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GL(4,R) is the group of all real nonsingular 4x4 matrices. What are the 4-dimensional complex representations of this group?

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The only 4-d representations are the obvious representation by matrix multiplication, its dual (matrix multiplication by the inverse transpose), and the trivial representation.

EDIT: Sorry, the list above is the 4-d representations of $SL(4;\mathbb R)$. So, if we have 4-dimensional representation of $GL(4;\mathbb R)$ will be an $SL(4; \mathbb R)$-represention. If $SL(4; \mathbb R)$ acts trivially on your representation, then it factors through $\mathbb R^{\times}$, and thus will always be of the form $(-1)^pe^t\mapsto \epsilon^p\operatorname{exp}(tA)$ for commuting matrices $A,\epsilon$ such that $\epsilon^2=I$, up to conjugacy.

If $SL(4;\mathbb R)$ acts by a non-trivial representation, then it could be the obvious extension by matrix multiplication; we could also multiply the action of $A$ by $\det(A)^{a}|\det(A)|^p$ for $a\in \{0,1\}$ and $p\in \mathbb{C}$ (where, as usual, $x^p=e^{p\log x}$ for $x$ real and positive). A little thinking about Clifford theory shows that this is all the possible extensions (it's true that $GL(4;\mathbb R)\cong \mathbb{R}^{\times}_{>0}\times \{A||\det(A)|=1\}$, and we can use Clifford theory on the extension of $SL(4;\mathbb{R})$ by the second factor).

As for why this is the classification for $SL_4$: every non-trivial irrep has a non-trivial action of $S_4$ on its weight spaces. The only non-trivial $S_4$ action with 4 or fewer elements is $S_4/S_3$, so if this action has more than 1 orbit, then the dimension is $>4$. Thus, the irrep must be minuscule, and the only minuscule weights with the stabilizer conjugate to $S_3$ are the ones I described.

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    $\begingroup$ You can probably also twist irreducible ones by a character (and also consider various non-irreducible ones with abelian image). $\endgroup$
    – YCor
    Commented Dec 12, 2022 at 0:57
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    $\begingroup$ Ok, and... what's the easy explanation? :) $\endgroup$ Commented Dec 12, 2022 at 1:05
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    $\begingroup$ I know that this is being picky, but it's not true that $\mathrm{GL}(4,\mathbb{R})\simeq \mathrm{SL}(4,\mathbb{R})\times \mathbb{R}^{\times}$. Where would $A = \mathrm{diag}(-1,1,1,1)$ go under such an isomorphism? $\endgroup$ Commented Dec 12, 2022 at 15:54
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    $\begingroup$ Oy vey, this is hard to get right for such an "easy" question. Of course you’re right… $\endgroup$
    – Ben Webster
    Commented Dec 13, 2022 at 16:24
  • $\begingroup$ @RobertBryant Now it's fixed; it actually doesn't change the answer, since the irrep $\mathbb{R}^4$ is already the restriction of an irrep of $GL(4;\mathbb R)$. $\endgroup$
    – Ben Webster
    Commented Dec 13, 2022 at 20:31

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