I am currently reading "[Schiffer variations and the generic Torelli theorem for hypersurfaces](https://arxiv.org/abs/2004.09310)" 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. 



Subgroup of $\mathrm{GL}_n$ stabilizing linear subspace skew-symmetric matrices
Here is the worst possible proof of all but one case, namely $m=r=3$ (but this is not included in your question as stated, though I think it is claimed in the paper.)


Let $r$ denote the dimension of a generic subspace $W$ of $\bigwedge^2\mathbb{C}^{2m}$. We assume $3\leq r\leq\frac{1}{2}\operatorname{dim}(\bigwedge^2\mathbb{C}^{2m})$ (Voisin assumes a possibly weaker upper bound, but this is OK by duality.) Our goal is to show that no nontrivial element of $\operatorname{PGL}\_{2m}$ stabilizies $W$.


The first observation we make is that if $g$ stabilizes $W$, then the semisimple and unipotent parts $g\_{ss},g\_{un}$ also stabilize $W$ (because they are polynomials in $g$). If $g\_{ss}$ is nontrivial, we can take some polynomial in it to get a semisimple element with exactly two eigenvalues. If $g\_{un}$ is nontrivial, we can take some shifted power $(g\_{un}-1)^k+1$ which is nontrivial and only has Jordan blocks of length $1$ and $2$. In conclusion: it suffices to show that $W$ is not stabilized by any semisimple element with two eigenvalues nor any unipotent element with Jordan blocks of length $\leq 2$. We treat these two cases separately.


In both cases, we will do an incidence correspondence argument. Namely, we look at the variety of pairs $(g,W)$, with $g$ nontrivial and semisimple w/ two eigenvalues (or unipotent w/ $(g-1)^2=0$) and show this variety has dimension less than the dimension of $\operatorname{Gr}(r,m(2m-1)).$ We do this by splitting this variety into many many (but finitely many) strata, and showing the dimension bound for each strata.


The semisimple case. Our semisimple element $g$ with two eigenvalues gives a partition $d+e=2m$, given by the dimensions of each eigenspace. The action of $g$ on $\bigwedge^2\mathbb{C}^{2m}$ decomposes it as a sum of eigenspaces $\mathbb{C}^{\binom{d}{2}}\oplus\mathbb{C}^{de}\oplus\mathbb{C}^{\binom{e}{2}}.$


A subspace $W$ is stabilized by $g$ iff it is a direct sum of subspaces in each component. Letting the dimension of these subspaces be given by $f\_1,f\_2,f\_3$, we see that we have $f\_1(\binom{d}{2}-f\_1)+f\_2(de-f\_2)+f\_3(\binom{e}{2}-f\_3)$ degrees of freedom to choose $W$.


Consider the strata given by fixing d,e, and the $f\_i$. We have a $2de+1$-dimensional space of choices of $g$; $de$ for a subspace of dimension $d$, another $de$ for a subspace of dimension $e$, and a final degree of freedom for the ratio between the two eigenvalues. So our desired inequality becomes 


$$2de+1+f\_1(\binom{d}{2}-f\_1)+f\_2(de-f\_2)+f\_3(\binom{e}{2}-f\_3)<r(m(2m-1)-r).$$


Assume this inequality is false. Consider the choice of $f\_i$ that maximizes the left hand side. As $f\_1+f\_2+f\_3=r\leq\frac{1}{2}m(2m-1),$ we must have $f\_1\leq\frac{1}{2}\binom{d}{2}$, $f\_2\leq\frac{1}{2}de$, $f\_3\leq\frac{1}{2}\binom{e}{2}$ (by the maximizing assumption). Now rewrite (the negation of) the inequality as:


$$2de+1\geq (f\_2+f\_3)(\binom{d}{2}-f\_1)+(f\_1+f\_3)(de-f\_2)+(f\_1+f\_2)(\binom{e}{2}-f\_3).$$


Assume WLOG that $d\geq e$. As $f\_2\leq\frac{1}{2}de$, we must have $f\_1+f\_3\leq 4.$ The right hand side contains $f\_2(\binom{d}{2}+\binom{e}{2}-f\_1-f\_3)$ as a summand, and because $m\geq 3$, we have $\binom{d}{2}+\binom{e}{2}\geq\frac{2}{3}de,$ so $\binom{d}{2}+\binom{e}{2}-f\_1-f\_3\geq\frac{1}{3}de,$ so $f\_2\leq 6$.


