Automorphism counts for nilpotent Jordan representations

For the Jordan-string basis $v_1,\dots ,v_r$ of $I_r$, we have $T{v}_1=0$ and $T{v}_j=v_{j-1}$ for $j>1$. Give $I_r$ the $k[T]$-module structure in which $T$ acts by its nilpotent operator. The linear map $k[T]\mathrm{/}(T^r)\to I_r$ sending $T^i$ to $v_{r-i}$ for $0\le i<r$ is a bijection commuting with $T$, hence a module isomorphism. Intertwining linear maps are exactly module homomorphisms. A module homomorphism $k[T]\mathrm{/}{T}^r\to k[T]\mathrm{/}{T}^h$ is determined by the image of $1$, which may be any vector killed by $T^r$; that space has dimension $\min (r,h)$. Thus the endomorphism algebra of $I_{\lambda }$ has dimension $D={\sum }_{r,h}{m}_r{m}_h\min (r,h)={\sum }_j({\lambda }_j^{\mathrm{\prime }}{)}^2$. Counting each column's ordered pairs of boxes gives ${\sum }_j({\lambda }_j^{\mathrm{\prime }}{)}^2=\mathrm{\mid }\lambda \mathrm{\mid }+2{\sum }_i(i-1){\lambda }_i$: in a column of height $c$, its square is $c+2{\sum }_{i=1}^c(i-1)$.

Consider the induced map of an endomorphism $f$ on $I_{\lambda }\mathrm{/}T{I}_{\lambda }$. Ordered by block length, its matrix is block triangular: a homomorphism from length $r$ to a larger length $h$ has image of $1$ divisible by $T^{h-r}$ and hence has zero reduction in the target head. Its diagonal blocks are arbitrary matrices $B_r\in {\mathrm{Mat}}_{m_r}(k)$; all choices occur independently by constant matrices between blocks of the same length. The linear map from the $D$-dimensional endomorphism space to these diagonal blocks is onto, with every fiber of size $q^{D-{\sum }_r{m}_r^2}$.

The map $f$ is invertible exactly when all $B_r$ are invertible. Necessity follows from invertibility on the quotient. Conversely those blocks make the triangular head map surjective, so $I_{\lambda }=f(I_{\lambda })+T{I}_{\lambda }$. Since $f$ commutes with $T$, iteration gives $I_{\lambda }=f(I_{\lambda })+T^N{I}_{\lambda }$ for every $N$, and nilpotence makes the last summand zero for large $N$. Thus $f$ is onto and, on a finite-dimensional vector space, bijective; its inverse commutes with $T$.

Successively choosing linearly independent columns gives $\mathrm{\mid }{\mathrm{GL}}_m(k)\mathrm{\mid }={\prod }_{i=0}^{m-1}(q^m-q^i)=q^{m^2}{\prod }_{j=1}^m(1-q^{-j})$, including $m=0$. Hence the two preceding steps count the automorphisms as $q^{D-{\sum }_r{m}_r^2}{\prod }_r\mathrm{\mid }{\mathrm{GL}}_{m_r}(k)\mathrm{\mid }=q^D{\prod }_r{\prod }_{j=1}^{m_r}(1-q^{-j})$. Substitute $D$ from step 1.1. For $\lambda =\mathrm{\varnothing }$ the space is zero and its sole automorphism is the identity.