# Identification of products in the tensor product of hilbert spaces

I'm relatively new to tensor products and particularly interested in tensor products of Hilbert spaces. I did read a short note on the explicit construction of theses spaces, which is also covered on Wikipedia (link below). However, I have some general question and hope someone can bring some light into my confusion.

Following the wikipedia article one can construct the tensor product of Hilbert spaces $$H_1$$ and $$H_2$$ as the space which is isometrically and linearly isomorphic to $$HS(H_{1}^{*},H_{2})$$, the space of Hilber-Schmidt operators from $$H_1^*$$ to $$H_2$$.

The idea is to identify to every tensor $$x_1\otimes x_2$$ with $$x_i\in H_i$$ and $$x^*\in H_1^*$$ the map

$$x^*\mapsto x^*(x_1)x_2$$

On the other hand one often defines the tensor product via a new inner product. For this the tensor product (in algebraic way) is first constructed. Then an inner product is defined. I quote Wikipedia:

Construct the tensor product of $$H_1$$ and $$H_2$$ as vector spaces as explained in the article on tensor products. We can turn this vector space tensor product into an inner product space by defining $${\displaystyle \langle \phi _{1}\otimes \phi _{2},\psi _{1}\otimes \psi _{2}\rangle =\langle \phi _{1},\psi _{1}\rangle _{1}\,\langle \phi _{2},\psi _{2}\rangle _{2}\quad {\mbox{for all }}\phi _{1},\psi _{1}\in H_{1}{\mbox{ and }}\phi _{2},\psi _{2}\in H_{2}}$$

and extending by linearity. That this inner product is the natural one is justified by the identification of scalar-valued bilinear maps on $$H_1 \times H_2$$ and linear functionals on their vector space tensor product. Finally, take the completion under this inner product. The resulting Hilbert space is the tensor product of $$H_1$$ and $$H_2$$.

question 1 In the last definition, if I read it correctly, we build first the tensor product in a algebraic way. Then an inner product is on that tensor product defined. Finally the completion is taken and one sees that this corresponds with the original (algebraic) tensor product? Or how does the algebraic and the one stemming from the completion related?

question 2 I often see that for Hilbert spaces, say of functions, tensor products are defined via multiplication. That is for functions $$f_i\in H_i$$ we have $$f_1\otimes f_2:=f_1(x)f_2(x)$$. How do we get this concrete example of "multiplication". I'm particular interested in Hilbert spaces of scalar valued functions, if that matters.

question 3. What is the relation of the explicit construction via Hilber-Schmidt operators to the multiplicative case mentioned in question 2.

answer 1: I am not exactly sure, what you mean by the "original (algebraic)" tensor product. Usually, for two Hilbert spaces $$H_1, H_2$$ the algebraic tensor product $$H_1 \otimes_{alg} H_2$$ is the space of all linear combinations of vectors $$\phi_1 \otimes \phi_2$$. As explained in the Wikipedia article, you define an inner product $$\langle \cdot , \cdot \rangle$$ on it by $$\langle \phi_1 \otimes \phi_2, \psi_1 \otimes \psi_2 \rangle := \langle \phi_1 , \psi_1 \rangle_1 \; \langle \phi_2, \psi_2 \rangle_2.$$

Now, $$H_1 \otimes_{alg} H_2$$ with this inner product is not a Hilbert space, since it is not complete. But its completion $$H_1 \otimes H_2 := \overline{H_1 \otimes_{alg} H_2}^{\langle \cdot , \cdot \rangle}$$ is a Hilbert space and is called the tensor product Hilbert space.

The space of Hilbert-Schmidt operators $$HS(H_1^*,H_2)$$ you mentioned above is isometrically isomorphic to $$H_1 \otimes H_2$$. You may identify all tensor products $$\phi_1 \otimes \phi_2$$ with the map $$x^* \mapsto x^*(\phi_1) \phi_2$$. This map is in $$HS(H_1^*,H_2)$$. Taking linear combinations and limits, this identification extends to an identification of $$H_1 \otimes H_2$$ and $$HS(H_1^*,H_2)$$, which can be proven to be isometric and isomorphic.

