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How would one classify the strata for the standard nilpotent cone for $GL_{k}(\mathbb{C})$, using the definition from Hesselink's paper "Desingularizations of Varieties of Nullforms"? I know that they correspond to partitions / nilpotent orbits etc, but from first principles why aren't two different nilpotent orbits possibly in the same strata - how would you prove that? (preferably using the definition of Hesselink)

I would like to classify the strata for the problem I'm working on, but don't completely understand how to do it for the more basic case (for which the method is probably well-known), I get stuck on the details, so that would be very helpful. Thanks.

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I linkified Hesselink's paper for you. Since that paper doesn't seem to be publicly available, it might be helpful for you to restate the definition here. – David Speyer Dec 3 '09 at 1:42
up vote 3 down vote accepted

I'm going to give a partial answer here for two reasons: (1) I am lazy and (2) this is starting to feel a little homeworky to me. Obviously, no one would assign this material as homework, but part of reading a math paper is taking the time to work out lots of simple examples and see how the definitions work. I feel like you are pushing the boundaries of how much of this work it is reasonable to ask other people to do. Not a major criticism, certainly not a vote to close the question, but my input.

On to the math. I've scanned the first 3 pages of Hesselink's paper. He make the following definitions. G acts on V, v is a point of V and $\star$ a chosen base point of V fixed by G. In your setting, G is $GL_n$, V is the $n \times n$ matrices where G acts by conjugation, and $\star$ is zero. Hesselink writes Y(G) for what is essentially $\mathrm{Hom}(\mathbb{C}^*, G)$. More precisely, Hesselink tensors with $\mathbb{Q}$, so that he can talk about maps like $t \mapsto \left( \begin{smallmatrix} t^{1/3} & 0 \\\\ 0 & t^{-2/7} \end{smallmatrix} \right)$. I'll ignore this detail.

For $\lambda \in Y(G)$, Hesselink defines a rational number $m(\lambda)$. We talked about this in your previous question. In this setting, where V is an $N$-dimensional vector space, Hesselink gives an explicit formula for m on the bottom of page 142/top of page 143: Diagonalize the action of $\lambda$ as $t \mapsto \mathrm{diag}(t^{m_1}, \cdots, t^{m_N})$ and write $v = \sum v_i e_i$.. Then $m(\lambda) = \min(m_i : v_i \neq 0)$ if this number is nonnegative, and is $- \infty$ if this minimum is negative.

Let's see what this definition means in your setting. We can conjugate any $\lambda$ into diagonal form as $t \mapsto \mathrm{diag}(t^{c_1}, \cdots, t^{c_n})$. I've replaced $m_i$ by $c_i$ to point out that these $c$'s are not the $m$'s of the previous paragraph. In our notation, the $N$ of the previous paragraph is $n^2$. The vector space $V$ has dimension $n^2$ with basis $e_{ij}$. The action of $\lambda(t)$ on $e_{ij}$ is by $t^{c_i - c_j}$. (Exercise!).

So $m(\lambda) > 0$ if and only if $c_i \leq c_j$ implies $v_{ij} =0$.

We may as well order our basis such that $c_1 \geq c_2 \geq \cdots \geq c_n$. If $c_1 > c_2 > \cdots >c_n$ then we see that $m(\lambda) > 0$ if and only if $v$ is a strictly upper triangular matrix. When there are some equalities among the $c$'s, you want $v$ to be strictly block upper triangular. For such a $v$, $m(\lambda) = \min(c_i - c_j : v_{ij} \neq 0)$. In particular, notice that there exists a $\lambda$ such that $m(\lambda) > 0$ if and only if $v$ is nilpotent.

Hesselink defines $\Lambda(v)$ to be the locus in $\{ \lambda : m(\lambda) = 1 \}$ where $q(\lambda)$ is minimized, where $q$ is the inner product from your previous question. What you want to show is that $\Lambda(v)$ determines the Jordan normal form of $v$.

I must admit that I haven't thought out how to prove this. But I hope this makes things explicit enough that you can attack it.

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thanks, that was more than enough detail to be of help! if you think my question is unreasonable, and if you have time to answer it briefly, please just answer concisely as possible and some of the major steps and I'll try to fill in. I didn't ask for a 100% mathematically rigorous answer. – Vinoth Dec 7 '09 at 6:28

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