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In one of the earliest posts on this blog, I talked about the ability to “arbitrage” a disparity of symmetry in an inequality, and in particular to “amplify” such an inequality into a stronger one. (The principle can apply to other mathematical statements than inequalities, with the “hypothesis” and “conclusion” of that statement generally playing the role of the “right-hand side” and “left-hand side” of an inequality, but for sake of discussion I will restrict attention here to inequalities.) One can formalise this principle as follows. Many inequalities in analysis can be expressed in the form

$\displaystyle A(f) \leq B(f) \ \ \ \ \ (1)$

for all ${f}$ in some space ${X}$ (in many cases ${X}$ will be a function space, and ${f}$ a function in that space), where ${A(f)}$ and ${B(f)}$ are some functionals of ${f}$ (that is to say, real-valued functions of ${f}$). For instance, ${B(f)}$ might be some function space norm of ${f}$ (e.g. an ${L^p}$ norm), and ${A(f)}$ might be some function space norm of some transform of ${f}$. In addition, we assume we have some group ${G}$ of symmetries ${T: X \rightarrow X}$ acting on the underlying space. For instance, if ${X}$ is a space of functions on some spatial domain, the group might consist of translations (e.g. ${Tf(x) = f(x-h)}$ for some shift ${h}$), or perhaps dilations with some normalisation (e.g. ${Tf(x) = \frac{1}{\lambda^\alpha} f(\frac{x}{\lambda})}$ for some dilation factor ${\lambda > 0}$ and some normalisation exponent ${\alpha \in {\bf R}}$, which can be thought of as the dimensionality of length one is assigning to ${f}$). If we have

$\displaystyle A(Tf) = A(f)$

for all symmetries ${T \in G}$ and all ${f \in X}$, we say that ${A}$ is invariant with respect to the symmetries in ${G}$; otherwise, it is not.

Suppose we know that the inequality (1) holds for all ${f \in X}$, but that there is an imbalance of symmetry: either ${A}$ is ${G}$-invariant and ${B}$ is not, or vice versa. Suppose first that ${A}$ is ${G}$-invariant and ${B}$ is not. Substituting ${f}$ by ${Tf}$ in (1) and taking infima, we can then amplify (1) to the stronger inequality

$\displaystyle A(f) \leq \inf_{T \in G} B(Tf).$

In particular, it is often the case that there is a way to send ${T}$ off to infinity in such a way that the functional ${B(Tf)}$ has a limit ${B_\infty(f)}$, in which case we obtain the amplification

$\displaystyle A(f) \leq B_\infty(f) \ \ \ \ \ (2)$

of (1). Note that these amplified inequalities will now be ${G}$-invariant on both sides (assuming that the way in which we take limits as ${T \rightarrow \infty}$ is itself ${G}$-invariant, which it often is in practice). Similarly, if ${B}$ is ${G}$-invariant but ${A}$ is not, we may instead amplify (1) to

$\displaystyle \sup_{T \in G} A(Tf) \leq B(f)$

and in particular (if ${A(Tf)}$ has a limit ${A_\infty(f)}$ as ${T \rightarrow \infty}$)

$\displaystyle A_\infty(f) \leq B(f). \ \ \ \ \ (3)$

If neither ${A(f)}$ nor ${B(f)}$ has a ${G}$-symmetry, one can still use the ${G}$-symmetry by replacing ${f}$ by ${Tf}$ and taking a limit to conclude that

$\displaystyle A_\infty(f) \leq B_\infty(f),$

though now this inequality is not obviously stronger than the original inequality (1) (for instance it could well be trivial). In some cases one can also average over ${G}$ instead of taking a limit as ${T \rightarrow \infty}$, thus averaging a non-invariant inequality into an invariant one.

