# Prove Harish-Chandra's theorem on central characters with algebraic-geometrical observation and the linkage relations among weights

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?

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.

• 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"? Nov 23, 2021 at 15:34
• ...... 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 $\Omega$. :) Nov 23, 2021 at 15:34
• I am going to add more details for Question 2. Nov 23, 2021 at 15:56
• 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 $\xi$. Nov 24, 2021 at 0:16
• 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.) Nov 24, 2021 at 15:46