Subgroup of $\mathrm{GL}_n$ stabilizing linear subspace skew-symmetric matrices I am currently reading "Schiffer variations and the generic Torelli theorem for hypersurfaces" by Voisin, where it is claimed that the subgroup of $\mathrm{SL}_{2m}$ ($m \geq 3$) which preserves a generic subspace of $\bigwedge^2 \mathbb{C}^{2m}$ of dimension bigger than $3$ must be finite. There are no references given in the paper, where it is claimed that this fact is easily checked. Unfortunately (for me), I am not able to prove it.
Here "preserves" means that $g\cdot w\cdot{\vphantom g^tg} \in W$, for any $w \in W$. Is this fact indeed well-known? I am looking for a reference of this fact, or a quick proof. 
 A: Here is an outline of the argument that shows that the $\mathrm{SL}_6(\mathbb{C})$-stabilizer of the generic $3$-plane $W\subset\Lambda^2(\mathbb{C}^6)$ has dimension $1$, not $0$, as (apparently) claimed.
First, note that the cone $C_2\subset \Lambda^2(\mathbb{C}^6)$ consisting of the elements $b\in \Lambda^2(\mathbb{C}^6)$ that satisfy $b^3 = 0$ is a hypersurface of degree $3$ in $\Lambda^2(\mathbb{C}^6)\simeq\mathbb{C}^{15}$.  It is irreducible but not smooth, as it is singular along the locus $C_1\subset C_2$ consisting of the elements that satisfy $b^2=0$, and $C_1$ itself is a smooth cone of dimension $9$ (i.e., its only singularity is the origin $b=0$).
Thus, a generic $3$-plane $W$ in $\Lambda^2(\mathbb{C}^6)$ will only meet $C_1$ at the origin and will not be tangent to $C_2$ anywhere.  Thus, the intersection $W\cap C_2$ will be a smooth $2$-dimensional cone that projectivizes to a smooth cubic curve in $\mathbb{P}W\simeq \mathbb{P}^2$. 
The group $G\subset \mathrm{SL}_6(\mathbb{C})$ that stabilizes $W$ must act on $\mathbb{P}W$ as symmetries of the nonsingular cubic curve and hence must act as a finite group on $\mathbb{P}W$. By passing to a subgroup $G'\subset G$ of finite index in $W$, we can assume that $G'$ acts trivially on $\mathbb{P}W$ and hence as a scalar multiple of the identity on $W$.  However, if we let $b\in W$ be an element that satisfies $b^3\not=0$, then $G'$ must preserve $b^3$ and hence, it can at most multiply $b$ by a nontrivial cube root of unity.  Again passing to a subgroup $G''\subset G'$ of index at most $3$, we can arrange that $G''$ acts trivially on $W$.
Let $P\subset W$ be a plane that intersects $W\cap C_2$ in three distinct lines.  This means that $P$ has a basis consisting of two elements $b_1\not\in C_2$ and $b_2$ such that the polynomial $p(\lambda)$ that satisfies
$$
(b_2-\lambda b_1)^3 = p(\lambda) b_1^3
$$ 
has $3$ distinct roots.  After a change of basis in $P$, I can assume that those roots are $1$, $2$, and $3$.  
Then, by an elementary argument, there is a basis $e_1,\ldots,e_6$ of $\mathbb{C}^6$ such that
$$
b_1 = e_1\wedge e_2 + e_3\wedge e_4 + e_5\wedge e_6
\quad\text{and}\quad
b_2 = e_1\wedge e_2 + 2e_3\wedge e_4 + 3e_5\wedge e_6
$$
Moreover, the fact that $G''$ fixes $b_1$ and $b_2$ implies that it also fixes the individual monomial terms $e_1\wedge e_2$, $e_3\wedge e_4$, and $e_5\wedge e_6$. 
