is the contour consisting of the line segment , the semicircle , and the line segment .
- (iii) Use (ii) to meromorphically continue to the entire complex plane .
- (iv) By shifting the contour to the contour for a large natural number and applying the residue theorem, show that again using the branch of the logarithm to define .
- (v) Establish the functional equation (5).

Exercise 14Use the formula from Exercise 12, together with the functional equation, to give yet another proof of the identity .

Exercise 15 (Relation between zeta function and Bernoulli numbers)

- (i) For any complex number with , use the Poisson summation formula (Proposition 3(v) from Notes 2) to establish the identity
- (ii) For as above and sufficiently small, show that Conclude that for any natural number , where the Bernoulli numbers are defined through the Taylor expansion Thus for instance , , and so forth.
- (iii) Show that for any odd natural number . (This identity can also be deduced from the Euler-Maclaurin formula, which generalises the approach in Exercise 12; see this previous post.)
- (iv) Use (14) and the residue theorem (now working inside the contour , rather than outside) to give an alternate proof of (15).

Exercise 16 (Convexity bounds)It is possible to improve the bounds (iii) in the region ; such improvements are known as

- (i) Establish the bounds for any and with .
- (ii) Establish the bounds for any and with . (Hint: use the functional equation.)
- (iii) Establish the bounds for any and with . (Hint: use the Phragmén-Lindelöf principle, Exercise 19 from Notes 2, after dealing somehow with the pole at .)
subconvexity estimates. For instance, it is currently known that for any and , a result of Bourgain; the Lindelöf hypothesis asserts that this bound in fact holds for all , although this remains unproven (it is however a consequence of the Riemann hypothesis).

Exercise 17 (Riemann-von Mangoldt formula)Show that for any , the number of zeroes of in the rectangle is equal to . (Hint:apply the argument principle to evaluated at a rectangle for some that is chosen so that the horizontal edges of the rectangle do not come too close to any of the zeroes (cf. the selection of the radii in the proof of the Hadamard factorisation theorem in Notes 1), and use the functional equation and Stirling’s formula to control the asymptotics for the horizontal edges.)We remark that the error term , due to von Mangoldt in 1905, has not been significantly improved despite over a century of effort. Even assuming the Riemann hypothesis, the error has only been reduced very slightly to (a result of Littlewood from 1924).

Remark 18Thanks to the functional equation and Rouche’s theorem, it is possible to numerically verify the Riemann hypothesis in any finite portion of the critical strip, so long as the zeroes in that strip are all simple. Indeed, if there was a zero off of the critical line , then an application of the argument principle (and Rouche’s theorem) in some small contour around but avoiding the critical line would be capable of numerically determining that there was a zero off of the line. Similarly, for each simple zero on the critical line, applying the argument principle for some small contour around that zero and symmetric around the critical line would numerically verify that there was exactly one zero within that contour, which by the functional equation would then have to lie exactly on that line. (In practice, more efficient methods are used to numerically verify the Riemann hypothesis over large finite portions of the strip, but we will not detail them here.)

** — 2. The explicit formula — **

We now prove Riemann-von Mangoldt explicit formula. Since is a non-trivial entire function of order , with zeroes at the non-trivial zeroes of (the trivial zeroes having been cancelled out by the Gamma function), we see from the Hadamard factorisation theorem (in the form of Exercise 35 from Notes 1) that

away from the zeroes of , where ranges over the non-trivial zeroes of (note from Exercise 11 that there is no zero at the origin), and is some constant. From (4) we can calculate while from Exercise 27 of Notes 1 we have and thus (after some rearranging) whereOne can compute the values of explicitly:

Exercise 19By inspecting both sides of (16) as , show that , and hence .

Jensen’s formula tells us that the number of non-trivial zeroes of in a disk is at most for any and . One can obtain a local version:

Exercise 20 (Local bound on zeroes)

- (i) Establish the upper bound whenever and with . (
Hint:use (10). More precise bounds are available with more effort, but will not be needed here.)- (ii) Establish the bounds uniformly in . (
Hint:use the Euler product.)- (iii) Show that for any , the number of non-trivial zeroes with imaginary part in is . (
Hint:use Jensen’s formula and the functional equation.)- (iv) For , , and , with not a zero of , show that (
Hint:use Exercise 9 of Notes 1.)

Meanwhile, from Perron’s formula (Exercise 12 of Notes 2) and (8) we see that for any non-integer , we have

We can compute individual terms here and then conclude the Riemann-von Mangoldt explicit formula:

Exercise 21 (Riemann-von Mangoldt explicit formula)Let and . Establish the following bounds:(

- (i) .
- (ii) .
- (iii) For any positive integer , we have
- (iv) For any non-trivial zero , we have
- (v) We have .
- (vi) We have .
Hint:for (i)-(iii), shift the contour to for an that gets sent to infinity, and using the residue theorem. The same argument works for (iv) except when is really close to , in which case a detour to the contour may be called for. For (vi), use Exercise 20 and partition the zeroes depending on what unit interval falls into.)

- (viii) Using the above estimates, conclude Theorem 8.

The explicit formula in Theorem 8 is completely exact, but turns out to be a little bit inconvenient for applications because it involves all the zeroes , and the series involving them converges very slowly (indeed the convergence is not even absolute). In practice it is preferable to work with a smoothed version of the formula. Here is one such smoothing:

Exercise 22 (Smoothed explicit formula)

- (i) Let be a smooth function compactly supported on . Show that is entire and obeys the bound (say) for some , all , and all .
- (ii) With as in (i), establish the identity with the summations being absolutely convergent by applying the Fourier inversion formula to , shifting the contour to frequencies for some , applying (8), and then shifting the contour again (using Exercise 20 and (i) to justify the contour shifting).
- (iii) Show that whenever is a smooth function, compactly supported in , with the summation being absolutely convergent.
- (iv) Explain why (iii) is
formallyconsistent with Theorem 8 when applied to the non-smooth function .

** — 3. Extending the zero free region, and the prime number theorem — **

We now show how Theorem 9 implies Theorem 6. Let be parameters to be chosen later. We will apply Exercise 22 to a function which equals one on , is supported on , and obeys the derivative estimates

for all and , and for all and . Such a function can be constructed by gluing together various rescaled versions of (antiderivatives of) standard bump functions. For such a function, we have On the other hand, we have and and hence We split into the two cases and , where is a parameter to be chosen later. For , there are only zeros, and all of them have real part strictly less than by Theorem 9. Hence there exists such that for all such zeroes. For each such zero, we have from the triangle inequality and so the total contribution of these zeroes to (17) is . For each zero with , we integrate parts twice to get some decay in : and from the triangle inequality and the fact that we conclude Since is convergent (this follows from Exercise 20 we conclude (for large enough depending on ) that the total contribution here is . Thus, after choosing suitably, we obtain the bound and thus whenever is sufficiently large depending on (since depends only on , which depends only on ). A similar argument (replacing by in the construction of ) gives the matching lower bound whenever is sufficiently large depending on . Sending , we obtain Theorem 6.

Exercise 23Assuming the Riemann hypothesis, show that for any and , and that for any and . Conversely, show that either of these two estimates are equivalent to the Riemann hypothesis. (Hint:find a holomorphic continuation of to the region in a manner similar to how was first holomorphically continued to the region ).

It remains to prove Theorem 9. The claim is clear for thanks to the simple pole of at , so we may assume . Suppose for contradiction that there was a zero of at , thus

for sufficiently close to . Taking logarithms, we see in particular that Using Lemma 5(v), we conclude that Note that the summands here are oscillatory due to the cosine term. To manage the oscillation, we use the simple pole at that gives for sufficiently close to one, and on taking logarithms as before we get These two estimates come close to being contradictory, but not quite (because we could have close to for most numbers that are weighted by . To get the contradiction, we use the analytic continuation of to to conclude that and hence Now we take advantage of the

Exercise 24Establish the inequality for any and .

Remark 25There are a number of ways to improve Theorem 9 that move a little closer in the direction of the Riemann hypothesis. Firstly, there are a number ofzero-free regionsfor the Riemann zeta function known that give lower bounds for (and in particular preclude the existence of zeros) a small amount inside the critical strip, and can be used to improve the error term in the prime number theorem; for instance, theclassical zero-free regionshows that there are no zeroes in the region for some sufficiently small absolute constant , and lets one improve the error term in Theorem 6 to (with a corresponding improvement in Theorem 1, provided that one replaces with the logarithmic integral ). A further improvement in the zero free region and in the prime number theorem error term was subsequently given by Vinogradov. We also mention a number of importantzero density estimateswhich provide non-trivial upper bounds for the number of zeroes in other, somewhat larger regions of the critical strip; the bounds are not strong enough to completely exclude zeroes as is the case with zero-free regions, but can at least limit the collective influence of such zeroes. For more discussion of these topics, see the various lecture notes to this previous course.

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We were saddened and horrified to learn that Ilya Dumanski, a brilliant young mathematician who has been admitted to our graduate programs at Stanford and MIT, has been imprisoned in Russia, along with several other mathematicians, for participation in a peaceful demonstration. Our thoughts are with them. We urge their rapid release, and failing that, that they be kept in humane conditions. A petition in their support has been started at

https://www.ipetitions.com/petition/a-call-for-immediate-release-of-arrested-students/

Signed,

Roman Bezrukavnikov (MIT)

Alexei Borodin (MIT)

Daniel Bump (Stanford)

Sourav Chatterjee (Stanford)

Otis Chodosh (Stanford)

Ralph Cohen (Stanford)

Henry Cohn (MIT)

Brian Conrad (Stanford)

Joern Dunkel (MIT)

Pavel Etingof (MIT)

Jacob Fox (Stanford)

Michel Goemans (MIT)

Eleny Ionel (Stanford)

Steven Kerckhoff (Stanford)

Jonathan Luk (Stanford)

Eugenia Malinnikova (Stanford)

Davesh Maulik (MIT)

Rafe Mazzeo (Stanford)

Haynes Miller (MIT)

Ankur Moitra (MIT)

Elchanan Mossel (MIT)

Tomasz Mrowka (MIT)

Bjorn Poonen (MIT)

Alex Postnikov (MIT)

Lenya Ryzhik (Stanford)

Paul Seidel (MIT)

Mike Sipser (MIT)

Kannan Soundararajan (Stanford)

Gigliola Staffilani (MIT)

Nike Sun (MIT)

Richard Taylor (Stanford)

Ravi Vakil (Stanford)

Andras Vasy (Stanford)

Jan Vondrak (Stanford)

Brian White (Stanford)

Zhiwei Yun (MIT)

On the real line, the quintessential examples of a periodic function are the (normalised) sine and cosine functions , , which are -periodic in the sense that

By taking various polynomial combinations of and we obtain more general trigonometric polynomials that are -periodic; and the theory of Fourier series tells us that all other -periodic functions (with reasonable integrability conditions) can be approximated in various senses by such polynomial combinations. Using Euler’s identity, one can use and in place of and as the basic generating functions here, provided of course one is willing to use complex coefficients instead of real ones. Of course, by rescaling one can also make similar statements for other periods than . -periodic functions can also be identified (by abuse of notation) with functions on the quotient space (known as the
What about periodic functions on the complex plane? We can start with *singly periodic functions* which obey a periodicity relationship for all in the domain and some period ; such functions can also be viewed as functions on the “additive cylinder” (or equivalently ). We can rescale as before. For holomorphic functions, we have the following characterisations:

Proposition 1 (Description of singly periodic holomorphic functions)In both cases, the coefficients can be recovered from by the Fourier inversion formula for any in (in case (i)) or (in case (ii)).

