# Surprisingly short or elegant proofs using Lie theory

Today, I was listening to someone give an exhausting proof of the fundamental theorem of algebra when I recalled that there was a short proof using Lie theory:

A finite extension $K$ of $\mathbb{C}$ forms a finite-dimensional vector space over $\mathbb{C}$, so the group of units $K^\times$ would be of the form $\mathbb{C}^n\setminus\{0\}$, which is simply connected for $n>1$. Since the operation on $K^\times$ is essentially just multiplication of polynomials over $\mathbb{C}$, it must be a Lie group. In sum, if $n>1$, then $K^\times$ is a simply connected abelian Lie group, thus isomorphic (as a Lie group) to $\mathbb{C}^n$, which is absurd (since $\mathbb{C}^n$ is torsion-free). Thus, $n=1$.

What other examples are there of theorems which yield such short or elegant proofs by appealing to Lie theory?

To clarify the criteria: I'm looking for (nontrivial) theorems that are usually stated in terms outside of Lie theory (e.g. the fundamental theorem of algebra) that can be proven in a particularly short or elegant way using Lie groups or Lie algebras.

• I am not an expert on Lie theory, but I am wondering about your example. Clearly, you need to use some properties of $\mathbb{R}$ to prove that $\mathbb{C}$ is algebraically closed. Is it clear that your argument is not cyclic and you are not using the fact that $\mathbb{C}$ is algebraically closed somewhere? – Yiftach Barnea Apr 5 '17 at 18:13
• @YiftachBarnea: The argument, due to Witt, is not circular. It really takes place entirely in the context of real Lie groups, not complex Lie groups (the unit group of a finite-dimensional associative $\mathbf{R}$-algebra $A$ with identity is a Lie group via its structure as an open subspace of the vector space $A$); complex Lie groups and complex analysis play no role in the proof. Moreover, setting up real Lie groups through the key fact that a connected Lie group that is simply connected is determined up to isomorphism by its Lie algebra does not make any use of $\mathbf{C}$. – nfdc23 Apr 6 '17 at 14:51
• @RobinGoodfellow: I think that to a non-expert in Lie theory, it is natural to wonder if the analytic techniques underlying making the Lie group to Lie algebra correspondence involve the same circle of ideas used in various analytic proofs of the Fundamental Theorem of Algebra (especially if the argument is phrased in terms of complex Lie groups as above); e.g., winding numbers or Cauchy's formula, etc. That is probably what Yiftach Barnea was wondering (and I also wondered very briefly when I first learned of this argument when I was a student). – nfdc23 Apr 6 '17 at 14:58
• @YiftachBarnea: The argument uses a lot from the basic theory of real Lie groups; e.g., exponential map (rests on flow along vector fields, so existence/uniqueness basics for first-order ODE's) and the Frobenius theorem on integrable subbundles of vector bundles (underlies the functorial dictionary between connected Lie groups that are simply connected and Lie algebras). Hence, lots of basic differential geometry is lurking in the proofs of the input. And at one key step to show closed subgroups are submanifolds it is used that $\mathbf{Q}$ is dense in $\mathbf{R}$ (false for $\mathbf{C}$)! – nfdc23 Apr 6 '17 at 20:44
• @YiftachBarnea: You're correct that the argument doesn't use that $\mathbf{C}$ contains a square root of $-1$. That is why I highlighted that one really doesn't need to (and for clarity should not) mention $\mathbf{C}$ at all: it suffices to show that there does not exist a field extension $F$ over $\mathbf{R}$ of finite degree $d> 2$. Given such an $F$, $F^{\times}=F-\{0\}=\mathbf{R}^d-\{0\}$ is simply connected (!), and then Lie theory implies this commutative Lie group is isomorphic to $\mathbf{R}^d$ and hence is torsion-free. But $-1 \in F^{\times}$ is non-trivial torsion, contradiction. – nfdc23 Apr 6 '17 at 20:48

The coefficients of the polynomial $$(1+x)(1+x^2)(1+x^3)\cdots(1+x^n)$$ are unimodal. This innocuous-looking fact is surprisingly hard to prove, and perhaps the most elegant proof uses the representation theory of semisimple Lie algebras. See Stanley's survey for further details and related examples.

