# Solving algebraic problems with topology

Often, topologists reduce a problem which is - in some sense - of geometric nature, into an algebraic question that is then (partiallly) solved to give back some understanding of the original problem.

Sometimes, it goes the other way, i.e. algebraic topology is used to attack some problem from algebra. Here are two examples I particularly like:

1) Schreiers theorem: Every subgroup of a free group is free: This reduces to study coverings of some wedge of circles.

2) Group cohomology: E.g. questions like "Does every finite group have nontrivial cohomology in infinitely many degrees?" - see Non-vanishing of group cohomology in sufficiently high degree. Such questions can be phrased purely algebraically, but still the best way to think about them is via topology.

I am interested in seeing more examples of this kind. What is your favourite application of topology in algebra?

• See the book: Ol'shanskii, A.Yu. "Geometry of Defining Relations in Groups" – TT_ Jun 6 '15 at 2:43

Theorem (Arnold - 1970): The algebraic function defined by the solutions of the equation $$\ \ z^n+a_1z^{n-1}+\cdots +a_{n-1}z+a_n=0\ \ \ ,$$ cannot be written as a composition of polynomial functions of any number of variables and algebraic functions of less than $\phi(n)$ variables, where $\phi(n)$ is $n$ minus the number of ones appearing in the binary representation of the number $n$.

The proof is essentially a clever application of the computation of the mod. 2 cohomology ring of the braid group $B_n$ by Fuchs.

(And I seem to remember Vershinin explaining that Arnold asked Fuchs to compute this ring for this very reason).

• How is this connected to Hilbert's 13th? And is the Bring quintic relevant to this theorem? We have $\phi(5) = 3$, but the general quintic can be reduced to a form with only one variable. – Tito Piezas III Jun 7 '15 at 4:48
• @TitoPiezasIII : I just found this paper math.toronto.edu/khesin/papers/ArnoldSecond.pdf that explains the whole story. See in particular pp.13-14 for the idea, and the text of Fuchs (the first one) for a historical account. – few_reps Jun 8 '15 at 0:10
• @few_reps what is the reference for the quoted theorem? – T.... Sep 6 '16 at 11:50
• @Turbo : reference 5 in the paper cited in the above comment, as indicated in the contribution by Fuchs. – few_reps Sep 6 '16 at 13:07

If $n \geq 2$ is even and there exists a continuous map $f \colon S^{2n-1} \to S^n$ with Hopf invariant $1$, then $n \in \{2,4,8\}$.

This can be used to prove that $\mathbb{R}^n$ has the structure of a real division algebra if and only if $n \in \{1,2,4,8\}$.

For a nice reference, have a look at Atiyah's book about K-theory (as Vincent mentions in the comments). There is also the book "General Cohomology and K-Theory" by Peter Hilton, which I really liked. This is book 1 in the LMS Lecture Note Series. The book project "Vector Bundles and K-Theory" by Allen Hatcher also covers the proof. If you want (german) notes of a course with an introduction to K-theory where this will be (hopefully) proven in the end, you can find them for example here.

• If I recall correctly this is also explained quite well in the K-theory book by Attyah. – Vincent Jun 5 '15 at 8:30
• As much as I love this theorem, it's worth pointing out that the algebraic corollary has a purely algebraic proof: en.wikipedia.org/wiki/…, en.wikipedia.org/wiki/… – Dylan Wilson Jun 15 '15 at 20:31
• (Moreover, unlike the case of a "subgroup of a free group is free", the algebraic proof is substantially easier than the topological one. But, in this case, the topological theorem is much stronger, so it's worth the extra work!) – Dylan Wilson Jun 15 '15 at 20:32
• @DylanWilson, the proof is simple only for associative algebras, while the topological result is true without restrictions. – Anton Fetisov Jun 15 '15 at 22:23

Here is one I like very much :

Let $F_s$ denote the free group on $s$ generators.

• If $F_s$ is a subgroup of finite index in $F_3$, then $s$ is odd.

More generally :

• assume we have an inclusion $F_s\subset F_t$ with $[F_t:F_s]=m$. Then $m=\frac{1-s}{1-t}$.

