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I am trying to get a good understanding of what a tensor product is and I am trying to understand one particular example.

Suppose that $H=L^2([0,1],\mathbb R)$ and take the inner-product $$ \langle f,g\rangle=\int_0^1f(x)g(x)dx $$ defined for each $f,g\in H$ as a bilinear map from $H\times H$ to $\mathbb R$. I want to investigate the tensor product $H\otimes H$. First of all, as far as I understand, there exists a universal bilinear map $\varphi:H\times H\to H\otimes H$ and then there exists a linear map $l:H\otimes H\to\mathbb R$ such that $\langle\cdot,\cdot\rangle=l\circ \varphi$, i.e. $\langle f,g\rangle=l(\varphi(f,g))=l(f\otimes g)$ for each $f,g\in H$.

My questions are as follow.

  1. As far as I understand (see also here), a tensor $f\otimes g$ is defined as $(f\otimes g)(x,y)=f(x)g(y)$ for each $f,g\in H$ and $x,y\in[0,1]$ . How can we deduce that tensors $f\otimes g$ look like this? How can we come up with this expression?

  2. The bilinear map $\varphi$ is given by $\varphi(f,g)=f\otimes g$. $H\otimes H$ also contains elements that are finite linear combinations of the tensors $f\otimes g$. So the map $\varphi$ is not necessarily surjective, the image of $H\times H$ is smaller than $H\otimes H$, right?

  3. The tensor product $H\otimes H$ is actually isomorphic to the space $L^2([0,1]^2,\mathbb R)$, is that correct?

  4. How can we find the linear map $l$ that maps tensors $f\otimes g$ to $\langle f,g\rangle$ for each $f,g\in H$?

Thanks a lot for your help!

user159517
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Cm7F7Bb
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1 Answers1

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We define $f \otimes g := f(x)g(y)$. For the moment, think of $f \otimes g$ like this and not as an abstract tensor product. Let

$$T := \text{span}{\{ f \otimes g: f,g \in H\}} \subseteq L^2([0,1]^2)$$ We distinguish between the topological tensor product and the algebraical tensor product, where the latter is the one satisfying the universal property you gave. I will show first that $T$ is isomorphic to the algebraical tensor product. Let $(e_i)_{i\in I}$ be a basis of $H$.

Claim: $(e_i \otimes e_j)_{(i,j)\in I\times I}$ is a Hamel basis of $T$.

It is easy to see that $(e_i \otimes e_j)_{(i,j)\in I\times I}$ generates $T$. For the linear independence, let $c_{ij} \in \mathbb{R}$ such that $$\sum_{i=1}^{n}\sum_{j=1}^{n} c_{ij} e_{i} \otimes e_{j} = \sum_{j=1}^{n} \left(\sum_{i=1}^{n} c_{ij}e_{i}\right) \otimes e_{j} = 0$$ Define $k_{j} := \sum_{i=1}^{n} c_{ij}e_{i}$. Let $z_1,\dots,z_n$ be orthonormal with the same span as $e_{1},\dots,e_{n}$. Thus, we may write $k_{j} = \sum_{l} \lambda_{jl}z_{l}$. We find

$$ 0 = \sum_{j} k_{j} \otimes e_{j} = \sum_{j} \left(\sum_{l} \lambda_{jl} z_{l}\right) \otimes e_{j} = \sum_{l} z_{l} \otimes \left(\sum_{j} \lambda_{jl} e_{j} \right)$$ Defining $y_l := \sum_{j} \lambda_{jl} e_{j}$ we may rewrite this as $$\sum_{l} z_{l} \otimes y_{l} = 0.$$ Now observe that for $l \neq m$ we have $$(z_l \otimes y_l, z_{m}\otimes y_{m})_{L^2([0,1]^2)} = (z_l , z_m)_{L^2([0,1])} \cdot (y_{l},y_{m})_{L^2([0,1])} = 0.$$ Thus, the terms in the above sum are orthogonal, and by Pythagoras we find $$ 0 = \left\|\sum_{l} z_{l} \otimes y_{l} \right\|^2 = \sum_{l} \| z_{l} \otimes y_{l}\|^2 = \sum_{l} \|y_{l}\|^2.$$ Thus, $y_1 = \dots = y_n = 0$, which by linear independence of the $e_j$ implies that all the $\lambda_{jl}$ are equal to $0$, which in turn implies $k_1 = \dots = k_n = 0$. Again by the linear independence of the $e_i$, this implies $c_{ij} = 0$ for all $i,j$.

Claim: $T$ satisfies the universal property

Let $\varphi(f,g) := f \otimes g$. Let $V$ be some $\mathbb{R}$-vector space and $B : H \times H \rightarrow V$ be a bilinear map. For $(i,j) \in I \times I$, define $$ l(e_i \otimes e_j) := B(e_i , e_j).$$ Due to the extension theorem for linear maps, as $(e_i \otimes e_j)_{(i,j)\in I\times I}$ is a Hamel basis of $T$, this uniquely defines a linear map $l: T \to V$. Note that any $l$ satisfying $l(f \otimes g) = B(f,g)$ for all $f,g \in H$ must satisfy the equation above, so the map $l$ is indeed unique.