- (i) Every -periodic entire function has an absolutely convergent expansion where is the nome , and the are complex coefficients such that Conversely, every doubly infinite sequence of coefficients obeying (2) gives rise to a -periodic entire function via the formula (1).
- (ii) Every bounded -periodic holomorphic function on the upper half-plane has an expansion where the are complex coefficients such that Conversely, every infinite sequence obeying (4) gives rise to a -periodic holomorphic function which is bounded away from the real axis (i.e., bounded on for every ).

*Proof:* If is -periodic, then it can be expressed as for some function on the “multiplicative cylinder” , since the fibres of the map are cosets of the integers , on which is constant by hypothesis. As the map is a covering map from to , we see that will be holomorphic if and only if is. Thus must have a Laurent series expansion with coefficients obeying (2), which gives (1), and the inversion formula (5) follows from the usual contour integration formula for Laurent series coefficients. The converse direction to (i) also follows by reversing the above arguments.

For part (ii), we observe that the map is also a covering map from to the punctured disk , so we can argue as before except that now is a bounded holomorphic function on the punctured disk. By the Riemann singularity removal theorem (Exercise 35 of 246A Notes 3) extends to be holomorphic on all of , and thus has a Taylor expansion for some coefficients obeying (4). The argument now proceeds as with part (i).

The additive cylinder and the multiplicative cylinder can both be identified (on the level of smooth manifolds, at least) with the geometric cylinder , but we will not use this identification here.

Now let us turn attention to *doubly periodic* functions of a complex variable , that is to say functions that obey two periodicity relations

Within the world of holomorphic functions, the collection of doubly periodic functions is boring:

Proposition 2Let be an entire doubly periodic function (with periods linearly independent over ). Then is constant.

In the language of Riemann surfaces, this proposition asserts that the torus is a non-hyperbolic Riemann surface; it cannot be holomorphically mapped non-trivially into a bounded subset of the complex plane.

*Proof:* The fundamental domain (up to boundary) enclosed by is compact, hence is bounded on this domain, hence bounded on all of by double periodicity. The claim now follows from Liouville’s theorem. (One could alternatively have argued here using the compactness of the torus .

To obtain more interesting examples of doubly periodic functions, one must therefore turn to the world of *meromorphic functions* – or equivalently, holomorphic functions into the Riemann sphere . As it turns out, a particularly fundamental example of such a function is the Weierstrass elliptic function

** — 1. Doubly periodic functions — **

Throughout this section we fix two complex numbers that are linearly independent over , which then generate a lattice .

We now study the doubly periodic meromorphic functions with respect to these periods that are not identically zero. We first observe some constraints on the poles of these functions. Of course, by periodicity, the poles will themselves be periodic, and thus the set of poles forms a finite union of disjoint cosets of the lattice . Similarly, the zeroes form a finite union of disjoint cosets . Using the residue theorem, we can obtain some further constraints:

Lemma 3 (Consequences of residue theorem)Let be a doubly periodic meromorphic function (not identically zero) with periods , poles at , and zeroes at .

- (i) The sum of residues at each (i.e., we sum one residue per coset) is equal to zero.
- (ii) The number of poles (counting multiplicity, but only counting once per coset) is equal to the number of zeroes (again counting multiplicity, and once per coset).
- (iii) The sum of the poles (counting multiplicity, and working in the group ) is equal to the sum of the zeroes .

*Proof:* For (i), we first apply a translation so that none of the pole cosets intersects the fundamental parallelogram boundary ; this of course does not affect the sum of residues. Then, by the residue theorem, the sum in (i) is equal to the expression

For part (iii), we again translate so that none of the pole or zero cosets intersects , noting from part (ii) that any such translation affects the sum of poles and sum of zeroes by the same amount. By the residue theorem, it now suffices to show that

lies in the lattice . But one can rewrite this using the double periodicity as so it suffices to show that is an integer for . But (a slight modification of) the argument principle shows that this number is precisely the winding number around the origin of the image of under the map , and the claim follows.This lemma severely limits the possible number of behaviors for the zeroes and poles of a meromorphic function. To formalise this, we introduce some general notation:

Definition 4 (Divisors)

- (i) A divisor on the torus is a formal integer linear combination , where ranges over a finite collection of points in the torus (i.e., a finite collection of cosets ), and are integers, with the obvious additive group structure; equivalently, the space of divisors is the free abelian group with generators for (with the convention ).
- (ii) The number is the
degreeof a divisor , the point is thesumof , and each is theorderof the divisor at (with the convention that the order is if does not appear in the sum). A divisor isnon-negative(oreffective) if for all . We write if is non-negative (i.e., the order of is greater than or equal to that of at every point , and if and .- (iii) Given a meromorphic function (or equivalently, a doubly periodic function ) that is not identically zero, the
principal divisoris the divisor , where ranges over the zeroes and poles of , and is the order of the zero (if is a zero) or negative the order of the pole (if is a pole).- (iv) Given a divisor , we define to be the space of all meromorphic functions that are either zero, or are such that . That is to say, consists of those meromorphic functions that have at most a pole of order at if is positive, or at least zero of order if is negative.

A divisor can be viewed as an abstraction of the concept of a set of zeroes and poles (counting multiplicity). Observe that principal divisors obey the laws , when are meromorphic and non-zero. In particular, the space of principal divisors is a subgroup of the space of all divisors. By Lemma 3(ii), all principal divisors have degree zero, and from Lemma 3(iii), all principal divisors have sum zero as well. Later on we shall establish the converse claim that every divisor of degree and sum zero is a principal divisor; see Exercise 7.

Remark 5One can define divisors on other Riemann surfaces, such as the complex plane . Observe from the fundamental theorem of algebra that if one has two non-zero polynomials , then if and only if divides as a polynomial. This may give some hint as to the origin of the terminology “divisor”. The machinery of divisors turns out to have a rich algebraic and topological structure when applied to more general Riemann surfaces than tori, for instance enabling one to associate an abelian variety (the Jacobian variety) to every algebraic curve; see these 246C notes for further discussion.

It is easy to see that is always a vector space. All non-zero meromorphic functions belong to at least one of the , namely , so to classify all the meromorphic functions on , it would suffice to understand what all the spaces are.

Liouville’s theorem (in the form of Proposition 2) tells us that all elements of – that is to say, the holomorphic functions on – are constant; thus is one-dimensional. If is a negative divisor, the elements of are thus constant and have at least one zero, thus in these cases is trivial.

Now we gradually work our way up to higher degree divisors . A basic fact, proven from elementary linear algebra, is that every time one adds a pole to , the dimension of the space only goes up by at most one:

Lemma 6For any divisor and any , is a subspace of of codimension at most one. In particular, is finite-dimensional for any .

*Proof:* It is clear that is a subspace of . If has order at , then there is a linear functional that assigns to each meromorphic function the coefficient of the Laurent expansion of at (note from periodicity that the exact choice of coset representative is not relevant. A little thought reveals that the kernel of is precisely , and the first claim follows. The second claim follows from iterating the first claim, noting that any divisor can be obtained from a suitable negative divisor by the addition of finitely many poles .

Now consider the space for some point . Lemma 6 tells us that the dimension of this space is either one or two, since was one-dimensional. The space consists of functions that possibly have a simple pole at most at , and no other poles. But Lemma 3(i) tells us that the residue at has to vanish, and so is in fact in and thus is constant. (One could also argue here using the other two parts of Lemma 2; how?) So is no larger than , and is thus also one-dimensional.

Now let us study the space – the space of meromorphic functions that have at most a double pole at and no other poles. Again, Lemma 6 tells us that this space is one or two dimensional. To figure out which, we can normalise to be the origin coset . The question is now whether there is a doubly periodic meromorphic function that has a double pole at each point of . A naive candidate for such a function would be the infinite series

however this series turns out to not be absolutely convergent. Somewhat in analogy with the discussion of the Weierstrass and Hadamard factorisation theorems in Notes 1, we then proceed instead by working with the normalised function defined by the formula (6). Let us first verify that the series in (6) is absolutely convergent for . There are only finitely many with , and all the summands are finite for , so we only need to establish convergence of the tail However, from the fundamental theorem of calculus we have so to demonstrate absolute convergence it suffices to show that But a simple volume packing argument (considering the areas of the translates of the fundamental domain ) shows that the number of lattice points in any disk , is , and so by dyadic decomposition as in Notes 1, the series is absolutely convergent. Further repetition of the arguments from Notes 1 shows that the series in (6) converges locally uniformly in , and thus is holomorphic on this set. Furthermore, for any , the same arguments show that stays bounded in a punctured neighbourhood of , thus by the Riemann singularity removal theorem is equal to plus a bounded holomorphic function in the neighbourhood of . Thus is meromorphic with double poles (and vanishing residue) at every lattice point , and no other poles.Now we show that is doubly periodic, thus and for . We just prove the first identity, as the second is analogous. From (6) we have

The series on the right is absolutely convergent, and on every coset of it telescopes to zero. The claim then follows by Fubini’s theorem.By construction, lies in , and is clearly non-constant. Thus is two-dimensional, being spanned by the constant function and . By translation, we see that is two-dimensional for any other point as well.

From (6) it is also clear that the function is even: . In particular, for any avoiding the half-lattice (so that and occupy different locations in the torus ), the function has a zero at both and . By Lemma 3(ii) there are no other zeroes of this function (and this claim is also consistent with Lemma 3(iii)); thus the divisor of this function is given by

If lies in the half-lattice but not in (thus, it lies in one of the

Exercise 7 (Classification of principal divisors)

- (i) Let be four points such that . Show that the divisor is a principal divisor. (
Hint:if are all distinct, use the function If some of the coincide, use some transformed version of the Weierstrass elliptic function instead.)- (ii) Show that every divisor of degree zero and sum zero is a principal divisor.
- (iii) Two divisors are said to be
equivalentif their difference is a principal divisor. Show that two divisors are equivalent if and only if they have the same degree and same sum.- (iv) Show that the quotient group (known as the
divisor class groupor Picard group) is isomorphic (as a group) to , and that the subgroup arising from degree zero divisors (also known as the Jacobian variety of ) is isomorphic to .

Now let us study the space , where we again normalise for sake of discussion. Lemma 6 tells us that this space is two or three dimensional, being spanned by , , and possibly one other function. Note that the derivative of the meromorphic function is also doubly periodic with a triple pole at , so it lies in and is not a linear combination of or (as these have a lower order singularity at ). Thus is three-dimensional, being spanned by . A formal term-by-term differentiation of (6) gives (7). To justify (7), observe that the arguments that demonstrated the meromorphicity of the right-hand side of (6) also show the meromorphicity of (7). From Fubini’s theorem, the fundamental theorem of calculus, and (6) we see that

for any contour in from one point to another , and the claim (7) now follows from another appeal to the fundamental theorem of calculus. Of course, will then also be three-dimensional for any other point on the torus. From (7) we also see that is odd; this also follows from the even nature of . From the oddness and periodicity has to have zeroes at the half-periods ; in particular, from Lemma 3(ii) there are no other zeroes, and the principal divisor is given byTurning now to , we could differentiate yet again to generate a doubly periodic function with a fourth order pole at the origin, but we can also work with the square of the Weierstrass function. From Lemma 6 we conclude that is four-dimensional and is spanned by . In a similar fashion, is a five-dimensional space spanned by .

Something interesting happens though at . Lemma 6 tells us that this space is the span of , and possibly one other function, which will have a pole of order six at the origin. Here we have *two* natural candidates for such a function: the cube of the Weierstrass function, and the square of its derivative. Both have a pole of order exactly six and lie in , and so must be a linear combination of . But since are even and are odd, must in fact just be a linear combination of . To work out the precise combination, we see by repeating the derivation of (7) that

Exercise 8Derive (8) directly from Proposition 2 by showing that the difference between the two sides is doubly periodic and holomorphic after removing singularities.

