up vote 13 down vote favorite
6

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.

share|cite|improve this question
up vote 7 down vote

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.

share|cite|improve this answer
up vote 6 down vote

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 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.

share|cite|improve this answer
up vote 2 down vote

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).

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.