Bilinear forms in compact/semisimple Lie group theory

If you look up the list of compact or semisimple Lie groups, you will see that three out of four infinite families (B, C and D) are defined in terms of a bilinear form on a vector space, either symmetric or skew-symmetric.

Are there any underlying reasons for this prominence of bilinear/quadratic forms in Lie group theory? Why do they, and not any other geometric objects, play such a fundamental role?

• If you know that simple Lie groups and simple algebraic groups are mostly the same things, then you know that these have to be defined by algebraic equations. These kind of forms certainly give algebraic equations. Other constructions in geometry can be more of the flavor, your object satisfies a differential equation, which are not going to be algebraic in the same way. Sep 12 '20 at 13:58

(Edit: I rewrote this answer. In the first draft I tried to take some shortcuts and found that they didn't work.)

Let $$G$$ be a compact Lie group acting faithfully on a f.d. vector space $$V$$ over $$\mathbb{C}$$. It's a nice exercise to show that every f.d. irreducible representation of $$G$$ appears in some tensor product $$V^{\otimes n} \otimes (V^*)^{\otimes m}$$ (see, for example, this old MO question). What this implies is that the entire structure of the category $$\operatorname{Rep}_\text f(G)$$ of f.d. representations of $$G$$ is contained in the data of the invariant tensors $$\operatorname{Hom}(V^{\otimes m}, V^{\otimes n})^G$$: more formally, these invariant tensors describe the subcategory of $$\operatorname{Rep}_\text f(G)$$ generated by $$V$$ under tensor product and dual and direct sum, and the nice exercise implies that $$\operatorname{Rep}_\text f(G)$$ is the idempotent completion of this subcategory.

Furthermore, the Tannaka half of Tannaka–Krein duality tells us that $$G$$ is determined by $$\operatorname{Rep}_\text f(G)$$ in a suitable sense, although depending on how you take "suitable sense" means you may instead recover the complexification $$G_{\mathbb{C}}$$. From here on I will blithely ignore the difference between $$G$$ and its complexification. (Really I should say something here about averaging over a compact group and $$\operatorname U(n)$$ being the maximal compact subgroup of $$\operatorname{GL}_n(\mathbb{C})$$.)

Taken together, these two results tell us that $$G$$ or maybe its complexification is determined as a subgroup of $$\operatorname{GL}(V)$$ by its $$G$$-invariant tensors $$\operatorname{Hom}(V^{\otimes m}, V^{\otimes n})^G$$. What this means is that we ought to be able to define various $$G$$ of interest by saying "the $$G$$ preserving such-and-such tensors," and we do.

Moreover, if we decompose a given space of tensors $$\operatorname{Hom}(V^{\otimes m}, V^{\otimes n})$$ into its irreducible components under the action of $$GL(V)$$, then $$G$$ preserves some tensor iff it preserves the projection of the tensor to each irreducible component, so we can restrict our attention to collections of "irreducible tensors."

The tensors of rank $$1$$ are not so interesting; the stabilizer of a nonzero vector $$v \in V$$ is a general affine group, so we don't get anything new. Next are the tensors of rank $$2$$. The tensors in $$\operatorname{Hom}(V, V)$$ are again not so interesting; generically their stabilizers look like products of $$\operatorname{GL}(V_i)$$ where $$V_i$$ are the eigenspaces of a diagonalizable $$T \in \operatorname{Hom}(V, V)$$, so we again don't get anything new. So the next candidate is bilinear forms, and since $$V^{\otimes 2} \cong \operatorname S^2(V) \oplus \bigwedge^2(V)$$ is the irreducible decomposition here, we are naturally led to considering the stabilizers of symmetric resp. skew-symmetric forms, hence to the orthogonal and symplectic groups.

It's not just bilinear stuff out there though; to get the special linear groups we have to go all the way to a tensor in $$V^{\otimes \dim V}$$, namely any choice of a nonzero element of $$\bigwedge^{\dim V}(V)$$, and we can get, for example, $$G_2$$ using trilinear forms. But bilinear stuff is the simplest stuff after linear stuff.

• Part of the mystery here concerns the relationship between compact groups and algebraic groups, for which see BCnrd's answer here: mathoverflow.net/a/16269/290 Sep 12 '20 at 22:20
• It should be possible to avoid the use of Tannaka's theorem above, which is what I tried to do in the first draft. What I want to show is that every Zariski closed subgroup $G \subset GL(V)$ is precisely the subgroup preserving the $G$-invariant tensors. I bet this has a simple proof but I'm not seeing it. Sep 12 '20 at 22:34
• I did some TeX tidying. In one place, you had $\operatorname{Hom}(V^{\otimes m}, V^{\otimes n})^*)^G$, which I assumed was supposed to be the fixed points in $\operatorname{Hom}(V^{\otimes m}, V^{\otimes n})$ as elsewhere, so I deleted the spare $*$. Sep 13 '20 at 3:46
• Whoops, thanks, thought I fixed that already. Sep 13 '20 at 3:51
• How hard is it to see that you actually need just finitely many $G$-invariant tensors? Sep 13 '20 at 9:12

According to a theorem of Serre, all semi-simple Lie groups are linear algebraic groups. See https://en.wikipedia.org/wiki/Complex_Lie_group for the precise statement and a reference.

That shows why we should look at "algebraic functions" when looking for complex semi-simple Lie groups. Natural place to start is subgroups of $$GL(n, \mathbb{C})$$ that preserve some linear forms. But then we get something isomorphic to $$GL(n, \mathbb{C})$$ as such subgroup has to preserve the kernel. Bilinear forms are the next best things and it turns out they provide plenty of examples. But we don't have to stop there! The complex Lie group $$F_4$$ can be defined as the subgroup of $$GL(26, \mathbb{C})$$ fixing a symmetric trilinear form. And the complex simple Lie group $$G_2$$ can be defined as the stabilizer of a generci $$3$$-form on $$\mathbb{C}^7.$$ There are similar descriptions for $$E$$-series. See e.g. this answer by Robert Bryant https://mathoverflow.net/a/99795/6818 (I think he wrote about this more explicitly somewhere else on MO, but I have trouble finding it. Anyway... this description of $$E$$-series goes back to Elie Cartan.)