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?


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. $k^{\times} \times k^{\times}$.

  • $\begingroup$ 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 Nov 8 '19 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 Nov 9 '19 at 16:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.