# Irreducible representations of $\mathrm{SL}_n(K)$, $K$ finite

Let $$\mathrm{SL}_n/K$$ ($$K$$ finite) be given with its natural action on an $$n$$-dimensional vector space $$V/K$$. Consider the action of $$\mathrm{SL}_n$$ on the $$m$$-fold tensor product $$V\otimes \dotsc \otimes V$$. What is the largest-dimensional irreducible subrepresentation?

(For $$m$$ much larger than the maximum of $$n$$ and the size of $$K$$, it cannot be the symmetric power, as its dimension becomes larger than the size of $$\mathrm{SL}_n(K)$$. Or am I missing something?)

The paper [https://archives.maths.anu.edu.au/people/Kovacs/K033.pdf] by Bryant and Kovacs is very relevant here, at a reasonably generic level. You have to be careful about the action of the centre of $${\rm SL}_{n}(K)$$: note that the centre has order $$d = {\rm gcd}(n,|K|-1).$$ The result of Bryant and Kovacs shows that if $$m$$ is large enough, then the regular module is a direct summand of $$V^{m} \oplus V^{m+1} \oplus V^{m + d-1}$$, where I let $$V^{j}$$ denote the $$j$$-fold tensor product of $$V$$ with itself. Note that each irreducible representation restricts to a multiple of a one-dimensional representation of the centre, and that for a given choice of one-dimensional representation $$\sigma$$ of $$Z({\rm SL}(n,K))$$, the irreducibe rpresentations (of $${\rm SL})$$ which lie over $$\sigma$$ occur in (the socle of) one and only one of the $$d$$ successive tensor powers of $$V$$.
• For large enough $m$, it would be the dimension of the largest simple module of ${\rm SL}(n,K)$ which lies over $\lambda^{m}$, where $\lambda$ is the one dimension representation of $K^{\times}$ afforded by the action of $Z({\rm SL}(n,K))$ on $V$. – Geoff Robinson Dec 31 '19 at 21:12
• and, for $K=\mathbb{R},\mathbb{C}$ or $K=\mathbb{F}_q$, that would be...? – H A Helfgott Jan 1 at 8:39
• The answers must be well-documented, but I am not sure about the best references. I guess that someone like Jim Humphreys would have the answer at his fingertips.In the case that $|K| = q,$ I would expect the largest dimension simple module in the case that $\lambda^{m}$ is the trivial character to be the Steinberg module, but I am unsure about other powers of $\lambda$. – Geoff Robinson Jan 1 at 9:08
• Actually, my answer was probably only valid for $K$ finite. – Geoff Robinson Jan 1 at 9:18