# Inner product on the tensor product of Hilbert spaces

Let $H_1$ and $H_2$ be Hilbert spaces with inner products $\langle\cdot,\cdot\rangle_1$ and $\langle\cdot,\cdot\rangle_2$, respectively. Then $H_1\otimes H_2$ is at least a pre-Hilbert space (we are not concerned with completeness here).
On this tensor product we can define an inner product by setting $$\langle v_1\otimes v_2,w_1\otimes w_2\rangle:=\langle v_1,w_1\rangle_1 \langle v_2,w_2\rangle_2$$ for $v_1,w_1\in H_1,\;v_2,w_2\in H_2$. This is how all textbooks define it.

However, I do not understand why this product is bilinear. To prove bilinearity, we would consider something like $$\langle \lambda u_1\otimes u_2+ v_1\otimes v_2,w_1\otimes w_2\rangle,$$ but in general it is not possible to write the sum $\lambda u_1\otimes u_2+ v_1\otimes v_2$ as one pure tensor product. Therefore it would not be possible to write this in the form of the definition which formally does not permit sums of tensor products.

In the literature on this topic, the above definition is the only information on the new inner product, so my question is: Are the textbooks missing the fact that bilinearity does not follow from the above definition (i.e. bilinearity should be part of the definition in order to get an inner product) or did I miss something and bilinearity is already clear from the definition?

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It extends to a sesquilinear form on the free vector space with basis $H_1 \times H_2$ and factors through the relations defining $H_1 \otimes H_2$. In other words, it is defined on pure products but extends to $H_1 \otimes H_2$. – WimC Jun 2 '12 at 9:04

The map $b:(v_1,v_2,w_1,w_2)\in (H_1\times H_2)\times (H_1\times H_2)\to \langle v_1,w_1\rangle_1\langle v_2,w_2\rangle_2\in\mathbb R$ is multinear.
So it factors through $$\begin{array}{cc}\otimes\times\otimes:&(H_1\times H_2)\times (H_1\times H_2)\to (H_1\otimes H_2)\times (H_1\otimes H_2)\\&(v_1,v_2,w_1,w_2)\mapsto(v_1\otimes v_2,w_1\otimes w_2)\end{array}$$ and a bilinear form $\langle\cdot,\cdot\rangle:(H_1\otimes H_2)\times (H_1\otimes H_2)\to\mathbb R.$
The latter is precisely your product.

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The statement you cited does provide enough information to completely determine the function $\langle \cdot, \cdot \rangle$. Sesquilinearity follows from the clause stating that it's an inner product, and any sesquilinear map is completely determined by its values on pairs of basis elements.

The challenge isn't to prove it's sesquilinear, but to prove it's well-defined.

In general, definitions aren't of the form

<thing to be defined> := <thing it is defined as>


it is rather more common to implicitly define it by providing sufficient information to uniquely determine it. It may help to digest this fact by noting that defining the function $f$ by

$$f(x) = x^2 + 2x + 1$$

is actually an implicit definition for $f$ -- you aren't defining $f$ directly, but instead you are providing an equation involving the evaluation operator applied to $f$ (and $x$).

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I see. In my particular case, what justifies the extension of the inner product on pure products to an inner product on all tensor products (i.e. sums of them)? – Deniz Jun 2 '12 at 10:43
The prose that says that you are defining "an inner product", and the fact that if you choose a basis for $H_1$ and $H_2$ and the corresponding basis for $H_1 \otimes H_2$, that the formula provides values for each pair of basis elements. As I said, the real challenge is to prove the sesquilinear form defined by that data is consistent with all of the other values claimed by the formula (but that's an easy calculation). – Hurkyl Jun 2 '12 at 10:50
Thanks, please note that at this point I cannot vote up, but will do so when I have enough reputation points. – Deniz Jun 2 '12 at 10:52
Now, if I see it correctly, extending the inner product to the whole space $H_1\otimes H_2$ is justified by the universal property of tensor products, right? – Deniz Jun 2 '12 at 11:20