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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?)

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  • $\begingroup$ Hi Harald, since V is faithful, every irrep becomes a component of V^m when m is large enough and especially the regular rep would eventually be a subrep. $\endgroup$
    – Zello
    Commented Mar 6, 2020 at 19:52
  • $\begingroup$ That's right - but what is the (dimension of the) largest-dimensional irreducible representation? $\endgroup$
    – H A Helfgott
    Commented Mar 7, 2020 at 20:04
  • $\begingroup$ The question has no short answer. To induce it from $V\otimes V$, the Clebsch-Gordan coefficients play. They are the multiplicities of irreps $W$ occurring in $V\otimes V$ for $V$ irreducible, which has character formula $\dfrac{1}{|G|}\sum_{g\in G}\chi_V(g)^2\bar{\chi_W(g)}. There are computations of these coefficients for linear groups in papers that I'm not familiar with. But for symmetric groups and their natural representation, their is a simple rule for general Young characters in section 2.9 of James-Kerber, The representations of symmetric groups. We can discuss it in person if needed. $\endgroup$
    – Zello
    Commented Mar 9, 2020 at 9:32
  • $\begingroup$ Sorry, for multiplicity in $V^n$ you just change the exponent to $n$ in the character formula. Note that the formula is for finite groups. $\endgroup$
    – Zello
    Commented Mar 9, 2020 at 10:06

1 Answer 1

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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 $V$.

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  • $\begingroup$ So what would the dimension of the largest-dimensional representation be, then? $\endgroup$
    – H A Helfgott
    Commented Dec 31, 2019 at 16:04
  • $\begingroup$ 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 $V$. $\endgroup$
    – Geoff Robinson
    Commented Dec 31, 2019 at 21:12
  • $\begingroup$ and, for $K=\mathbb{R},\mathbb{C}$ or $K=\mathbb{F}_q$, that would be...? $\endgroup$
    – H A Helfgott
    Commented Jan 1, 2020 at 8:39
  • $\begingroup$ 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 $\lambda$. $\endgroup$
    – Geoff Robinson
    Commented Jan 1, 2020 at 9:08
  • $\begingroup$ Actually, my answer was probably only valid for $K$ finite. $\endgroup$
    – Geoff Robinson
    Commented Jan 1, 2020 at 9:18

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