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.

## Lie algebra representations

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- The Sleeping Tyrant
**Posts:**533**Joined:**Tue Jan 23, 2007 2:49 am UTC**Location:**Ont., Canada-
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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.

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
**Posts:**533**Joined:**Tue Jan 23, 2007 2:49 am UTC**Location:**Ont., Canada-
**Contact:**

### 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.

Thanks for your help.

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.

Thanks for your help.

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