Continuous tensor products

Question

Is there a notion of infinite tensor product Hilbert spaces? Let $I$ be an infinite set, for example taking $I= \mathbb{R}$, then is the space $\mathcal{H} = \bigotimes_{i \in I} \mathcal{H}_{I}$ well-defined? Is it separable if each $\mathcal{H}_{I}$ is?

Answer 1

First of all, let's construct such a notion for vector spaces. Let $(V_i)_{i\in I}$ be an infinite family of vector spaces, such that for each $i\in I$, there is a choice of a nonzero vector $v_i\in V_i$. Then, there exists a vector space ${\bigotimes}'_{i\in I}V_i$ of linear combinations of vectors $\otimes_{i\in I}w_i$ such that for all but finitely many $i\in I$, we have $w_i=v_i$. This is a well-defined vector space, with the usual bilinearity relations.

To extend this to Hilbert spaces, let $(\mathcal H_i)_{i\in I}$ be an infinite family of Hilbert spaces, and again fix a choice of nonzero $v_i\in\mathcal H_i$ for each index $i\in I$. Then, we can again define the vector space ${\bigotimes}'_{i\in I}\mathcal H_i$. The Hilbert space structure is given by the inner product $$\langle \otimes_{i\in I}u_i,\otimes_{i\in I}w_i\rangle=\prod_{i\in I}\frac{\langle u_i,w_i\rangle}{\langle v_i,v_i\rangle}=\prod_{i\in I}\frac{\langle u_i,w_i\rangle}{\|v_i\|^2}.$$ Here, the product is well-defined since for all but finitely many $i$ we have $u_i=w_i=v_i$, so $\frac{\langle u_i,w_i\rangle}{\langle v_i,v_i\rangle}=1$. Thus, the seemingly infinite product is actually already finite.

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A comment about separability: suppose each $\mathcal H_i$ is separable and $I$ is a countable set. Then the infinite tensor product $\mathcal H:={\bigotimes}'_{i\in I}\mathcal H_i$ is again separable. Indeed, for each $i\in I$, there is a countable dense subset $D_i\subset \mathcal H_i$, where we may assume $v_i\in D_i$. Then let ${\bigotimes}'_{i\in I}D_i$ be the set of $\otimes_{i\in I}w_i$, where each $w_i\in D_i$, and all but finitely many $w_i$ are equal to $v_i$. This is a countable set (this is an exercise). Moreover, let $D$ be the set of finite linear combinations of elements of ${\bigotimes}'_{i\in I}D_i$. This is also countable, and you can check $D$ is a dense subset of $\mathcal H$.