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Let $H$ be a subgroup of $G$ with Lie algebras $\mathfrak{h}$ and $\mathfrak{g}$ respectively. If I have 2 representations $V, W$ of $\mathfrak{h}$ equipped with a $\mathfrak{h}$ invariant inner product. I can use a generalized Verma module to induce up to representations on $\mathfrak{g}$ (this is from Kac's book "Infinite dimentional Lie Algebras").

$$ M_\mathfrak{g} (V) = \mathfrak{U}({\mathfrak{g}})\otimes_{\mathfrak{U}({\mathfrak{b}}_- + \mathfrak{h})\ }V$$

where $\mathfrak{b}_-$ is the negative Borel sub-algebra with $\mathfrak{b}_-/ \mathfrak{h}$ acting trivially on $V$. This has a maximal subalgebra $m_V$ so that $ M_\mathfrak{g} (V)/m_V$ is an $\mathfrak{g}$ representation. My question is there an induced inner product on $ M_\mathfrak{g} (V)/m_V$ that is $\mathfrak{g}$ invariant and if so then is this construction natural with respect to isometries?

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  • $\begingroup$ I am not familiar with your notation ($\mathfrak{b}_-$?) and it seems to me that $m_V$ should be a submodule not a subalgebra. But isn't "Shapovalov form" the thing you are looking for? $\endgroup$ Nov 14, 2013 at 23:09
  • $\begingroup$ I think that you're a bit confused about the underlying ideas here; what on earth does it mean that $\mathfrak{b}_-+\mathfrak{h}$ acts with $\mathfrak{b}_/\mathfrak{h}$ acting trivially. It doesn't make sense to ask a quotient to act trivially, and if $\mathfrak{b}_-+\mathfrak{h}$ is a subalgebra, it's a parabolic, and if $\mathfrak{b}_-$ acts trivially, the module must be trivial. $\endgroup$
    – Ben Webster
    Nov 15, 2013 at 4:44
  • $\begingroup$ Sorry, I was wording this too generally and too quickly. What I wrote admittingly does not make sense. I wantthe vectors in $\mathfrak{h}$ to act as normal and the vectors in $\mathfrak{b}_-$ that don't occor in $\mathfrak{h}$ to act trivially. I just need this for loop groups, where $H$ is the subgroup of constant loops and $G$ is the affine loop group. $\endgroup$ Nov 20, 2013 at 20:39
  • $\begingroup$ This is not true, btw. I have a counterexample but it's quite long and involved. $\endgroup$ Jul 9, 2014 at 16:41

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