• ice example. Come to think of it, this may be a special case of the answer I gave: when $q$ is the size of a finite field $k$, the formula $(1+q)(1+q^2)(1+q^3) \cdots (1+q^n)$ gives the number of maximal isotropic spaces for a symplectic form on $k^{2n}$, which is the number of $k$-rational points on a symplectic Grassmannian $G = {\rm Sp}_{2n}(k) / P$; so the $x^i$ coefficient of $(1+x)(1+x^2)(1+x^3) \cdots (1+x^n)$ must be the Betti number $b_{2i}(G) = \dim H^{2i}(G)$, and then the $\mathfrak{sl}_2$ structure yields the unimodality of the sequence $\{b_{2i}\}_{i=0}^{\dim G}$. – Noam D. Elkies Apr 6 '17 at 0:49
• [Naturally I meant to start with "Nice example", not "ice example" . . .] – Noam D. Elkies Apr 6 '17 at 1:12
• @NoamD.Elkies, it's just how the cool kids say "Cool example", I assume. – LSpice Apr 7 '17 at 18:00
• Another example along these lines -- one can prove that the number of (isomorphism classes) of unlabelled graphs on $n$ vertices and $k$ edges is unimodal in $k$, using $\mathfrak{sl}_2$-representation theory. See e.g. here for an exposition. – Daniel Litt Apr 11 '17 at 18:00

A famous example is the proof of the "Hard Lefschetz theorem" via finite-dimensional representations of $\mathfrak{sl}_2$. For example (http://relaunch.hcm.uni-bonn.de/fileadmin/perrin/chap10.pdf):

Example 10.4.5 Let $X$ be a compact Kähler manifold of complex dimension $n$ (say for example a compact [smooth] projective variety). Then Hodge theory defines endomorphisms $L$ and $\Lambda$ on $H^*(X,{\mathbb C})$. Set $X = L$ and $Y = \Lambda$ and $H(v) = (n-p)v$ for $v \in H_p(X,{\mathbb C})$. Then one can prove that this defines a $\mathfrak{sl}_2$-representation structure on $H^*(X,{\mathbb C})$. Then Corollary 10.4.4 (ii) for $V = H^*(X,{\mathbb C})$ is called the Hard Lefschetz Theorem. Of course the difficulty here is to construct the endomorphisms $L$ and $\Lambda$ and prove that they satisfy the correct commuting relations.

Likewise J.-P. Serre, in Complex Semisimple Lie Algebras (Springer 1966, tr. 1987 by G. A. Jones), Remark 2 at the end of Section 5 of "IV. The Algebra $\mathfrak{sl}_2$ and Its Representations":

Here is an example of an application of Theoremes 3 and 4, independent of the interpretation of $\mathfrak{sl}_2$ as the Lie algebra of ${\rm SL}_2$:

Let $U$ be a compact Kähler variety of complex dimension $n$, and let $V$ be the cohomology algebra $H^*(U,{\bf C})$. Hodge theory associates endomorphisms $\Lambda$ and $L$ of $V$ with the kählerian structure on $U$ (cf. A. Weil, Variétés kähleriennes, Chap. IV); let us take $X$ and $Y$ to be these endomorphisms, and define $H$ by the relation $Hx = (n-p)x$ if $x \in H^p(U,{\bf C})$. Then one can check (Weil, loc. cit.) that $V$ becomes a $\mathfrak{g}$-module. By applying Theorems 3 and 4 to this module, one retrieves Hodge's theorem on "primitive" cohomology classes.

Let $\Gamma$ be an arbitrary group, and $(V, \rho), (V', \rho')$ two semisimple finite-dimensional linear representations of $\Gamma$ over a field $k$ of characteristic 0. The tensor product representation $V \otimes V'$ is typically not irreducible when $\rho$ and $\rho'$, but is it at least semisimple? Note that there are no "finiteness" hypotheses on $\Gamma$ at all.

The affirmative answer is a classic result due to Chevalley, the statement of which does not mention Lie theory at all, but the only known proof (as far as I'm aware) goes through applying the structure theory of linear algebraic groups to the (possibly disconnected) Zariski closure of $\Gamma$ in ${\rm{GL}}(V)$ and ${\rm{GL}}(V')$ to ultimately reduce to the semisimplicity of finite-dimensional representations of semisimple Lie algebras (i.e., those with vanishing radical) in characteristic 0.

The proof is not short (if one is not familiar with the structure theory of linear algebraic groups), but it is very elegant and more importantly (for the question posed) it really does use Lie theory in an essential way (but Lie algebras, not Lie groups).

• This is amazing. Is it written anywhere more modern than the Chevalley corpus? – LSpice Apr 7 '17 at 18:01
• Serre discusses this in a number of places, e.g. arxiv.org/abs/math/0305257, part 2, lecture 1. – Denis Chaperon de Lauzières Apr 7 '17 at 18:20
• @LSpice: The above arxiv link to the Serre lecture notes illustrates my statement that the proof is not short if one is not familiar with the structure theory of linear algebraic groups: Serre assumes his audience knows that theory, so his proof is a model of efficiency (as always), but if one doesn't know it then the proof becomes somewhat longer because it isn't obvious from the raw definition of "semisimple group" (via geometric solvable radical) that such groups have completely reducible representation theory in characteristic 0. Anyway, Chevalley's result & proof are indeed amazing. – nfdc23 Apr 8 '17 at 19:57
• Thanks, @DenisChaperondeLauzières and nfdc23. I am reasonably familiar with the structure theory, so I have no problem with assuming it. I had also somehow never seen the linked Serre lecture notes before, so that's a bonus. – LSpice Apr 8 '17 at 20:11

Does there exist a nonvanishing vector field on $S^3$?

