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?



Bilinear forms in compact/semisimple Lie group theory
(**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](https://mathoverflow.net/questions/58633/does-every-irreducible-representation-of-a-compact-group-occur-in-tensor-product)). 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](https://ncatlab.org/nlab/show/Karoubi+envelope) of this subcategory.


Furthermore, the Tannaka half of [Tannaka–Krein duality](https://en.wikipedia.org/wiki/Tannaka%E2%80%93Krein_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](https://en.wikipedia.org/wiki/G2_(mathematics)#History). But bilinear stuff is the simplest stuff after linear stuff.


