# Determining the isomorphism classes of these symmetry groups (exterior algebra)

Let $V$ be a real $d$-dimensional vector space. Let $\omega \in \bigwedge^kV$ be a fixed non-zero multivector for some $1 < k < d$.

Define $G_{\omega}=\{ T \in \text{Aut}(V) \, | \, (\bigwedge^k T)(\omega)=\omega\}$ to be the subgroup of linear automorphisms of $V$ which preserve $\omega$.

(Here $\bigwedge^k V$ is the $k$-th exterior power of $V$, and $\bigwedge^k T \in \text{End}(\bigwedge^k V)$ is the exterior power of $T$).

Question: What are the different isomorphism classes of the $\{G_{\omega}\}_{\omega \in \bigwedge^k V}$?

If that is too hard, what are the possible dimensions of these groups?

Commnets:

$G_{\omega}$ is the isotropy group of the action of $\text{GL}(V)$ on $\bigwedge^K V$, defined by $(T,\omega) \to (\bigwedge^k T) \omega$. Thus, for every two elements in the same orbit $\omega,\omega'$, we have $G_{\omega} \simeq G_{\omega'}$.

(Thus it might be helpful to classify the orbits of this action).

In particular, this is true for every two non-zero decomposable elements.

In the case where $\omega$ is decomposable, it is easy to see that $$\dim(G_{\omega})=d^2-k(d-k)-1. \tag{1}$$

Indeed, suppose that $\omega=v_1 \wedge \dots \wedge v_k$, we can complete $v_1,\dots,v_k$ into a basis $v_1,\dots,v_d$ of $V$. Then we have $$Tv_1 \wedge \dots \wedge Tv_k=v_1 \wedge \dots \wedge v_k \neq 0,$$

so $\text{span}(Tv_1,\dots,Tv_k)=\text{span}(v_1,\dots,v_k)$. Denote $W=\text{span}(v_1,\dots,v_k)$. Then $T \in G_{\omega}$ implies $T|_{W} \in \text{SL}(W) \simeq \text{SL}(k)$. On a direct summand of $W$, $T$ can do whatever it wants, as long as it stays invertbile. Thus $$\dim(G_{\omega})=\dim \text{SL}(k)+ (d-k)\cdot d=k^2-1+(d-k)\cdot d.$$

Let us consider for a moment the (simplest?) non-decomposable case:

$\omega = e_1\wedge e_2 + e_3\wedge e_4\in \Lambda^2(\Bbb R^4)$. In that case $G_{\omega}$ clearly contains a copy of $\text{SL}(2) \times \text{SL}(2)$, corresponding to actions on the two planes $\text{span}\{ e_1,e_2\},\text{span}\{ e_3,e_4\}$. Thus, $$\dim(G_{\omega}) \ge 2\dim \text{SL}(2)=6.$$

Is $\dim(G_{\omega})=6$? (As commented by Ted Shifrin, we can also switch between the two planes, but I don't think this particular option increases the dimension of the group).

Note that for $d=4,k=2$, the formula $(1)$ gives $\dim(G_{\omega})=11$.

• This doesn't seem right. What if $\omega = e_1\wedge e_2 + e_3\wedge e_4\in \Lambda^2(\Bbb R^4)$? Then any rotation leaving the $e_1e_2$- and $e_3e_4$-planes fixed (as planes) works. So that's $SO(2)\times SO(2)$. And then you can switch the two planes, as well. – Ted Shifrin Jul 7 '18 at 1:01
• The group won't be "trivial" in general. I haven't checked the details, but I'm pretty such that if you take $V = W^{*}$ and $\omega \in \Lambda^2(W)$ to be a non-degenerate two-form and then identify $\operatorname{Aut}(V)$ with $\operatorname{Aut}(W)$ using the dual map, you just get the symplectic group (I guess this what happens in Ted Shifrin's example). In general, the dimension of the group will depend on $\omega$. By differentiating $\Lambda^k(T + tS)(\omega) = \omega$ with respect to $t$, you can write down an "explicit" description for the Lie algebra. – levap Jul 7 '18 at 13:05
• @levap Thanks, you are right of course. Are you sure that the dimension of the group will depend on $\omega$? For any two non-zero decomposable $k$-multivectors, the dimension would be the same I think: The two groups will be conjugate, via a change of basis. That is, if $\omega_1=v_1 \wedge \dots v_k,\omega_2=w_1 \wedge \dots w_k$, you can complete the $v_i,w_i$ into bases for $V$. Then one group is obtained from the other by conjugation with the map $v_i \to w_i$. – Asaf Shachar Jul 7 '18 at 21:38
• @AsafShachar: Your group is the isotropy group of $\omega$ under the action of $\operatorname{GL}(V)$ on $\Lambda^k(V)$. If $\omega, \omega'$ lie in the same orbit then the groups will be conjugate which is what happens if $\omega,\omega'$ are decomposable but there's no reason this will happen in general. For two-forms, I expect the dimension to depend on the rank of the associated bilinear-form (with the case of full rank corresponding to the symplectic group). – levap Jul 8 '18 at 4:48