# Optimal reference for tensor product of symmetric bilinear forms?

This is just a reference request on a relatively elementary level (for which I apologize in advance), but every time I bump into this question I suspect I'm missing the "correct" conceptual setting. In the simplest case, one is given two vector spaces $V, W$ over a field of characteristic 0, each endowed with a symmetric bilinear form. Then the tensor product $V \otimes W$ inherits an obvious symmetric bilinear form. A natural result is that nondegeneracy of the given forms implies nondegeneracy of the new form, though the proof seems to require somewhat messy manipulation of bases and indices. Even if the vector spaces are infinite dimensional, the same principle seems valid. Then there is the possibility of working over a field of prime characteristic, as well as passing to free modules over commutative rings, etc.

Where in the textbook literature can one find the most definitive treatment of nondegeneracy of symmetric bilinear forms on tensor products? (Preferably with few indices to keep track of.)

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[An index-free argument.] Lemma: if $A\to B$ is injective, then $A\otimes C \to B\otimes C$ is too. Think of a bilinear form on $V$ as a map $V \to V^\ast$. Proof of desired theorem: compose $V \otimes W \to V^\ast \otimes W \to V^\ast \otimes W^\ast$ to get an injective map $V\otimes W \to V^\ast \otimes W^\ast$, the desired bilinear forms. The only space I see finite-dimensionality being relevant is to know the 1:1 maps are also onto ("strong nondegeneracy"). –  Allen Knutson May 24 '11 at 0:43
@Allen: This makes good sense, up to a point, but I'm especially concerned with treatment of infinite dimensional vector spaces where it gets risky to bring in dual spaces this way (?) –  Jim Humphreys May 24 '11 at 14:40
In the infinite-dimensional case, are you working with topological vector spaces and continuous (in some topology) bilinear forms? Or just purely algebraic vector spaces? –  Marty May 24 '11 at 16:16
@Marty: All of this is in principle purely algebraic (from my point of view). –  Jim Humphreys May 24 '11 at 17:18
I really don't see the risk here. From an inner product $\langle,\rangle$, define $\phi:V\to V^*$ by $\vec v \mapsto \langle \vec v,\bullet \rangle$. In general (and without Choice), I am ready to believe that e.g. $V^*$ might be just $0$, but then $V$ isn't going to have an inner product, either. I'm not using that the natural map $V\to (V^*)^*$ is 1:1 anywhere. –  Allen Knutson May 25 '11 at 2:36