# Contravariant forms and tensor products

In Humphreys's book "Representations of semi-simple Lie algebras in the BGG-category O", section 3.14 deals with contravariant forms on highest weight modules. I wanted to define a map by a bilinear form and came across the following problem.

Let $$V$$ and $$W$$ be two irreducible $$\mathfrak{g}$$-modules, where $$\mathfrak{g}$$ is a semi-simple Lie algebra.

Then there exists a unique (up to scalar) contravariant non-degenerate form on $$V$$ and $$W$$ giving rise to a contravariant non-degenerate form on $$V \otimes W$$.

On the other hand, decomposing $$V\otimes W$$ into irreducible $$\mathfrak{g}$$-modules, we also have contravariant non-degenerate forms on each summand, yielding a contravariant non-degenerate form on the sum.

Is there any chance, that these two bilinear forms on $$V \otimes W$$ are the same?

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$$\textbf{Edit:}$$ Here is my special case of the above question, which may be easier to understand. Let $$\mathfrak{g}$$ be a simple Lie algebra and let $$\theta$$ be the highest root, such that $$\mathfrak{g} = V(\theta)$$ as irreducible $$\mathfrak{g}$$-modules.

$$V(k\theta)$$ is a submodule of $$\mathfrak{g}^{\otimes k}$$ by sending the highest weight vector to $$e_\theta^{\otimes k}$$. Hence also $$V(k\theta) \otimes V(l\theta)$$ can be viewed as a submodule of $$\mathfrak{g}^{\otimes k+l}$$. $$V(k\theta) \otimes V(l\theta)$$ admits a contraviariant bilinar form $$\beta$$ coming from the forms on $$V(k\theta)$$ and $$V(l\theta)$$. I would like to show, that the form on $$\mathfrak{g}^{\otimes k+l}$$ coming from the Killing form restricts to $$\beta$$ on $$V(k\theta) \otimes V(l\theta)$$.

• These forms are all defined only up to scalar factor. Now, $V \otimes W$ will usually split into a sum of multiple irreducible modules, and there is a choice of scalar for each addend. It gets worse when there are multiplicities, because then even the choice of decomposition becomes non-canonical. When you say "the two forms are the same", are you saying "there is a way to make all these choices so that the two forms are the same"? – darij grinberg Jan 18 at 11:20
• Yes, indeed. I think, that starting with a bilinear form on $V \otimes W$, one can decompose $V \otimes W$ into irreducible submodules and obtain bilinear forms on the summands. They are unique (up to scalar), so they can be chosen to coincide with the initial bilinear forms on the summands. The problem is, that this only works for the decomposition induced by the bilinear form on $V \otimes W$, but not on any decomposition into irreducible, as you pointed out. I realised, that may initial problem may be easier to solve, therefore I will edit the initial question in a couple of minutes. – dura10 Jan 21 at 10:46