(Topological proof : we have a covering $F_t/F_s\to BF_s\to BF_t$ and we compute the Euler characterstics.)

Even more generally :

• Let $\Gamma$ be a torsion-free group of finite homological type (i.e. it has finite cohomological dimension, and a finite index subgroup of $\Gamma$ has finitely generated integral cohomology - any torsion-free arithmetic group satisfies these hypotheses). If $\Gamma'$ is a torsion-free group containing $\Gamma$ as a subgroup of finite index $m$, then $m$ divides $\chi(\Gamma)$.

But the most spectacular in this vein seems to be

• If moreover $\Gamma$ is normal in $\Gamma'$, and $m$ is a prime power $p^r$ with $\text{gcd}(p,\chi(\Gamma))=1$, then $\Gamma\to\Gamma'\to Q$ is split !

This last theorem is due to K.S. Brown., and, applying it to $F_3$ we get : any extension $F_3\to\Gamma'\to Q$ with $Q$ a $p$-group of odd order is split.

The fundamental theorem of algebra, any non-constant polynomial with complex coefficients has at least one root.

The whole field of geometric group theory very much uses geometric and topological arguments to prove group theory facts. Stallings for example used topological arguments to prove that only free groups have cohomological dimension 1. He also gave a topological proof of Grushko's theorem on generators of free products.

Gromov's proof that only virtually nilpotent groups have polynomial growth uses geometry and topology.

The train track proof of Scott's conjecture that the fixed point set of a free group automorphism has rank n is topological.

Van Kampen diagram proofs line Lyndons theorem on cohomlogical dimension of one relator groups tend to use some topology plus properties of planar graphs.

There are many applications of numerical invariants (LS-category) to differential geometry, but let me give one pretty example to resolution of polynomial equations.

In the paper

"On the equation of degree 6" Comment. Math. Helv. 79 (2004) 605–617

C. De Concini, C. Procesi and M. Salvetti prove using the schwartz genus of fibrations that 5 holomorphic functions suffice to compute the roots of a polynomial of degree 6.

.......

Definition: For a fibration $p:A\rightarrow B$ the Schwarz genus $g(p)$ is the minimum number for which one can cover the base with open sets $U_i$ so that the fibration, restricted to $U_i$ is trivial.

Let $\mathcal{P}_n$ be the space of of monic polynomials of degree $n$ over $\mathbb{C}$ with distinct roots. Let $\Delta$ be the big diagonal $$\Delta=\{(z_1,\ldots,z_n)\in \mathbb{C}^n:z_i\neq z_j, \forall i\neq j\}$$ then $\mathcal{P}_n=(\mathbb{C}^n-\Delta)/S_n$ (where $S_n$ is the symmetric group). And let consider the covering: $$\gamma_n:\mathbb{C}^n-\Delta\rightarrow \mathcal{P}_n$$ Concini, Procesi and Salvetti prove that $g(\gamma_6)=5$.

Let me add that the genus of $\gamma_n$ is related to topological complexity of algorithms that should compute the roots of a polynomial (S. Smale, "On the topology of algorithms I", Journal of complexity 3 (1987), 81–89.).

There are cool examples already for finite group: Dijkgraaf and Witten recover and generalize a combinatorial formula due to Burnside using the TFT formalism.

It's probably worth elaborating on this. Let $\Sigma_g$ denote a compact orientable surface of genus $g$ and let $G$ denote a finite group. The problem is to compute $|\text{Hom}(\pi_1(\Sigma_g), G)|$; for example, when $g = 1$ this is the number of pairs of commuting elements of $G$, which is $|G|$ times the number of conjugacy classes of $G$, or equivalently $|G|$ times the number of irreducible representations. The formula, which I know as Mednykh's formula, gives the answer as

$$\frac{|\text{Hom}(\pi_1(\Sigma_g), G)|}{|G|} = \sum_V \left( \frac{\dim V}{|G|} \right)^{\chi(\Sigma_g)}.$$

where the sum ranges over all irreducible representations $V$ of $G$. Note that when $g = 1$ we recover the above result and when $g = 0$ we recover the fact that $|G| = \sum_V (\dim V)^2$. This result was used here to show that $\pi_1(\Sigma_g)$ is not a free group ($g \ge 1$).

