TITLE: What is the hidden symmetry behind four generic planes in $\mathbb{R}^4$?
QUESTION [10 upvotes]: 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?

REPLY [12 votes]: 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. $k^{\times} \times k^{\times}$.