4
$\begingroup$

For representations of semisimple Lie algebras $L$, we have the following theorem on central characters:

Theorem (Harish-Chandra): Let $\lambda, \mu \in H^{\ast}$. Then $\chi_{\lambda} = \chi_{\mu}$ if and only if $\lambda \sim \mu$.

Notation issue: Let $L$ be a semisimple Lie algebra, $\mathcal{U}(L)$ be its universal enveloping algebra, $\mathcal{Z}(L)$ be the center of $\mathcal{U}(L)$. Let $\lambda \in H^{\ast}$. The action of $\mathcal{Z}(L)$ on the Verma module $Z(\lambda)$ of $L$ induces a central character (or called the infinitesimal character) $\chi_{\lambda} : \mathcal{Z}(L) \rightarrow \mathbb{F}$.

Linkage: We say $\lambda, \mu \in H^{\ast}$ are linked (written as $\lambda \sim \mu$) if $\lambda + \delta$ and $\mu + \delta$ are $\mathcal{W}$-conjugate, where $\mathcal{W}$ is the Weyl group of the root system of $L$ (after the CSA $H$ is chosen).

My two questions below are both related to understanding the notion of Linkage:

Exercise 23.8 Prove that the weight lattice $\Lambda$ is Zariski dense in $H^{\ast}$. Use this to prove the if part of the above quoted theorem.

My question 1 is how to solve the above exercise?

My attempt 1: I have proved that the weight lattice $\Lambda$ is indeed Zariski dense by identifying $\Lambda$ with $\mathbb{Z}^{\ell}$ and $H^{\ast}$ with $\mathbb{C}^{\ell}$. Then $\mathbb{Z}^{\ell}$ is indeed dense in $\mathbb{C}^{\ell}$ by noting $$ \overline{\mathbb{Z}^{\ell}} = Z(I(\mathbb{Z}^{\ell})) = Z(0) = \mathbb{C}^{\ell}. $$ But I got stuck on how to carry on to show $\chi_{\lambda} = \chi_{\mu}$ for linked $\lambda$ and $\mu$.

Here is another exercise in Humphrey:

Exercise 23.6: If $\Lambda \in \Lambda^{+}$, prove that all $\mu$ linked to $\lambda$ satisfy $\mu \prec \lambda$, hence that all such $\mu$ occurs as weights of $Z(\lambda)$.

My question 2 is how to show the "hence that" part?

My attempts 2: I have managed to show the first part: Since $\delta$ is strongly dominant, we see $\lambda + \delta$ is strongly dominant. Since $\mu$ is linked to $\lambda$, $\mu + \delta$ is $\mathcal{W}$-conjugate to $\lambda + \delta$. Hence by Lemma 13.2A of Humphrey's book, $\mu + \delta \prec \lambda + \delta$, and hence $\mu \prec \lambda$. But I got stuck on the "hence that" part. It seems that this part is a direct corollary of $\mu \prec \lambda$? But I cannot figure it out, as I know few on the weights of the Verma module $Z(\lambda)$. The only thing relevent that I can think of is Proposition 21.3 on the weights of $V(\lambda)$, the irreducible highest weight representation. Is this related to the Harish-Chandra theorem quoted above?

As the two are both related to the Linkage relation and the Verma module $Z(\lambda)$, I listed the two questions in one post. Sorry for such a long post and thank you all for your answers and comments!


EDIT after @Erica's hints on Question 2: Following @Erica's hints on Question 2, I have solved Exercise 21.4 in Humphrey's book using Lemma 21.2, this is also Proposition 23.2:

Let $\lambda \in \Lambda, \alpha \in \Delta$. If the integer $m = \langle \lambda, \alpha \rangle$ is nonnegative, then the coset of $y_{\alpha}^{m+1}$ in $Z(\lambda)$ is a maximal vector of weight $\lambda - (m+1)\alpha$.

So for any $\mu \prec \lambda$, to show that $\mu$ is a weight of $Z(\lambda)$, it suffices to find a simple root $\alpha \in \Delta$ such that $\mu = \lambda - (m+1)\alpha $. In the proof of the Corollary 23.2, we have seen that $$\lambda - (m+1)\alpha = \sigma_{\alpha}(\lambda + \delta) - \delta.$$ So in other words, we shall find some simple reflection $\sigma_{\alpha} \in \mathcal{W}$ such that $\mu + \delta = \sigma_{\alpha}(\lambda + \delta)$. But it seems that the only thing we know from the linkage of $\lambda$ and $\mu$ is that there exist $\sigma \in \mathcal{W}$ such that $\mu + \delta = \sigma(\lambda + \delta)$, but how can we make $\sigma$ to be a simple reflection (i.e. $\sigma = \sigma_{\alpha}$ for some $\alpha \in \Delta$)?

  • Note that $\mathcal{W}$ is generated by simple reflections, we obtain $\sigma = \sigma_{\alpha_1} \cdots \sigma_{\alpha_r}$ for some $\alpha_1, \ldots, \alpha_r \in \Delta$. But I got stuck here. I have tried to show "by induction on $r$" but failed.

  • We haven't invoke $\mu \prec \lambda$ or $\lambda \in \Lambda^{+}$ in this part. Will these conditions helpful?

