# Contraction of the maximal submodule in a Verma module

Suppose $\mathfrak{g}$ is a real semisimple Lie algebra, $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p}$ is a Cartan decomposition, and $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{k}$. Assume $\mathfrak{g}$ and $\mathfrak{k}$ have equal ranks, so that $\mathfrak{h}$ is also a Cartan subalgebra of $\mathfrak{g}$.

Fix $\lambda$ in $\mathfrak{h}^\star$, write $M(\lambda)$ for the Verma module of highest weight $\lambda$, and $N(\lambda)$ for its maximal proper submodule. Suppose $\lambda$ is such that the quotient is finite-dimensional.

Now, if $t$ is a nonzero real number, define $\varphi_{t} \in GL(\mathfrak{g})$ by $\varphi_{t}(X_{\mathfrak{k}}+X_{\mathfrak{p}}) = X_{\mathfrak{k}}+ t \cdot X_{\mathfrak{p}}$ when $X_{\mathfrak{k}}$ is in $\mathfrak{k}$ and $X_{\mathfrak{p}}$ in $\mathfrak{p}$. Going through the definitions, $\varphi_{t}$ induces a linear map, say $\tilde{\varphi}_{t}$, from $M(\lambda)$ to itself.

Could it be that $N(\lambda)$ is $\tilde{\varphi}_{t}$-stable ?

For $\mathfrak{g} = \mathfrak{sl}(2, \mathbb{R})$ it is if I'm not mistaken, but I do not understand the general situation at all.

(For $\mathfrak{sl}(2, \mathbb{R})$, because there is only one root, $\tilde{\varphi}_{t}$ turns out to act as a multiple of the identity on each weight subspace $N(\lambda)_{\mu}$ ; but in general when a product of negative root vectors is in $N(\lambda)_{\mu}$ it's only the parity of the number of noncompact roots, with multiplicity, which is determined by $\mu$, so this argument does not generalize).

Remark : the question might seem mysterious, so here is the context. Define a Lie algebra $\mathfrak{g}_{t}$ (with underlying vector space $\mathfrak{g}$) by transferring the structure through $\varphi_{t}$. Then $F_{t} = M(\lambda) / \tilde{\varphi}_{t}(N(\lambda))$ carries an irreducible, finite-dimensional representation of $\mathfrak{g}_{t, \mathbb{C}}$, and I'd like to understand how things behave as $t$ goes to zero. If the answer to my question is yes, then $\tilde{\varphi}_{t}$ defines an "intertwining" map from $M(\lambda)/N(\lambda)$ to itself, $\tilde{\varphi}_{t} v$ has a limit for each $v$ as $t$ goes to zero, and the space of limits thus obtained should turn out to be an irreducible $\mathfrak{k}_{\mathbb{C}}$-module with highest weight $\lambda$. If the answer to my question is no, I would welcome any help on giving a meaning to "the limit of the $F_{t}$s" which would preserve the conclusion.

Thanks !

• Could you explain in more detail how you are supposed to extend this map to the Verma module? I don't believe you should be able to even extend it to the universal enveloping algebra since $\mathfrak p$ isn't a Lie subalgebra. – Ben Webster Apr 30 '15 at 11:58
• Thanks for your question, I should have been clearer. Once you choose a basis of $\mathfrak{g}$ and order it, you can extend $\varphi_t$ to the universal enveloping algebra by saying how it acts on the associated (PBW) basis of the enveloping algebra. There is a natural basis here with vectors in the root spaces or in $\mathfrak{h}$, and a natural ordering once a system of positive roots (and an ordering on $\mathfrak{h}$) has been chosen. If a basis element has $p$ noncompact roots counting multiplicities, decide $\varphi_t$ multiplies it with $t^p$. – Alexandre Afgoustidis Apr 30 '15 at 17:37
• This extension to the enveloping algebra turns out to descend to $M(\lambda)$, in a manner compatible with the vector space isomorphism between U(sum of negative root spaces) and $M(\lambda)$. So the extension can be defined directly through that isomorphism and the natural PBW basis of U(sum of negative root spaces). Of course it does not respect the $\mathfrak{g}\u{\mathbb{C}}$ action, otherwise my question would be really stupid ! – Alexandre Afgoustidis Apr 30 '15 at 17:42