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Shapovalov and Jantzen showed us how to construct a nice inner product on finite dimensional representations of a semi-simple Lie algebra, by simply giving the highest weight vector inner product 1 with itself and making the upper and lower halves adjoint.

The result I need is an extension of this to tensor products. Roughly, I would like a statement like:

There is a unique system of $U_q(\mathfrak{g})$-invariant Hermitian inner products on all tensor products $V_{\lambda_1}\otimes \cdots \otimes V_{\lambda_\ell}$ such that

  1. On $V_\lambda$, it is the Shapovalov form.
  2. The action of $E_i$ and $F_i$ are biadjoint (up to some powers of $q$).
  3. For any $j<\ell$, the natural map $V_{\lambda_1}\otimes\cdots\otimes V_{\lambda_j}\hookrightarrow V_{\lambda_1}\otimes \cdots \otimes V_{\lambda_\ell}$ is an isometric embedding.

I said to myself, "Self, it would be silly to post this question on MathOverflow. You are in a math library, just feet away from Lusztig's book. Surely it is in there." However, I've had no luck finding it in Lusztig's book, which is sadly lacking in index. Is this actually written down anywhere?

EDIT: Jim asks for more motivation. I feel like this is the sort of question where motivation will not be very helpful in actually finding an answer, but there's no harm in saying a little (and it will allow me to put off real work).

One of the foundational principles of categorification is that things with nice categorifications have nice inner products (since Grothendieck groups have a nice inner product given by Euler characteristic of the Ext's between objects). I'm working right now on categorifying tensor products of representations, so it would be rather convenient for me to find some earlier references that used this form.

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More context and motivation would help here. You are probably far away from the original motivation of both Shapovalov and Jantzen: find a nonzero symmetric bilinear form on a highest weight module for a semisimple Lie algebra so distinct weight spaces are orthogonal and the radical of the form is the unique maximal submodule. Thus the form is nondegenerate on the simple quotient module. Jantzen worked over the integers in order to study the behavior of f.d. "Weyl modules" mod $p$. Both of them found a miraculous determinant formula on each weight space. –  Jim Humphreys Mar 22 '10 at 21:57
    
Edit request: "... a nice inner product on finite dimensional representations of a SEMISIMPLE Lie algebra ..." At least, I can't imagine interpreting the rest of the post without this extra word. –  Theo Johnson-Freyd Mar 23 '10 at 2:12
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2 Answers

up vote 2 down vote accepted

I know a couple of ways to get a Shapovalov type form on a tensor product. The details of what I say depends on the exact conventions you use for quantum groups. I will follow Chari and Pressley's book.

The first method is to alter the adjoint slightly. If you choose a * involution that is also a coalgebra automorphism, you can just take the form on a tensor product to be the product of the form on each factor, and the result is contravariant with respect to *. There is a unique such involution up to some fairly trivial modifications (like multiplying $E_i$ by $z$ and $F_i$ by $z^{-1}$). It is given by: $$ *E_i = F_i K_i, \quad *F_i=K_i^{-1}E_i, \quad *K_i=K_i, $$ The resulting forms are Hermitian if $q$ is taken to be real, and will certainly satisfy your conditions 1) ad 3). Since the $K_i$s only act on weight vectors as powers of $q$, it almost satisfies 2).

The second method is in case you really want * to interchange $E_i$ with exactly $F_i$. This is roughly contained in this http://www.ams.org/mathscinet-getitem?mr=1470857 paper by Wenzl, which I actually originally looked at when it was suggested in an answer to one of your previous questions.

It is absolutely essential that a * involution be an algebra-antiautomorphism. However, if it is a coalgebra anti-automorphism instead of a coalgebra automorphism there is a work around to get a form on a tensor product. There is again an essentially unique such involution, given by

$$ *E_i=F_i, \quad *F_i=E_i, \quad *K_i=K_i^{-1}, \quad *q=q^{-1}. $$

Note that $q$ is inverted, so for this form one should think of $q$ as being a complex number of the unit circle. By the same argument as you use to get the Shapovalov form, then is a unique sesquilinear *-contravariant form on each irreducible representation $V_\lambda$, up to overall rescaling.

To get a form on $V_\lambda \otimes V_\mu$, one should define $$(v_1 \otimes w_1, v_2 \otimes w_2)$$ to be the product of the form on each factor applied to $v_1 \otimes w_1$ and $R( v_2 \otimes w_2)$, where $R$ is the universal $R$ matrix. It is then straightforward to see that the result is *-contravariant, using the fact that $R \Delta(a) R^{-1} =\Delta^{op}(a).$

If you want to work with a larger tensor product, I believe you replace $R$ by the unique endomorphism $E$ on $\otimes_k V_{\lambda_k}$ such that $w_0 \circ E$ is the braid group element $T_{w_0}$ which reverses the order of the tensor factors, using the minimal possible number of positive crossings. Here $w_0$ is the symmetric group element that reverses the order of the the tensor factors.

The resulting form is *-contravariant, but is not Hermitian. In Wenzl's paper he discusses how to fix this.

Now 1) and 2) on your wish list hold. As for 3): It is clear from standard formulas for the $R$-matrix (e.g. Chari-Pressley Theorem 8.3.9) that $R$ acts on a vector of the form $b_\lambda \otimes c \in V_\lambda \otimes V_\mu$ as multiplication by $q^{(\lambda, wt(c))}$. Thus if you embed $V_\mu$ into $V_\lambda \otimes V_\mu$ as $w \rightarrow b_\lambda \otimes w$, the result is isometric up to an overall scaling by a power of $q$. This extends to the type of embedding you want (up to scaling by powers of $q$), only with the order reversed. I don't seem to understand what happen when you embed $V_\lambda$ is $V_\lambda \otimes V_\mu$, which confuses me, and I don't see your exact embeddings.

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I have come across this for the tensor products of two highest weight modules but not more.

In this case start with the tensor product of a highest weight vector with a lowest weight vector and proceed as before.

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