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?

  • 1
    $\begingroup$ 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? $\endgroup$ – LSpice Oct 21 at 18:24
  • $\begingroup$ @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? $\endgroup$ – user839372 Oct 21 at 18:35
  • $\begingroup$ Just out of curiosity: you rejected my edit that inserted the < missing from the title, and then inserted it yourself. Why? $\endgroup$ – LSpice Oct 21 at 22:04
  • $\begingroup$ 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? $\endgroup$ – LSpice Oct 21 at 22:07
  • 1
    $\begingroup$ 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. $\endgroup$ – 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.

| cite | improve this answer | |
  • $\begingroup$ Definitely harder than I thought! Many thanks! I will go in detail through it soon! $\endgroup$ – user839372 Oct 28 at 19:32
  • 1
    $\begingroup$ You are welcome. And please fell free to ask for any clarification if needed! $\endgroup$ – Ruy Oct 28 at 19:47
  • $\begingroup$ Thanks! That's very kind of you! $\endgroup$ – user839372 Oct 28 at 19:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.