Exercise 9 (Classification of doubly periodic meromorphic functions)

- (i) For any , show that has dimension , and every element of this space is a polynomial combination of .
- (ii) Show that every doubly periodic meromorphic function is a rational function of .

We have an alternate form of (8):

Exercise 10Define the roots , , .

- (i) Show that are distinct, and that for all . (
Hint:use (10).) Conclude in particular that , , and .- (ii) Show that the modular discriminant is equal to , and is in particular non-zero.

If we now define the elliptic curve

to be the union of a certain cubic curve in the complex plane together with the point at infinity (where the notions of “curve” and “plane” are relative to the underlying complex field rather than the more familiar real field ), then we have a map from to , with the convention that the origin is mapped to the point at infinity . For instance, the half-periods , , are mapped to the points of respectively.

Lemma 11The map defined by (11) is a bijection between and .

Among other things, this lemma implies that the elliptic curve is topologically equivalent (i.e., homeomorphic to) a torus, which is not an entirely obvious fact (though if one squints hard enough, the real analogue of an elliptic curve does resemble a distorted slice of a torus embedded in ).

*Proof:* Clearly is the only point that maps to , and (from (10)) the half-periods are the only points that map to . It remains to show that all the other points arise via from exactly one element of . The function has exactly two zeroes by Lemma 3(ii), which lie at for some as is even; since , is not equal to , hence is not a half-period. As is odd, the map (11) must therefore map to the two points of the elliptic curve that lie above , and the claim follows.

Analogously to the Riemann sphere , the elliptic curve can be given the structure of a Riemann surface, by prescribing the following charts:

- (i) When is a point in other than or , then locally is the graph of a holomorphic branch of the square root of near , and one can use as a coordinate function in a sufficiently small neighbourhood of .
- (ii) In the neighbourhood of for some , the function has a simple zero at and so has a local inverse that maps a neighbourhood of to a neighbourhood of , and a point sufficiently near can be parameterised by . One can then use as a coordinate function in a neighbourhood of .
- (iii) A neighbourhood of consists of and the points in the remaining portion of with sufficiently large; then is asymptotic to a square root of , so in particular and should both go to zero as goes to infinity in . We rewrite the defining equation of the curve in terms of and as . The function has a simple zero at zero and thus has a holomorphic local inverse that maps to , and we have in a neighbourhood of infinity. We can then use as a coordinate function in a neighbourhood of , with the convention that this coordinate function vanishes at infinity.

It is then a tedious but routine matter to check that has the structure of a Riemann surface. We then claim that the bijection defined by (11) is holomorphic, and thus a complex diffeomorphism of Riemann surfaces. In the neighbourhood of any point of the torus other than the origin , maps to a neighbourhood of finite point of , including the three points , the holomorphicity is a routine consequence of composing together the various local holomorphic functions and their inverses. In the neighbourhood of the origin , maps for small to a point of with a Laurent expansion

from the Laurent expansions of , so in particular the coordinate takes the form where the error term is holomorphic, with mapping to . In particular the map is a local complex diffeomorphism here and again we have holomorphicity. We thus conclude that the elliptic curve is complex diffeomorphic to the torus using the map . From Exercise 9, the meromorphic functions on may be identified with the rational functions on .While we have shown that all tori are complex diffeomorphic to elliptic curves, the converse statement that all elliptic curves are diffeomorphic to tori will have to wait until the next section for a proof, once we have set up the machinery of modular forms.

Exercise 12 (Group law on elliptic curves)

- (i) Let be three distinct elements of the torus that are not equal to the origin . Show that if and only if the three points , , are collinear in , in the sense that they lie on a common complex line for some complex numbers with not both zero.
- (ii) What happens in (i) if (say) and agree? What about if ?
- (iii) Using (i), (ii), give a purely geometric definition of a group addition law on the elliptic curve which is compatible with the group addition law on the torus via (11). (We remark that the associativity property of this law is not obvious from a purely geometric perspective, and is related to the Cayley-Bacharach theorem in classical geometry; see this previous blog post.)

Exercise 13 (Addition law)Show that for any lying in distinct cosets of , one has

Exercise 14 (Special case of Riemann-Roch)

- (i) Show that if two divisors are equivalent (in the sense of Exercise 7(iii)), then the vector spaces and are isomorphic (in particular, they have the same dimension).
- (ii) If is a divisor of some degree , show that the dimension of the space is zero if , equal to if , equal to if and has non-zero sum, and equal to if and has zero sum. (
Hint:use Exercise 7(iii) and part (i) to replace with an equivalent divisor of a simple form.)- (iii) Verify the identity for any divisor . This is a special case of the more general Riemann-Roch theorem, discussed in these 246C notes.

Exercise 15 (Elliptic integrals)

Remark 16The integral is an example of an elliptic integral; many other elliptic integrals (such as the integral arising when computing the perimeter of an ellipse) can be transformed into this form (or into a closely related integral) by various elementary substitutions. Thus the Weierstrass elliptic function can be employed to evaluate elliptic integrals, which may help explain the terminology “elliptic” that occurs throughout these notes. In 246C notes we will introduce the notion of a meromorphic -form on a Riemann surface. The identity (12) can then be interpreted in this language as the differential form identity , where are the standard coordinates on the elliptic curve ; the meromorphic -form is initially only defined on outside of the four points , but this identity in fact reveals that the form extends holomorphically to all of ; it is an example of what is known as an Abelian differential of the first kind.

Remark 17The elliptic curve (for various choices of parameters ) can be defined in other fields than the complex numbers (though some technicalities arise in characteristic two and three due to the pathological behaviour of the discriminant in those cases). On the other hand, the Weierstrass elliptic function is a transcendental function which only exists in complex analysis and does not have a direct analogue in other fields. So this connection between elliptic curves and tori is specific to the complex field. Nevertheless, many facts about elliptic curves that were initially discovered over the complex numbers through this complex-analytic link to tori, were then reproven by purely algebraic means, so that they could be extended without much difficulty to many other fields than the complex numbers, such as finite fields. (For instance, the role of the complex torus can be replaced by the Jacobian variety, which was briefly introduced in Exercise 7.) Elliptic curves over such fields are of major importance in number theory (and cryptography), but we will not discuss these topics further here.

** — 2. Modular functions and modular forms — **

In Exercise 32 of 246A Notes 5, it was shown that two tori and are complex diffeomorphic if and only if one has

for some integers with . From this it is not difficult to see that if are two lattices in , then and are diffeomorphic if and only if for some , i.e., the lattices are complex dilations of each other.
Let us write for the set of all tori quotiented by the equivalence relation of complex diffeomorphism; this is the (classical, level one, noncompactified) modular curve. By the above discussion, this set can also be identified with the set of pairs of linearly independent (over ) complex numbers quotiented by the equivalence relation given implicitly by (13). One can simplify this a little by observing that any pair is equivalent to for some in the upper half-plane , namely either or depending on the relative phases of and ; this quantity is known as the *period ratio*. From (13) (swapping the roles of as necessary), we then see that two pairs are equivalent if one has

If we use the relation to write

we see that approaches the real line as if is non-zero; also, if is zero, then from we must have , and will either have imaginary part going off to infinity (if goes to infinity) or real part going to infinity (if is bounded and goes to infinity). In all cases we then conclude that goes to infinity as goes to infinity, uniformly for in any fixed compact subset of , which makes the action of on proper (for any compact set , one has intersecting for at most finitely many . If acted freely on (i.e., any element of other than the identity and negative identity has no fixed points in ), then the quotient would be a Riemann surface by the discussion in Section 2 of 246A Notes 5. Unfortunately, this is not quite true. For instance, the point is fixed by the Möbius transformation coming from the rotation matrix of , and the point is similarly fixed by the transformation coming from the matrix of . Geometrically, these fixed points come from the fact that the Gaussian integgers are invariant with respect to rotation by , while the Eisenstein integers are invariant with respect to rotation by . On the other hand, these are basically the only two places where the action is not free:

Exercise 18Suppose that is an element of which is fixed by some element of which is not the identity or negative identity. Let be the lattice .

- (i) Show that obeys a dilation invariance for some complex number which is not real.
- (ii) Show that the dilation in part (i) must have magnitude one. (Hint: look at a non-zero element of of minimal magnitude.)
- (iii) Show that there is no rotation invariance with . (Hint: again, work with a non-zero element of of minimal magnitude, and use the fact that is closed under addition and subtraction. It may help to think geometrically and draw plenty of pictures.)
- (iv) Show that is equivalent to either the Gaussian lattice or the Eisenstein lattice , and conclude that the period ratio is equivalent to either or .

Remark 19The conformal map on the complex numbers preserves the Gaussian integers and thus descends to a conformal map from the Gaussian torus to itself; similarly the conformal map preserves the Eisenstein integers and thus descends to a conformal map from the Eisenstein torus to itself. These rare examples of complex tori equipped with additional conformal automorphisms are examples of tori (or elliptic curves) endowed with complex multiplication. There are additional examples of elliptic curves endowed with conformalendomorphismsthat are still considered to have complex multiplication, and have a particularly nice algebraic number theory structure, but we will not pursue this topic further here.

Remark 20The fact that the action of on lattices contains fixed points is somewhat annoying, as it prevents one from immediately viewing the modular curve as a Riemann surface. However by passing to a suitable finite index subgroup of , one can remove these fixed points, leading to a theory that is cleaner in some respects. For instance, one can work with the congruence group , which roughly speaking amounts to decorating the lattices (or their tori ) with an additional “-marking” that eliminates the fixed points. This leads to a modification of the theory which is for instance well suited for studying theta functions; the role of the -invariant in the discussion below is then played by the modular lambda function , which also gives a uniformisation of the twice-punctured complex plane . However we will not develop this parallel theory further here.

If we let be the elements of not equivalent to or , and the equivalence class of tori not equivalent to the Gaussian torus or the Eisenstein torus , then can be viewed as the quotient of the Riemann surface by the free and proper action of , so it has the structure of a Riemann surface; can thus be thought of as the Riemann surface with two additional points added. Later on we will also add a third point (known as the *cusp*) to the Riemann surface to compactify it to .

A function on the modular curve can be thought of, equivalently, as a function that is -invariant in the sense that for all and , or equivalently that one has the identity

whenever and are integers with . Similarly if takes values in the Riemann sphere rather than . If is holomorphic (resp. meromorphic) on , this will in particular define a holomorphic (resp. meromorphic) function on , and morally to all of as well (although we have not yet defined a Riemann structure on all of ).We define a modular function to be a meromorphic function on that obeys the condition (15), and which also has at most polynomial growth at the cusp in the sense that one has a bound of the form

for all with sufficiently large imaginary part, and some constants (this bound is needed for technical reasons to ensure “meromorphic” behaviour at the cusp , as opposed to an essential singularity). Specialising to the matrices we see that the condition (15) in particular implies the -periodicity and the inversion law for all . Conversely, these two special cases of (15) imply the general case:

Exercise 21

- (i) Let be two elements of with . Show that it is possible to transform the quadruplet to the quadruplet after a finite number of applications of the moves and ({
Hint:use the principle of infinite descent, applying the moves in a suitable order to decrease the lengths of and when the dot product is not too small, taking advantage of the Lagrange identity to determine when this procedure terminates. It may help to think geometrically and draw plenty of pictures.) Conclude that the two matrices (17) generate all of .- (ii) Show that a function obeys (15) if and only if it obeys both (18) and (19).

