2
$\begingroup$

In the book "Representation of Semisimple Lie Algebra in the BGG Category $\mathcal{O}$".

Exercise 1.13. Suppose $\lambda\not\in\Lambda$, so the linkage class $W\cdot\lambda$ is the disjoint union of its nonempty intersections with various cosets of $\Lambda_r$ in $\mathbb{h}^*$. Prove that each $M\in\mathcal{O}_{\chi_\lambda}$ has a corresponding direct sum decomposition $M=\oplus_i M_i$, in which all weights of $M_i$ lies in a single coset.

My attempt:

$M=\bigoplus_{\nu\in\mathfrak{h}^*} M_\nu =\bigoplus_{[\nu]\in\mathfrak{h}^*/\Lambda_r} M^{[\nu]}$.

Since $\mathfrak{g}_\alpha\cdot M_\mu\subseteq M_{\mu+\alpha}$ for all $\alpha\in \Phi$, we get $U(\mathfrak{n})\cdot M^{[\nu]}\subseteq M^{[\nu]}$, $U(\mathfrak{h})\cdot M^{[\nu]}\subseteq M^{[\nu]}$ and $U(\mathfrak{n}^-)\cdot M^{[\nu]}\subseteq M^{[\nu]}$.

Since $U(\mathfrak{g})=U(\mathfrak{n}^-)U(\mathfrak{h})U(\mathfrak{n})$, we get $U(\mathfrak{g})\cdot M^{[\nu]}\subseteq M^{[\nu]}$. Hence $M^{[\nu]}$ is a $U(\mathfrak{g})$-submodule of $M$.

Since $M\in\mathcal{O}$, $M$ is finitely generated as a $U(\mathfrak{g})$-module. Therefore, $M=\bigoplus_{i=1}^n M^{[\nu_i]}$.

Now, let $W\cdot\lambda=\{\eta_1,\cdots,\eta_k\}$. Consider $\{\eta_{i_1},\cdots, \eta_{i_r}\}\subseteq \{\eta_1,\cdots,\eta_k\}$ such that $[\eta_{i_1}],\cdots, [\eta_{i_r}]$ are distinct and $\{[\eta_{i_1}],\cdots, [\eta_{i_r}]\}=\{[\eta_1],\cdots,[\eta_k]\}$. It is clearly that $W\cdot\lambda\cap[\eta]\neq \emptyset\implies [\eta]\in \{ [\eta_{i_1}],\cdots, [\eta_{i_r}]\}$.

Then $W\cdot \lambda =\bigcup_{\eta\in\mathfrak{h}^*} W\cdot\lambda \cap[\eta] =\bigsqcup_{j=1}^{r}W\cdot\lambda \cap[\eta_{i_j}] =\bigsqcup_{j=1}^{r}W\cdot \eta_{i_j} \cap[\eta_{i_j}] =\bigsqcup_{j=1}^{r}W_{[\eta_{i_j}]}\cdot \eta_{i_j}$

I would like to know whether the corresponding direct sum decomposition means $M=\bigoplus_{j=1}^r M^{[\eta_{i_j}]}$. If so, how to prove it? Also, am I on the right track?

$\endgroup$

2 Answers 2

3
$\begingroup$

I think you're overcomplicating things. You have showed that: (a) As a vector space we have the following decomposition: $$ M = \bigoplus_{[\nu]\in\mathfrak{h}^*/\Lambda_r} M^{[\nu]}. $$ (b) Furthermore, each $M^{[\nu]}$ is a submodule of $M$.

From this it follows directly that the direct sum above is a decomposition of modules.

Edit to clarify: Note that just before Exercise 1.13, Humphreys writes "The following easy exercise gives a refinement of the linkage classes in the nonintegral case..." (emphasis mine). Chapter 1.1 has just introduced Category $\mathcal{O}$, and Exercise 1.1(b) shows that $\mathcal{O}$ decomposes as

$$\mathcal{O}=\bigoplus_{[\nu]∈\mathfrak{h}^∗/\Lambda_r}\mathcal{O}^{[\nu]}.$$

Chapter 1.12 shows that $\mathcal{O}$ also decomposes by central character as

$$\mathcal{O}=\bigoplus_{\chi}\mathcal{O}_\chi.$$

Chapter 1.13 introduces the notion of a block of a category, and shows that if $\lambda$ is integral then $\mathcal{O}_{\chi_\lambda}$ is a block. Exercise 1.13 is about what happens when $\lambda$ is not integral, and is a straightforward application of the result of Exercise 1.1(b). In this case $[w\cdot\lambda]$ is not a single coset of $\Lambda_r$ in $\mathfrak{h}^*$, but several, say $[w\cdot\lambda] = \{[\nu_1], \dots, [\nu_k]\}$. Exercise 1.1(b) now tells us that

$$\mathcal{O}_{\chi_\lambda}=\bigoplus_{i=1}^k\mathcal{O}_{\chi\lambda}^{[\nu_i]},$$

so when $\lambda$ is not integral then $\mathcal{O}_{\chi_\lambda}$ splits into several blocks, each corresponding to one of the cosets in $[w\cdot\lambda]$.

