Subgroup of O(Q) Let $(V,Q)$ be a quadratic space over some field $K$, where $V$ is a $K$-vector space of dimension eight. I would like to understand the structure of the subgroup of the group $O(Q)$ of orthogonal transformations which fix a given three-dimensional subspace $W$ of $V$. Can anyone give a concrete description of this subgroup (there might be different possibilities depending on $Q$ and/or $W$)? I find it hard to do this kind of concrete calculations, any hints would be welcome...   
(I am already very happy with a description of what happens when $W$ is totally isotropic for $Q$.)
 A: All right, here's a solution that ought to work for any algebraically closed field of characteristic $0$ in the case where $Q$ is nondegenerate.  This is all based on a bunch of matrix calculations, so I don't have proofs for some of it, but it ought to be correct unless I've made a calculation mistake.  The obvious way to check it would be to prove these statements rigorously.
Let $W$ be an arbitrary subspace of $V$, and let $G \leq O(V)$ be the group of orthogonal transformations that leave $W$ invariant.  Define
$$
R \;=\; W \cap W^\perp \qquad\text{and}\qquad S \;=\; W + W^\perp.
$$
Note that $W^\perp$ is invariant under the action of $G$, and therefore $R$ and $S$ are also invariant.
The structure of $G$ is as follows.  First, there is a short exact sequence
$$
1 \;\;\to\;\; A \;\;\to\;\; G \;\;\to\;\; O(S) \;\;\to\;\; 1
$$
where $A$ is an abelian group isomorphic to $K^d$ for some value of $d$.  The homomorphism $G \to O(S)$ is surjective because of Witt's theorem.
The group $O(S)$ is a semidirect product.  Specifically,
$$
O(S) \;\cong\; \bigl( \text{Lin}(R,W/R) \times \text{Lin}(R,W^\perp/R) \bigr) \;\rtimes\; \bigl(O(R) \times O(W/R) \times O(W^\perp/R) \bigr).
$$
Here $\text{Lin}(R,W/R)$ is the additive abelian group of all linear functions $R\to W/R$, and $\text{Lin}(R,W^\perp/R)$ is similarly an additive abelian group.  Since $Q$ restricts to the null quadratic form on $R$, the orthogonal group $O(R)$ is the same as $GL(R)$.  Moreover, since $Q$ is null on $R$, the quotients $W/R$ and $W^\perp/R$ are quadratic spaces, and $O(W/R)$ and $O(W^\perp/R)$ are the corresponding orthogonal groups.  Note also that the quadratic forms on $W/R$ and $W^\perp/R$ are nondegenerate.
As for the question of connectivity: since the kernels $A$ and $\text{Lin}(R,W/R)\times\text{Lin}(R,W^\perp/R)$ are both connected, the group $G$ will be connected if and only if the groups $O(R)$, $O(W/R)$, and $O(W^\perp/R)$ are connected.  Note that $O(R) \cong GL(R)$, while $O(W/R)$ and $O(W^\perp/R)$ are orthogonal groups for nondegenerate quadratic spaces.  (Perhaps it is obvious to you whether these are connected -- I do not know much about the connectedness of orthogonal groups over arbitrary fields.)
Edit:  Here is a bit more information on the kernel $A$ of the epimorphism $G \to O(S)$.  Since $Q|_{R\times S} = 0$, the quadratic form $Q$ defines a bilinear map $B \colon R \times (V/S) \to K$,
and it is not hard to show that $B$ is a perfect pairing.  It follows that the action of an element $g\in G$ on  $V/S$ is entirely determined by the action of $g$ on $R$.  In particular, every element of $A$ acts trivially on $V/S$.  Therefore, every element $g\in A$ has the form
$$
g(v) \;=\; v + \varphi(\pi(v))
$$
where $\pi\colon V \to V/S$ is the quotient map, and $\varphi\colon V/S \to S$ is a linear map.  Thus $A$ is isomorphic to some subgroup of the abelian group $\text{Lin}(V/S,S)$.
To be specific, $A$ is isomorphic to the group of all linear maps $\varphi\colon V/S \to S$ that
satisfy the following conditions:


*

*The range of $\varphi$ lies in $R$.

*The map $\varphi$ is "antisymmetric" with respect to $B$ in the sense that
$$
B(\varphi(u),v) + B(\varphi(v),u) = 0
$$
for all $u,v \in V/S$.
In particular, $A$ is isomorphic to the additive group of all $m\times m$ antisymmetric matrices over $K$, where $m = \dim(R)$.
A: Okay, the most basic case is when $W^\perp = V$.  In this case, $Q$ induces a quadratic form $Q/W$ on $V/W$, and the group has the following structure:
$$
\text{Lin}(W,V/W) \rtimes \bigl(GL(W) \times O(Q/W)\bigr).
$$
Here $\text{Lin}(W,V/W)$ is the additive group of all linear functions $W\to V/W$, and $GL(W)\times O(Q/W)$ acts on $\text{Lin}(W,V/W)$ in the obvious way.
Is this the kind of information you're looking for?  I'm not sure exactly what you mean by "concrete description".  Would you prefer something in terms of matrices?