$\endgroup$

1 Answer 1

2
+50
$\begingroup$

Question 1: Humphreys already proved the if part for $\lambda,\mu$ in the weight lattice $\Lambda$. Let $\mu=\sigma(\lambda+\delta)-\delta$ for some $\sigma$ in the Weyl group. Consider the subset $$\Omega=\{\lambda\in H^{*}:\chi_{\lambda}=\chi_{\sigma(\lambda+\delta)-\delta}\},$$ which contains the weight lattice $\Lambda$. Notice that the map $$H^{*}\rightarrow \mathrm{Hom}_{alg}(Z(U(L)),F),\lambda\mapsto \chi_{\lambda}$$ can be viewed as the $F$-scheme morphism induced by $Z(U(L))\rightarrow \operatorname{Sym}(H)$. So the set $\Omega$ is Zariski closed and thus $\Omega=H^{*}$ since $\Lambda$ is Zariski dense.

Question 2: I think Lemma 21.2 in Humphreys' book can help you show that any $\mu<\lambda$ occurs as a weight of the standard cyclic module $Z(\lambda).$

Let $\{\alpha_{1},\cdots,\alpha_{r}\}$ be a basis of the root system. Firstly we review the construction of $Z(\lambda).$ Let $D_{\lambda}=Fv$ be a $B$-module, where $B$ is the Borel subalgebra corresponds to our choice of basis, defined by $$(h+\sum_{\alpha>0}x_{\alpha}).v=h.v=\lambda(h)v.$$ The standard cyclic module $Z(\lambda)$ is the tensor product $U(L)\otimes_{U(B)}D_{\lambda}$.

For each $\alpha_{j}$, we choose a nonzero $x_{j}\in L_{\alpha_{j}}$ and $y_{j}\in L_{-\alpha_{j}}$. Lemma 21.2 tells us $[h,y_{j}^{k}]=-k\alpha_{j}(h)y_{j}^{k}$ for $h\in H$.

Suppose that $\lambda=\mu+\sum_{j}k_{j}\alpha_{j}$, where the coefficients $k_{j}\in\mathbb{N}.$ Consider the element $$v_{\mu}:= (\prod_{j=1}^{r}y_{j}^{k_{j}}).v.$$

Then we can see \begin{align*} h.v_{\mu}&=hy_{1}^{k_{1}}\cdots y_{r}^{k_{r}}.v\\ &=[h,y_{1}^{k_{1}}]y_{2}^{k_{2}}\cdots y_{r}^{k_{r}}.v+y_{1}^{k_{1}}hy_{2}^{k_{2}}\cdots y_{r}^{k_{r}}.v\\ &=[h,y_{1}^{k_{1}}]y_{2}^{k_{2}}\cdots y_{r}^{k_{r}}.v+y_{1}^{k_{1}}[h,y_{2}^{k_{2}}]y_{3}^{k_{3}}\cdots y_{r}^{k_{r}}.v+\cdots+y_{1}^{k_{1}}\cdots y_{r-1}^{k_{r-1}}[h,y_{r}^{k_{r}}].v+y_{1}^{k_{1}}\cdots y_{r}^{k_{r}}h.v\\ &=\big(-k_{1}\alpha_{1}(h)-\cdots-k_{r}\alpha_{r}(h)+\lambda(h)\big)y_{1}^{k_{1}}\cdots y_{r}^{k_{r}}.v\\ &=\mu(h)v_{\mu}. \end{align*} Hence $\mu$ is a weight of $Z(\lambda)$.

I hope there is no mistake in my proof because I haven't used them for a long time.

$\endgroup$
7
  • $\begingroup$ Thank you so much for your help again! I have tried on Question 2 following your hint. Yet I still got stuck on some "final steps". I have added my attempts in this part in the post. Could you please explain more on these "final steps"? $\endgroup$
    – Hetong Xu
    Nov 23, 2021 at 15:34
  • $\begingroup$ ...... I have also gone through your hints on Question 1 and they are again quite helpful and inspiring! But I still need to think more on the "notice that the map ..." part for a while, especially how these lead to the closedness of $\Omega$. :) $\endgroup$
    – Hetong Xu
    Nov 23, 2021 at 15:34
  • $\begingroup$ I am going to add more details for Question 2. $\endgroup$
    – Erica
    Nov 23, 2021 at 15:56
  • $\begingroup$ About Question 1, I have a look at Humphreys' book tonight, he mentions this interpretation in the beginning of section 23.3. I think the algebra homomorphism is what he denotes by $\xi$. $\endgroup$
    – Erica
    Nov 24, 2021 at 0:16
  • $\begingroup$ Thank you for your further edits! These explainations can also work for many exercises including Exercise 20.9. That helps a lot. On Question 1, I'm still quite confused on why the map $\xi^{\ast}: H^{\ast} \rightarrow \mathrm{Hom}_{F-\mathrm{alg}}(Z(U(L)), F)$ is a $F$-scheme morphism and how this implies that $\Omega$ is closed. (Though I saw that this $\xi^{\ast}$ is induced by composing $\lambda: \mathrm{Sym}(H)=U(H) \rightarrow F$ with $\xi: Z(U(L)) \rightarrow \mathrm{Sym}(H)$ by your previous comment.) $\endgroup$
    – Hetong Xu
    Nov 24, 2021 at 15:46

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.