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Jan 1, 2020 at 10:07 comment added H A Helfgott Sure, let us assume $K$ is finite.
Jan 1, 2020 at 9:18 comment added Geoff Robinson Actually, my answer was probably only valid for $K$ finite.
Jan 1, 2020 at 9:08 comment added Geoff Robinson 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 $\lambda$.
Jan 1, 2020 at 8:39 comment added H A Helfgott and, for $K=\mathbb{R},\mathbb{C}$ or $K=\mathbb{F}_q$, that would be...?
Dec 31, 2019 at 21:12 comment added Geoff Robinson 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 $V$.
Dec 31, 2019 at 16:04 comment added H A Helfgott So what would the dimension of the largest-dimensional representation be, then?
Nov 27, 2019 at 23:05 history edited Geoff Robinson CC BY-SA 4.0
Added a little clarification
Nov 27, 2019 at 9:59 history answered Geoff Robinson CC BY-SA 4.0