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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.

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    $\begingroup$ This doesn't directly answer your question, but for the analogous thing with skew-symmetric forms instead of symmetric, these "intermediate symplectic groups" were studied by Prcotor in his paper "Odd symplectic groups": doi.org/10.1007/BF01404455. They have a highest weight character theory similar to that of semisimple Lie groups. $\endgroup$ Dec 13, 2020 at 17:36
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    $\begingroup$ $\DeclareMathOperator\rad{rad}$Let $b$ be the quadratic form. Such a matrix must preserve the—I think it's called—radical $\rad_b(V)$ (the subspace of vectors $v$ for which $b(v, V) = 0$), and must induce an orthogonal transformation of $V/\rad_b(V)$. Thus, at least if we replace $\operatorname{SL}$ by $\operatorname{GL}$, we get a sequence $1 \to \operatorname{Hom}(V, \rad_b(V)) \to \operatorname O_b(V) \to \operatorname O_b(V/\rad_b(V)) \times \operatorname{GL}(\rad_b(V) \to 1$. (@YCor's description below says this is split.) $\endgroup$
    – LSpice
    Dec 13, 2020 at 17:36
  • $\begingroup$ Wouldn't you denote your example $SO(k,0,n-k)$? The fact that you need three parameters, corresponding to positive, negative and zero, is to do with Sylvester's law of inertia $\endgroup$
    – wlad
    Dec 13, 2020 at 17:37
  • $\begingroup$ The left-hand term in my exact sequence should be $\DeclareMathOperator\rad{rad}\DeclareMathOperator\Hom{Hom}$$\Hom(V/\rad_b(V), \rad_b(V))$, not just $\Hom(V, \rad_b(V))$. $\endgroup$
    – LSpice
    Dec 13, 2020 at 19:21
  • $\begingroup$ Thank you for the answer. Just a question: here are you considering $Hom(V/rad_b(V), rad_b(V))$ as a group with respect to the sum? In this case shouldn't the first $1$ in the above exact sequence be a $0$? $\endgroup$
    – user82886
    Dec 14, 2020 at 16:21

1 Answer 1

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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.

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  • $\begingroup$ As mentioned, the description for skew-symmetric forms appears in Proctor's "Odd symplectic group" paper: doi.org/10.1007/BF01404455. He also writes it as a semidirect product $(Sp_{2n}\times GL_m)\ltimes \mathbb{C}_{+}^{2nm}$, and notes that it is neither semisimple nor reductive. It would be interesting if the combinatorics of the highest weight character theory worked similarly in this "intermediate orthogonal group" case as well. $\endgroup$ Dec 13, 2020 at 17:51
  • $\begingroup$ @SamHopkins, no need for "neither semisimple nor reductive"—semisimple implies reductive. In fact @‍YCor's presentation gives a clear Levi decomposition, with reductive component the block-diagonal matrices and unipotent radical the block lower-unitriangular matrices. $\endgroup$
    – LSpice
    Dec 13, 2020 at 17:53
  • $\begingroup$ This paper is from 1988, but I'd bet it was known to Elie Cartan... it's quite straightforward. A basis-free description is: the stabilizer of a bilinear form on a space $V$ with kernel $W$ (intersection of left and right kernel) is the set of operators preserving $W$ and inducing on $V/W$ an operator preserving the induced bilinear form on $V/W$. $\endgroup$
    – YCor
    Dec 13, 2020 at 17:58
  • $\begingroup$ @YCor, for what it's worth, this is the description in my comment. $\endgroup$
    – LSpice
    Dec 13, 2020 at 18:15
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    $\begingroup$ @LSpice sure, I claim no ownership :) I was actually the currently single upvoter of your comment. $\endgroup$
    – YCor
    Dec 13, 2020 at 18:21

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