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One day I was reading an article on the infinitude of prime numbers in the Proof Wiki. The article introduced a proof that used only topology to prove the infinitude of primes, and I found it very interesting and satisfying. I'm wondering, if there are similar proofs that use topology where it's not obvious that it can be applied. I'm sure that seeing such proofs could also strengthen my intuition with topology.

So my question is: "Which theorems, not directly linked with topology, have interesting proofs that use topology?".

Thanks in advance!

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    $\begingroup$ In my opinion, all proofs of the infinitude of primes rely on the same core mathematical content, and the rest is just fluff. Take your cited topological proof, for example. At one point it uses the existence of lowest common multiple of two integers, and at another point it uses the fact (proven by induction) that inclusion of empty set and closure under intersection implies closure under finite intersection (for the proof to even justify that the topological axioms satisfied). But existence of LCM and induction are the key ingredients of the usual proof of infinitude of primes! $\endgroup$ – user21820 Apr 17 '18 at 14:19
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    $\begingroup$ (At least, equivalent.) The standard proof involves the ability to construct the product of all primes no larger than a given integer. There is a slightly simpler proof by considering $r = p!+1$ (defined inductively) where $p$ is a largest prime, and then if $r$ is not prime then there is (by induction) a smallest integer $q \mid r$ such that $q>1$ and $q$ must be prime so $\gcd(q,r) = \gcd(q,p!+1) = \gcd(q,1) = 1$. I purposely wrote it this way to show the link between Euclid's lemma and existence of LCM. $\endgroup$ – user21820 Apr 17 '18 at 14:25
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    $\begingroup$ Hochster used the hairy ball theorem to prove the existence of two non-isomorphic commutative rings $R \not \cong S$ such that $R[t] \cong S[t]$. $\endgroup$ – Watson Apr 18 '18 at 6:48
  • $\begingroup$ @Watson reference? $\endgroup$ – Andres Mejia Apr 18 '18 at 13:37
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    $\begingroup$ @Andrew Mejia ams.org/journals/proc/1972-034-01/S0002-9939-1972-0294325-3/… $\endgroup$ – Teddy38 Apr 18 '18 at 14:32

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One of the richest sources for this is geometric group theory, and especially the parts of it that construct topological spaces corresponding to particular groups (schreier graphs, cayley complexes, BG spaces etc.) For example, the proof that any subgroup of a free group is free (Nielsen Schreier) epitomizes how topological arguments greatly simplify some parts of group theory.

There are proof methods for theorems from discrete geometry that utilize methods from equivariant topology (such as the ham-sandwich theorem via Borsuk ulam.) These use the CS/TM paradigm, which studies maps between the space of "geometric configurations" to the "Space of viable solutions."

See here for a general reference and wikipedia for a shorter exposition.

I have a blogpost, where I note some "weird proofs in algebra" using algebraic topology: blogpost

These include:

  1. Bezout's identity via separating curves (assumes the euclidian algorithm in disguise.)

  2. Fundamental Theorem of Algebra (resp. existence of eigenvalues) via lefschetz fixed point theorem

and of course, there is the use of "Denseness arguments" in algebraic geometry (such as cayley-hamilton).

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How about the Nielsen-Schreier theorem? The statement of the theorem is entirely algebraic:

Every subgroup of a free group is free.

To prove it, we use algebraic topology: the free group on $n$ generators is the fundamental group of the bouquet of $n$ circles, and any subgroup of the fundamental group is the fundamental group of some covering space of this bouquet. Then it's a matter of using topological arguments to show that every covering space of the bouquet of $n$ circles is itself homotopy equivalent to some bouquet of circles.

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One of the proofs of the Fundamental Theorem of Algebra uses topology.