Exercise 22 (Standard fundamental domain)Define thestandard fundamental domainfor to be the set

- (i) Show that every lattice is equivalent (up to dilations) to a lattice with , with unique except when it lies on the boundary of , in which case the lack of uniqueness comes either from the pair for some , or from the pair for some . (
Hint:arrange so that is a non-zero element of of minimal magnitude.)- (ii) Show that can be identified with the fundamental domain after identifying with for , and with for . Show also that the set is then formed the same way, but first deleting the points from .

We will give some examples of modular functions (beyond the trivial example of constant functions) shortly, but let us first observe that when one differentiates a modular function one gets a more general class of function, known as a modular form. In more detail, observe from (14) that the derivative of the Möbius transformation is , and hence by the chain rule and (15) the derivative of a modular function would obey the variant law

Motivated by this, we can define a (weakly)

Exercise 23Let be a natural number. Show that a function obeys (20) if and only if it is -periodic in the sense of (18) and obeys the law for all .

Exercise 24 (Lattice interpretation of modular forms)Let be a modular form of weight . Show that there is a unique function from lattices to complex numbers such that for all , and such that one has the homogeneity relation for any lattice and non-zero complex number .

Observe that the product of a modular form of weight and a modular form of weight is a modular form of weight , and that the ratio of two modular forms of weight will be a modular function (if the denominator is not identically zero). Also, the space of modular forms of a given weight is a vector space, as is the space of modular functions. This suggests a way to generate non-trivial modular functions, by first locating some modular forms and then taking suitable rational combinations of these forms.

Somewhat analogously to how we used Lemma 3 to investigate the spaces for divisors on a torus, we will investigate the space of modular forms via the following basic formula:

Theorem 25 (Valence formula)Let be a modular form of weight , not identically zero. Then we have where is the order of vanishing of at , is the order of vanishing of (i.e., viewed as a function of the nome ) at , and ranges over the zeroes of that are not equivalent to , with just one zero counted per equivalence class. (This will be a finite sum.)

Informally, this formula asserts that the point only “deserves” to be counted in with multiplicity due to its order stabiliser, while the point only “deserves” to be counted in with multiplicity due to its order stabiliser. (The cusp has an infinite stabiliser, but this is compensated for by taking the order with respect to the nome variable rather than the period ratio variable .) The general philosophy of weighting points by the reciprocal of the order of their stabiliser occurs throughout mathematics; see this blog post for more discussion.

*Proof:* Firstly, from Exercise 22, we can place all the zeroes in the fundamental domain . When parameterised in terms of the nome , this domain is compact, hence has only finitely many zeros, so the sum in (22) is finite.

As in the proof of Lemma 3(ii), we use the residue theorem. For simplicity, let us first suppose that there are no zeroes on the boundary of the fundamental domain except possibly at the cusp . Then for large enough, we have from the residue theorem that

where is the closed contour consisting of the polygonal path concatenated with the circular arc . From the -periodicity, the contribution of the two vertical edges and cancel each other out. The contribution of the horizontal edge can be written using the change of variables as which by the residue theorem is equal to . Finally, using the modularity (21), one calculates that the contribution of the left arc is equal to minus the contribution of the right arc . This gives the proof of the valence theorem in the case that there are no zeroes on the boundary of .Suppose now that there is a zero on the right edge of , and hence also on the left edge by periodicity, for some . One can account for this zero by perturbing the contour to make a little detour to the right of (e.g., by a circular arc), and a matching detour to the right of . One can then verify that the same argument as before continues to work, with this boundary zero being counted exactly once. Similarly, if there is a zero on the left arc for some , and hence also at by modularity, one can make a detour slightly above and slightly below (with the two detours being related by the transform to ensure cancellation), and again we can argue as before. If instead there is a zero at , one makes an (approximately) semicircular detour above ; in this case the detour does not cancel out, but instead contributes a factor of in the limit as the radius of the detour goes to zero. Finally, if there is a zero at (and hence also at ), one makes detours by two arcs of angle approximately at these two points; these two (approximate) sixth-circles end up contributing a factor of in the limit, giving the claim.

Exercise 26 (Quick applications of the valence formula)

- (i) Let be a modular form of weight , not identically zero. Show that is equal to or an even number that is at least .
- (ii) (Liouville theorem for ) If is a modular form of weight zero, show that it is constant. (Hint: apply the valence theorem to various shifts of by constant.)
- (iii) For , show that the vector space of modular forms of weight is at most one dimensional. (Hint: in these cases, there are a very limited number of solutions to the equation with natural numbers.)
- (iv) Show that there are no cusp forms of weight when or , and for the space of cusp forms of weight is at most one dimensional.
- (v) Show that for any , the space of cusp forms of weight is a subspace of the space of modular forms of weight of codimension at most one, and that both spaces are finite-dimensional.

A basic example of modular forms are provided by the Eisenstein series

that we have already encountered for even integers greater than two (we ignore the odd Eisenstein series as they vanish). We can view this as a function on by the formula Observe that if are integers with , then using the matrix inverse in . Inserting this into (23), (24) we conclude that (Compare also with Exercise 24.) Also, from (23), (24) we have The series here is locally uniformly convergent for , so is holomorphic. Also, using the bounds for non-zero , while where is the famous Riemann zeta function we conclude on summing in and using the hypothesis that In particular, is bounded at infinity. Summarising, we have established that the Eisenstein series is a modular form of weight , which is not identically zero (since it approaches the non-zero value at the cusp ). Combining this with Exercise 26(iii), we see that we have completely classified the modular forms of weight for , namely they are the scalar multiples of . For instance, the coefficients and appearing in the previous section are modular forms of weight and weight respectively, and the modular discriminant from Exercise 10 is a modular form of weight . From that exercise, this modular form never vanishes on , hence by the valence formula it must have a simple zero at , and in particular is a cusp form. From Exercise 26 it is the unique cusp form of weight , up to constants.

Exercise 27Give an alternate proof that is a cusp form, not using the valence identity, by first establishing that and .

We can now create our first non-trivial modular function, the -invariant

The factor of is traditional, as it gives a nice normalisation at , as we shall see later. One can take advantage of complex multiplication to compute two special values immediately:

Lemma 28We have and .

*Proof:* Using the rotation symmetry we see that , hence which implies that and hence . Similarly, using the rotation symmetry we have , hence . (One can also use the valence formulae to get the vanishing ).

Being modular, we can think of as a map from to . We have the following fundamental fact:

Proposition 29The map is a bijection.

*Proof:* Note that for any , if and only if is a zero of . It thus suffices to show that for every , the zeroes of the function in consist of precisely one orbit of . This function is a modular form of weight that does not vanish at infinity (since does not vanish while does). By the valence formula, we thus have

- has a simple zero at precisely one -orbit, not equivalent to or .
- has a double zero at (and equivalent points), and no other zeroes.
- has a triple zero at (and equivalent points), and no other zeroes.

Note that this proof also shows that has a double zero at and has a triple zero at , but that has a simple zero for any not equivalent to or .

We can now give the entire modular curve the structure of a Riemann surface by declaring to be the coordinate function. This is compatible with the existing Riemann surface structure on since was already holomorphic on this portion of the curve. Any modular function can then factor as for some meromorphic function that is initially defined on the punctured complex plane ; but from meromorphicity of on and at infinity we see that blows up at an at most polynomial rate as one approaches , , or , and so is in fact a meromorphic function on the entire Riemann sphere and is thus a rational function (Exercise 19 of 246A Notes 4). We conclude

Proposition 30Every modular function is a rational function of the -invariant .

Conversely, it is clear that every rational function of is modular, thus giving a satisfactory description of the modular functions.

Exercise 31Show that every modular function is the ratio of two modular forms of equal weight (with the denominator not identically zero).

Exercise 32 (All elliptic curves are tori)Let be two complex numbers with . Show that there is a lattice such that and , so in particular the elliptic curve is complex diffeomorphic to a torus .

Remark 33By applying some elementary algebraic geometry transformations one can show that any (smooth, irreducible) cubic plane curve generated by a polynomial of degree is a Riemann surface complex diffeomorphic to a torus after adding some finite number of points at infinity; also, some degree curves such as can also be placed in this form. However we will not detail the required transformations here.

A famous application of the theory of the -invariant is to give a short Riemann surface-based proof of the the little Picard theorem (first proven in Theorem 55 of 246A Notes 4):

Theorem 34 (Little Picard theorem)Let be entire and non-constant. Then omits at most one point of .

*Proof:* Suppose for contradiction that omits at least two points of . By applying a linear transformation, we may assume that omits the points and . Then is a holomorphic function from to . Since the domain is simply connected, lifts to a holomorphic function from to . Since is complex diffeomorphic to a disk, this lift must be constant by Liouville’s theorem, hence is constant as required. (This is essentially Picard’s original proof of this theorem.)

The great Picard theorem can also be proven by a more sophisticated version of these methods, but it requires some study of the possible behavior of elements of ; see Exercise 37 below.

All modular forms are -periodic, and hence by Proposition 1 should have a Fourier expansion, which is also a Laurent expansion in the nome. As it turns out, the Fourier coefficients often have a highly number-theoretic interpretation. This can be illustrated with the Eisenstein series ; here we follow the treatment in Stein-Shakarchi. To compute the Fourier coefficients we first need a computation:

Exercise 35Let and , and let be the nome. Establish the identity in two different ways:

- (i) By applying the Poisson summation formula (Proposition 3(v) of Notes 2).
- (ii) By first establishing the identity by applying Proposition 1 to the difference of the two sides, and differentiating in . (It is also possible to establish (27) from differentiating and then manipulating the identities in Exercises 25 or 27 of Notes 1.)

From (25), (26) (and symmetry) one has

and hence by the above exercise Since it is not difficult to show that the double sum here is absolutey convergent and can be rearranged as we please. If we group the terms based on the product we thus have the Fourier expansion where the divisor function is defined by where the sum is over those natural numbers that divide . Thus for instance and so after some calculation and therefore thus the factor of in the definition of the -invariant normalises the “residue” of at infinity to equal .

Remark 36If one expands out a few more terms in the above expansions, one can calculate The various coefficients in here have several remarkable algebraic properties. For instance, applying this expansion at for a natural number , so that , one obtains the approximation Now for certain values of , most famously , the torus admits a complex multiplication that allows for computation of the -invariant by algebraic means (think of this as a more advanced version of Lemma 28; it is closely related to the fact that the ring of algebraic integers in admit unique factorisation, see these previous notes for some related discussion). For instance, one can eventually establish that which eventually leads to the famous approximation (first observed by Hermite, but also attributed to Ramanujan via an April Fools’ joke of Martin Gardner) which is accurate to twelve decimal places. The remaining coefficients have a remarkable interpretation as dimensions of components of a certain representation of the monster group known as the moonshine module, a phenomenon known as monstrous moonshine. For instance, the smallest irreducible representation of the monster group has dimension , precisely one less than the coefficient of . The Fourier coefficients of the (normalised) modular discriminant, form a sequence known as the Ramanujan function and obeys many remarkable properties. For instance, there is the totally non-obvious fact that this function is multiplicative in the sense that whenever are coprime; see Exercise 43.

