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In all that follows, we are working over $\mathbb{C}$. Let $B \subseteq P \subseteq {\rm GL}(n)$ be a parabolic subgroup. Can you say anything in general about the representations of $P$? I suspect the answer is no because I couldn't find anything about this in the standard books or using google.

If $P = B$ then everything is easy: any rep breaks up as a sum of 1 dimensional subreps on which the torus acts by some character and the unipotent radical acts as the identity.

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    $\begingroup$ It depends on what you mean by "easy", I think. Any semisimple representation of $B$ will of course be a direct sum of one dimensional subrepresentations. But arbitrary representations need not be. $\endgroup$
    – Stephen
    Commented Aug 26, 2014 at 1:00
  • $\begingroup$ Ohh, the representations of B are not semisimple? $\endgroup$
    – Daniel Barter
    Commented Aug 26, 2014 at 1:03
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    $\begingroup$ Definitely not! A simple example is the standard $n$-dimensional representation of $GL(n)$ restricted to $B$. $\endgroup$
    – Steven Sam
    Commented Aug 26, 2014 at 2:03
  • $\begingroup$ The problem is that $B$ is not a reductive group, so some of its finite-dimensional reps are not semisimple. $\endgroup$
    – Peter Crooks
    Commented Aug 26, 2014 at 2:39
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    $\begingroup$ I hope it's OK to appreciate 5 years later @StevenSam's simple example of a non-semisimple representation. :-) $\endgroup$
    – LSpice
    Commented Sep 9, 2019 at 14:15

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I think the idea is to write $P$ as a semidirect product of its unipotent radical $N$ and its (maximal reductive) Levi subgroup $M$. (If $P$ is a Borel, then $M$ is a maximal torus.) You can then restrict a complex representation of $P$ to $M$ and decompose it into irreducibles.

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