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    $\begingroup$ 1.7 here is one such proof. $\endgroup$ – Andres Mejia Apr 16 '18 at 20:22
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    $\begingroup$ But that isn't unexpected. The real numbers are developed from the rationals by topological considerations. They cannot be obtained strictly algebraicly. Thus one would expect that important properties of $\Bbb R$ and its extension $\Bbb C$ that are not true for $\Bbb Q$ would involve topology somewhere in the proof. $\endgroup$ – Paul Sinclair Apr 17 '18 at 16:45
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    $\begingroup$ @PaulSinclair All such opinions are subjective. Both that of mine and that of yours. :) $\endgroup$ – szw1710 Apr 17 '18 at 19:06
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    $\begingroup$ Actually, I should have said "one should expect" rather than "one would expect" - after all, just because something should be, doesn't mean it will be - but other than that, there is nothing subjective in my earlier comment. Any proof of the FTA has to involve properties of $\Bbb C$ that are not shared by $\Bbb Q+i\Bbb Q$, and those properties are topological in nature. $\endgroup$ – Paul Sinclair Apr 17 '18 at 23:18
  • $\begingroup$ @PaulSinclair I agree that it's not surprising that the proof is topological in nature. What's surprising is the topological method used, which indeed only makes sense to speak of on connected spaces, but is relatively simple technically. $\endgroup$ – Andres Mejia Apr 18 '18 at 21:07
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Existence of real nowhere-differentiable continous functions can be proved using Baire's theorem.

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The compactness theorem of propositional logic (that a set of sentences has a model as soon as every finite subset hast a model) can be proved by using Tychonoff's theorem on an appropriate Stone Space. It can be proved without topology as a consequence of Gödel's completeness theorem, but the topological proof is more elegant and in fact gives the theorem its name. As far as your stated goal of using such things to improve your intuition on topology, I recommend Steven Vicker's excellent book Topology via Logic.

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Here is a paper using the Borsuk-Ulam theorem to deal with a combinatorial segmentation problem. It got popularized again by this video of the YouTube channel 3Blue1Brown.

Maybe you also like this related video which uses Sperner's lemma to solve another discrete division problems. The interesting thing is that the discrete "Sperner's lemma" is closely related to the topological "Brouwer's fixed point theorem". Each one can be used to prove the other one.

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The proof of Picard–Lindelöf Theorem on existence and uniqueness of a solution of a Cauchy problem $y'=f(x,y)$, $y(x_0)=y_0$ uses the Banach Contraction Principle, which is, of course, of topological nature. Of course, we assume continuity of the right hand side, which is a topology, but differential equations are not immanantly connected with topology.

In my 1st scientific paper together with Nikodem we have proved the affine separation theorem. In the proof we applied Helly's Theorem on nonempty intersection of a certain family of compact sets. Here is a link: https://eudml.org/doc/182478

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The multitude of proofs that go through the Baire Category Theorem. For a list of such proofs, see Your favourite application of the Baire Category Theorem .

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  • $\begingroup$ BCT is awesome. $\endgroup$ – nomen Apr 18 '18 at 20:08
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I don't know much about Wiles' proof of Fermat's last theorem, but from what I have read, it uses topology.

Let me cite a part of that article (p. 7)

Any finite set of Diophantine equations in several variables defines a scheme, actually a special case called a spectrum, and general schemes are gotten by patching together compatible spectra just as a differential manifold patches together parts of n-dimensional real coordinate space $\mathbb{R}^n$. When a scheme $X$ is presented as a topological space plus some algebraic structure then the points correspond to specialized forms of the equations for $X$. Notably, given integer polynomial equations the scheme organizes the corresponding equations on rational numbers and the corresponding equations modulo $p$ for each prime number $p$. The algebra and topology of $X$ capture all information about the equations including their solutions $-$in a form beautifully revealed by cohomology.

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  • $\begingroup$ I would like to learn some new things by being informed about what's wrong with my answer, and I am not saying this sarcastically. So I'd appreciate very much if the downvoter explains why they didn't like this one. $\endgroup$ – polfosol Apr 18 '18 at 16:58
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The first proof of the chromatic number of the Kneser graph, due to Lovász, used the Borsuk–Ulam theorem. Since then, a purely combinatorial proof has been found by Matoušek, but even his proof is topological in spirit: it uses Tucker's lemma, which is a combinatorial analog of the Borsuk–Ulam theorem.