As discussed in the previous post, this use of amplification gives rise to a general principle about inequalities: the most efficient inequalities are those in which the left-hand side and right-hand side enjoy the same symmetries. It is certainly possible to have true inequalities that have an imbalance of symmetry, but as shown above, such inequalities can always be amplified to more efficient and more symmetric inequalities. In the case when limits such as ${A_\infty}$ and ${B_\infty}$ exist, the limiting functionals ${A_\infty(f)}$ and ${B_\infty(f)}$ are often simpler in form, or more tractable analytically, than their non-limiting counterparts ${A(f)}$ and ${B(f)}$ (this is one of the main reasons why we take limits at infinity in the first place!), and so in many applications there is really no reason to use the weaker and more complicated inequality (1), when stronger, simpler, and more symmetric inequalities such as (2), (3) are available. Among other things, this explains why many of the most useful and natural inequalities one sees in analysis are dimensionally consistent.

One often tries to prove inequalities (1) by directly chaining together simpler inequalities. For instance, one might attempt to prove (1) by by first bounding ${A(f)}$ by some auxiliary quantity ${C(f)}$, and then bounding ${C(f)}$ by ${B(f)}$, thus obtaining (1) by chaining together two inequalities

$\displaystyle A(f) \leq C(f) \leq B(f). \ \ \ \ \ (4)$

A variant of the above principle then asserts that when proving inequalities by such direct methods, one should, whenever possible, try to maintain the symmetries that are present in both sides of the inequality. Why? Well, suppose that we ignored this principle and tried to prove (1) by establishing (4) for some ${C}$ that is not ${G}$-invariant. Assuming for sake of argument that (4) were actually true, we could amplify the first half ${A(f) \leq C(f)}$ of this inequality to conclude that

$\displaystyle A(f) \leq \inf_{T \in G} C(Tf)$

and also amplify the second half ${C(f) \leq B(f)}$ of the inequality to conclude that

$\displaystyle \sup_{T \in G} C(Tf) \leq B(f)$

and hence (4) amplifies to

$\displaystyle A(f) \leq \inf_{T \in G} C(Tf) \leq \sup_{T \in G} C(Tf) \leq B(f). \ \ \ \ \ (5)$

Let’s say for sake of argument that all the quantities involved here are positive numbers (which is often the case in analysis). Then we see in particular that

$\displaystyle \frac{\sup_{T \in G} C(Tf)}{\inf_{T \in G} C(Tf)} \leq \frac{B(f)}{A(f)}. \ \ \ \ \ (6)$

Informally, (6) asserts that in order for the strategy (4) used to prove (1) to work, the extent to which ${C}$ fails to be ${G}$-invariant cannot exceed the amount of “room” present in (1). In particular, when dealing with those “extremal” ${f}$ for which the left and right-hand sides of (1) are comparable to each other, one can only have a bounded amount of non-${G}$-invariance in the functional ${C}$. If ${C}$ fails so badly to be ${G}$-invariant that one does not expect the left-hand side of (6) to be at all bounded in such extremal situations, then the strategy of proving (1) using the intermediate quantity ${C}$ is doomed to failure – even if one has already produced some clever proof of one of the two inequalities ${A(f) \leq C(f)}$ or ${C(f) \leq B(f)}$ needed to make this strategy work. And even if it did work, one could amplify (4) to a simpler inequality

$\displaystyle A(f) \leq C_\infty(f) \leq B(f) \ \ \ \ \ (7)$

(assuming that the appropriate limit ${C_\infty(f) = \lim_{T \rightarrow \infty} C(Tf)}$ existed) which would likely also be easier to prove (one can take whatever proofs one had in mind of the inequalities in (4), conjugate them by ${T}$, and take a limit as ${T \rightarrow \infty}$ to extract a proof of (7)).