Topological Tensor Product

We've shown that $T$ is isomorphic to the algebraical tensor product. As you correctly concluded, $\varphi: H \rightarrow T$ is not surjective. Also, we do not have $T = L^2([0,1]^2)$, but we do have that $T$ is dense in $L^2([0,1]^2)$. Clearly, it would be nice if the operation of taking tensor products would be 'closed' for Hilbert spaces, meaning that $H \otimes H$ should again be a Hilbert space. Thus, in this context, one would usually define $H \otimes H$ to be the closure of the algebraical tensor product, so the topological tensor product is in fact isomorphic to $L^2([0,1]^2)$. This tensor product of Hilbert spaces satisfies a different universal property, as described in the link.

Addendum: Motivating the definition of $f \otimes g$

Let us consider the finite-dimensional case, say we want to gain some intuition on what the tensor product of $\mathbb{R}^n \otimes \mathbb{R}^m$ should be. Let $e_i, f_j$ be bases of $\mathbb{R}^n, \mathbb{R}^m$ respectively. Let $x = \sum_{i} x_i e_i$ and $y = \sum_{j} y_j f_j$.For any bilinear form $B: \mathbb{R}^n \times \mathbb{R}^m \to V$ and $\mathbb{R}$-vector space $V$, we have $$B(x,y) = \sum_{i}\sum_{j} x_i y_j B(e_i , f_j).$$ Due to the universal property, we must have $l(e_i \otimes e_j) = B(e_i, f_j)$, and we see that $$l(x \otimes y) = \sum_{i}\sum_{j} x_i y_j l(e_i \otimes e_j) = \sum_{i}\sum_{j} x_i y_j B(e_i , f_j).$$ Now we want $l$ to act linearly on elements of the form $x \otimes y$. We can see by the above considerations that this is in fact achieved if we set $x \otimes y$ to be the matrix of tuples $x_i y_j$, i.e. $x^{T}y$ if we let $e_i, f_j$ be the standard canonical bases. This should clarify the finite-dimensional case.

Now for the infinite-dimensional case: You can think of $\mathbb{R}^n$ as the set of all functions from $\{1,\dots,n\}$ to $\mathbb{R}$. Taking the tensor product of that with the set of all functions from $\{1,\dots, m\}$ to $\mathbb{R}$, we saw that $$ (x \otimes y)(i,j) = x(i)y(j), \quad (i,j) \in \{1,\dots,n\} \times \{1,\dots,m\}.$$ So it seems natural to define $(f \otimes g) (x,y) := f(x)g(y)$ for functions $f,g$ from $[0,1]$ to $\mathbb{R}$.

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user159517
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  • Thank you very much for your answer (+1). I am still trying to understand the example with $L^2([0,1],\mathbb R)$ better. I would be grateful if you could clarify a few things. (1) How do we know how the tensor $f\otimes g$ looks like, i.e. we have that $(f\otimes g)(x,y)=f(x)g(y)$, but how can we come up with this expression? (2) Hamel basis means that any element can be expressed as a finite linear combination, right? (3) Is there a way to obtain an explicit expression of the linear map that maps $f\otimes g$ to $\langle f,g\rangle$? – Cm7F7Bb Aug 04 '17 at 08:37
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    (1) Do you understand what $x \otimes y$ is if $x,y \in \mathbb{R}^n$? (2) Yes (3) In this case, it is easy to see that the map $$ f \otimes g \mapsto \int_{0}^{1} (f\otimes g) (x,x)~\mathrm{d}x$$ does the job. – user159517 Aug 04 '17 at 14:56
  • (1) That's a good point, I cannot say that I understand. If $x,y\in\mathbb R^n$, $x\otimes y=xy^{\mathrm T}$, right? How can we conclude that $x\otimes y=xy^\mathrm T$? – Cm7F7Bb Aug 04 '17 at 15:04
  • Why is that not correct? It is the same thing as you wrote, right? If $x,y\in\mathbb R^n$, then $xy^\mathrm T$ is an $n\times n$ matrix, i.e. an element of $\mathbb R^{n\times n}$. So I have seen this definition, but I am not sure how we can conclude that the elements of $\mathbb R^n\otimes\mathbb R^n$ look like this. – Cm7F7Bb Aug 05 '17 at 07:46
  • @Cm7F7Bb Oh, I'm sorry, I read $x^T y$ for some reason. You can argue similarly as I did: Show that $e_i \otimes e_j$ is a basis of $\mathbb{R}^{n\times n}$ and thus $\mathbb{R}^{n\times n}$ satisfies the universal property which implies that $\mathbb{R}^{n\times n}$ is isomorphic to $\mathbb{R}^n \otimes \mathbb{R}^n$ – user159517 Aug 05 '17 at 10:29
  • @Cm7F7Bb I made my answer even longer to give a little more motivation. – user159517 Aug 05 '17 at 11:30