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I would appreciate very much if you can point to me some references on the following:

1) Representations of the linear group $GL(n,2)$ over $F_2$.

2) Representations of $GL(n,2)$ over an algebraic closure of $F_2$.

3) Representations of $GL(n,2^f)$ over $F_{2^f}$.

Here $F_q$ is the Galois field of size $q$.

Thank you in advanced!

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  • $\begingroup$ What sort of things should be covered, how advanced knowledge can it assume, and does it have to focus on this specific case? If the answers to the latter two are "quite" and "no" then the book by Humphreys titled "Representations on finite groups of Lie type" should cover pretty much any immediate need. $\endgroup$
    – Tobias Kildetoft
    Commented Oct 9, 2018 at 15:54
  • $\begingroup$ To Tobias: I would like to know what are the irreducible representations and their degrees? Is there any description or construction of irreducible representations of GL(n,2) over F_2? $\endgroup$
    – Uep
    Commented Oct 9, 2018 at 16:17
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    $\begingroup$ Once $n$ becomes large enough, I don't think the dimensions of the irreducible representations are known, even in characteristic $2$. The case of $SL_n$ is a bit easier to describe (though passing to $GL_n$ does not change much, as it just adds the possibility of twisting by powers of the determinant representation). Here there are $p^{n-1}$ irreducible representations, indexed by restricted weights. There is also a description due to Jantzen of which of these irreducibles have their dimension given by Weyl's formula, but this will be only some of them. $\endgroup$
    – Tobias Kildetoft
    Commented Oct 9, 2018 at 16:50
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    $\begingroup$ Just to add a bit more detail: For the case in 2) we have that $GL_n = SL_n$ so we don't need to worry about twisting with the determinant at all. The irreducible representations are labelled by $(n-1)$-tuples of $0$s and $1$s (by writing the weight in the fundamental basis), and denoted $L(a_1,a_2,\dots, a_{n-1})$. We then have the following dimensions for $n\leq 4$: $\dim(L(0)) = 1$, $\dim(L(1)) = 2$, $\dim(L(0,0)) = 1$, $\dim(L(1,0)) = \dim(L(0,1)) = 3$, $\dim(L(1,1)) = 8$ $\endgroup$
    – Tobias Kildetoft
    Commented Oct 12, 2018 at 11:06
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    $\begingroup$ $\dim(L(0,0,0)) = 1$, $\dim(L(1,0,0)) = \dim(L(0,0,1)) = 4$, $\dim(L(0,1,0)) = 6$, $\dim(L(1,1,0)) = \dim(L(0,1,1)) = 20$, $\dim(L(1,0,1)) = 14$, $\dim(L(1,1,1)) = 64$. Here the dimensions of all except $L(1,0,1)$ are given by Weyl's dimension formula, while the exception is one smaller than that due to the appearance of a copy of the trivial module in the induced module with that highest weight. $\endgroup$
    – Tobias Kildetoft
    Commented Oct 12, 2018 at 11:09

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