Here are some simple (but somewhat contrived) examples to illustrate these points. Suppose one wishes to prove the inequality

$\displaystyle xy \leq x^2 + y^2 \ \ \ \ \ (8)$

for all ${x,y>0}$. Both sides of this inequality are invariant with respect to interchanging ${x}$ with ${y}$, so the principle suggests that when proving this inequality directly, one should only use sub-inequalities that are also invariant with respect to this interchange. However, in this particular case there is enough “room” in the inequality that it is possible (though somewhat unnatural) to violate this principle. For instance, one could decide (for whatever reason) to start with the inequality

$\displaystyle 0 \leq (x - y/2)^2 = x^2 - xy + y^2/4$

to conclude that

$\displaystyle xy \leq x^2 + y^2/4$

and then use the obvious inequality ${x^2 + y^2/4 \leq x^2+y^2}$ to conclude the proof. Here, the intermediate quantity ${x^2 + y^2/4}$ is not invariant with respect to interchange of ${x}$ and ${y}$, but the failure is fairly mild (changing ${x}$ and ${y}$ only modifies the quantity ${x^2 + y^2/4}$ by a multiplicative factor of ${4}$ at most), and disappears completely in the most extremal case ${x=y}$, which helps explain why one could get away with using this quantity in the proof here. But it would be significantly harder (though still not impossible) to use non-symmetric intermediaries to prove the sharp version

$\displaystyle xy \leq \frac{x^2 + y^2}{2}$

of (8) (that is to say, the arithmetic mean-geometric mean inequality). Try it!

Similarly, consider the task of proving the triangle inequality

$\displaystyle |z+w| \leq |z| + |w| \ \ \ \ \ (9)$

for complex numbers ${z, w}$. One could try to leverage the triangle inequality ${|x+y| \leq |x| + |y|}$ for real numbers by using the crude estimate

$\displaystyle |z+w| \leq |\hbox{Re}(z+w)| + |\hbox{Im}(z+w)|$

and then use the real triangle inequality to obtain

$\displaystyle |\hbox{Re}(z+w)| \leq |\hbox{Re}(z)| + |\hbox{Re}(w)|$

and

$\displaystyle |\hbox{Im}(z+w)| \leq |\hbox{Im}(z)| + |\hbox{Im}(w)|$

and then finally use the inequalities

$\displaystyle |\hbox{Re}(z)|, |\hbox{Im}(z)| \leq |z| \ \ \ \ \ (10)$

and

$\displaystyle |\hbox{Re}(w)|, |\hbox{Im}(w)| \leq |w| \ \ \ \ \ (11)$

but when one puts this all together at the end of the day, one loses a factor of two:

$\displaystyle |z+w| \leq 2(|z| + |w|).$

One can “blame” this loss on the fact that while the original inequality (9) was invariant with respect to phase rotation ${(z,w) \mapsto (e^{i\theta} z, e^{i\theta} w)}$, the intermediate expressions we tried to use when proving it were not, leading to inefficient estimates. One can try to be smarter than this by using Pythagoras’ theorem ${|z|^2 = |\hbox{Re}(z)|^2 + |\hbox{Im}(z)|^2}$; this reduces the loss from ${2}$ to ${\sqrt{2}}$ but does not eliminate it completely, which is to be expected as one is still using non-invariant estimates in the proof. But one can remove the loss completely by using amplification; see the previous blog post for details (we also give a reformulation of this amplification below).

Here is a slight variant of the above example. Suppose that you had just learned in class to prove the triangle inequality

$\displaystyle (\sum_{n=1}^\infty |a_n+b_n|^2)^{1/2} \leq (\sum_{n=1}^\infty |a_n|^2)^{1/2} + (\sum_{n=1}^\infty |b_n|^2)^{1/2} \ \ \ \ \ (12)$

for (say) real square-summable sequences ${(a_n)_{n=1}^\infty}$, ${(b_n)_{n=1}^\infty}$, and was tasked to conclude the corresponding inequality

