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.



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


