# Inner product on $V_{-\rho}$?

Prologue. To $M^n$ a compact real manifold with frame bundle $F$ (a principal $GL_n$ bundle), we associate a line bundle using the representation $M\mapsto \sqrt{|\det M|}$, the bundle of half-densities. Then the space of sections of this has a canonical inner product (multiply then integrate), invariant under all diffeomorphisms of $M$. If $M$ is oriented then one can use the representation $M\mapsto \sqrt{\det M}$, the "half-form" bundle.

Present day. I was thinking about "parabolic category $\mathcal O$" and realized that the most parabolic is the one containing (only) the Verma module $L(-\rho)$ with highest weight $-\rho$. The corresponding Borel-Weil line bundle is the half-form bundle, so its space of sections carries a canonical inner product... except there are no holomorphic ones. Or indeed any sheaf cohomology.

What special structures do the very special representation $L(-\rho)$ bear? Is there any shadow of the inner product that the finite-dimensional irrep would have, if it existed?

EDIT: "If it existed" was a lousy way to talk about that finite-dimensional irrep. As Ben points out, Borel-Weil-Bott makes clear that it does exist but is $0$.

• This representation is isomorphic to its contragredient. So being irreducible, it has a nondegenerate invariant form. Over the reals, you can ask whether this form is definite. I think that for $\frak{sl}_2$ the answer is affirmative, by an explicit calculation. – Victor Protsak Dec 1 '16 at 6:02
• Isn't its contragredient a lowest-weight, not highest-weight, representation? – Allen Knutson Dec 2 '16 at 15:51
• @Allen: Starting with a semisimple Lie algebra over $\mathbb{C}$, note that a Verma module with an integral highest weight is irreducible just when the weight is "antidominant"; so $M(-\rho) = L(-\rho)$ is a basic example. Here the most standard symmetric bilinear form is the contravariant form, which is nondegenerate precisely in such cases (Jantzen, Shapovalov). But all of this is over $\mathbb{C}$, so it probably doesn't fit your situation. [P.S. The first question in your highlighted part lacks a word at the end.] – Jim Humphreys Dec 2 '16 at 17:42
• @AllenKnutson The obvious one, yes. I think in the context of category O, people often use "contragredient" to mean the obvious dual twisted by the Cartan involution, so you end up back in category O. – Ben Webster Dec 3 '16 at 1:40
• Allen, @Ben: Right, contragredient preserves the highest weight property, whereas the dual maps highest weight modules to lowest weight, and vice-versa. – Victor Protsak Dec 3 '16 at 2:25

I'm not totally sure what will satisfy you on this account. Of course you pointedly restricted yourself to the reals, but some complex manifolds do have holomorphic half-form bundles, and the flag manifold is one of them. Of course, the half-form line bundle $\omega_X^{1/2}$ "should" have sections $V(-\rho)$ (actually, this should be the cohomology in all degrees) but in order to make sense of this you have to declare that $V(-\rho)=0$. It's still true that multiplying two sections of this bundle gives a (now holomorphic) top form, but there just are no global sections.
One manifestation of this that is of interest to geometric representation theorists is that the ring and sheaf $D_{\omega^{1/2}}$ of holomorphic differential operators twisted in half-forms on any complex manifold (which make sense even if half-forms are not a well-defined line bundle) is self-opposite (the sheaf never is for other twists, sometimes the ring is various global reasons). This makes it often the most natural twist to work with, and it plays a important role in D-module theory.