Exercise 37 (Great Picard theorem)

- (i) Show that every fractional linear transformation on with , is either of finite order (elliptic case), conjugate to a translation for some after conjugating by another fractional linear transformation (parabolic case), or conjugate to a dilation for some after conjugating by another fractional linear transformation (hyperbolic case). (
Hint:study the eigenvalues and eigenvectors of , based on the value of the trace and in particular whether the magnitude of the trace is less than two, equal to two, or greater than two. Note that the trace also has to be an integer.)- (ii) Let be holomorphic. Show that there exists a holomorphic function such that for all , as well as a fractional linear transformation with and such that for all .
- (iii) If the transformation in (ii) is in the elliptic case of (i), show that is bounded in a neighbourhood of , and hence has a removable singularity at the origin. (
Hint:will have some finite period and can thus be studied using Proposition 1 after applying a Möbius transform to map to a disk.)- (iv) If the transformation in (ii) is in the hyperbolic case of (i), show that is bounded in a neighbourhood of , and hence has a removable singularity at the origin. (
Hint:The standard branch of maps to an annulus, and is invariant with respect to the dilation action . Use this to create a bounded -periodic holomorphic function on .)- (v) If the transformation in (ii) is in the parabolic case of (i), show that exhibits at most polynomial growth as one approaches , and hence has at most a pole at the origin. (
Hint:If for instance , then is -periodic and takes values in , and one can now repeat the arguments of (iii). Also use the expansion (28).)- (vi) Use the previous parts of this exercise to give another proof of the great Picard theorem (Theorem 56 of 245A Notes 4): the image of a holomorphic function in a punctured disk with an essential singularity at omits at most one value of .

Exercise 38 (Dimension of space of modular forms)

- (i) If is an even natural number, show that the dimension of the space of modular forms of weight is equal to except when is equal to mod , in which case it is equal to . (
Hint:for this follows from Exercise 26; to cover the larger ranges of , use the modular discriminant to show that the space of cusp forms of weight is isomorphic to the space of modular forms of weight .- (ii) If is an even natural number, show that a basis for the space of modular forms of weight is provided by the powers where range over natural numbers (including zero) with .

Thus far we have constructed modular forms and modular functions starting from Eisenstein series . There is another important, and seemingly quite different, way to generate modular forms coming from theta functions. Typically these functions are not *quite* modular in the sense given in these notes, but are close enough that after some manipulation one can transform theta functions into modular forms. The simplest example of a theta function is the Jacobi theta function

Exercise 39Define the Dedekind eta function by the formula or in terms of the nome where is one of the roots of .

- (i) Establish the modified -periodicity and the modified modularity using the standard branch of the square root. (
Hint:a direct application of Poisson summation applied to gives a sum that looks somewhat like but with different numerical constants (in particular, one sees terms like instead of arising). Split the index of summation into three components , , based on the residue classes modulo and rearrange each component separately.)- (ii) Establish the identity (
Hint:show that both sides are cusp forms of weight that vanish like near the cusp.)

Remark 40The relationship between and the power of the eta function can be interpreted (after some additional effort) as a relation between the modular discriminant and the theta function of a certain highly symmetric -dimensional lattice known as the Leech lattice, but we will not pursue this connection further here.

The function has a remarkable factorisation coming from Euler’s pentagonal number theorem

so that There are many proofs of the pentagonal number theorem in the literature. One approach is to first establish the more general Jacobi triple product identity:

Theorem 41 (Jacobi triple product identity)For any and , one has

Observe that by replacing by and with we have

and this gives the identity (31) after splitting the integers into the three residue classes modulo . One can obtain many further identities of this type by other substitutions; for instance, by setting in the triple product identity, one obtains
*Proof:* Let us denote the left-hand side and right-hand side of (33) by and respectively. For fixed , both sides are clearly holomorphic in , with . Our strategy in showing that and agree (following Stein-Shakarchi) is to first observe that they have many of the same periodicity properties. We clearly have -periodicity

Remark 42Another equivalent form of (32) is where is the partition function of – the number of ways to represent as the sum of positive integers (up to rearrangement). Among other things, this formula can be used to ascertain the asymptotic growth of (which turns out to roughly be of the order of , as famously established by Hardy and Ramanujan).

Theta functions can be used to encode various number-theoretic quantities involving quadratic forms, such as sums of squares. For instance, from (30) and collecting terms one obtains the formula

for any natural number , where denotes the number of ways to express a natural number as the sum of squares of integers. From Fourier inversion (Proposition 1 and a rescaling) one then has a representation for any , which allows one to obtain asymptotics for when is large through estimation of the theta function (this is an example of the circle method); moreover, explicit identities relating the theta function to other near-modular forms (such as the Eisenstein series and their relatives) can be used to obtain exact formulae for for small values of that can be used for instance to establish the famous Lagrange four-square theorem that all natural numbers are the sum of four squares. We refer the reader to the Stein-Shakarchi text for an exposition of this connection.

Exercise 43 (Hecke operators)Let be a natural number.Simultaneous eigenfunctions of the Hecke operators are known as

- (i) If is a modular form of weight , and is the corresponding function on lattices given by Exercise 24, and is a positive natural number, show that there is a unique modular form of weight whose corresponding function on lattices is related to by the formula where the sum ranges over all sublattices of whose index is equal to . Show that is a linear operator on the space of weight modular forms that also maps the space of weight cusp forms to itself; this operator is known as a Hecke operator.
- (ii) Give the more explicit formula
- (iii) Show that the Hecke operators all commute with each other, thus whenever is a modular form of weight and are positive natural numbers. Furthermore show that if are coprime.
- (iv) If is a modular form of weight with Fourier expansion , show that for any coprime positive integers that the coefficient of is equal to .
- (v) Establish the multiplicativity of the Ramanujan tau function (the Fourier coefficients of the modular discriminant). (
Hint:use the one-dimensionality of the space of cusp forms of weight to conclude that is a simultaneous eigenfunction of the Hecke operators.)Hecke eigenfunctionsand are of major importance in number theory.

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In Exercise 5 (and Lemma 1) of 246A Notes 4 we already observed some links between complex analysis on the disk (or annulus) and Fourier series on the unit circle:

- (i) Functions that are holomorphic on a disk are expressed by a convergent Fourier series (and also Taylor series) for (so in particular ), where conversely, every infinite sequence of coefficients obeying (1) arises from such a function .
- (ii) Functions that are holomorphic on an annulus are expressed by a convergent Fourier series (and also Laurent series) , where conversely, every doubly infinite sequence of coefficients obeying (2) arises from such a function .
- (iii) In the situation of (ii), there is a unique decomposition where extends holomorphically to , and extends holomorphically to and goes to zero at infinity, and are given by the formulae where is any anticlockwise contour in enclosing , and and where is any anticlockwise contour in enclosing but not .

This connection lets us interpret various facts about Fourier series through the lens of complex analysis, at least for some special classes of Fourier series. For instance, the Fourier inversion formula becomes the Cauchy-type formula for the Laurent or Taylor coefficients of , in the event that the coefficients are doubly infinite and obey (2) for some , or singly infinite and obey (1) for some .

It turns out that there are similar links between complex analysis on a half-plane (or strip) and Fourier *integrals* on the real line, which we will explore in these notes.

We first fix a normalisation for the Fourier transform. If is an absolutely integrable function on the real line, we define its Fourier transform by the formula

From the dominated convergence theorem will be a bounded continuous function; from the Riemann-Lebesgue lemma it also decays to zero as . My choice to place the in the exponent is a personal preference (it is slightly more convenient for some harmonic analysis formulae such as the identities (4), (5), (6) below), though in the complex analysis and PDE literature there are also some slight advantages in omitting this factor. In any event it is not difficult to adapt the discussion in this notes for other choices of normalisation. It is of interest to extend the Fourier transform beyond the class into other function spaces, such as or the space of tempered distributions, but we will not pursue this direction here; see for instance these lecture notes of mine for a treatment.

Exercise 1 (Fourier transform of Gaussian)If is a coplex number with and is the Gaussian function , show that the Fourier transform is given by the Gaussian , where we use the standard branch for .

The Fourier transform has many remarkable properties. On the one hand, as long as the function is sufficiently “reasonable”, the Fourier transform enjoys a number of very useful identities, such as the Fourier inversion formula

the Plancherel identity and the Poisson summation formula On the other hand, the Fourier transform also intertwines various

Exercise 2 (Decay of implies regularity of )Let be an absolutely integrable function.

- (i) If has super-exponential decay in the sense that for all and (that is to say one has for some finite quantity depending only on ), then extends uniquely to an entire function . Furthermore, this function continues to be defined by (3).
- (ii) If is supported on a compact interval then the entire function from (i) obeys the bounds for . In particular, if is supported in then .
- (iii) If obeys the bound for all and some , then extends uniquely to a holomorphic function on the horizontal strip , and obeys the bound in this strip. Furthermore, this function continues to be defined by (3).
- (iv) If is supported on (resp. ), then there is a unique continuous extension of to the lower half-plane (resp. the upper half-plane which is holomorphic in the interior of this half-plane, and such that uniformly as (resp. ). Furthermore, this function continues to be defined by (3).
Hint:to establish holomorphicity in each of these cases, use Morera’s theorem and the Fubini-Tonelli theorem. For uniqueness, use analytic continuation, or (for part (iv)) the Cauchy integral formula.

Later in these notes we will give a partial converse to part (ii) of this exercise, known as the Paley-Wiener theorem; there are also partial converses to the other parts of this exercise.

From (3) we observe the following intertwining property between multiplication by an exponential and complex translation: if is a complex number and is an absolutely integrable function such that the modulated function is also absolutely integrable, then we have the identity

whenever is a complex number such that at least one of the two sides of the equation in (7) is well defined. Thus, multiplication of a function by an exponential weight corresponds (formally, at least) to translation of its Fourier transform. By using contour shifting, we will also obtain a dual relationship: under suitable holomorphicity and decay conditions on , translation by a complex shift will correspond to multiplication of the Fourier transform by an exponential weight. It turns out to be possible to exploit this property to derive many Fourier-analytic identities, such as the inversion formula (4) and the Poisson summation formula (6), which we do later in these notes. (The Plancherel theorem can also be established by complex analytic methods, but this requires a little more effort; see Exercise 8.)The material in these notes is loosely adapted from Chapter 4 of Stein-Shakarchi’s “Complex Analysis”.

** — 1. The inversion and Poisson summation formulae — **

We now explore how the Fourier transform of a function behaves when extends holomorphically to a strip. For technical reasons we will also impose a fairly mild decay condition on at infinity to ensure integrability. As we shall shortly see, the method of contour shifting then allows us to insert various exponentially decaying factors into Fourier integrals that make the justification of identities such as the Fourier inversion formula straightforward.

Proposition 3 (Fourier transform of functions holomorphic in a strip)Let , and suppose that is a holomorphic function on the strip which obeys a decay bound of the form for all , , , and some and (or in asymptotic notation, one has whenever and ).

- (i) (Translation intertwines with modulation) For any in the strip , the Fourier transform of the function is .
- (ii) (Exponential decay of Fourier transform) For any , there is a quantity such that for all (or in asymptotic notation, one has for and ).
- (iii) (Partial Fourier inversion) For any and , one has and
- (iv) (Full Fourier inversion) For any , the identity (4) holds for this function .
- (v) (Poisson summation formula) The identity (6) holds for this function .

*Proof:* We begin with (i), which is a standard application of contour shifting. Applying the definition (3) of the Fourier transform, our task is to show that

For (ii), we apply (i) with to observe that the Fourier transform of is . Applying (8) and the triangle inequality we conclude that

for both choices of sign and all , giving the claim.For the first part of (iii), we write . By part (i), we have , so we can rewrite the desired identity as

By (3) and Fubini’s theorem (taking advantage of (8) and the exponential decay of as ) the left-hand side can be written as But a routine calculation shows that giving the claim. The second part of (iii) is proven similarly.To prove (iv), it suffices in light of (iii) to show that

for any . The left-hand side can be written after a change of variables as On the other hand, from dominated convergence as in the proof of (i) we have while from the Cauchy integral formula one has giving the claim.Now we prove (v). Let . From (i) we have

and for any . If we sum the first identity for we see from the geometric series formula and Fubini’s theorem that and similarly if we sum the second identity for we have Adding these two identities and changing variables, we conclude that We would like to use the residue theorem to evaluate the right-hand side, but we need to take a little care to avoid the poles of the integrand , which are at the integers. Hence we shall restrict to be a half-integer . In this case, a routine application of the residue theorem shows that Noting that stays bounded for in or when is a half-integer, we also see from dominated convergence as before that The claim follows.