Yes: $S^3$ is the unit quaternions and hence a Lie group. Thus there exist left invariant (never zero!) vector fields.

• And you can say a bit more: $S^3 \simeq SU_2({\bf C})$ is parallelizable. On the other hand, $S^7$ is parallelizable because it is the set of unit octonions. Yet this set is not a group due to the lack of associativity, so this is not really the group structure that is at hand there. – coudy Apr 5 '17 at 9:36
• I am with coudy here: the real prize is parallelizability. Existence of a nonvanishing vector field on odd-dimensional spheres is straightforward, due to a fixed point free action of the circle $|z|=1$ on the unit sphere in $\Bbb {C}^n$ by multiplication. – Victor Protsak Apr 5 '17 at 15:49
• And I agree with you two :). (I used this [parallelizability] fact just yesterday!) – Artur Jackson Apr 5 '17 at 23:50

Many results in hyperbolic geometry are proven using the group of isometries of the hyperbolic space in an essential way.

My personal favorite is the Ergodicity of the geodesic flow$^*$.

Consider a compact hyperbolic manifold $M$. Given a point $x\in M$, a unit vector $v\in T_xM$ and a time $t\in \mathbb{R}$, when moving time $t$ in a unit speed along the geodesic emanating from $x$ in direction $v$, one reaches a new point $t\cdot x\in M$ pointing in direction $t\cdot v \in T_{t\cdot x}M$. This process, which is an action of $\mathbb{R}$ on $T^1M$, is known as the geodesic flow on $M$.

Now, the compact space $T^1M$ carries a natural volume form which gives rise to a finite measure, invariant under the $\mathbb{R}$-action. It is an important fact that this action is ergodic (in fact, mixing). This is a result of Hopf from 1939. This result was reproved by Mautner in 1957 using Lie groups reasoning. From a more modern point of view it is seen as a special application of a general result regarding vanishing at infinity of matrix coefficients of unitary representation of semi-simple Lie groups. The latter result is due to Howe-Moore and it is rather easy to prove.

* You should appreciate this theorem even if you have no interest in ergodicity per-se. It has many outside applications, e.g for counting and equidistribution results. Let me mention in particular that it is an essential ingridant in the celebrated Wise-Agol result (previously known as the virtually fibered conjecture). It enters its proof via the work of Kahn--Markovic.

I originally intended to write this as a comment but it turned out to be too long. The following is a proof of the fundamental theorem of algebra similar to OP's but which does not rely on the correspondence between connected, simply-connected Lie groups and Lie algebras but on somewhat easier results on Lie theory and some topological facts.

Suppose $K$ is a non-trivial field extension of $\Bbb{R}$ of degree $n$. Then, $K^\times$ is a connected, abelian real Lie group of dimension $n$. Commutativity implies that the exponential map $\mathrm{exp}:\Bbb{R}^n\to K^\times$ is a group morphism, and by connectedness we get that it is surjective and thus a covering map. The kernel $\Gamma$ of the exponential map is then a discrete subgroup of $\Bbb{R}^n$, isomorphic to $\Bbb{Z}^d$, and so $K^\times \simeq \Bbb{R}^{n-d}\times (S^1)^d$.

We know that $d\geq 1$ since there is torsion in $K^\times$, and in fact we must have $d=1$, for otherwise there would be $n^d$ $n$-th roots of unity in $K$. Therefore $K^\times \simeq \Bbb{R}^{n-1}\times S^1$ but also $K^\times \simeq \Bbb{R}^n\setminus\{0\}$, and we get that $n=2$ for otherwise $K^\times$ would be simply-connected.

• How does this differ from what I wrote? – Robin Goodfellow Jan 18 '18 at 19:15
• I'm not sure if I misread/misunderstood your argument but it seems to rely on the equivalence between connected, simply connected Lie groups and Lie algebras (when you state that $K^\times\simeq\Bbb{C}^n$ because it is a simply connected abelian Lie group). My argument relies only on basic facts about the exponential and the topology of the spheres. – Fernando Martin Jan 18 '18 at 19:27
• (This is along the lines of what Uri Bader suggests in his comment). – Fernando Martin Jan 18 '18 at 19:38
• Ah. I see the confusion. The covering property of the exponential is my reason that they are isomorphic. Essentially, from my point of view, you just wrote what I did but expanded upon what I had considered to be trivialities. Of course, we could instead use that simply connected locally isomorphic Lie groups are isomorphic (it isn't that much more difficult), but part of my point in the OP was that the proof from the exponential covering perspective is genuinely elementary. – Robin Goodfellow Jan 18 '18 at 19:50