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Let $L$ a Lie Algebra. I need prove that that every Verma module $\Delta(\lambda)$ admits a composition series, i.e a series of submodules with simple factors. I found a proof that is quite short in this scripts, at Proposition 5.5. At the end of the proof, when builds a concrete series, the term $M_i$ is given and is considered $M_i$ a maximal sub-module of $M$. I am not too much sure how it is obvious that such a maximal sub-module exists.

I thought to two ways to proceed, the first is to consider that $M$ is a sub-module of a Verma module, that the weights of $M$ are bounded by above and deduce (in someway, if it is true) that there is a finite number of maximal weights of $M$, say $\mu_1 \dots \mu_n $. Then since $\text{dim}$ $ M_{\mu_i} < \infty$ we can choose a basis $B_i$ of $M_{\mu_i}$. If $M$ is generated by all $v \in B_i$ for all $i$, then we can conclude using the theory of finitely generated modules (in particular, this question).

The second way is to consider a maximal weight $\mu$ of $M$ and $v \in M_\mu$, then try to prove that $M'=\sum N $, where the sum is over all the submodules of $M'$ that don't contain $v$, is maximal in $M$.

Then, how can I prove that the factors (and their occurrence) are independent on the choice of the Jordan-Holder chain?

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  • $\begingroup$ Ok, I just realized that the first way doesn't work. If M is the maximal submodule of a verma module, then sometimes is not possible to cover it with the sum of the verma modules that it contains, as J. Humphreys explained on math overflow. mathoverflow.net/questions/150803/… $\endgroup$ – Lorban Feb 21 '16 at 10:39
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Ok, I found a proof of the fact that every submodule of highest weight module have a maximal proper submodule. According to this question on math overflow, the universal enveloping algebra $U(L)$ of a lie algebra $L$ is noetherian. Now an $L$-module $M$ that is a highest weight module is a finitely generated $U(L)$-module (actually is $1$-generated by a maximal vector). Recall

If $R$ is a noetherian ring and $M$ a finitely generated left $R$-module, then $M$ is a noetherian module

The proof of that can be found in "Basic Algebra II" of Jacobson, Theorem 3.4. Then a highest weight module $M$ is a noetherian module.

Now remains to prove that a noetherian module $M$ has a maximal proper submodule. Consider $\mathcal{S}$ the set of proper submodules of $M$, that is non empty since contains the zero module. Suppose that a maximal submodule doesn't exists, so for each $N \in\mathcal{S}$ we can define the non empty set

\begin{gather} \mathcal{S}(N):=\{L \in \mathcal{S} \colon N < L\} \end{gather} By the choice axiom for each $N \in \mathcal{S}$ we can choose a $L_N \in \mathcal{S}(N)$, then the mapping $N \to L_N$ define a map $f$ from $\mathcal{S}$ to itself such that $f(N)>N$ for each $N \in \mathcal{S}$. Pick $N_1 \in \mathcal{S}$, then define $N_2=f(N_1)>N_1$, $N_3=f(N_2)>N_2$ and so on. Then $N_1<N_2<N_3 \dots $ is an infinite ascending chain in $M$ noetherian, that is impossible. This is inspired to Isaacs, Algebra - A graduate course, lemma 11.2.

Take a highest weight module $M$, then take $M_1$ the unique maximal proper submodule, since $M$ in noetherian, $M_1$ it is. Then by above $M_1$ contains a maximal proper submodule $M_2$, still noetherian. Iterating this process can build then a chain $M=M_0>M_1>M_2 \dots$ of submodules with simple factors. For ending the proof that a highest weight module admits a composition sequence I should prove that the sequence $M_i$ is bounded from above, but I have no idea at the moment.

EDIT:ok, I'm writing to myself and seems that nobody have considered this question. Nevertheless I think that an answer can be useful for someone in future, so I write how I found out a complete (I hope) solution.

Suppose that a highest weight module $M$ has a descending series $M=M_0>M_1>M_2> \dots$ with simple factors. From the general theory of highest weight modules, a simple factor $M_i/M_{i+1}$ must be isomorphic to $L(\mu)$, i.e. the Verma module $\Delta(\mu)$ over its maximal (and unique) submodule. Studying the value of the Casimir operator, we conclude that the number of $\mu$ such that $L(\mu)$ appears as simple factor in the series above is finite.

Suppose now that the series $M=M_0>M_1>M_2> \ldots$ is infinite, by the considerations above so there exists a $\mu$ such that $L(\mu)$ appears infinitely many times. Then there exists a subsequence $M_{i_j}$ of $M_i$ such that $M_{i_j}/M_{i_j+1}$ is isomorphic to $L(\mu)$. For each $j \in \mathbb{N}$ choose $v_j \in (M_{i_j})_{\mu}$ such that $v_j$ projects to a generator on the factor.. The set $\{v_1,v_2,v_3 \dots\}$ is and infinite set of independent eigenvectors of the same eigenvalue $\mu$, that is impossible since the eigenspaces of $M$ are finite dimensional.

There is a much quicker and easier way that is sketched in the amazing book "Introduction of Lie Algebras and Representation Theory" of J. Humphreys. It is in the appendix at the end of chapter 24. This appendix was added in the later editions, so if someone wants to read it should take care to get a enough recent edition of the book.

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