There's some leeway in terms of how you juggle the powers of $|G|$ here, but I prefer this statement because the LHS is the groupoid cardinality of the "moduli groupoid" of principal $G$-bundles on $\Sigma_g$; this is the number that directly appears in the TFT story, as the value of "untwisted" Dijkgraaf-Witten theory on $\Sigma_g$.

The TFT proof can be found, for example, here, although that document commits the terrible sin of having no diagrams. The important diagram to draw here involves breaking up a surface of genus $g$ as a composite, in the cobordism category, of a cup, $g$ copies of the "tube of genus $1$," and a cap, and the important computation involves applying a TFT to this composite.

The TFT proof generalizes directly to the case of "twisted" Dijkgraaf-Witten theory, where a class $\alpha \in H^2(G, \mathbb{C}^{\times})$ appears. The LHS is now a sum over homomorphisms $\pi_1(\Sigma_g) \to G$ weighted in a manner determined by $\alpha$ while the RHS is now a sum over the projective irreducible representations corresponding to $\alpha$. I don't know if this more general result has a character-theoretic proof.

Jiří Matoušek's book, Using the Borsuk-Ulam Theorem, gives applications of a topological theorem (that for every continuous map $f\colon\mathbf S_n\to\mathbf R^n$, there exists a point $x\in\mathbf S_n$ such that $f(x)=f(-x)$) to geometry and combinatorics.

• I think the question was about applications to algebra. – Omar Antolín-Camarena Jul 3 at 1:26

Special values of Dedekind zeta functions can frequently be expressed (and then calculated) in topological terms, e.g. from the Euler characteristics of certain manifolds/orbifolds.

An easy-to-formulate example is that of zeta functions of totally real quadratic fields $K$: the value of $\zeta_K(2)$ can be computed from the orbifold Euler characteristics of the Hilbert modular surface $$SL(2,O_K)\backslash (H^2\times H^2),$$ where $SL_2(O_K)$ is embedded into $SL(2,R)\times SL(2,R)\subset Isom^+(H^2\times H^2)$ via the two different embeddings $K\to R$.

More involved examples can be found in http://people.mpim-bonn.mpg.de/zagier/files/scanned/ValuesofZetaFunsAndApplications/fulltext.pdf and its references.

In a similar vein, the class number of $O_K$ equals the number of cusps (= the number of ends) of the Hilbert modular surface, which (I hope) might be easier to compute.

• Not sure about your last hope ... the identification of cusps with classes in $\mathcal O_K$ is usually used in the other direction, to find the cusps. (But I'd be happy to see an application as in your hopes). – few_reps Jun 1 '15 at 13:21

The Brouwer fixed point theorem can be used to prove a particular case of the Perron-Frobenius theorem; specifically, one can prove that a matrix $A\in\mathbb{R}^{n\times n}$ with positive entries has a real, positive eigenvalue.

• And, more generally, the entire Perron--Frobenius theorem can be deduced from Banach's contraction mapping theorem, using the Hilbert metric on the positive orthant. – HJRW Jun 6 '15 at 9:34
• The Brouwer fixpoin theorem can also be used to prove Nash's theorem about equilibria in game theory. Unfortunately, this is not an algebraic statement. However, if you talk about this to high school students or undergraduates, and you go all the way through combinatorics and elementary probability to state the problem, it is usually a surprise that the proof comes from topology. – Sebastian Goette Oct 21 '15 at 11:14

Dylan Thurston gave a new proof of the Duflo isomorphism using ideas from knot theory (in particular the Kontsevich integral).

Not the same league as the other answers but still nice: One can prove the simplicity of $PSU(2)$ (and I mean in the algebraic sense of not having non-obvious normal subgroups, not in the Lie-group sense of not having non-obvious normal closed subgroups) by topological methods.

