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Consider the action of $\operatorname{GL}(\mathbb{R}^4)$ on the Grassmannian of 2-dimensional subspaces of $\mathbb{R}^4$. In experiments, I observe that four randomly drawn points in this space are simultaneously stabilized by nontrivial members of $\operatorname{GL}(\mathbb{R}^4)$. This violates my naive attempts at counting dimensions:

The stabilizer subgroup of any given plane is a generic subset of a 12-dimensional subspace of $\mathbb{R}^{4\times 4}$. If I draw three planes at random, the intersection of the corresponding subspaces is 4-dimensional, as expected. If I draw a fourth plane at random, I expect the resulting intersection to be the span of the identity matrix, but alas, I consistently obtain a 2-dimensional intersection. Only after I draw a fifth plane does the resulting intersection equal the span of the identity matrix.

What's going on here? More generally, when should I expect subsets of Grassmannian spaces to have trivial stabilizer?

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Work over any field $k$. Taking two generic planes in 4-dimensions, we can get them to our favourites by linear transformation $k^4=k^2\oplus k^2$, reducing $GL_4$ to $GL_2 \times GL_2$. A third plane, generically, is a graph of a unique linear map from one to the other $y=Ax$. The group action is by matrix similarity. We normalize to get $A=I$, reducing to $GL_2$. A 4th plane, generically, is $y=Ax$ but with $A$ having distinct nonzero eigenvalues. The action of $GL_2$ on $A$ is by conjugation. The group preserving all 4 planes is now reduced to the subgroup preserving a splitting of the plane into eigenspaces of $A$, i.e. $k^{\times} \times k^{\times}$.

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  • $\begingroup$ To be more precise, over $\mathbb{Z}/2\mathbb{Z}$ you can't get distinct nonzero eigenvalues, so for any field but that one the argument works. $\endgroup$
    – Ben McKay
    Commented Nov 8, 2019 at 10:58
  • $\begingroup$ Ah, put another way, four generic planes can be linearly transformed to take the form $\operatorname{span}\{e_1,e_2\}$, $\operatorname{span}\{e_3,e_4\}$, $\operatorname{span}\{e_1+e_3,e_2+e_4\}$, and $\operatorname{span}\{e_1+ae_3,e_2+be_4\}$ for some distinct nonzero $a$ and $b$. These planes are preserved by transforms of the form $\operatorname{diag}(x,y,x,y)$ for nonzero $x$ and $y$. $\endgroup$
    – Dustin G. Mixon
    Commented Nov 9, 2019 at 16:46

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