I have a question regarding highest-weight modules:

Let be $\mathfrak{g}$ a Lie algebra, $\mathfrak{b}$ a Borel subalgebra, $\mathfrak{h}$ a Cartan subalgebra and $U(\mathfrak{g})$ its universal enveloping algebra. Let be $M$ a $\mathfrak{b}$-module and $\text{Ind}^{\mathfrak{g}}_{\mathfrak{b}} M = U(\mathfrak{g}) \otimes_{U(\mathfrak{b})} M$ the induced module on $\mathfrak{g}$.

My question is: If $M=M(\lambda)$ is a highest-weight module to the weight $\lambda \in \mathfrak{h}^*$, is then $\text{Ind}^{\mathfrak{g}}_{\mathfrak{b}} M(\lambda)$ a highest-weight module with weight $\lambda$, too?

An example would be Verma modules, but is this correct in general? If yes, how would one proof it?

Thank you! Andreas


The answer to our question is "no, but in a sense, almost yes".

In general, if you take a highest weight $\mathfrak{b}$-module $M(\lambda)$ (which is one dimensional), and take the induced $\mathfrak{g}$-module $M'(\lambda)$, then $M'(\lambda)$ is not necessarily irreducible but has a unique simple quotient, which you can prove using the fact that $M'(\lambda)$ is a weight module in the following manner:

By the Poincare-Birkhoff-Witt Theorem, if $\mathfrak{n}$ is the lower nilpotent corresponding to $\mathfrak{b}$, then $$M'(\lambda) \cong U(\mathfrak{n}) \otimes M(\lambda).$$

So, if we take a proper submodule of $M'(\lambda)$, since it has to be a weight submodule, it cannot contain $1 \otimes v$ for $v \not = 0$ as this generates all of $M'(\lambda)$. This property is preserved by taking sums of proper submodules and hence the sum of all proper submodules is itself a proper submodule and is hence the unique maximal submodule of $M'(\lambda).$ Taking a quotient then gives us the unique simple submodule $L(\lambda).$

It is not, however, the case that $L(\lambda) = M'(\lambda)$. In fact, if $\lambda$ is dominant, then $L(\lambda)$ is finite dimensional but $M'(\lambda)$ is not. So, in such a case, $M'(\lambda)$ is not a highest-weight module because it is not irreducible.

It should be noted that in all cases (that is, no matter what $\lambda$ is), $M'(\lambda)$ does contain the highest-weight module with weight $\lambda$ as a submodule.


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