The key is to show that the conjugacy classes in $SU(2)=S^3=\{h\in\mathbb{H} \mid \|h\|=1\}$ are intersections of $S^3$ with hyperplanes (parallel to the hyperplane of purely imaginary quaternions). These intersections are either single points (north and south pole of the 3-sphere, corresponding to the elements 1 and -1) or whole 2-spheres.

Now consider a normal subgroup $N\notin \{ 1, \{\pm 1\}\}$. Then $N$ contains a 2-sphere which is a path-connected subset. Therefore there also exists a path from 1 to an element $\neq 1$ which lies completely in $N$. This path must intersect many conjugacy classes which therefore also lie in $N$. Their union contains a neighbourhood of $1$. The only subgroup of a connected topological group which contains an open subset is the whole group. Therefore $N=SU(2)$ which proves simplicity of $PSU(2)$.

Here is a simple (to state) linear algebra problem: given $n$ Hermitian matrices with known spectra, what sets of real numbers can be the spectrum of their sum?

In this beautiful paper the authors address this question using Schubert calculus. (Admittedly, I'm not sure on which side of the boundary between algebraic topology and algebraic geometry this falls.)

Quillen's first construction of higher algebraic $K$-theory (the plus construction) was purely topological. It is related to group cohomology as mentioned in the question. The later constructions (Quillen's $Q$-construction, Waldhausen $K$-theory etc.) involve manipulations with categories, but in the end one always takes the homotopy groups of the classifying space. I am not sure if there is a construction now that throws out all topology. See Weibel's book for more details.

There are many examples coming from topological field theories, though I'm not quite sure that it's the sort of example you're looking for.

At first it seems that TFT's are about using algebraic structures to produce topological invariants. In fact, quite often those invariants are themselves complicated objects which are meaningful in representation theory. Hence, computing the invariant of two knots or surfaces which you know are equivalent gives interesting identities.

There are cool examples already for finite group: Dijkgraaf and Witten recover and generalize a combinatorial formula due to Burnside using the TFT formalism. For quantum groups, you get identities for multiplicities of decomposition of tensor product of modules, or for Macdonald's polynomial.

Last but not least, it turns out that in some sense those invariants have another construction using a direct analytical approach. You can then use this invariant to reconstruct the algebraic structure you would start from in the previous paragraph. Long story short, this analytically defined knot invariant, a variant of the Kontsevich integral, can be used to prove Kontsevich theorem that every Poisson manifold can be quantized, for example.

Bezout's theorem says that the intersection number (defined algebraically as the sum of the intersection multiplicities) of two algebraic curves $C$ and $D$ in $\mathbb{P}^2$ is equal to $\deg(C)\cdot \deg(D)$. The fastest way to prove this is to show that the intersection number is equal to the topological intersection number $[C]\cdot [D]$ of the fundamental classes of $C$ and $D$ in $H^2(\mathbb{P}^2,\,\mathbb{Z})$. It is clear that $[C]\cdot \ell=\deg(C)$ from the definition of degree and using that $\ell\cdot \ell=1$, we conclude that $[C]=\deg(C)\ell$. Bezout's theorem follows.

A perhaps slicker rephrasing which highlights the way that topology is used to help out algebra: A set of two polynomial equations in two variables of degrees $c$ and $d$ has at most $cd$ solutions.

Using Priestley duality for distributive lattices and compact, totally disconnected ordered topological spaces, many purely algebraic questions have been solved using quite simple topological tools. For instance the fact that free products of affine complete lattices are affine complete, boils down to a topological argument, as it has been done in section 3 of this article.

In principle, purely algebraic questions about commensurated subgroups of a group $G$ can be rephrased in terms of homomorphisms $\phi: G \rightarrow H$ such that $H$ is a totally disconnected, locally compact (t.d.l.c.) group and the image of $\phi$ is dense: specifically, commensurated subgroups of $G$ show up as preimages of compact open subgroups of $H$, and all commensurated subgroups of $G$ will arise in this way. The most sophisticated example I have seen of such an application is the paper 'Commensurated subgroups of arithmetic groups, totally disconnected groups and adelic rigidity' by Y. Shalom and G. A. Willis, where the authors use the theory of t.d.l.c. groups to tackle a conjecture of Margulis and Zimmer about commensurated subgroups of arithmetic groups. The theory of t.d.l.c. groups is not as deep from a topological perspective as that of Lie groups, because the structure of a t.d.l.c. group as a topological space is much less informative, but topological arguments (especially compactness arguments and the Baire category theorem) inevitably come into it.

