Let $\mathfrak{g}$ be a semisimple Lie algebra over $\mathbb{C}$ with root system $\Phi$, Weyl group $W$ and Cartan decomposition $\mathfrak{g}=\mathfrak{h}\oplus \bigoplus_{\alpha \in \Phi} \mathfrak{g}_\alpha $. Fix a set of positive roots $\Phi^+ \subset \Phi$ and simple roots $\Delta \subset \Phi$. Then $I \subset \Delta$, defines a root system $\Phi_I \subset \Phi$ with positive roots $\Phi_I^+ \subset \Phi^+$ and a Weyl group $W_I \subset W$. Furthermore be $\mathfrak{n}_I^-=\bigoplus_{\alpha \in -\Phi_I^+}\mathfrak{g}_\alpha$.

In Humphreys "Representations of Semisimple Lie algebras in the BGG Category $\mathcal{O}$" we have the following Lemma 9.3:

Let $M \in \mathcal{O}$ have the set of weights $\Pi(M)$. The following conditions are equivalent:

  1. $M$ is locally $\mathfrak{n}_I^-$-finite.
  2. For all $\alpha \in I$ and $\mu \in \Pi(M)$ we have $\dim M_\mu=\dim M_{s_{\alpha}\mu}$.
  3. For all $w \in W_I$ and $\mu \in \Pi(M)$ we have $\dim M_\mu=\dim M_{w\mu}$.
  4. $\Pi(M)$ is stable under $W_I$.

Now I struggle with the main parts of the proof.

In "$1. \Rightarrow 2.$" he argues the following way. For $\alpha \in I$ fixed, consider the action of the subalgebra generated by $x_\alpha \in \mathfrak{g}_\alpha$ and $y_\alpha \in \mathfrak{g}_{-\alpha}$, which is isomorphic to $\mathfrak{sl} (2,\mathbb{C})$, on $M_\mu$. This gives a finite dimensional submodule $N$ of $M$, which is stable under $\mathfrak{h}$. Then the standard theory for finite dimensional representations of $\mathfrak{sl} (2,\mathbb{C})$ yields $2.$

Questions that arise for me:

  1. Does he consider $N$ as a finite dimensional $\mathfrak{sl} (2,\mathbb{C})$-module?
  2. Why is $N$ $\mathfrak{h}$-stable and why is this necessary?
  3. I know how the simple $\mathfrak{sl} (2,\mathbb{C})$-modules look like but how does this imply 2. ?

In "$4. \Rightarrow 1.$" He constructed a weight $\mu'(=w_I\mu)$, such that only finitely many $\mathbb{Z}^+$-linear combinations of $-\Phi_I^+$ can be added to $\mu'$ to lie in $\Pi(M)$. Then he says that $\mu'$ is a typical weight of $M$, forcing $M$ to be locally $\mathfrak{n}_I^-$-finite.

  1. How does this weight forces $M$ to be locally $\mathfrak{n}_I^-$-finite? Have the impression there is kind of standard argument used.
  • 1
    $\begingroup$ (1) Yes. (2) $h x_\alpha m_\mu = (\mu(h) - \alpha(h))x_\alpha m_\mu$, or something like that, and similarly for $y_\alpha$. (3) By decomposing into irreducible $\mathfrak{sl}$-submodules that intersect $M_\mu$, one sees that the dimension computation can be done for an irreducible submodule. For each such module, one knows that the dimension of the weight spaces behaves as desired. (4) The $\mathfrak n^-$-submodule of $M$ spanned by $M_{\mu'}$ is spanned by the weight spaces as you say. There are finitely many of those and each is finite dimensional. $\endgroup$ – LSpice Aug 26 at 12:55
  • 1
    $\begingroup$ (My argument for (3) is a little sketchy, but I think it basically works.) $\endgroup$ – LSpice Aug 26 at 13:01
  • $\begingroup$ Thanks for the fast answer! (2) Yeah right. But why is it necessary to have $\mathfrak{h}$-stability? (4) Then the argument is that every weight of $M$ can be expressed as such a $\mu'$? Otherwise I don't see how I get it locally everywhere. My definition of locally $\mathfrak{n}_I^-$-finite is that for every $v \in M$ is $U(\mathfrak{n}_I^-)v$ is finite dimensional. $\endgroup$ – CJS Aug 26 at 13:29
  • 1
    $\begingroup$ math.berkeley.edu/~reb/courses/261/35.pdf Excercise 283 let me believe something different. But okay I will think on my own. Thanks for all your comments. $\endgroup$ – CJS Sep 1 at 13:43
  • 1
    $\begingroup$ After your doubts I went through some basic textbook and you are right- $\langle \mu, \alpha \rangle \in \mathbb{Z}$. Can be found for example in Humphreys "Lie algebras and Representations Theory" section 7.2. $\endgroup$ – CJS Sep 9 at 16:08

Thanks to the outstanding help of LSpice I present a version of more detailed proof of the two parts above. Do not hesitate to point out mistakes.

