maximal subgroup of the general linear group

Question

Maybe I'm being silly.

As algebraic groups over an algebraically closed field $K$ of characteristic not $2$, is the conformal orthogonal group $CO_{2n}$ a maximal subgroup of $GL_{2n}$(or $CO_{2n+1}$ a maximal subgroup of $GL_{2n+1}$)?

Answer 1

Let $V$ be a finite-dimensional $K$-vector space, and let $Q \in \text{Sym}^2(V^*)$ be a nondegenerate symmetric bilinear form on $V$. Then we set $$ \mathrm{O}(V) = \{g \in \mathrm{GL}(V): Q(g(v),g(w)) = Q(v,w), \forall v, w \in V\} $$ and $\mathrm{GO}(V)= Z(\mathrm{GL}(V)).\mathrm{O}(V)$ (the subgroup of $\mathrm{GL}(V)$ which preserves $Q$ up to a scalar multiple).

If $\mathrm{GL}(V)$ contains a subgroup $H$ with

$$ \mathrm{GO}(V) = N_{\mathrm{GL}(V)}(\mathrm{O}(V))<H\leq \mathrm{GL}(V), $$ then at the level of Lie algebras, $\mathrm{Lie}(H)$ would be an $\mathrm{O}(V)$-subrepresentation of $\mathfrak{gl}(V)$ which strictly contained $\mathrm{Lie}(\mathrm{O}(V))\oplus \mathsf k.I$, and therefore (identifying $\mathfrak{gl}(V)$ with $V\otimes V$ as a $\mathrm{O}(V)$-representation, it would have to be all of $V\otimes V$, since I think $V\otimes V$ decomposes as a direct sum of $3$ irreducible representations $\bigwedge^2(V)\oplus \mathsf k.Q \oplus H$ of $\mathrm{O}(V)$.

I'm not 100% sure that works outside characteristic zero, but I think the weights are small enough that it has a good chance!