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?


1 Answer 1


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.

  • $\begingroup$ 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

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