MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.)

share|cite|improve this question
[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
up vote 2 down vote accepted

A rather satisfactory treatment (at least in my opinion) can be found in Greub's book "Multilinear Algebra".

In the copy I have (the 1967 edition) the nondegeneracy of bilinear forms on tensor products is considered in Chapter I, Section 7, Subsection 1.22 (p. 31).

The arguments involved are index-free and work in any characteristic.

share|cite|improve this answer
I'll check this reference, but meanwhile I should repeat my comment to Allen about wanting to include infinite dimensional vector spaces. (This comes up naturally in a representation theory context.) I'm not familiar enough with Greub's book to see how general he gets, but the reference is probably the right one for the classical case. – Jim Humphreys May 24 '11 at 14:42
P.S. Both the original Springer edition and the expanded Universitext edition are technically out of print, so hang onto the copy you own. One or two copies of the older edition (but not the newer one) seem to be in Five College libraries, which I will visit in a day or so. – Jim Humphreys May 24 '11 at 17:23
Thanks for this reference, which I wasn't at all familiar with. The advantage is that Greub takes full advantage of universal properties and avoids computations with bases, treating vector spaces of arbitary dimension over any field. The main disadvantage is that the argument for non-degeneracy is fairly long and formal, with some reliance on his earlier book on linear algebra. I've seen the principle applied without further comment and was therefore motivated to check the details concretely, which seems to be straightforward but is unpleasant. – Jim Humphreys May 26 '11 at 17:30
You are welcome. You are right, Greub's book is unfortunately out of print. However, it is possible to buy a (used) copy on or spending less than 50 dollars – Francesco Polizzi May 26 '11 at 22:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.