Invariant planes of a nilpotent matrix with two Jordan blocks of size two

Describe all the invariant 2-dimensional subspaces of $\mathbb{C}^4$ (or $\mathbb{R}^4$) of the nilpotent map $$N = \begin{pmatrix} 0 & 1 & & \\ 0 & 0 & & \\ & & 0 & 1 \\ & & 0 & 0 \\ \end{pmatrix}$$

Equivalently, describe all fixed points of the unipotent map $u = \exp(N)$ acting on the grassmanian of planes $\mathrm{Gr}(2,4)$ of $\mathbb{C}^4$ (or $\mathbb{R}^4$).

This seems like an easy Linear Algebra problem, but I don´t see how to solve it by easy arguments. We have, for example, the invariant planes $[e_1,e_2]$, $[e_3,e_4]$ and $[e_1,e_3]$ spanned by the standard basis. But there are many more, in fact, a connected continuum of them!

This is because this is an example of a partial Springer fiber (perhaps the simplest nontrival example). Such a space can be defined in the more general context of reductive algebraic groups and is known to be connected (see, for example, Chapter 6 of the 1995 AMS book Conjugacy Classes in Semisimple Algebraic Groups, by Jim Humphreys). Altough there are many deep results about the topology of these spaces (they are paved by affines, their cohomology is related to representations of the Weyl group and so on) I have not been able to use these results to give a concrete description of this simple example.

I would like to list all invariant planes of $N$ and, if possible, see how they are glued together. Perhaps the theory and approach one uses to solve this example can be used to describe concretely other higher dimensional examples.

• Take any vector $v$ not belonging to the kernel of $N$ and consider the space $\langle v,Nv\rangle$. This family and the kernel of $N$ are all spaces you ask about. Aug 25, 2014 at 0:55
• Thanks @AntonKlyachko, but I would like a description of all of them! PS - The kernel of $N$ is $[e_1,e_3]$. Aug 25, 2014 at 0:57
• This is the description of all of them, because any invariant subspace containing a vector $v$ shoud contain $Nv$. Aug 25, 2014 at 1:00
• That´s great @AntonKlyachko, thanks a lot! I think that from this it follows that they are glued together to form a pinched torus (in the $\mathbb{R}$ case). Aug 25, 2014 at 1:21
• The 2-dimensional subspaces are parametrised by the points on a quadric $Q$ in $P^5$, via Plucker coordinates. You can write down how $N$ acts there, and take the fixed vectors of linear map which satisfy the quadratic relation defining $Q$. Thus you will have them parametrised by the intersection of $Q$ and the linear subspace of fixed points. Aug 25, 2014 at 8:14

I think it is easiest to just do it in Plucker coordinates. Write a $2$ plane in $4$ space as the row span of $\begin{pmatrix} w_1 & x_1 & y_1 & z_1 \\ w_2 & x_2 & y_2 & z_2 \end{pmatrix}$. Then the unipotent group action is: $$\begin{pmatrix} w_1 & x_1 & y_1 & z_1 \\ w_2 & x_2 & y_2 & z_2 \end{pmatrix} \mapsto \begin{pmatrix} w_1 + t x_1 & x_1 & y_1 + t z_1 & z_1 \\ w_2 + t x_2 & x_2 & y_2 + t z_2 & z_2 \end{pmatrix}.$$ Written in Plucker coordinates, that is: $$(p_{12}, p_{13}, p_{14}, p_{23}, p_{24}, p_{34}) \mapsto (p_{12}, p_{13} + t(p_{14}+p_{23}) + t^2 p_{24}, p_{14} + t p_{24}, p_{23} + t p_{24}, p_{24}, p_{34}).$$ So we have a fixed point if and only if $p_{24} = p_{14}+p_{23} = 0$. Subbing into the Plucker relation $p_{12} p_{34} - p_{13} p_{24} + p_{14} p_{23} =0$, we get $p_{12} p_{34} = p_{23}^2$, with the variable $p_{13}$ unconstrained. In $\mathbb{P}^2$, the equation $xz=y^2$ is a conic; adding one more variable takes the cone over that conic.
You are correct that, topologically, this is a pinched torus. There are $4$ possible sign patterns for the $p$'s (we have $\mathrm{sign}(p_{12}) = \mathrm{sign}(p_{34})$ since their product is square, $p_{23} = - p_{14}$ and $p_{24}=0$, and remember to mod out by a global sign change because these are homogenous coordinates). Each fixed sign pattern forms a triangle, and you can check that the triangles glue into a pinched torus.
One useful way to think about this is think about the structure of flags of a line and a 2-space, both of them invariant. The line has to be in the kernel, so we have a $\mathbb{P}^1$ worth of them. Over each one of these, there is a $\mathbb{P}^1$ of 2-spaces whose image lies in that line. In real terms, $\mathbb{P}^1$ is just a circle, so this is the torus you're seeing. What's not visible in real terms is that this is not a trivial $\mathbb{P}^1$ bundle; I believe it's actually isomorphic to $\mathbb{P}^2$ blown-up at a point.
That is, we have the lines $ae_1+be_3$ (where the line only depends on point $[a:b]$ in $\mathbb{P}^1$). The preimage of this line is spanned by $\{ae_2+be_4,e_1,e_3\}$. The fact that this is not a trivial $\mathbb{P}^1$ bundle exactly makes it impossible to parametrize all the fibers uniformly. The description $\mathrm{span}(ae_1+be_3, c(ae_2+be_4)+de_1)$ works unless $b=0$, and $\mathrm{span}(ae_1+be_3, c(ae_2+be_4)+de_1)$ works unless $a=0$.
However, there's a redundancy in this description; the kernel shows up $\mathbb{P}^1$-many times, so the space itself is $\mathbb{P}^2$ blown-up at a point and blown down along a different $\mathbb{P}^1$. I don't think that the result is $\mathbb{P}^2$, since the $\mathbb{P}^1$ you blow down doesn't have self-intersection $-1$, but I don't swear it.