• In Shalom-Willis the whole argument is algebraic insofar as I understand it: bounded generation by distorted unipotents, the Schlichting-(Bergman-Lenstra) theorem. You don't need to introduce locally compact completions to perform it. Rather, the algebraic result provides a corollary in terms of homomorphisms to locally compact groups. – YCor Jun 5 '15 at 20:35

Fundamental theorem of algebra for Quaternions. The proof for $\mathbb{H}$ uses the degree of the maps to $S^3$, very similar to the Gauss's proof for $\mathbb{C}$ with the degree of map to $S^1$.

Taras Banakh today told me about a solution of a group theory problem by means of topological algebra. If I remembered it right, it is the following.

For a cardinal $$\kappa$$ as $$S(\kappa)$$ we denote the group of all permutations of $$\kappa$$. Prove that if $$\kappa<\lambda$$ then the group $$S(\lambda)$$ is not embeddable into the group $$S(\kappa)$$. If $$| S(\lambda)|=2^\lambda>2^\kappa=| S(\kappa)|$$ then the non-embeddability follows from the size comparison, which implies the proof under GCH. But in order to obtain a proof in ZFC, Taras Banakh introduced a cardinal invariant $$w(G)$$ of a group $$G$$, which equals to a minimal weight of a Hausdorff group topology on the group $$G$$. Clearly, if $$H$$ is a subgroup of the group $$G$$ then $$w(H)\le w(G)$$. Banakh and Mildenberger proved that $$w(G)=\kappa$$ for any group $$G\subset S(\kappa)$$ containing the subgroup $$Alt(\kappa)$$ consisting of finitely supported even permutations of an infinite cardinal $$\kappa$$. So, for any infinite cardinals $$\kappa<\lambda$$ the inequality $$w(S(\kappa))=\kappa<\lambda=w(\mathrm{Alt}(\kappa))$$ implies that $$\mathrm{Alt}(\lambda)$$ is not isomorphic to a subgroup of $$S(\kappa)$$.

• That for $\kappa<\lambda$, the subgroup of finitely supported permutations in $S(\lambda)$ does not embed as a subgroup of $S(\kappa)$, was first obtained by N de Bruijn (Theorem 5.1 in Embedding theorems for infinite groups. Indag. Math. 19 1957 560--569. Erratum: Indag. Math. 26 1964 594--595.) It was extended by R. McKenzie: A note on subgroups of infinite symmetric groups. Indag. Math. 33 (1971), 53--58 who showed that the direct sum of $\lambda$ non-abelian groups (which can be found in $\mathrm{Alt}(\lambda)$ can't embed into $S(\kappa)$. The methods are not topological. – YCor Oct 21 '18 at 8:27
• Actually it's even slightly earlier: G. Higman proved (On infinite simple permutation groups. Publ. Math. Debrecen 3 (1954), 221–226 (1955)) that for any infinite cardinal $\lambda$, $\mathrm{Alt}(\lambda)$ has no subgroup of index $<\lambda$. This appears as Theorem 11.2.5 in W.R. Scott's book *Group Theory" (1964). – YCor Oct 21 '18 at 9:30
• Scott's book: (1987 reprint of 1964 original) hep.fcfm.buap.mx/ecursos/TTG/libros/… – YCor Oct 21 '18 at 9:31

One can use the hairy ball theorem to give examples of stably free modules that are not free. See the proof of proposition 1.1 in these notes for example.

