# What is the hidden symmetry behind four generic planes in $\mathbb{R}^4$?

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. $$k^{\times} \times k^{\times}$$.
• 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. – Ben McKay Nov 8 '19 at 10:58
• 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 $y$. – Dustin G. Mixon Nov 9 '19 at 16:46