We see in particular that $2de+1\geq f\_2(\binom{d}{2}+\binom{e}{2}-4)+(f\_1+f\_3)(de-6),$ and since $(f\_2+f\_3)+(f\_1+f\_3)\geq r\geq 3$, either $\binom{d}{2}+\binom{e}{2}-4$ or $de-6$ is $\leq\frac{2de+1}{3}$. If $de-6\leq\frac{2de+1}{3}$, then $de\leq 19$ and $m \leq 10$. while if $\frac{2de+1}{3}\geq\binom{d}{2}+\binom{e}{2}-4,$ then we have $\frac{2m^2+1}{3}\geq\frac{2de+1}{3}\geq\binom{d}{2}+\binom{e}{2}-4=\frac{d^2+e^2}{2}-m-4\geq m^2-m-4,$ so $m\leq 5.$


In summary, we have $m\leq 10.$ Using a computer to check all such cases (as I said, this is the worst possible proof), we see that the only possible case is $f\_1=0,f\_2=3,f\_3=0,d=e=m=3$, but this is exactly the case I excluded at the start.


Now let do the same argument for unipotent $g$ with $(g-1)^2=0$. Let $n$ denote the number of nontrivial Jordan blocks, or equivalently, the dimension of $\operatorname{im}(g-1).$ Then the action of $g$ on $\bigwedge^2\mathbb{C}^{2m}$ has $n(2m-n-1)$ nontrivial blocks (each of length $2$). 


The data of a preserved subspace $W\subseteq\bigwedge^2\mathbb{C}^{2m}$ is equivalent to the data of a subspace $W'\cong W\cap\operatorname{im}(g-1)\subseteq\operatorname{im}(g-1)$ and a subspace $W/W'\subseteq (g-1)^{-1}W'/W'.$ If the dimension of $W'$ is given by $d$, then there are $d(n(2m-n-1)-d)+(r-d)(m(2m-1)+d-n(2m-n-1)-r)$ degrees of freedom. Meanwhile, the data of a choice of $g$ is equivalent to the choice of the subspace $\operatorname{im}(g-1)$ and of a map $\mathbb{C}^{2m}/\operatorname{im}(g-1)\rightarrow \operatorname{im}(g-1)$, which gives $n(2m-n)+n(2m-n)=2n(2m-n)$ degrees of freedom. So the inequality we want is


$$d(n(2m-n-1)-d)+(r-d)(m(2m-1)+d-n(2m-n-1)-r)+2n(2m-n)< r(m(2m-1)-r).$$


We can rearrange this to


$$2d^2+(m(2m-1)-2r-2n(2m-n-1))d-2n(2m-n)+rn(2m-n-1) > 0.$$


Assume otherwise. Note that $\frac{2m-n-1}{2m-n}\geq\frac{2}{3}$, so $-2n(2m-n)+rn(2m-n-1)\geq 0,$ with equality only when $m=n=r=3$ (and so we land in the excluded case again.) So we must have $m(2m-1)-2r-2n(2m-n-1)<0$ and, by the discriminant test,


$$8(rm(2m-n-1)-2n(2m-n))\leq(m(2m-1)-2r-2n(2m-n-1))^2.$$


Note that we have $m(2m-1)-2r-2n(2m-n-1)\geq m-2r$ and $m(2m-1)-2r-2n(2m-n-1)\geq -2n(2m-n-1)$. Applying these inequalities, we see that


$$8(rm(2m-n-1)-2n(2m-n))\leq(2r-m)2n(2m-n-1)$$.


Again applying $2(2m-n)\leq 3(2m-n-1),$ we get


$$8(rm-3n)(2m-n-1)\leq(2r-m)2n(2m-n-1)$$


or


$$8(rm-3n)\leq(2r-m)2n.$$


This implies


$$8(rn-3n)\leq(2r-m)2n$$


and


$$8r-24\leq 4r-2m$$


$$4r\leq 24-2m$$


so $m\leq 6$. Again, we check all these cases by computer. All of them either have $m=r=3$ or have equality in the dimension bound. But in this case, equality suffices, because all subgroups of $G$ containing a unipotent element have positive dimension.