Exercise 4 (Hilbert transform and Plemelj formula)Let be as in Proposition 3. Define the Cauchy-Stieltjes transform by the formula

- (i) Show that is holomorphic on and has the Fourier representation in the upper half-plane and in the lower half-plane .
- (ii) Establish the Plemelj formulae and uniformly for any , where the Hilbert transform of is defined by the principal value integral
- (iii) Show that is the unique holomorphic function on that obeys the decay bound and solves the (very simple special case of the) Riemann-Hilbert problem uniformly for all , with both limits existing uniformly in .
- (iv) Establish the identity where the signum function is defined to equal for , for , and for .
- (v) Assume now that has mean zero (i.e., ). Show that extends holomorphically to the strip and obeys the bound (8) (but possibly with a different constant , and with replaced by a slightly smaller quantity ), with the identity holding for . ({
Hint:To exploit the mean value hypothesis to get good decay bounds on , write as the sum of and and use the mean value hypothesis to manage the first term. For the contribution of the second term, take advantage of contour shifting to avoid the singularity at . One may have to divide the integrals one encounters into a couple of pieces and estimate each piece separately.)- (vi) Continue to assume that has mean zero. Establish the identities and (
Hint:for the latter inequality, square both sides of (9) and use (iii).)

Exercise 5 (Kramers-Kronig relations)Let be a continuous function on the upper half-plane which is holomorphic on the interior of this half-plane, and obeys the bound for all non-zero in this half-plane and some . Establish the Kramers-Kronig relations and relating the real and imaginary parts of to each other.

Exercise 6

- (i) By applying the Poisson summation formula to the function , establish the identity for any positive real number . Explain why this is consistent with Exercise 24 from Notes 1.
- (ii) By carefully taking limits of (i) as , establish yet another alternate proof of Euler’s identity

Exercise 7For in the upper half-plane , define the theta function . Use Exercise 1 and the Poisson summation formula to establish the modular identity for such , where one takes the standard branch of the square root.

Exercise 8 (Fourier proof of Plancherel identity)Let be smooth and compactly supported. For any with , define the quantityRemarkably, this proof of the Plancherel identity generalises to a nonlinear version involving a trace formula for the scattering transform for either Schrödinger or Dirac operators. For Schrödinger operators this was first obtained (implicitly) by Buslaev and Faddeev, and later more explicitly by by Deift and Killip. The version for Dirac operators more closely resembles the linear Plancherel identity; see for instance the appendix to this paper of Muscalu, Thiele, and myself. The quantity is a component of a nonlinear quantity known as the

- (i) When is real, show that . (
Hint:find a way to rearrange the expression .)- (ii) For non-zero, show that , where the implied constant in the notation can depend on . (
Hint:integrate by parts several times.)- (iii) Establish the Plancherel identity (5).
transmission coefficientof a Dirac operator with potential and spectral parameter (or , depending on normalisations).

The Fourier inversion formula was only established in Proposition 3 for functions that had a suitable holomorphic extension to a strip, but one can relax the hypotheses by a limiting argument. Here is one such example of this:

Exercise 9 (More general Fourier inversion formula)Let be continuous and obey the bound for all and some . Suppose that the Fourier transform is absolutely integrable.

Exercise 10 (Laplace inversion formula)Let be a continuously twice differentiable function, obeying the bounds for all and some .The Laplace-Mellin inversion formula in fact holds under more relaxed decay and regularity hypotheses than the ones given in this exercise, but we will not pursue these generalisations here. The limiting integral in (10) is also known as the

- (i) Show that the Fourier transform obeys the asymptotic for any non-zero .
- (ii) Establish the principal value inversion formula for any positive real . (
Hint:modify the proof of Exercise 9(ii).) What happens when is negative? zero?- (iii) Define the Laplace transform of for by the formula Show that is continuous on the half-plane , holomorphic on the interior of this half-plane, and obeys the
Laplace-Mellin inversion formulafor any and , where is the line segment contour from to . Conclude in particular that the Laplace transform is injective on this class of functions .Bromwich integral, and often written (with a slight abuse of notation) as . The Laplace transform is a close cousin of the Fourier transform that has many uses; for instance, it is a popular tool for analysing ordinary differential equations on half-lines such as .

Exercise 11 (Mellin inversion formula)Let be a continuous function that is compactly supported in . Define theMellin transformby the formula Show that is entire and one has theMellin inversion formulafor any and . The regularity and support hypotheses on can be relaxed significantly, but we will not pursue this direction here.

Exercise 12 (Perron’s formula)Let be a function which is of subpolynomial growth in the sense that for all and , where depends on (and ). For in the half-plane , form the Dirichlet series For any non-integer and any , establish Perron’s formula What happens when is an integer? (The Perron formula and its many variants are of great utility in analytic number theory; see these previous lecture notes for further discussion.)

Exercise 13 (Solution to Schrödinger equation)Let be as in Proposition 3. Define the function by the formula \{ u(t,x) := \int_R\hat f(\xi) e^{2\pi i x \xi – 4 \pi^2 i \xi^2 t} d\xi.\}

- (i) Show that is a smooth function of that obeys the Schrödinger equation with initial condition for .
- (ii) Establish the formula for and , where we use the standard branch of the square root.

** — 2. Phragmén-Lindelöf and Paley-Wiener — **

The maximum modulus principle (Exercise 26 from 246A Notes 1) for holomorphic functions asserts that if a function continuous on a compact subset of the plane and holomorphic on the interior of that set is bounded in magnitude by a bound on the boundary , then it is also bounded by on the interior. This principle does not directly apply for noncompact domains : for instance, on the entire complex plane , there is no boundary whatsoever and the bound is clearly vacuous. On the half-plane , the holomorphic function (for instance) is bounded in magnitude by on the boundary of the half-plane, but grows exponentially in the interior. Similarly, in the strip , the holomorphic function is bounded in magnitude by on the boundary of the strip, but is grows double-exponentially in the interior of the strip. However, if one does not have such absurdly high growth, one can recover a form of the maximum principle, known as the Phragmén-Lindelöf principle. Here is one formulation of this principle:

Theorem 14 (Lindelöf’s theorem)Let be a continuous function on a strip for some , which is holomorphic in the interior of the strip and obeys the bound for all and some constants . Suppose also that and for all and some . Then we have for all and .

Remark 15The hypothesis (12) is a qualitative hypothesis rather than a quantitative one, since the exact values of do not show up in the conclusion. It is quite a mild condition; any function of exponential growth in , or even with such super-exponential growth as or , will obey (12). The principle however fails without this hypothesis, as discussed previously.

*Proof:* By shifting and dilating (adjusting as necessary) we can reduce to the case , , and by multiplying by a constant we can also normalise .

Suppose we temporarily assume that as . Then on a sufficiently large rectangle , we have on the boundary of the rectangle, hence on the interior by the maximum modulus principle. Sending , we obtain the claim.

To remove the assumption that goes to zero at infinity, we use the trick of giving ourselves an epsilon of room. Namely, we multiply by the holomorphic function for some . A little complex arithmetic shows that the function goes to zero at infinity in . Applying the previous case to this function, then taking limits as , we obtain the claim.

Corollary 16 (Phragmén-Lindelöf principle in a sector)Let be a continuous function on a sector for some , which is holomorphic on the interior of the sector and obeys the bound for some and . Suppose also that on the boundary of the sector for some . Then one also has in the interior.

*Proof:* Apply Theorem 14 to the function on the strip .

Exercise 17With the notation and hypotheses of Theorem 14, show that the function is log-convex on .

Exercise 18 (Hadamard three-circles theorem)Let be a holomorphic function on an annulus . Show that the function is log-convex on .

Exercise 19 (Phragmén-Lindelöf principle)Let be as in Theorem 14 with , but with the hypotheses after “Suppose also” in that theorem replaced instead by the bounds and for all and some exponents and a constant . Show that one has for all and some constant (which is allowed to depend on the constants in (12), as well as ). (Hint: it is convenient to work first in a half-strip such as for some large . Then multiply by something like for some suitable branch of the logarithm and apply a variant of Theorem 14 for the half-strip. A more refined estimate in this regard is due to Rademacher.) This particular version of the principle gives theconvexity boundfor Dirichlet series such as the Riemann zeta function. Bounds which exploit the deeper properties of these functions to improve upon the convexity bound are known assubconvexity boundsand are of major importance in analytic number theory, which is of course well outside the scope of this course.

Now we can establish a remarkable converse of sorts to Exercise 2(ii) known as the Paley-Wiener theorem, that links the exponential growth of (the analytic continuation) of a function with the support of its Fourier transform:

Theorem 20 (Paley-Wiener theorem)Let be a continuous function obeying the decay condition for all and some . Let . Then the following are equivalent:

- (i) The Fourier transform is supported on .
- (ii) extends analytically to an entire function that obeys the bound for some .
- (iii) extends analytically to an entire function that obeys the bound for some .

The continuity and decay hypotheses on can be relaxed, but we will not explore such generalisations here.

*Proof:* If (i) holds, then by Exercise 9, we have the inversion formula (4), and the claim (iii) then holds by a slight modification of Exercise 2(ii). Also, the claim (iii) clearly implies (ii).

Now we see why (iii) implies (i). We first assume that we have the stronger bound

for . Then we can apply Proposition 3 for any , and conclude in particular that for any and . Applying (14) and the triangle inequality, we see that If , we can then send and conclude that ; similarly for we can send and again conclude . This establishes (i) in this case.Now suppose we only have the weaker bound on assumed in (iii). We again use the epsilon of room trick. For any , we consider the modified function . This is still holomorphic on the lower half-plane and obeys a bound of the form (14) on this half-plane. An inspection of the previous arguments shows that we can still show that for despite no longer having holomorphicity on the entire upper half-plane; sending using dominated convergence we conclude that for . A similar argument (now using in place of shows that for . This proves (i).

Finally, we show that (ii) implies (iii). The function is entire, bounded on the real axis by (13), bounded on the upper imaginary axis by (iii), and has exponential growth. By Corollary 16, it is also bounded on the upper half-plane, which gives (iii) in the upper half-plane as well. A similar argument (using in place of ) also yields (iii) in the lower half-plane.

** — 3. The Hardy uncertainty principle — **

Informally speaking, the uncertainty principle for the Fourier transform asserts that a function and its Fourier transform cannot simultaneously be strongly localised, except in the degenerate case when is identically zero. There are many rigorous formulations of this principle. Perhaps the best known is the *Heisenberg uncertainty principle*

Another manifestation of the uncertainty principle is the following simple fact:

Lemma 21

- (i) If is an integrable function that has exponential decay in the sense that one has for all and some , then the Fourier transform is either identically zero, or only has isolated zeroes (that is to say, the set is discrete.
- (ii) If is a compactly supported continuous function such that is also compactly supported, then is identically zero.

*Proof:* For (i), we observe from Exercise 2(iii) that extends holomorphically to a strip around the real axis, and the claim follows since non-zero holomorphic functions have isolated zeroes. For (ii), we observe from (i) that must be identically zero, and the claim now follows from the Fourier inversion formula (Exercise 9).

