# Surprising applications of topology

Today in class we got to see how to use the Brouwer Fixed Point theorem for $D^2$ to prove that a $3 \times 3$ matrix $M$ with positive real entries has an eigenvector with a positive eigenvalue. The idea is like this: consider $T = \{ (x,y,z) \mid x + y + z = 1, x, y, z \geq 0 \}$. This is a triangle in $\mathbb R^3$. Take a point $\overline x \in T$, and consider $\lambda_x M \overline x \in T$, for some $\lambda_x \in \mathbb R$. This is a vector which is equal to some $y \in T$. In particular, $\lambda_x M$ is a homeomorphism $T \to T$. Hence it has a fixed point $\overline x$. So $\lambda_x M \overline x =\overline x \implies M \overline x = \frac{1}{\lambda_x} \overline x$. So $\overline x$ is an eigenvector with eigenvalue $\frac{1}{\lambda _x}$, which is certainly positive.

I was extremely surprised when this question came up, as we are studying fundamental groups at the moment and that doesn't seem the least bit related to eigenvalues at first. My question is: what are some other example of surprising applications of topology?

• – Najib Idrissi Jan 27 '15 at 8:22

## 10 Answers

Francis Su described in 1999 ("Rental Harmony: Sperner's Lemma in Fair Division", Amer. Math. Monthly, 106, 1999, 930-42) how to apply Sperner's Lemma---which says that every so-called Sperner coloring of a triangulation of an $n$-simplex contains a cell colored with a complete set of colors---to produce a list of variously sized rents for rooms in a shared house that are fair in a certain sense that accounts for all roommates' preferences. See the column "To Divide the Rent, Start With a Triangle", (New York Times 2014 April 28) for an interesting interactive tool that illustrates the algorithm that exploits the lemma.

How about Furstenberg's proof of the infinitude of prime numbers?

• Thank you, that one is really cool. I hadn't seen it before :) – Johanna Jan 27 '15 at 4:09
• @Johanna Be aware that many don't consider this to be innately topological, e.g. see the MO discussion here and the comments here. – Bill Dubuque Jan 27 '15 at 6:00

The proof of the Cayley–Hamilton theorem in the case of different eigenvalues is very easy. The extension to general case in any field is possible using the Zariski topology.

The Nielsen–Schreier theorem (a subgroup of a free group is itself free) can be proven using methods of elementary algebraic topology.

My personal favourite is the pretty deep result that every finite dimensional divison algebra over $\mathbf{R}$ has dimension $1,2,4$, or $8$. This result seems to be due Kervaire and Milnor; the proof uses methods of advanced algebraic topology. An accessible one-page proof using K- theory was given by Adams and Atiyah, see here (they actually prove a more general result concerning the hopf invariant one problem, of which the statement I mention is a corollary). At present no purely algebraic proof is known.

• Do you know any specific book / paper where I can find this proof? – Johanna Jan 27 '15 at 13:09
• @Johanna This is proved in Hatcher in section 1.A. See also Trees by Serre or this article. – André 3000 Jan 27 '15 at 20:12
• @Johanna I would have mentioned Hatcher, but this was already done by SpamIAm. – Mister Benjamin Dover Jan 27 '15 at 20:29

The Necklace Splitting problem in combinatorics has been beautifully solved by Alon and West using the Borsuk-Ulam theorem.

http://en.wikipedia.org/wiki/Necklace_splitting_problem

Hindman's Finite Sums Theorem: If you partition $\mathbb N$ into finitely many classes, there is an infinite sequence $a_1\lt a_2\lt a_3\lt\cdots$ in $\mathbb N$ such that all of the finite sums $a_{i_1}+a_{i_2}+\cdots+a_{i_k}$, where $i_1\lt i_2\lt\cdots\lt i_k$ and $k\in\mathbb N$, belong to the same partition class. The proof uses some kind of topological algebra on the Stone-Čech compactification of $\mathbb N$.

The Borsuk-Ulam Theorem states that for any continuous map $f : \mathbb{S}^n \to \mathbb{R}^n$, there are two antipodal (opposite) points of $\mathbb{S}^n$ that $f$ maps to the same value. (This follows from the fact that every antipodes-preserving map from the sphere to itself has odd degree.)

For example, in the case $n = 2$, we can conclude that there are always two antipodal points on Earth which have the same temperature and barometric pressure (assuming we idealize the Earth's surface in a suitable way, and regard temperature and pressure as continuous functions of position).

A corollary of the result is the amusingly named Ham Sandwich Theorem, which says that given $n$ objects (measurable sets) in $n$-dimensional space, one can always cut all three of them in half with a single stroke of a knife: More precisely, there is a hyperplane that simultaneously partitions each object into two sets of equal measure.

The fundamental theorem of algebra

Every nonconstant polynomial with coefficients in $\mathbb{C}$ has a root in $\mathbb{C}$

can be shown using algebraic topology. See for example Hatcher (http://www.math.cornell.edu/~hatcher/AT/AT.pdf) Page 31. Theorem 1.8.

Sketch of the proof:

Take a polynomial: $p(z) = z^n +a_1 z^{n−1} +\cdots+a_n$ . Then construct a homotopy:

$$f_r(s) = \frac{p(re^{2πis})/p(r)}{|p(re^{2πis})/p(r)|}$$ ...

I was really surprised about the following theorem. (General Question, sharp result)

Theorem: If a compact Lie Group $G$ acts freely on a sphere, then $G$ is either finite or isomorphic to $S^1$, $S^3$ or the Normalizer of $S^1$ in $S^3$. (As usual i write $S^1=SO(2)$ and $S^3=SU(2)$.

The proof of this results is an easy (If you know a tiny bit of Lie theory) consequence of the following

Lemma For every $n>1$ the group $\mathbb{Z}_n\times \mathbb{Z}_n$ cannot act freely on a sphere.

However this result can be proven by topological methods as can be seen in

https://mathoverflow.net/questions/140703/pseudofree-t2-actions-on-spheres/141132#141132

• Did you mean $S^3$ everywhere you wrote $S^2$? – Jason DeVito Jan 27 '15 at 15:17
• Yes, thank you very much. – O. Straser Jan 27 '15 at 16:26
• It may also be worth mentioning that there is a complete characterization (in terms of presentations) of which finite groups can act freely on $S^n$. This is only hard when $n = 4k-1$: When $n$ is even, only $\mathbb{Z}_2$ can act freely, and if $n = 4k+1$, only cyclic groups act freely. (Actually, I'm not 100% positive about the $4k+1$ case. Certainly all finite cyclic groups act freely, but conversely...?) – Jason DeVito Jan 27 '15 at 16:59
• Yes, I remember that. It was proven in a book of Joseph A. Wolf. However, if I recall it correctly the proof was horrible technical. Anyway, you have also a similiar result, if you replace free by one orbit type and require the action to be smooth. – O. Straser Jan 27 '15 at 18:01

Here is a nice application in the "real world" of engineering. The use of industrial robots is standard in many manufacturing processes, for example to produce cars. Robot arms resemble a human arm and are able to move to (more or less arbitrary) spatial positions. However, there is the big problem of singular positions. See for example https://blog.robotiq.com/why-singularities-can-ruin-your-day. To overcome this problem, engineers increased the number of joints to gain more flexibility. But to their surprise it was not successful: Singular positions proved to be persistent.

In fact, topology shows that it is impossible to avoid singular positions by increasing the number of joints. This was proved in

Gottlieb, Daniel H. "Topology and the robot arm." Acta Applicandae Mathematica 11.2 (1988): 117-121.