This question comes from Exercise 4.1 of Lectures of geometric constructions, Kamnitzer – arXiv link.
In this exercise, we are given that $X: \mathbb{C}^N \to \mathbb{C}^N$ is a nilpotent matrix with $X^n = 0$. Associated to it is the partition $ \mu = (\mu_1, \dots, \mu_n) $ with $$\mu_i = \dim \ker (X^i) – \dim\ker(X^{i-1}).$$
To $X$ we can associate also associate the partition $\nu = (\nu_1, \dots, \nu_m)$ where each $\nu_i$ is the size of the $i$-th Jordan block of $X$, imposing an ordering so the 1st Jordan block is of largest size, and so on. The Young diagram of $\nu$ has a conjugate partition $\lambda$, where each $\lambda_i$ is the number of $j$ such that $\nu_j \geq i$ (i.e. it is the number of Jordan blocks of size greater than or equal to $i$).
The first part is to show that for every $k$, $$\mu_1 + \dots + \mu_k \leq \lambda_1 + \dots + \lambda_k.$$ Following the hint (and using a linear algebra result), $$ \dim \ker X^k = \text{number of Jordan blocks of size }(\geq k) + \text{number of Jordan blocks of size } k-1,$$ and inductively this is equal to the sum $\lambda_1 + \dots + \lambda_k$.
On the other hand, by construction, $$ \mu_1 + \dots + \mu_k = \dim\ker(X^k).$$ So shouldn't the inequality actually become an equality here? What part have I missed?
Best Answer
We know via the hint that $\lambda_1 + \dots + \lambda_k = \dim \ker (X^k)$.
The partition $\mu$ is an arbitrary partition not related to that of $X$. In particular, it describes flags of the form $$ 0 \subset F_1 \subset F_2 \subset \dots$$ where $\mu_i = \dim F_i - \dim F_{i-1}$. In other words, $\mu$ controls the dimension jump of each subspace of the flag, and it is clear that $\dim F_k = \mu_1 + \dots + \mu_k$. Following the notation in the paper, this must be a flag inside $\text{Fl}_\mu (\mathbb{C}^N)^X$, which by definition forces $F_k \subseteq \ker (X^k)$, from which the desired inequality follows.