$\endgroup$
5
  • $\begingroup$ I do not think I am overcomplicating thing since what you have suggested is the exercise 1.1 in Humphreys' Category $\mathcal{O}$ book. But what I am doing is exercise 1.13. I fairly sure the word "corresponding" has something to do with the decomposition of $W\cdot\lambda$ into disjoint union. $\endgroup$ Sep 21, 2018 at 11:58
  • $\begingroup$ I've edited my answer, hopefully this clears things up? $\endgroup$ Sep 24, 2018 at 9:06
  • $\begingroup$ Now it is much clearer but I still DO NOT understand how to get the decomposition $\mathcal{O}_{\chi_\lambda}=\bigoplus_{i=1}^k\mathcal{O}_{\chi\lambda}^{[\nu_i]}$. Since $M$ itself (NOT its composition factor) may not have weight in the the form of $w\cdot\lambda$. $\endgroup$ Sep 24, 2018 at 15:53
  • $\begingroup$ M is assumed to be in $\mathcal{O}_{\chi_\lambda}$, so by the Harish-Chandra Theorem (b) (Humphreys section 1.10) all composition factors of $M$ are on the form $L(\mu)$ where $\mu = w\cdot \lambda$ for some $w\in W$. Is that enough? $\endgroup$ Sep 25, 2018 at 20:59
  • $\begingroup$ By corollary 1.2 in Humphreys' book, what you have said indeed tell $M\in\mathcal{O}_{\chi_\lambda}$ and also $M^{[\nu]}\in\mathcal{O}_{\chi_\lambda}$ have weight of form $w\cdot\lambda$, for some $w\in W$. Yes. I think your proof also works as well as my proof below. $\endgroup$ Sep 26, 2018 at 7:43
0
$\begingroup$

In Humphreys' book, I quote "A formal argument (which we leave to the reader) shows that M decomposes uniquely as a direct sum of submodules, each belonging to a single block. In particular, each indecomposable module belongs to a single block. (See for example Jantzen [153, II.7.1]; his finite dimensionality assumption can be replaced here by the chain conditions on modules.)"

Remark: Now if $M$ is arbitrary, we say it belongs to a block if all its composition factors do.

The blocks of $\mathcal{O}$ are precisely the subcategories consisting of modules whose composition factors all have highest weights linked by $W_{[\lambda]}$ to an antidominant weight $\lambda$. We denote the block associated to an antidominant weight by $\mathcal{O}_\lambda$.

By mimicking the proof in Jantzen [153, II.7.1]:

Let us denote the set of blocks by $\mathcal{B}$. For any $M\in \mathcal{O}_{\chi_\lambda}$, let $M_b$ be the sum of all submodules $M'$ of $M$ such that all composition factors of $M'$ belong to $b$. Then $M_b$ is the largest submodule with this property. By the remark above, we get $M_b\in b$.

Claim $M=\bigoplus_{b\in\mathcal{B}}M_b$.

The sum is clearly direct. It suffices to show $N=\sum_{b\in\mathcal{B}}M_b$ is all of $M$.

Suppose $N\neq M$. Then $\{0\}\neq M/N\in \mathcal{O}$, by corollary 1.2. in Humphreys' book, there is a chain $\{0\}\subseteq N'/N\subseteq M/N$ such that $N'/N=L$ is simple.

Equivalently, there is a submodule $N'\subseteq M$ such that $N\subset N'$ such that $N'/N$ is simple. There is a block $b$ with $L\in b$.

Set $N''=\bigoplus_{b'\neq b}M_{b'}\subseteq N\subset N'$. Then $\{0\}\subseteq N/N''\subseteq N'/N''$.

All composition factors of $N'/N''$ belong to $b$ since they are either $L\cong (N'/N'')/(N/N'')$ or composition factors of $N/N''\cong M_b$.

By the definition of blocks, the exact sequence $0\to N''\to N'\to N'/N''\to 0$ splits. So there is a submodule $E$ of $N'$ with $N'=N''\oplus E$. We have $E\cong N'/N''$, hence $E\subseteq M_b$. It follows that $N'\subseteq \bigoplus_{b'}M_{b'}=N$, a contradiction. Therefore, the claim follows.

Now claim all weights of $M_b\in b$ lies in a single coset.

If $M_b$ is indecomposable, by exercise 1.1 (b) in Humphreys' book, we are done.

If $M_b$ is decomposable, then $M_b=M_b^1\oplus\cdots\oplus M_b^k$, where $M_b^i$ is indecomposable. Note that $M_b^i\in b$ since $M_b\in b$.

Let $b=\mathcal{O}_\lambda$. By corollary 1.2., there is a composition series for $M_b^i$: $\{0\}\subseteq M_1^i\subseteq \cdots\subseteq M_b^i$. Then $M_1^i$ have a highest weight in the form $w_i\cdot\lambda$ with $w_i\in W_{[\lambda]}$. This implies $M_b^i$ contains a weight in the form $w_i\cdot\lambda$ with $w_i\in W_{[\lambda]}$. By exercise 1.1 (b), all weights of $M_b^i$ lie in $[w_i\cdot\lambda]$ for each $i$. Since $w_i\in W_{[\lambda]}$, we get $w_i\lambda-w_j\lambda\in\Lambda_r$ and then $[w_i\cdot\lambda]=[w_j\cdot\lambda]$ for any $i,j$. Hence all weight of $M_b$ lie in $[w\cdot\lambda]$ where $w\in W_{[\lambda]}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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