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Suppose we have a Borel subgroup $B$ of a linear algebraic group and a 1-dimensional representation $\pi:B\rightarrow \mathbb{C}_\lambda$, where $\lambda\in Hom(\mathbb T,\mathbb{C}^*)$ with $\mathbb T$ the maximal torus.

Choose some simple root $\alpha$, consider the minimal parabolic subgroup $P_\alpha$ containing $B$.

My question is when the representation can be extended to a representation of $P_\alpha$. Will the answer be $\langle\lambda,\alpha^\vee \rangle=0$?

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Yes. Write $P_\alpha = (SL_2) \ltimes ((\ker \alpha) Rad(P_\alpha))$ where $\ker\alpha$ is the subgroup of $T$. The necessity of your condition is about getting the $SL_2$ to act. Once you impose it, then the group to the right of the $\ltimes$ is acting trivially, so the question has reduced to the $SL_2$ case.

This is reasonably elementary structure theory of algebraic groups and I don't consider it a research-level question.

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  • $\begingroup$ I agree that the quetion is not research-level. It's worth comparing the deeper questions about extensions of representations in papers by Cline-Parshall-Scott Induced modules and extensions of representations, Invent. Math. 47 (1978), no. 1, 41–51, Induced modules and extensions of representations II, J. London Math. Soc. (2) 20 (1979), no. 3, 403–414, and by H.H. Andersen Vanishing theorems and induced representations, J. Algebra 62 (1980), no. 1, 86–100. $\endgroup$
    – Jim Humphreys
    Commented Jan 30, 2014 at 14:35
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    $\begingroup$ P.S. Strictly speaking, your semidirect product might be an "almost" semidirect product if the Levi subgroup has derived group $SL_2$ rather than $PGL_2$. But that has little effect on the situation. $\endgroup$
    – Jim Humphreys
    Commented Jan 30, 2014 at 14:43
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    $\begingroup$ I don't agree that this question would not be research-level. This depends a lot on the background of the OP, and it is not unlikely that he came across this problem during his research in a different context but he does not have the required familiarity with algebraic groups, which is why he asked it on MO. This is a very good reason to use MO, in my humble opinion. $\endgroup$
    – Tom De Medts
    Commented Jan 30, 2014 at 17:45
  • $\begingroup$ I agree with Tom De Medts. I'm certainly no expert and my first guess would be $\langle \lambda, \alpha^\vee \rangle \in\mathbb{N}$. $\endgroup$
    – Vít Tuček
    Commented Jan 30, 2014 at 17:52

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