Lemma 21(ii) rules out the existence of a bump function whose Fourier transform is also a bump function, which would have been a rather useful tool to have in harmonic analysis over the reals. (Such functions do exist however in some non-archimedean domains, such as the -adics.) On the other hand, from Exercise 1 we see that we do at least have gaussian functions whose Fourier transform also decays as a gaussian. Unfortunately this is basically the best one can do:

Theorem 22 (Hardy uncertainty principle)Let be a continuous function which obeys the bound for all and some . Suppose also that for all and some . Then is a scalar multiple of the gaussian , that is to say one has for some .

*Proof:* By replacing with the rescaled version , which replaces with the rescaled version , we may normalise . By multiplying by a small constant we may also normalise .

From Exercise 2(i), extends to an entire function. By the triangle inequality, we can bound

for any . Completing the square and using , we conclude the bound In particular, if we introduce the normalised function then In particular, is bounded by on the imaginary axis. On the other hand, from hypothesis is also bounded by on the real axis. We can nowOne corollary of this theorem is that if is continuous and decays like or better, then cannot decay any faster than without vanishing identically. This is a stronger version of Lemma 21(ii). There is a more general tradeoff known as the Gel’fand-Shilov uncertainty principle, which roughly speaking asserts that if decays like then cannot decay faster than without vanishing identically, whenever are dual exponents in the sense that , and is large enough (the precise threshold was established in work of Morgan). See for instance this article of Nazarov for further discussion of these variants.

Exercise 23If is continuous and obeys the bound for some and and all , and obeys the bound for some and all , show that is of the form for some polynomial of degree at most .

Exercise 24In this exercise we develop an alternate proof of (a special case of) the Hardy uncertainty principle, which can be found in the original paper of Hardy. Let the hypotheses be as in Theorem 22.It is possible to adapt this argument to also cover the case of general that are not required to be even; see the paper of Hardy for details.

- (i) Show that the function is holomorphic on the region and obeys the bound in this region, where we use the standard branch of the square root.
- (ii) Show that the function is holomorphic on the region and obeys the bound in this region.
- (iii) Show that and agree on their common domain of definition.
- (iv) Show that the functions are constant. (You may find Exercise 13 from 246A Notes 4 to be useful.)
- (v) Use the above to give an alternate proof of Theorem 22 in the case when is even. (
Hint:subtract a constant multiple of a gaussian from to make vanish, and conclude on Taylor expansion around the origin that all the even moments vanish. Conclude that the Taylor series coefficients of around the origin all vanish.)

Exercise 25(This problem is due to Tom Liggett; see this previous post.) Let be a sequence of complex numbers bounded in magnitude by some bound , and suppose that the power series obeys the bound for all and some .

- (i) Show that the Laplace transform extends holomorphically to the region and obeys the bound in this region.
- (ii) Show that the function is holomorphic in the region , obeys the bound in this region, and agrees with on the common domain of definition.
- (iii) Show that is a constant multiple of .
- (iv) Show that the sequence is a constant multiple of the sequence .

Remark 26There are many further variants of the Hardy uncertainty principle. For instance we have the following uncertainty principle of Beurling, which we state in a strengthened form due to Bonami, Demange, and Jaming: if is a square-integrable function such that , then is equal (almost everywhere) to a polynomial times a gaussian; it is not difficult to show that this implies Theorem 22 and Exercise 23, as well as the Gel’fand-Shilov uncertainty principle. In recent years, PDE-based proofs of the Hardy uncertainty principle have been established, which have then been generalised to establish uncertainty principles for various Schrödinger type equations; see for instance this review article of Kenig. I also have some older notes on the Hardy uncertainty principle in this blog post. Finally, we mention theBeurling-Malliavin theorem, which provides a precise description of the possible decay rates of a function whose Fourier transform is compactly supported; see for instance this paper of Mashregi, Nazarov, and Khavin for a modern treatment.

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Suppose one has an eigenfunction

of a Schrödinger operator , where is the Laplacian on , is a potential, and is an energy. Where would one expect the eigenfunction to be concentrated? If the potential is smooth and slowly varying, the correspondence principle suggests that the eigenfunction should be mostly concentrated in the potential energy wells , with an exponentially decaying amount of tunnelling between the wells. One way to rigorously establish such an exponential decay is through an argument of Agmon, which we will sketch later in this post, which gives an exponentially decaying upper bound (in an sense) of eigenfunctions in terms of the distance to the wells in terms of a certain “Agmon metric” on determined by the potential and energy level (or any upper bound on this energy). Similar exponential decay results can also be obtained for discrete Schrödinger matrix models, in which the domain is replaced with a discrete set such as the lattice , and the Laplacian is replaced by a discrete analogue such as a graph Laplacian.
When the potential is very “rough”, as occurs for instance in the random potentials arising in the theory of Anderson localisation, the Agmon bounds, while still true, become very weak because the wells are dispersed in a fairly dense fashion throughout the domain , and the eigenfunction can tunnel relatively easily between different wells. However, as was first discovered in 2012 by my two coauthors, in these situations one can replace the rough potential by a smoother *effective potential* , with the eigenfunctions typically localised to a single connected component of the effective wells . In fact, a good choice of effective potential comes from locating the *landscape function* , which is the solution to the equation with reasonable behavior at infinity, and which is non-negative from the maximum principle, and then the reciprocal of this landscape function serves as an effective potential.

There are now several explanations for why this particular choice is a good effective potential. Perhaps the simplest (as found for instance in this recent paper of Arnold, David, Jerison, and my two coauthors) is the following observation: if is an eigenvector for with energy , then is an eigenvector for with the same energy , thus the original Schrödinger operator is conjugate to a (variable coefficient, but still in divergence form) Schrödinger operator with potential instead of . Closely related to this, we have the integration by parts identity

for any reasonable function , thus again highlighting the emergence of the effective potential .These particular explanations seem rather specific to the Schrödinger equation (continuous or discrete); we have for instance not been able to find similar identities to explain an effective potential for the bi-Schrödinger operator .

In this paper, we demonstrate the (perhaps surprising) fact that effective potentials continue to exist for operators that bear very little resemblance to Schrödinger operators. Our chosen model is that of an -matrix: self-adjoint positive definite matrices whose off-diagonal entries are negative. This model includes discrete Schrödinger operators (with non-negative potentials) but can allow for significantly more non-local interactions. The analogue of the landscape function would then be the vector , where denotes the vector with all entries . Our main result, roughly speaking, asserts that an eigenvector of will then be exponentially localised to the “potential wells” , where denotes the coordinates of the landscape function . In particular, we establish the inequality

if is normalised in , where theOur approach is based on Agmon’s methods, which we interpret as a double commutator method, and in particular relying on exploiting the negative definiteness of certain double commutator operators. In the case of Schrödinger operators , this negative definiteness is provided by the identity

for any sufficiently reasonable functions , where we view (like ) as a multiplier operator. To exploit this, we use the commutator identity valid for any after a brief calculation. The double commutator identity then tells us that If we choose to be a non-negative weight and let for an eigenfunction , then we can write and we conclude that We have considerable freedom in this inequality to select the functions . If we select , we obtain the clean inequality If we take to be a function which equals on the wells but increases exponentially away from these wells, in such a way that outside of the wells, we can obtain the estimate which then gives an exponential type decay of away from the wells. This is basically the classic exponential decay estimate of Agmon; one can basically take to be the distance to the wells with respect to the Euclidean metric conformally weighted by a suitably normalised version of . If we instead select to be the landscape function , (3) then gives and by selecting appropriately this gives an exponential decay estimate away from the effective wells , using a metric weighted by .It turns out that this argument extends without much difficulty to the -matrix setting. The analogue of the crucial double commutator identity (2) is

for any diagonal matrix . The remainder of the Agmon type arguments go through after making the natural modifications.Numerically we have also found some aspects of the landscape theory to persist beyond the -matrix setting, even though the double commutators cease being negative definite, so this may not yet be the end of the story, but it does at least demonstrate that utility the landscape does not purely rely on identities such as (1).

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** — 1. Jensen’s formula — **

Suppose is a non-zero rational function , then by the fundamental theorem of algebra one can write

for some non-zero constant , where ranges over the zeroes of (counting multiplicity) and ranges over the zeroes of (counting multiplicity), and assuming avoids the zeroes of . Taking absolute values and then logarithms, we arrive at the formula as long as avoids the zeroes of both and . (In this set of notes we use for the natural logarithm when applied to a positive real number, and for the standard branch of the complex logarithm (which extends ); the multi-valued complex logarithm will only be used in passing.) Alternatively, taking logarithmic derivatives, we arrive at the closely related formula again for avoiding the zeroes of both and . Thus we see that the zeroes and poles of a rational function describe the behaviour of that rational function, as well as close relatives of that function such as the log-magnitude and log-derivative . We have already seen these sorts of formulae arise in our treatment of the argument principle in 246A Notes 4.

Exercise 1Let be a complex polynomial of degree .

- (i) (Gauss-Lucas theorem) Show that the complex roots of are contained in the closed convex hull of the complex roots of .
- (ii) (Laguerre separation theorem) If all the complex roots of are contained in a disk , and , then all the complex roots of are also contained in . (
Hint:apply a suitable Möbius transformation to move to infinity, and then apply part (i) to a polynomial that emerges after applying this transformation.)

There are a number of useful ways to extend these formulae to more general meromorphic functions than rational functions. Firstly there is a very handy “local” variant of (1) known as Jensen’s formula:

Theorem 2 (Jensen’s formula)Let be a meromorphic function on an open neighbourhood of a disk , with all removable singularities removed. Then, if is neither a zero nor a pole of , we have where and range over the zeroes and poles of respectively (counting multiplicity) in the disk .

One can view (3) as a truncated (or localised) variant of (1). Note also that the summands are always non-positive.

*Proof:* By perturbing slightly if necessary, we may assume that none of the zeroes or poles of (which form a discrete set) lie on the boundary circle . By translating and rescaling, we may then normalise and , thus our task is now to show that

An important special case of Jensen’s formula arises when is holomorphic in a neighborhood of , in which case there are no contributions from poles and one simply has

This is quite a useful formula, mainly because the summands are non-negative; it can be viewed as a more precise assertion of the subharmonicity of (see Exercises 60(ix) and 61 of 246A Notes 5). Here are some quick applications of this formula:

Exercise 3Use (6) to give another proof of Liouville’s theorem: a bounded holomorphic function on the entire complex plane is necessarily constant.

Exercise 4Use Jensen’s formula to prove the fundamental theorem of algebra: a complex polynomial of degree has exactly complex zeroes (counting multiplicity), and can thus be factored as for some complex numbers with . (Note that the fundamental theorem was invoked previously in this section, but only for motivational purposes, so the proof here is non-circular.)

Exercise 5 (Shifted Jensen’s formula)Let be a meromorphic function on an open neighbourhood of a disk , with all removable singularities removed. Show that for all in the open disk that are not zeroes or poles of , where and . (The function appearing in the integrand is sometimes known as the Poisson kernel, particularly if one normalises so that and .)

Exercise 6 (Bounded type)

- (i) If is a holomorphic function on that is not identically zero, show that .
- (ii) If is a meromorphic function on that is the ratio of two bounded holomorphic functions that are not identically zero, show that . (Functions of this form are said to be of bounded type and lie in the
Nevanlinna classfor the unit disk .)

Exercise 7 (Smoothed out Jensen formula)Let be a meromorphic function on an open set , and let be a smooth compactly supported function. Show that where range over the zeroes and poles of (respectively) in the support of . Informally argue why this identity is consistent with Jensen’s formula.