$\displaystyle (\sum_{n \in {\bf Z}} |a_n+b_n|^2)^{1/2} \leq (\sum_{n \in {\bf Z}} |a_n|^2)^{1/2} + (\sum_{n \in {\bf Z}} |b_n|^2)^{1/2} \ \ \ \ \ (13)$

for doubly infinite square-summable sequences ${(a_n)_{n \in {\bf Z}}, (b_n)_{n \in {\bf Z}}}$. The quickest way to do this is of course to exploit a bijection between the natural numbers ${1,2,\dots}$ and the integers, but let us say for sake of argument that one was unaware of such a bijection. One could then proceed instead by splitting the integers into the positive integers and the non-positive integers, and use (12) on each component separately; this is very similar to the strategy of proving (9) by splitting a complex number into real and imaginary parts, and will similarly lose a factor of ${2}$ or ${\sqrt{2}}$. In this case, one can “blame” this loss on the abandonment of translation invariance: both sides of the inequality (13) are invariant with respect to shifting the sequences ${(a_n)_{n \in {\bf Z}}}$, ${(b_n)_{n \in {\bf Z}}}$ by some shift ${h}$ to arrive at ${(a_{n-h})_{n \in {\bf Z}}, (b_{n-h})_{n \in {\bf Z}}}$, but the intermediate quantities caused by splitting the integers into two subsets are not invariant. Another way of thinking about this is that the splitting of the integers gives a privileged role to the origin ${n=0}$, whereas the inequality (13) treats all values of ${n}$ equally thanks to the translation invariance, and so using such a splitting is unnatural and not likely to lead to optimal estimates. On the other hand, one can deduce (13) from (12) by sending this symmetry to infinity; indeed, after applying a shift to (12) we see that

$\displaystyle (\sum_{n=-N}^\infty |a_n+b_n|^2)^{1/2} \leq (\sum_{n=-N}^\infty |a_n|^2)^{1/2} + (\sum_{n=-N}^\infty |b_n|^2)^{1/2}$

for any ${N}$, and on sending ${N \rightarrow \infty}$ we obtain (13) (one could invoke the monotone convergence theorem here to justify the limit, though in this case it is simple enough that one can just use first principles).

Note that the principle of preserving symmetry only applies to direct approaches to proving inequalities such as (1). There is a complementary approach, discussed for instance in this previous post, which is to spend the symmetry to place the variable ${f}$ “without loss of generality” in a “normal form”, “convenient coordinate system”, or a “good gauge”. Abstractly: suppose that there is some subset ${Y}$ of ${X}$ with the property that every ${f \in X}$ can be expressed in the form ${f = Tg}$ for some ${T \in G}$ and ${g \in Y}$ (that is to say, ${X = GY}$). Then, if one wishes to prove an inequality (1) for all ${f \in X}$, and one knows that both sides ${A(f), B(f)}$ of this inequality are ${G}$-invariant, then it suffices to check (1) just for those ${f}$ in ${Y}$, as this together with the ${G}$-invariance will imply the same inequality (1) for all ${f}$ in ${GY=X}$. By restricting to those ${f}$ in ${Y}$, one has given up (or spent) the ${G}$-invariance, as the set ${Y}$ will in typical not be preserved by the group action ${G}$. But by the same token, by eliminating the invariance, one also eliminates the prohibition on using non-invariant proof techniques, and one is now free to use a wider range of inequalities in order to try to establish (1). Of course, such inequalities should make crucial use of the restriction ${f \in Y}$, for if they did not, then the arguments would work in the more general setting ${f \in X}$, and then the previous principle would again kick in and warn us that the use of non-invariant inequalities would be inefficient. Thus one should “spend” the symmetry wisely to “buy” a restriction ${f \in Y}$ that will be of maximal utility in calculations (for instance by setting as many annoying factors and terms in one’s analysis to be ${0}$ or ${1}$ as possible).

As a simple example of this, let us revisit the complex triangle inequality (9). As already noted, both sides of this inequality are invariant with respect to the phase rotation symmetry ${(z,w) \mapsto (e^{i\theta} z, e^{i\theta} w)}$. This seems to limit one to using phase-rotation-invariant techniques to establish the inequality, in particular ruling out the use of real and imaginary parts as discussed previously. However, we can instead spend the phase rotation symmetry to restrict to a special class of ${z}$ and ${w}$. It turns out that the most efficient way to spend the symmetry is to achieve the normalisation of ${z+w}$ being a nonnegative real; this is of course possible since any complex number ${z+w}$ can be turned into a nonnegative real by multiplying by an appropriate phase ${e^{i\theta}}$. Once ${z+w}$ is a nonnegative real, the imaginary part disappears and we have

$\displaystyle |z+w| = \hbox{Re}(z+w) = \hbox{Re}(z) + \hbox{Re}(w),$

and the triangle inequality (9) is now an immediate consequence of (10), (11). (But note that if one had unwisely spent the symmetry to normalise, say, ${z}$ to be a non-negative real, then one is no closer to establishing (9) than before one had spent the symmetry.)