Let $Q_i\subset\mathbb{C}^6$ be the $2$-plane spanned by $e_{2i-1},e_{2i}$.  Then $G''\subset \mathrm{SL}(Q_1) \times\mathrm{SL}(Q_2)\times \mathrm{SL}(Q_3)$, and so, of course, $Q_i\simeq Q_i^*$ as $G$-modules.  Now let $b_3\in W$ be linearly independent from $b_1$ and $b_2$.  Using the $G''$-module decomposition
$$
\begin{aligned}
\Lambda^2(\mathbb{C}^6) &= \Lambda^2(Q_1\oplus Q_2\oplus Q_3) \\
&= \Lambda^2(Q_1)\oplus\Lambda^2(Q_2)\oplus\Lambda^2(Q_3)\oplus Q_1{\otimes}Q_2\oplus Q_2{\otimes}Q_3\oplus Q_3{\otimes}Q_1\\
&\simeq\mathbb{C}\oplus\mathbb{C}\oplus\mathbb{C}\oplus Q_1{\otimes}Q_2^*\oplus Q_2{\otimes}Q_3^*\oplus Q_3{\otimes}Q_1^*,
\end{aligned}
$$
one can decompose $b_3$ into a sum of the three basic monomials that occur in $b_1$ and $b_2$ plus a triple of linear maps $L_i:Q_{i+1}\to Q_i$.  For generic $W$ with basis $b_1$, $b_2$, and $b_3$ as above, these $L_i$ will be isomorphisms, and they will have to be $G''$-module isomorphisms in order for $b_3$ to be fixed by $G''$, whose elements can be thought of as triples of elements $(g_1,g_2,g_3)$ where $g_i\in\mathrm{SL}(Q_i)$.  However, the relations $L_ig_{i+1}=g_iL_i$ (necessary for $G''$ to fix $b_3$) will then determine $g_2$ and $g_3$ in terms of $g_1$ and the $L_i$.  Moreover, $L = L_1L_2L_3:Q_1\to Q_1$ must commute with $g_1$. Conversely, if $g_1$ commutes with $L$ then it determines an element of $G''$.  However, $L$ will always have a positive dimensional commutator in $\mathrm{SL}(Q_1)$ (generically of dimension $1$), so $G''$ always has dimension at least $1$.
A: In the following, a linear group is a (closed) subgroup $G$ of some $GL(V)$. Now the stabilizer of a generic $d$-space in $V$ is the same as the stabilizer in $G\times GL(d)$ of a generic vector of $V\otimes\mathbb C^d$. Thus, in our case we have to investigate the generic stabilizer of $GL(n)\times GL(d)$ acting on $\wedge^2\mathbb C^n\otimes\mathbb C^ d$ with $1\le d\le n(n-1)/4$ and determine when it is infinite.
Determining the generic isotropy group $H$ for a linear action of a reductive group $G$ is a classical problem of invariant theory. It was pretty much settled by the Vinberg school approx. 50 years ago. More specifically: In
Andreev, E. M.; Vinberg, È. B.; Èlašvili, A. G.: Orbits of highest dimension of semisimple linear Lie groups. (Russian) Funkcional. Anal. i Priložen. 1 1967 no. 4, 3–7
the authors derive a numerical criterion for $H$ to be infinite (the first theorem of that paper). The argument is really ingeneous and I recommend reading the paper. From their criterion they derive the classification of irreducible simple linear groups with infinite $H$. The table became later notorious because the very same table showed up in various different classification projects (see, e.g., the appendix to Mumford-Fogarty(-Kirwan)).
Probably, the numerical criterion alone is already enough to settle the finiteness of $H$ in the case at hand. But that is not necessary. In the follow-up paper
Èlašvili, A. G. Stationary subalgebras of points of general position for irreducible linear Lie groups. (Russian) Funkcional. Anal. i Priložen. 6 (1972), no. 2, 65–78
Èlašvili (almost, see below) settles the case of arbitrary irreducible linear groups. The answer for non-simple groups is given in Tables 5 and 6. The cases $\wedge^2\mathbb C^n\otimes\mathbb C^ d$ with infinite $H$ are $(n,d)=(n,1), (n,2), (4,3), (5,3)$, and $(6,3)$. So $d\ge4$ does not occur, as claimed.
There is more to say. Sifting through the tables one will notece that Robert Bryant's case $(n,d)=(6,3)$ is missing in Table 5. This case along with three more missing cases was pointed out in
Popov, A. M. Irreducible semisimple linear Lie groups with finite stationary subgroups of general position. (Russian) Funktsional. Anal. i Prilozhen. 12 (1978), no. 2, 91–92.
See the second paragraph. As for reliability of these results, the whole classification was recovered in
Knop, Friedrich; Littelmann, Peter Der Grad erzeugender Funktionen von Invariantenringen. (German) [The degree of generating functions of rings of invariants] Math. Z. 196 (1987), no. 2, 211–229
There we dealt with a slightly broader classification problem where we nevertheless had to do all calculations from scratch. So we were not using Èlašvili's tables directly. That's why I am pretty positive that the tables are complete.