"$(1) \Rightarrow (2)$": Fix $\alpha \in I$ and $\mu \in \Pi(M)$. Observe that for $\mu(h_\alpha)=0$, we have $s_\alpha\mu=\mu-\langle \mu, \alpha^{\vee}\rangle \alpha = \mu - \mu(h_\alpha)\alpha=\mu$ and $(2)$ follows trivially. Hence we can assume $\mu(h_\alpha) \neq 0$. Then by assumption and as $M\in \mathcal{O}$ the action on $M_\mu$ of the subalgebra $\mathfrak{s}_\alpha \cong \mathfrak{sl}(2,\mathbb{C})$, generated by $x_\alpha \in \mathfrak{g}_\alpha$, $y_\alpha \in \mathfrak{g}_{-\alpha}$, produces a finite dimensional $U(\mathfrak{sl}(2,\mathbb{C}))$-submodule $N \subset M$.

For $v \in M_\mu$, we have that $N \ni h_\alpha.v=\mu(h_\alpha)v$. Hence $M_\mu \subset N$. For $x_\alpha,y_\alpha$ exists $n_\alpha \in \mathbb{N}$ such that $x_\alpha^{n_\alpha+1}.N=y_\alpha^{n_\alpha+1}.N=0.$ Define \begin{align*} \exp(x_\alpha)&:=\sum_{k=0}^{n_\alpha} x_\alpha^k/k! \in U(\mathfrak{sl},(2,\mathbb{C}))\\ \exp(y_\alpha)&:=\sum_{k=0}^{n_\alpha} y_\alpha^k/k! \in U(\mathfrak{sl}(2,\mathbb{C})),\\ s&:=\exp(x_\alpha)\exp(-y_\alpha)\exp(x_\alpha) \in U(\mathfrak{sl}(2,\mathbb{C})) \end{align*} Then as in the proof of Theorem 21.2 of Humphreys "Lie algebras and Representations Theory", we additionally have that $M_{s_\alpha\mu}=s.M_\mu \subset N$.

As $N$ is finite dimensional, $N=\bigoplus N_i$ with $N_i$ simple $U(\mathfrak{sl}(2,\mathbb{C}))$-module. Observe that \begin{align*} h.(x_\alpha.v)&=(x_\alpha h+\alpha(h)x_\alpha).v=(\mu(h)+\alpha(h))x_\alpha.v, \\ h.(y_\alpha.v)&=(x_\alpha h-\alpha(h)x_\alpha).v=(\mu(h)-\alpha(h))y_\alpha.v, \end{align*} for $v \in M_\mu, h \in \mathfrak{h}$. Thus $N$ is $\mathfrak{h}$-stable and $M_\mu=\bigoplus (N_i)_\mu$ resp. $M_{s_\alpha\mu}=\bigoplus (N_i)_{s_\alpha\mu}$ follows. But for simple $U(\mathfrak{sl}(2,\mathbb{C}))$-module $N_i$ we know that $\dim((N_i)_{s_\alpha\mu})=\dim((N_i)_\mu)$ and the claim follows.

"$(4) \Rightarrow (1)$": We want to show that $U(\mathfrak{n}_I^-).v$ is finite dimensional for every $v \in M$. As $M$ is $\mathfrak{h}$-semisimple, we can assume that $v \in M_\mu$ for some $\mu \in \Pi(M)$. Then by assumption $w_I\mu \in \Pi(M)$ and only finitely many $\mathbb{Z}^+$-linear combinations of $\Phi_I^+$ can be added to get a weight of $M$.Thus, as $\mu=w_I(w_I\mu)$ and $w_I$ interchange $\Phi_I^+$ and $\Phi_I^-$, only finitely many $\mathbb{Z}^+$-linear combinations of $\Phi_I^-$ can be added to $\mu$ to lie in $\Pi(M)$. Hence $U(\mathfrak{n}_I^-)M_\mu$ is finite dimensional, especially $U(\mathfrak{n}_I^-).v$.

| cite | improve this answer | |
  • $\begingroup$ By the way, although it is confusing to the extent of being highly regrettable, I think that the usual notation is to use $\mathfrak u_\alpha$ for the root space and $U_\alpha$ for the root group, and then $\mathfrak g_\alpha$ for the Lie algebra of the group $G_\alpha \mathrel{:=} \langle U_\alpha, U_{-\alpha}\rangle$. $\endgroup$ – LSpice Sep 9 at 16:11
  • $\begingroup$ Mmmh where have you seen this? For example in Humphreys "Representations of Semisimple Lie algebras in the BGG Category $\mathcal{O}$" it is as I used it and I consider Humphreys as an expert of this topic. I cannot remember to have ever seen $\mathfrak{u}_\alpha$ for the root space. $\endgroup$ – CJS Sep 9 at 16:33
  • $\begingroup$ There's a holy trinity of reductive-group books: Borel, Humphreys, and Springer. Milne is a recent addition to the canon. I find as you say that all use $\mathfrak g_\alpha$ on the Lie algebra—to my astonishment, because they use $U_\alpha$ and $G_\alpha$ as I say, so $\mathfrak g_\alpha$ is not the Lie alg. of $G_\alpha$! $\endgroup$ – LSpice Sep 9 at 16:43
  • 1
    $\begingroup$ Alright that's consistent with what I know. $\endgroup$ – CJS Sep 9 at 16:50
  • $\begingroup$ I find to my semi-relief that Conrad, Gabber, and Prasad, at least in the 1st edition, while not adopting my suggestion of using $\mathfrak u_\alpha$ for the root space, at least, as far as I can tell, only ever refer to the space in words without ever giving it a symbolic name. $\endgroup$ – LSpice Sep 9 at 17:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.