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Dec 14, 2022 at 18:09 history edited Ben Webster CC BY-SA 4.0
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Dec 14, 2022 at 18:09 comment added Ben Webster @RobertBryant You're right; I'd dealt with that previously, and then messed it up fixing another mistake. Sigh.
Dec 13, 2022 at 22:07 comment added Robert Bryant @BenWebster: Yah, I didn't think it would really change the answer, but I didn't want the OP to be confused. Thanks for fixing it. Also, for some reason the OP asked for 4-dimensional complex representations, so I guess that you'd want to include representations such as $\rho(A) = e^{\lambda\,\log|\det(A)|} A$ where $\lambda$ is a complex number with nonzero imaginary part.
Dec 13, 2022 at 20:31 comment added Ben Webster @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 $GL(4;\mathbb R)$.
Dec 13, 2022 at 20:30 history edited Ben Webster CC BY-SA 4.0
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Dec 13, 2022 at 16:24 comment added Ben Webster Oy vey, this is hard to get right for such an "easy" question. Of course you’re right…
Dec 12, 2022 at 15:54 comment added Robert Bryant 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?
Dec 12, 2022 at 1:56 history edited Ben Webster CC BY-SA 4.0
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Dec 12, 2022 at 1:30 history edited Ben Webster CC BY-SA 4.0
added 299 characters in body
Dec 12, 2022 at 1:05 comment added paul garrett Ok, and... what's the easy explanation? :)
Dec 12, 2022 at 0:57 comment added YCor You can probably also twist irreducible ones by a character (and also consider various non-irreducible ones with abelian image).
Dec 12, 2022 at 0:35 history answered Ben Webster CC BY-SA 4.0