answer 2: For Hilbert spaces of functions, such as $$L^2(\mathbb{R})$$, you would build a tensor product as $$L^2(\mathbb{R}) \otimes L^2(\mathbb{R}) = L^2(\mathbb{R}^2), \qquad (\phi_1 \otimes \phi_2)(x_1,x_2) = \phi_1(x_1) \phi_2(x_2).$$ Note that the 2 arguments $$x_1, x_2$$ are different. So rather than multiplying the functions $$\phi_1, \phi_2$$ in an $$x$$-$$y$$-diagram, you go to a 3-dimensional $$x_1$$-$$x_2$$-$$y$$-diagram and multiply the functions $$f_1(x_1,x_2) = \phi_1(x_1) \; 1(x_2) \qquad \text{and} \qquad f_2(x_1,x_2) = 1(x_1) \; \phi_2(x_2)$$ (meaning $$(\phi_1 \otimes \phi_2)(x_1,x_2) = f_1(x_1,x_2) f_2(x_1,x_2)$$).

More generally, $$L^2(\mathbb{R}^n) \otimes L^2(\mathbb{R}^m) = L^2(\mathbb{R}^{n+m})$$ for any $$n,m \in \mathbb{N}$$, so the tensor product basically "adds the parameter dimensions of your functions". You may also use other Hilbert spaces, such as $$L^2(\Omega)$$ with $$\Omega \subset \mathbb{R}^n$$ being a nice set or even the Sobolev spaces $$H^s(\Omega), s \ge 0$$

answer 3: Each Hilbert-Schmidt operator $$K \in HS(L^2(\mathbb{R})^*, L^2(\mathbb{R}))$$ can be identified with a kernel $$k(x_1,x_2)$$ with $$k \in L^2(\mathbb{R}) \otimes L^2(\mathbb{R}) = L^2(\mathbb{R}^2)$$.

That means, for $$\phi^*_1 \in L^2(\mathbb{R})^*$$, the function $$K \phi^*_1 \in L^2(\mathbb{R})$$ is given by $$(K \phi^*_1)(x_2) = \int_\mathbb{R} k(x_1,x_2) \phi^*_1(x_1) \; dx_1$$

The identification of kernels $$k \in L^2(\mathbb{R}) \otimes L^2(\mathbb{R}) = L^2(\mathbb{R}^2)$$ and Hilbert-Schmidt operators is one-to-one. So you may also construct $$L^2(\mathbb{R}) \otimes L^2(\mathbb{R})$$ using Hilbert-Schmidt operators on such function spaces.

• thanks for your answer. Couple of follow up questions. For answer 1 we looked at multiplying the two inner products while for answer 2 we looked at multiplication of two functions (different inputs). Why do they refer to the same tensor product? In the case of $L^2$ using your answer 1 I would defin $\langle f_1\otimes g_1, f_2\otimes g_2\rangle = \int f_1 f_2\int g_1 g_2$ and take the completion of that space. Why do they end up with the same space? Can we always find a explicit definition of $\phi_1\otimes\phi_2$ as in the multiplicative function case? Do you have some good references on it?
– math
Jul 19, 2021 at 18:34
• That's a fair question. For $L^2$, concerning answer 1, the scalar product you wrote down is just the $L^2(\mathbb{R}^2)$ scalar product. So if you the the completion, you just get all $L^2$-integrable functions , i.e., $L^2(\mathbb{R}^2)$. Which is exactly the space in answer 2. The results agree, because you take the closure with respect to the same scalar product: the $L^2$ scalar product. Jul 23, 2021 at 17:23
• In general, the expression $\phi_1 \otimes \phi_2$ is just abstract, so it does not need to be a product of functions. In fact, your Hilbert space might also be represented by finite-dimensional vectors or infinite-dimensional sequences (as the $l^2$ sequence space). But often, one uses functions within Hilbert spaces and then you often have a product as above. Jul 23, 2021 at 17:24
• If $H_1, H_2$ are replaced by non-Hilbert spaces (e.g. test functions or distributions), then the tensor product is a lot more difficult to define, since you don't have a scalar product for a completion. A good reference is the Wikipedia article en.wikipedia.org/wiki/Topological_tensor_product. It also contains a short section with a link on the Hilbert space tensor product. Jul 23, 2021 at 17:30