When applied to entire functions , Jensen’s formula relates the order of growth of near infinity with the density of zeroes of . Here is a typical result:

Proposition 8Let be an entire function, not identically zero, that obeys a growth bound for some and all . Then there exists a constant such that has at most zeroes (counting multiplicity) for any .

Entire functions that obey a growth bound of the form for every and (where depends on ) are said to be of order at most . The above theorem shows that for such functions that are not identically zero, the number of zeroes in a disk of radius does not grow much faster than . This is often a useful preliminary upper bound on the zeroes of entire functions, as the order of an entire function tends to be relatively easy to compute in practice.

*Proof:* First suppose that is non-zero. From (6) applied with and one has

Just as (3) and (7) give truncated variants of (1), we can create truncated versions of (2). The following crude truncation is adequate for many applications:

Theorem 9 (Truncated formula for log-derivative)Let be a holomorphic function on an open neighbourhood of a disk that is not identically zero on this disk. Suppose that one has a bound of the form for some and all on the circle . Let be constants. Then one has the approximate formula for all in the disk other than zeroes of . Furthermore, the number of zeroes in the above sum is .

*Proof:* To abbreviate notation, we allow all implied constants in this proof to depend on .

We mimic the proof of Jensen’s formula. Firstly, we may translate and rescale so that and , so we have when , and our main task is to show that

for . Note that if then vanishes on the unit circle and hence (by the maximum principle) vanishes identically on the disk, a contradiction, so we may assume . From hypothesis we then have on the unit circle, and so from Jensen’s formula (3) we see that In particular we see that the number of zeroes with is , as claimed.Suppose has a zero with . If we factor , where is the Blaschke product (5), then

Observe from Taylor expansion that the distance between and is , and hence for . Thus we see from (9) that we may use Blaschke products to remove all the zeroes in the annulus while only affecting the left-hand side of (8) by ; also, removing the Blaschke products does not affect on the unit circle, and only affects by thanks to (9). Thus we may assume without loss of generality that there are no zeroes in this annulus.Similarly, given a zero with , we have , so using Blaschke products to remove all of these zeroes also only affects the left-hand side of (8) by (since the number of zeroes here is ), with also modified by at most . Thus we may assume in fact that has no zeroes whatsoever within the unit disk. We may then also normalise , then for all . By Jensen’s formula again, we have

and thus (by using the identity for any real ) On the other hand, from (7) we have which implies from (10) that and its first derivatives are on the disk . But recall from the proof of Jensen’s formula that is the derivative of a logarithm of , whose real part is . By the Cauchy-Riemann equations for , we conclude that on the disk , as required.

Exercise 10

- (i) (Borel-Carathéodory theorem) If is analytic on an open neighborhood of a disk , show that (
Hint:one can normalise , , , and . Now maps the unit disk to the half-plane . Use a Möbius transformation to map the half-plane to the unit disk and then use the Schwarz lemma.)- (ii) Use (i) to give an alternate way to conclude the proof of Theorem 9.

A variant of the above argument allows one to make precise the heuristic that holomorphic functions locally look like polynomials:

Exercise 11 (Local Weierstrass factorisation)Let the notation and hypotheses be as in Theorem 9. Then show that for all in the disk , where is a polynomial whose zeroes are precisely the zeroes of in (counting multiplicity) and is a holomorphic function on of magnitude and first derivative on this disk. Furthermore, show that the degree of is .

Exercise 12 (Preliminary Beurling factorisation)Let denote the space of bounded analytic functions on the unit disk; this is a normed vector space with norm

- (i) If is not identically zero, and denote the zeroes of in counting multiplicity, show that and
- (ii) Let the notation be as in (i). If we define the Blaschke product where is the order of vanishing of at zero, show that this product converges absolutely to a holomorphic function on , and that for all . (It may be easier to work with finite Blaschke products first to obtain this bound.)
- (iii) Continuing the notation from (i), establish a factorisation for some holomorphic function with for all .
- (iv) (Theorem of F. and M. Riesz, special case) If extends continuously to the boundary , show that the set has zero measure.

Remark 13The factorisation (iii) can be refined further, with being the Poisson integral of some finite measure on the unit circle. Using the Lebesgue decomposition of this finite measure into absolutely continuous parts one ends up factorising functions into “outer functions” and “inner functions”, giving the Beurling factorisation of . There are also extensions to larger spaces than (which are to as is to ), known as Hardy spaces. We will not discuss this topic further here, but see for instance this text of Garnett for a treatment.

Exercise 14 (Littlewood’s lemma)Let be holomorphic on an open neighbourhood of a rectangle for some and , with non-vanishing on the boundary of the rectangle. Show that where ranges over the zeroes of inside (counting multiplicity) and one uses a branch of which is continuous on the upper, lower, and right edges of . (This lemma is a popular tool to explore the zeroes of Dirichlet series such as the Riemann zeta function.)

** — 2. The Weierstrass factorisation theorem — **

The fundamental theorem of algebra shows that every polynomial of degree comes with complex zeroes (counting multiplicity). In the converse direction, given any complex numbers (again allowing multiplicity), one can form a degree polynomial with precisely these zeroes by the formula

where is an arbitrary non-zero constant, and by the factor theorem this is the complete set of polynomials with this set of zeroes (counting multiplicity). Thus, except for the freedom to multiply polynomials by non-zero constants, one has a one-to-one correspondence between polynomials (excluding the zero polynomial as a degenerate case) and finite (multi-)sets of complex numbers.As discussed earlier in this set of notes, one can think of a entire function as a sort of infinite degree analogue of a polynomial. One can then ask what the analogue of the above correspondence is for entire functions are – can one identify entire functions (not identically zero, and up to constants) by their sets of zeroes?

There are two obstructions to this. Firstly there are a number of non-trivial entire functions with no zeroes whatsoever. Most prominently, we have the exponential function which has no zeroes despite being non-constant. More generally, if is an entire function, then clearly is an entire function with no zeroes. In particular one can multiply (or divide) any other entire function by without affecting the location and order of the zeroes.

Secondly, we know (see Corollary 24 of 246A Notes 3) that the set of zeroes of an entire function (that is not identically zero) must be isolated; in particular, in any compact set there can only be finitely many zeroes. Thus, by covering the complex plane by an increasing sequence of compact sets (e.g., the disks ), one can index the zeroes (counting multiplicity) by a sequence of complex numbers (possibly with repetition) that is either finite, or goes to infinity.

Now we turn to the Weierstrass factorisation theorem, which asserts that once one accounts for these two obstructions, we recover a correspondence between entire functions and sequences of zeroes.

Theorem 15 (Weierstrass factorization theorem)Let be a sequence of complex numbers that is either finite or going to infinity. Then there exists an entire function that has zeroes precisely at , with the order of zero of at each equal to the number of times appears in the sequence. Furthermore, this entire function is unique up to multiplication by exponentials of entire functions; that is to say, if are entire functions that are both of the above form, then for some entire function .

We now establish this theorem. We begin with the easier uniqueness part of the theorem. If are entire functions with the same locations and orders of zeroes, then the ratio is a meromorphic function on which only has removable singularities, and becomes an entire function with no zeroes once the singularities are removed. Since the domain of an entire function is simply connected, we can then take a branch of the complex logarithm (see Exercise 46 of 246A Notes 4) to write for an entire function (after removing singularities), giving the uniqueness claim.

Now we turn to existence. If the sequence is finite, we can simply use the formula (11) to produce the required entire function (setting to equal , say). So now suppose that the sequence is infinite. Naively, one might try to replicate the formula (11) and set

Here we encounter a serious problem: the infinite product is likely to be divergent (that is to say, the partial products ) fail to converge, given that the factors go off to infinity. On the other hand, we do have this freedom to multiply by a constant (or more generally, the exponential of an entire function). One can try to use this freedom to “renormalise” the factors to make them more likely to converge. Much as an infinite series is more likely to converge when its summands converge rapidly to zero, an infinite series is more likely to converge when its factors converge rapidly to . Here is one formalisation of this principle:

Lemma 16 (Absolutely convergent products)Let be a sequence of complex numbers such that . Then the product converges. Furthermore, this product vanishes if and only if one of the factors vanishes.

Products covered by this lemma are known as *absolutely convergent products*. It is possible for products to converge without being absolutely convergent, but such “conditionally convergent products” are infrequently used in mathematics.

*Proof:* By the zero test, , thus converges to . In particular, all but finitely many of the lie in the disk . We can then factor , where is such that for , and we see that it will suffice to show that the infinite product converges to a non-zero number. But on using the standard branch of the complex logarithm on we can write . By Taylor expansion we have , hence the series is absolutely convergent. From the properties of the complex exponential we then see that the product converges to , giving the claim.

It is well known that absolutely convergent series are preserved by rearrangement, and the same is true for absolutely convergent products:

Exercise 17If is an absolutely convergent product of complex numbers , show that any permutation of the leads to the same absolutely convergent product, thus for any permutation of the positive integers .

Exercise 18

- (i) Let be a sequence of real numbers with for all . Show that converges if and only if converges.
- (ii) Let be a sequence of complex numbers. Show that is absolutely convergent if and only if is convergent.

To try to use Lemma 16, we can divide each factor by the constant to make it closer to in the limit . Since

we can thus attempt to construct the desired entire function using the formula Immediately there is the objection that this product is undefined if one or more of the putative zeroes is located at the origin, this objection is easily dealt with since the origin can only occur finitely many (say ) times, so if we remove the copies of the origin from the sequence of zeroes , apply the Weierstrass factorisation theorem to the remaining zeroes, and then multiply the resulting entire function by , we can reduce to the case where the origin is not one of the zeroes.In order to apply Lemma 16 to make this product converge, we would need to converge for every , or equivalently that

This is basically a requirement that the converge to infinity sufficiently quickly. Not all sequences covered by Theorem 15 obey this condition, but let us begin with this case for sake of argument. Lemma 16 now tells us that is well-defined for every , and vanishes if and only if is equal to one of the . However, we need to establish holomorphicity. This can be accomplished by the following product form of the Weierstrass -test.

Exercise 19 (Product Weierstrass -test)Let be a set, and for any natural number , let be a bounded function. If the sum for some finite , show that the products converge uniformly to on . (In particular, if is a topological space and all the are continuous, then is continuous also.)

Using this exercise, we see that (under the assumption (13)) that the partial products converge locally uniformly to the infinite product in (12). Since each of the partial products are entire, and the (locally) uniform limit of holomorphic functions is holomorphic (Theorem 34 of 246A Notes 3), we conclude that the function (12) is entire. Finally, if a certain zero appears times in the sequence, then after factoring out copies of we see that is the product of with an entire function that is non-vanishing at , and thus has a zero of order exactly at . This establishes the Weierstrass approximation theorem under the additional hypothesis that (13) holds.

What if (13) does not hold? The problem now is that our renormalized factors do not converge fast enough for Lemma 16 or Exercise 19 to apply. So we need to renormalize further, taking advantage of our ability to not just multiply by constants, but also by exponentials of entire functions. Observe that if is fixed and is large enough, then lies in and we can write

The function is not entire, but its Taylor approximation is. So it is natural to split We now discard the factors (which do not affect the zeroes) and now propose the new candidate entire function The point here is that for large enough, Taylor expansion gives and thus on exponentiating Repeating the previous arguments, we can then verify that is an entire function with the required properties as long as we have the hypothesis which is weaker than (13) (for instance it is obeyed when , whereas (13) fails in this case).This suggests the way forward to the general case of the Weierstrass factorisation theorem, by using increasingly accurate Taylor expansions of

when