Today I’d like to discuss a beautiful inequality in graph theory, namely the crossing number inequality. This inequality gives a useful bound on how far a given graph is from being planar, and has a number of applications, for instance to sum-product estimates. Its proof is also an excellent example of the amplification trick in action; here the main source of amplification is the freedom to pass to subobjects, which is a freedom which I didn’t touch upon in the previous post on amplification. The crossing number inequality (and its proof) is well known among graph theorists but perhaps not among the wider mathematical community, so I thought I would publicise it here.

In this post, when I talk about a graph, I mean an abstract collection of vertices V, together with some abstract edges E joining pairs of vertices together. We will assume that the graph is undirected (the edges do not have a preferred orientation), loop-free (an edge cannot begin and start at the same vertex), and multiplicity-free (any pair of vertices is joined by at most one edge). More formally, we can model all this by viewing E as a subset of $\binom{V}{2} := \{ e \subset V: |e|=2 \}$, the set of 2-element subsets of V, and we view the graph G as an ordered pair G = (V,E). (The notation is set up so that $|\binom{V}{2}| = \binom{|V|}{2}$.)

Now one of the great features of graphs, as opposed to some other abstract maths concepts, is that they are easy to draw: the abstract vertices become dots on a plane, while the edges become line segments or curves connecting these dots. [To avoid some technicalities we do not allow these curves to pass through the dots, except if the curve is terminating at that dot.] Let us informally refer to such a concrete representation D of a graph G as a drawing of that graph. Clearly, any non-trivial graph is going to have an infinite number of possible drawings. In some of these drawings, a pair of edges might cross each other; in other drawings, all edges might be disjoint (except of course at the vertices, where edges with a common endpoint are obliged to meet). If G has a drawing D of the latter type, we say that the graph G is planar.

Given an abstract graph G, or a drawing thereof, it is not always obvious as to whether that graph is planar; just because the drawing that you currently possess of G contains crossings, does not necessarily mean that all drawings of G do. The wonderful little web game “Planarity” illustrates this point excellently. Nevertheless, there are definitely graphs which are not planar; in particular the complete graph $K_5$ on five vertices, and the complete bipartite graph $K_{3,3}$ on two sets of three vertices, are non-planar.

There is in fact a famous theorem of Kuratowski that says that these two graphs are the only “source” of non-planarity, in the sense that any non-planar graph contains (a subdivision of) one of these graphs as a subgraph. (There is of course the even more famous four-colour theorem that asserts that every planar graphs is four-colourable, but this is not the topic of my post today.)

Intuitively, if we fix the number of vertices |V|, and increase the number of edges |E|, then the graph should become “increasingly non-planar”; conversely, if we keep the same number of edges |E| but spread them amongst a greater number of vertices |V|, then the graph should become “increasingly planar”. Is there a quantitative way to measure the “non-planarity” of a graph, and to formalise the above intuition as some sort of inequality?
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It occurred to me recently that the mathematical blog medium may be a good venue not just for expository “short stories” on mathematical concepts or results, but also for more technical discussions of individual mathematical “tricks”, which would otherwise not be significant enough to warrant a publication-length (and publication-quality) article. So I thought today that I would discuss the amplification trick in harmonic analysis and combinatorics (and in particular, in the study of estimates); this trick takes an established estimate involving an arbitrary object (such as a function f), and obtains a stronger (or amplified) estimate by transforming the object in a well-chosen manner (often involving some new parameters) into a new object, applying the estimate to that new object, and seeing what that estimate says about the original object (after optimising the parameters or taking a limit). The amplification trick works particularly well for estimates which enjoy some sort of symmetry on one side of the estimate that is not represented on the other side; indeed, it can be viewed as a way to “arbitrage” differing amounts of symmetry between the left- and right-hand sides of an estimate. It can also be used in the contrapositive, amplifying a weak counterexample to an estimate into a strong counterexample. This trick also sheds some light as to why dimensional analysis works; an estimate which is not dimensionally consistent can often be amplified into a stronger estimate which is dimensionally consistent; in many cases, this new estimate is so strong that it cannot in fact be true, and thus dimensionally inconsistent inequalities tend to be either false or inefficient, which is why we rarely see them. (More generally, any inequality on which a group acts on either the left or right-hand side can often be “decomposed” into the “isotypic components” of the group action, either by the amplification trick or by other related tools, such as Fourier analysis.)

