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.



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.


