## Lie algebra representations

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The Sleeping Tyrant
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### Lie algebra representations

I'm reading through Humphrey's text on category O (alongside his text on Lie algebras), and I'm having trouble understanding something.

Let [imath]L[/imath] be a semisimple Lie algebra, fix your simple roots [imath]\alpha_1,\dots,\alpha_t[/imath], and choose [imath]h_i,x_i,y_i[/imath] the standard basis for the isomorphic copy of SL(2,C) in L corresponding to the [imath]\alpha_i[/imath] weight space. Let [imath]L(\lambda)[/imath] be the simple L-module of highest weight [imath]\lambda[/imath] and let [imath]\phi[/imath] be the corresponding representation. Define [imath]r_i = exp (\phi(x_i)) exp(\phi(-y_i))exp(\phi(x_i))[/imath], and let [imath]s_i[/imath] be the simple reflection of the root system sending [imath]\alpha_i[/imath] to it's negative.

He claims that for any weight [imath]\mu[/imath] of [imath]L(\lambda)[/imath], that [imath]r_iL(\lambda)_\mu = L(\lambda)_{s_i\mu}[/imath]. I can see that it does send the [imath]\alpha_i[/imath] weight space to the [imath]-\alpha_i[/imath] weight space, but I don't see how acting by r_i permutes the other positive roots.

Jyrki
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### Re: Lie algebra representations

In my somewhat, but not too distant youth I think I was fully conversant with Humphreys' GTM series Lie algebra text. Category O I'm much less familiar with, but let's try anyway.

I'm fairly sure that you have left out something. The claim does not hold for all the simple roots [imath]\alpha_i[/imath], unless the module is finite dimensional, i.e. unless [imath]\lambda[/imath] is a dominant weight. For example, in the case L is [imath]sl_2(C)[/imath] and [imath]\lambda[/imath] non-dominant, the Verma module of highest weight [imath]\lambda[/imath] has no dominant weight spaces at all.

A further point is that unless [imath]L(\lambda)[/imath] is finite dimensional, then the action of [imath]\phi(-y_i)[/imath] is not locally finite, and we may have trouble defining [imath]\exp(\phi(-y_i))[/imath]. Mind you, this is a non-problem, if we know that [imath]L(\lambda)[/imath] is a sum of finite dimensional [imath]L_i[/imath]-modules, where by [imath]L_i[/imath] I mean the copy of [imath]sl_2(C)[/imath] spenned by [imath]x_i,y_i,h_i[/imath]. May be you are at a point, where it is known that you are working inside a finite dimensional [imath]L_i[/imath]-module?

Anyway the claimed relation only concerns an [imath]L_i[/imath]-submodule of [imath]L(\lambda)[/imath], where all the weights are of the form [imath]\mu+k\alpha_i[/imath] for some integer [imath]k[/imath]. This is the space the linear mapping [imath]r_i[/imath] acts on. It doesn't act on the root spaces at all, because it doesn't on the Lie algebra itself. In other words, I don't understand, why [imath]r_i[/imath] should act on the roots?

To see the claim, all you need to do is to show that this holds in the case of a finite dimensional [imath]sl_2(C)[/imath]-module. This may have been done earlier in the book(s). Handling [imath]sl_2[/imath] is usually done first. I don't remember the details of that calculation, because in the classical f.d. case you don't need to use these operators [imath]r_i[/imath] to reach this conclusion. May be it is a category O trick?

Edit: I guess all I mean is that you have left out an essential bit of context.

The Sleeping Tyrant
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### Re: Lie algebra representations

Actually, the issue is arising in a proof that [imath]L(\lambda)[/imath] is finite dimensional if [imath]\lambda[/imath] is dominant integral, so I can't rely on that. However, we do know that each [imath]x_i[/imath] and [imath]y_i[/imath] acts as a locally nilpotent operator.

But I think you're right about it only concerning an [imath]L_i[/imath]-submodule, and then refering to the weights of that action. I think that's what was confusing me.