qa.quantum algebra - Reference for the existence of a Shapovalov-type form on the tensor product of integrable modules

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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