# $H \odot K$ complete implies $\dim H$ or $\dim K <\infty$

Let $$H,K$$ be non-zero Hilbert spaces. Then the algebraic tensor product $$H \odot K$$ is an inner product space for the unique inner product determined by $$\langle h \otimes k, h' \otimes k'\rangle = \langle h, h'\rangle_H \langle k, k' \rangle_K$$

I am trying to prove that $$H \odot K$$ is a Hilbert space (i.e. $$H \odot K$$ is complete for the norm induced by this inner product) implies that $$\dim H <\infty$$ or $$\dim K < \infty$$.

Let us prove the contrapositive, i.e. so suppose that $$H$$ and $$K$$ are both infinite dimensional. How can I show that there is a Cauchy sequence in $$H \odot K$$ that does not converge?

• The statement is false unless $H \ne 0 \ne K$, so you probably want to assume that. \\ It is clear (when $K \ne 0$) that $H$ infinite dimensional implies $H \otimes K$ infinite dimensional (why $H \odot K$?). Surely you mean that you want to show that, if $H$ is infinite dimensional, then $H \otimes K$ is not complete? In that case, why not see if you can use the infinite dimensionality of $H$ to manufacture a non-convergent Cauchy sequence? – LSpice Oct 21 at 18:24
• @LSpice Thanks for the comment. I made some edits. Finding this non-convergent Cauchy sequence is exactly what I can't. Do you have any hints for this? – user839372 Oct 21 at 18:35
• Just out of curiosity: you rejected my edit that inserted the < missing from the title, and then inserted it yourself. Why? – LSpice Oct 21 at 22:04
• I didn't have a specific sequence in mind, and, in fact, I think that the result is still not true. If $K$ is 1-dimensional, then $H \otimes K$ is isometric to $H$, so is complete even if $H$ is infinite dimensional; and, more generally, $H \otimes \mathbb C^n$ is isometric to $H^{\oplus n}$, which is complete for $n < \infty$. Are you sure that you don't mean to prove that $\dim(H)$ or $\dim(K)$ is finite? – LSpice Oct 21 at 22:07
• As per en.wikipedia.org/wiki/… you just need an infinite matrix with finite sum of squares of the entries but not of finite rank; a diagonal one will do. – Max Oct 28 at 0:16

It is very easy to exhibit a non-converging Cauchy sequence in $$H\odot K$$, but actually proving that it does not converge is a bit more difficult. The approach suggested by @Max, although somewhat sophisticated, is perhaps the easiest way to pin down all of the details. Here is a pedestrian way to describe this method.

For each pair of vectors $$(x, y)\in H\times K$$, consider the bounded linear operator $$T_{x, y}:H\to K$$ given by $$T_{x, y}(z) = \langle x,z\rangle y, \quad\forall z\in H.$$ (For simplicity I am assuming that we are working with real vector spaces but this can be fixed in the complex case by seeing $$T_{x, y}$$ as an operator defined on $$\bar H$$, namely the Hilbert space made out of $$H$$ by adopting the new scalar multiplication operation $$\lambda \cdot x:= \bar \lambda x$$).

It is easy to see that the map $$(x, y)\in H\times K\mapsto T_{x, y}\in \mathscr {B}(H,K)$$ is bilinear so, by the universal property of (algebraic) tensor products, there exists a unique linear map $$T :H\odot K\to \mathscr {B}(H,K),$$ such that $$T (x\otimes y) = T_{x,y}$$. We next claim that $$\langle T (\xi )x,y\rangle = \langle \xi ,x\otimes y\rangle , \tag {1}$$ for every $$\xi$$ in $$H\odot K$$, $$x\in H$$, and $$y\in K$$. To prove this write $$\xi =\sum_{i=1}^nx_i\otimes y_i$$, and notice that $$\langle T (\xi )x,y\rangle = \sum_{i=1}^n \langle T_{x_i,y_i}(x),y\rangle =$$ $$= \sum_{i=1}^n \langle \langle x_i,x\rangle y_i,y\rangle = \sum_{i=1}^n \langle x_i,x\rangle \langle y_i,y\rangle =$$ $$= \sum_{i=1}^n \langle x_i\otimes y_i,x\otimes y\rangle = \langle \xi ,x\otimes y\rangle,$$ proving the claim.

Note that the range of each $$T_{x,y}$$ is the one-dimensional space spanned by $$y$$, so $$T_{x,y}$$ has rank 1. Furthermore, every $$\xi$$ in $$H\odot K$$ may be writen as $$\xi =\sum_{i=1}^nx_i\otimes y_i$$, so $$T (\xi ) = \sum_{i=1}^n T_{x_i,y_i},$$ so we see that $$T (\xi )$$ has rank at most $$n$$, hence finite, for every $$\xi$$ in $$H\odot K$$.

The final contradiction will be achieved by proving that, if $$H$$ and $$K$$ are infinite dimensional, and $$H\odot K$$ is complete, there exists a vector $$\xi$$ in $$H\odot K$$ such that $$T (\xi )$$ has infinite rank.

Let us therefore assume from now on that $$H$$ and $$K$$ are both infinite dimensional, so we may find (necessarily infinite) orthonormal bases $$\{e_i\}_{i\in I}$$, and $$\{f_j\}_{j\in J}$$, for $$H$$ and $$K$$ respectively, and it is well known that $$\{e_i\otimes f_j\}_{(i, j)\in I\times J}$$ is an orthonormal basis for $$H\otimes K$$.

Assuming by contradiction that $$H\odot K$$ is complete, observe that $$H\odot K$$ coincides with its completion, so $$H\odot K = H\otimes K$$.

Choose infinite subsets $$\{i_n:n\in {\mathbb N}\}\subseteq I$$, and $$\{j_n:n\in {\mathbb N}\}\subseteq J$$, and let $$\xi = \sum_{n=1}^\infty {1\over n^2} (e_{i_n}\otimes f_{j_n}).$$ This is well defined because the series is clearly absolutely convergent and we are working in a complete space!

Using (1) we have that $$T (\xi )e_{i_n} = \sum_{j\in J} \langle T (\xi )e_{i_n}, f_j\rangle f_j = \sum_{j\in J} \langle \xi , e_{i_n}\otimes f_j\rangle f_j = \langle \xi , e_{i_n}\otimes f_{j_n}\rangle f_{j_n} = {1\over n^2} f_{j_n}.$$ This shows that $${1\over n^2} f_{j_n}$$ lies in the range of $$T (\xi )$$, and hence that $$T (\xi )$$ has infinite rank, a contradiction.

• Definitely harder than I thought! Many thanks! I will go in detail through it soon! – user839372 Oct 28 at 19:32
• You are welcome. And please fell free to ask for any clarification if needed! – Ruy Oct 28 at 19:47
• Thanks! That's very kind of you! – user839372 Oct 28 at 19:48