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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.]