Analogue of the special orthogonal group for singular quadratic forms The special orthogonal group $SO(n)$ is the subgroup of the special linear group $SL(n)$ of $n\times n$ matrices with determinant one that preserve a non-degenerate symmetric bilinear form. If such a bilinear form is taken to be the one associated to the identity matrix, then
$$SO(n):=\{M\in SL(n)\: | \: MM^{t} = I\}$$
Is there a nice description of the group preserving a singular symmetric bilinear form? For instance, if we take $I_{k}=(a_{i,j})$ to be the matrix with $a_{i,i} = 1$ for $1\leq i\leq k$, $a_{i,i} = 0$ for $k+1\leq i\leq n$, and $a_{i,j} = 0$ for $i\neq j$, then how can we describe the group
$$SO_{I_k}(n):=\{M\in SL(n)\: | \: MI_kM^{t} = I_k\}$$
of determinant one matrices preserving $I_k$?
Thank you very much.
 A: Yes, it's quite immediate in general, over an arbitrary field (say with $0\neq 2$). Let $m$ be the dimension of the kernel and fix a supplement subspace.
Then under this decomposition, the quadratic form $q$ writes as $\begin{pmatrix}q_0 & 0\\ 0 & 0\end{pmatrix}$, with $q_0$ non-degenerate. Then the orthogonal group is
$$\begin{pmatrix}\mathrm{O}(q_0) & 0\\ \mathrm{Mat}_{m,n-m} & \mathrm{GL}_m\end{pmatrix}.$$
In particular, $\mathrm{SO}(q)$ consists of those matrices of determinant $1$, i.e. the diagonal blocks have both determinant $1$ or both $-1$ (the latter being possible if both blocks are nonzero, i.e., $q\neq 0$ and $q$ is degenerate: in this case, $\mathrm{SO}(q)$ has 2 components as algebraic group, while for $q=0$ or $q$ non-degenerate, it has a single component).

There's a similar description for alternating forms, the orthogonal group $\mathrm{O}(q_0)$ being replaced with a symplectic group. The symplectic group already being of determinant $1$, the determinant 1 group of an alternating form is then connected in all cases.

Other consequences of the description:
It also follows that the unipotent radical ($\mathrm{Mat}_{n,m-n}$) of $\mathrm{SO}(q)$ is contained in its the derived subgroup; it's in the derived subgroup of the connected component $\mathrm{SO}(q)^\circ$ unless $(n-m,m)=(1,1)$. Also if $\min(n-m,m)\ge 2$, we see that $\mathrm{SO}(q)^\circ$ is perfect.