Since Lovász's proof, topological methods have entered combinatorics. Indeed, Matoušek wrote a monograph on the topic, Using the Borsuk–Ulam theorem.

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Real division algebras

$\mathbb{R}^n$ is a division algebra only for $n\in\{1,\, 2,\, 4, \,8\}$.

This is a consequence of the properties of the mapping induced by multiplication map on cohomology groups. For details see Hatcher's Algebraic topology, 4B. For a similar version of this fact see Theorem 2B.5 from Hatcher's book (it only uses homology).

Application in data analysis (dimensionality reduction)

On fiber diameters of continuous maps is a paper that applies Borsuk-Ulam theorem (it's rather trivial application) to show that every map from $f: \mathbb{R}^n \to \mathbb{R}^m$ where $m \leq n$ has fibers (preimages of points) of arbitrarily large radius. This is something useful to know if you're doing data analysis, because it tells you something about the limits of dimensionality reduction.

Functional analysis

Banach-Alaoglu theorem is a theorem that states that any normed space $V$ is isometrically isomorphic to a subspace of $C(X) $ for some $X$. Concretely it states that about $V$'s closed unit ball with $^*$-weak topology. The proof is rather straightforward, it only requires showing that unit ball is compact in this topology (it follows from Tikhonov theorem on products). Also see Gelfand representation for a related fact on Banach algebras.

General remarks:

Many questions mentioned Borsuk-Ulam theorem. For someone interested in discrete math/computer science I'd suggest checking out Matousek's book on using the theorem.

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I was blown away by the fact that de Finetti's Theorem central to Bayesian inference, has different proof approaches and corollaries that use topological spaces. I think just a mere fact that a lot of things we do in Bayesian stats in an applied context on a daily basis are predominantly based on measure theory and then topology as well is just mind-boggling.

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This video discusses a topological proof of the following theorem:

Given a Jordan curve $C$, there exist four points on $C$ that are the vertices of a rectangle.

I'm not aware of any non-topological proof.

I'll sketch the proof. Let's begin with an equivalent restatement:

There exist distinct unordered pairs of points on $C$ of the same length and midpoint (namely, the rectangle's diagonals).

(If we wanted to find a square in $C$, the diagonals would need to be orthogonal, which isn't a property they have in common so much as a binary relation between them. This explains why, as the video notes, we haven't been able to adapt the techniques below to proving the always-a-square conjecture.)

Assume $C$ lives in the plane $z=0$, and consider the function $f:C\times C\to \mathbb{R}^3$ defined by $f(a,\,b):=(\frac{a_x+b_x}{2},\,\frac{a_y+b_y}{2},\,|b-a|)^T$ so $f(a,\,a)=(a_x,\,a_y,\,0)=a\in C$, and $S:=\mathrm{rang} f$ is a 2D surface in $\mathbb{R}^2\times [0,\,\infty)$ with $\partial S = C$. The claimed theorem can now be equivalently restated again:

The surface $S$ is self-intersecting, just like the Möbius strip.

In fact, we'll prove these two surfaces are topologically equivalent.

Endow $C$ with an affine parameter $\lambda$ running from $0$ to $1$, both extrema applying to the same point $p\in C$, with all other $q\in C$ having a unique value of $\lambda$. We can identify $(a,\,b)\in C^2$ with points on a square, with parallel edges joined as in "Space Invaders" topology. By contrast, an unordered pair $\{ a,\,b\}\subset C$ is identified with a point in the shape obtained as follows:

  • Fold the square along $y=x$ to form a triangle, so $\{ a,\,b\}$ isn't distinguished from $\{ b,\,a\}$.
  • We can't join the edges as before in an orientation-preserving way yet, so first cut the square along its other diagonal, $y=1-x$.
  • Rotate the triangles so the orientation-preserving joins can be made after all. But this introduces a half-twist; unsurprisingly, the result is the Möbius strip. (Feel free to try it with paper, scissors & glue.)
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You can prove Arrow's Impossibility Theorem topologically.

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