Watson also gave an application of the same theorem to prove that there are distinct rings that become isomorphic after adjoining a variable

• Really, this is just the image under Serre–Swan of the observation that the tangent bundle of $S^2$ is non-trivial yet stably trivial. – Branimir Ćaćić Jun 11 at 2:32
• Yep. I guess even then, the hairy ball theorem is a good way to show that it is not trivial. – Andres Mejia Jun 11 at 2:33
• Speaking of Serre–Swan, this is also giving you two very important (if morally identical) examples of a commutative monoid with non-injective canonical map to the corresponding Grothendieck group: two non-isomorphic fgp modules over $\mathbb{R}[x,y,z]/\langle x^2+y^2+z^2-1\rangle$ giving rise to the same class in algebraic $K$-theory, and two non-isomorphic real vector bundles over $S^2$ giving rise to the same class in topological $K$-theory. – Branimir Ćaćić Jun 11 at 2:56
• [All of which is to say that this is actually a really important example from the standpoint of $K$-theory.] – Branimir Ćaćić Jun 11 at 3:08

The following paper and its references contains some algebraic consequences of vector bundle theory.

Vakhtang Lomadze, Applications of vector bundles to factorization of rational matrices, Linear Algebra and its Applications 288 (1999) pp249–258, doi:10.1016/S0024-3795(98)10220-3

• I put in the reference to the paper, rather than just a ScienceDirect url – David Roberts Feb 4 '16 at 11:42
• @DavidRoberts thanks very much for your revision. – Ali Taghavi Feb 4 '16 at 14:36

My friends Jacob Mostovoy and Christopher Roque told me about a nice algebraic lemma they proved:

Let $$f_i : G_i \to H$$ be surjective group homomorphisms with kernel $$K_i$$ for $$i=1,2$$. Then the kernel $$K\require{AMScd}$$ of the induced homomorphism $$f : G_1 \ast G_2 \to H$$ from the free product is given by $$K_1 \ast K_2 \ast F_{|H|-1}$$, where $$F_{|H|-1}$$ is the free group on $$|H|-1$$ generators (if $$H$$ is infinite, here $$|H|-1 = |H|$$).

I told them you can prove this with classifying spaces. Since $$B(G_1 \ast G_2) \simeq BG_1 \vee BG_2$$, there is a homotopy pushout square: $$\begin{CD} \mathsf{pt} @>>> BG_1 \\ @VVV @VVV \\\ BG_2 @>>> B(G_1 \ast G_2) \\ \end{CD}$$

You can consider this a square of spaces over $$BH$$ by using the base point inclusion, $$Bf_1$$, $$Bf_2$$ and $$Bf$$ respectively. Taking homotopy pullback along the base point inclusion $$\mathsf{pt} \to BH$$ produces a new homotopy pushout square: $$\begin{CD} \Omega BH \simeq H @>>> BK_1 \\ @VVV @VVV \\\ BK_2 @>>> BK \\ \end{CD}$$

This means that $$BK$$ is homotopy equivalent to the space you get by connecting $$BK_1$$ and $$BK_2$$ by $$|H|$$ intervals, so $$BK \simeq BK_1 \vee BK_2 \vee \bigvee^{|H|-1} (S^1)$$.

You can find out what they use this lemma for and read both that homotopy-theoretical proof as well as their algebraic proof in their lovely short paper: Planar pure braids on six strands. The paper has other applications of algebraic topology to algebra: Proposition 1 makes short work of computing the pure planar braid groups on fewer than 6 strands using topological arguments. For an algebraic computation of the pure braid groups on up to 4 strands see Structural aspects of twin and pure twin groups by Valeriy Bardakov, Mahender Singh, Andrei Vesnin, which also conjectures that the one on 5 strands is $$F_{31}$$ as shown by Mostovoy and Roque in the first paper I mentioned.

• Hi, this also follows from the theory of graphs of groups. Take the canonical action of $G_1*G_2$ of a tree and restrict it to Ker(f). Now we need to compute the quotient graph of groups, which has two vertices with stabilizer groups $K_1$ and $K_2$ and $H$ edges connecting them. Then the fundamental theorem for graphs of groups tells us that $Ker(f)$ can be obtained as the fundamental group of that graph of groups, e.g. it is exactly given as above. – HenrikRüping Jul 3 at 7:38
• Hi @HenrikRüping! Thanks for that argument. As you know, for me the theory of graphs of groups is more exotic than classifying spaces, but I really should learn it some day. :P – Omar Antolín-Camarena Jul 3 at 15:32