The amplification trick is a deceptively simple one, but it can become particularly powerful when one is arbitraging an unintuitive symmetry, such as symmetry under tensor powers. Indeed, the “tensor power trick”, which can eliminate constants and even logarithms in an almost magical manner, can lead to some interesting proofs of sharp inequalities, which are difficult to establish by more direct means.

The most familiar example of the amplification trick in action is probably the textbook proof of the Cauchy-Schwarz inequality

$|\langle v, w \rangle| \leq \|v\| \|w\|$ (1)

for vectors v, w in a complex Hilbert space. To prove this inequality, one might start by exploiting the obvious inequality

$\|v-w\|^2 \geq 0$ (2)

but after expanding everything out, one only gets the weaker inequality

$\hbox{Re} \langle v, w \rangle \leq \frac{1}{2} \|v\|^2 + \frac{1}{2} \|w\|^2$. (3)

Now (3) is weaker than (1) for two reasons; the left-hand side is smaller, and the right-hand side is larger (thanks to the arithmetic mean-geometric mean inequality). However, we can amplify (3) by arbitraging some symmetry imbalances. Firstly, observe that the phase rotation symmetry $v \mapsto e^{i\theta} v$ preserves the RHS of (3) but not the LHS. We exploit this by replacing v by $e^{i\theta} v$ in (3) for some phase $\theta$ to be chosen later, to obtain

$\hbox{Re} e^{i\theta} \langle v, w \rangle \leq \frac{1}{2} \|v\|^2 + \frac{1}{2} \|w\|^2$.

Now we are free to choose $\theta$ at will (as long as it is real, of course), so it is natural to choose $\theta$ to optimise the inequality, which in this case means to make the left-hand side as large as possible. This is achieved by choosing $e^{i\theta}$ to cancel the phase of $\langle v, w \rangle$, and we obtain

$|\langle v, w \rangle| \leq \frac{1}{2} \|v\|^2 + \frac{1}{2} \|w\|^2$ (4)

This is closer to (1); we have fixed the left-hand side, but the right-hand side is still too weak. But we can amplify further, by exploiting an imbalance in a different symmetry, namely the homogenisation symmetry $(v,w) \mapsto (\lambda v, \frac{1}{\lambda} w)$ for a scalar $\lambda > 0$, which preserves the left-hand side but not the right. Inserting this transform into (4) we conclude that

$|\langle v, w \rangle| \leq \frac{\lambda^2}{2} \|v\|^2 + \frac{1}{2\lambda^2} \|w\|^2$

where $\lambda > 0$ is at our disposal to choose. We can optimise in $\lambda$ by minimising the right-hand side, and indeed one easily sees that the minimum (or infimum, if one of v and w vanishes) is $\|v\| \|w\|$ (which is achieved when $\lambda = \sqrt{\|w\|/\|v\|}$ when $v,w$ are non-zero, or in an asymptotic limit $\lambda \to 0$ or $\lambda \to \infty$ in the degenerate cases), and so we have amplified our way to the Cauchy-Schwarz inequality (1). [See also this discussion by Tim Gowers on the Cauchy-Schwarz inequality.]