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I am following this link under "definitions"

I need to see why the suggested inner product on the pre-Hilbert space $H_1$ tensor $H_2$ is well defined. Recall that the fundamental tensors are a spanning set, but not a linearly independent one, hence not free in the category of vector spaces. How do I know that this definition is consistent no matter what representation of an element of the tensor space I give?

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Is invoking a Hilbert basis of pure tensors possible/cheating? –  anon May 27 '12 at 6:31
    
So I just realized that you could use an algebraic basis to show that this possibly inconsistent definition is consistent with one defined on such a basis. Another thing I only just now remembered is the tensor product has a universal property. I'm an analysis student, algebra is slower for me. But I think I got it. –  Jeff May 27 '12 at 6:33
    
Yes I think it'd have to be cheating because isn't it circular? What you can do is the same trick, but instead of saying Hilbert basis, say Hamel basis without orthogonality. –  Jeff Aug 4 '12 at 0:40
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up vote 3 down vote accepted

As was pointed out in a comment, probably the simplest way to handle the technical details is to do everything in terms of chosen orthonormal bases for $H_1$ and $H_2$ (to lessen the number of subscripts, I will write $H$ and $K$ instead of $H_1$ and $H_2$ from now on).

But there are other ways to do it, as there are other ways to handle the details of the purely algebraic vector space tensor product (which I will denote by $\otimes_a$, to distinguish it from the Hilbert space tensor product).

One construction of $H \otimes_a K$ is as a quotient $V/W$, where $V$ is the free vector space with basis $H \times K$ and the subspace $W$ the span of an explicit set of vectors (including, e.g. all vectors of the form $(h_1 + h_2, k) - (h_1, k) - (h_2, k)$ with $h_1$ and $h_2$ in $H$ and $k$ in $K$). In this realization, the "elementary tensor" $h \otimes k$ is simply the image of $(h,k)$ in the quotient $V/W$, and the problem of defining an inner product on $H \otimes_a K$ is a special case of the problem of defining a positive definite Hermitian sesquilinear form on a quotient vector space $V/W$ in terms of something on $V$. So one can do this as a special case of a more general algebraic problem.

Suppose $S: V \times V \to \mathbb{C}$ is a positive (not necessarily positive definite) Hermitian sesquilinear form on a complex vector space $V$. Define $$ N = \{x \in V: S(x,y) = 0 \text{ for all } y \in V\}. $$ Note that $N$ is obviously a subspace of $V$.

Fix any subspace $W$ of $V$, and let $q: V \to V/W$ denote the quotient map. It is easy to verify that

  1. Then there is a positive Hermitian sesquilinear form $S'$ on $V/W$ satisfying $$ S'(q(x), q(y)) = S(x,y), \qquad x, y \in V, $$ if and only if $W \subseteq N$.

  2. In the case that $W \subseteq N$, the form $S'$ on $V/W$ from part 1. is positive definite if and only if $W = N$.

[In proving 2. it helps to recall that the Cauchy-Schwarz inequality for $S$ (which requires only positivity, not definiteness, of $S$) gives the alternate description $N = \{x \in V: S(x,x) = 0\}$ of $N$.]

So we have a general program to follow in the application of interest.

We take $V$ to be the free vector space on $H \times K$ and $W$ the subspace spanned by the elements representing the relations one wants in the tensor product.

  • It is clear (from freeness) that the recipe $$ S((\xi_1,\eta_1), (\xi_2,\eta_2)) = \langle\xi_1,\xi_2\rangle_H \langle\eta_1,\eta_2\rangle_K, \qquad \xi_1, \xi_2 \in H, \quad \eta_1, \eta_2 \in K, $$ gives rise to a well-defined sesquilinear form on all of $V$.

  • To know that $S$ is positive on $V$, one must know that for all $n$ and all $\xi_1, \dots, \xi_n \in H$ and $\eta_1, \dots, \eta_n$ the number $$ \sum_{i,j=1}^n \langle\xi_i,\xi_j\rangle_H \langle\eta_i,\eta_j\rangle_K $$ is nonnegative. One way to do this is to recall that the "Gram matrix" $m = (\langle\xi_i,\xi_j\rangle_H)_{i,j=1}^n \in M_n(\mathbb{C})$ of any finite sequence of $n$ vectors in a Hilbert space is positive semidefinite. So there is $b = (b_{kl})_{k,l=1}^n \in M_n(\mathbb{C})$ with $b^* b = m$, and hence $$ \sum_{i,j=1}^n \langle\xi_i,\xi_j\rangle_H \langle\eta_i,\eta_j\rangle_K = \sum_{i,j,p=1}^n \overline{b_{pi}} b_{pj} \langle\eta_i, \eta_j\rangle_K = \sum_{p=1}^n \left\|\sum_{k=1}^n \overline{b_{pk}} \eta_k\right\|_K^2 \geq 0. $$

  • It clear that the subspace $W$ one quotients with to get $H \otimes_a K$ is contained in the space "$N$" of the form (just check that each generator of $W$ is in $N$, which is immediate from the sesquilinearity of the inner products on $H$ and $K$).

    So the recipe does give, for rather general reasons, a well defined positive sesquilinear form $S'$ on $H \otimes_a K$.

The detail that is not immediately clear in this realization is why the form $S'$ is definite--- ie, why if $x \in V$ satisfies $S(x,x) = 0$, then $x \in W$. There are several ways to verify this. One is to note that (since any fixed $x \in V$ is a finite sum $\sum_{j=1}^n (s_j, t_j)$) it suffices to prove this assertion for $H$ and $K$ finite dimensional. In this case, letting $h = \dim H$ and $k = \dim K$ one can use orthonormal bases of $H$ and $K$ to observe that the quotient $V/N$ contains $hk$ mutually orthogonal vectors and hence has dimension at least $hk$. Since for purely linear algebraic reasons one knows that $V/W = H \otimes_a K$ has dimension $hk$, it follows that $\dim(N) \leq \dim(W)$, and hence $N = W$ as desired. (I think you can also tweak the above proof of positivity of $S$ into a proof of definiteness, if you understand the matrix algebraic consequences of an element of $V$ being $0$ in the tensor product.)

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Thank you for your response. I'm having trouble accepting the answer because of some account difficulties. Also, I should note that I am okay with the solution that was sort of being suggested in the comments, but that it's not correct logically as far as I can see. You can't talk about what an orthonormal basis on the tensor product is, neither algebraically in the algebraic tensor product (because it may not exist) nor Hilbert in the completion of the tensor product (because the inner product is as yet undefined.) What you can do is use a definition on an algebraic general basis. –  Jeff Aug 4 '12 at 0:36
    
I am a fan of the answers you've given to my questions. I haven't responded to two of them because of these account troubles (I apparently have multiple accounts from before I was registered), and also they're completely clear so no further questions have come to mind yet. –  Jeff Aug 4 '12 at 0:39
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