Let $K$ be an algebraically closed field with characteristic $0$. I consider the Jordan decomposition of a NILPOTENT matrix: $A=diag(J\_{r\_1},\cdots,J\_{r\_s})\in M\_n(K)$ where $J\_k$ is the nilpotent Jordan block of dimension $k$ and with $r\_1\geq\cdots\geq r\_s$. Let $S(A)=\{P^{-1}AP|P\in GL\_n(K)\}$ be the similarity class of $A$ and $d(A)=f(r\_1,\cdots,r\_s)$ be the dimension of the algebraic variety $S(A)$ (I hope that the previous assertion is true). To find $d(A)$ is equivalent to find the Hilbert dimension of the algebraic variety $\{X\in SL\_n(K)|XA=AX\}$. With Maple, we can do actually the calculation until $n=7$.


For instance, for a fixed $n$, $\sup\_A d(A)=n^2-n$ and this sup is reached only for $J\_n$.


Does there exist an explicit formula for $f(r\_1,\cdots,r\_s)$, or, an algorithm that does the job within a reasonable time ? Is this problem linked to the Young tableaux ?


Thanks in advance



Dimension of a similarity class
There is an explicit formula. Let $r'\_1\geq \ldots \geq r'\_t$ be the transpose partition of $[r\_1,\ldots ,r\_s]$ (obtained by exchanging rows and columns in the Young diagram). Then $\ d(A)=n^2-\sum r'\_i{}^2$ . See for instance Collingwood-McGovern, *Nilpotent orbits in semisimple Lie algebras*, Cor. 7.2.4.


