There are two exercises about projective tensor products.

1) If either $X$ or $Y$ is finite dimensional, then the projective tensor product $X\otimes Y$ is complete.

2) If both $X$ and $Y$ are infinite dimensional, then $X\otimes Y$ is never complete. (As a result, $X\otimes Y\ne X\otimes_{\pi}Y$)

If both $X$ and $Y$ are finite dimensional, then $\dim X\otimes Y=(\dim X)(\dim Y)$ is finite, and hence the norm space $X\otimes Y$ is complete and is just $X\otimes_{\pi}Y$ itself.

But I can't make it for only just either of $X$ or $Y$ is finite dimensional.

Any idea?

Note that here both $X$ and $Y$ are Banach spaces, and the projective tensor product norm defined on $X\otimes Y$ is \begin{align*} \pi(u)=\inf\left\{\sum_{i=1}^{n}\|x_{i}\|\|y_{i}\|: u=\sum_{i=1}^{n}x_{i}\otimes y_{i}\right\}, \end{align*} where $u=\displaystyle\sum_{i=1}^{n}x_{i}\otimes y_{i}$ is any representation of $u$ in $X\otimes Y$. The space $X\otimes_{\pi}Y$ is the completion of $X\otimes Y$ under this norm.

Maybe the first one can be dealt in this way:

I suspect that $(X\otimes Y)/Y\cong X$ where $Y$ is assumed to be finite dimensional, so $(X\otimes Y)/X$ is complete, as $X$ is also complete, then $X\otimes Y$ is complete by the three-space property of Banach spaces.

  • $\begingroup$ How do you embed $Y$ into $X \otimes Y$? You need to fix some vector $x_0 \in X$. Either way, the last quotient cannot hold. Just by counting dimensions $dim(X \otimes Y)/Y = n m - m = n(m-1)$, where $dim(X) = n$ and $dim(Y) = m$. $\endgroup$ – Adrián González-Pérez Oct 21 '19 at 11:29

It is not necessarily true that $(X \otimes Y)/Y$ is isomorphic to $X$; it may help verify that even in the finite-dimensional case, the dimensions of these spaces do not match.

I think that the easiest approach for the case where $Y$ is finite dimensional is to note that if $Y$ has dimension $n$ (so that $Y$ is isomorphic to $\Bbb C^n$ under some norm), we have $$ X \otimes Y \cong X \otimes \Bbb C^n \cong \overbrace{X \oplus \cdots \oplus X}^{n \text{ times}}. $$ In the infinite-dimensional case, suppose that $\{x_i\} \subset X$ and $\{y_i\} \subset Y$ are linearly independent sequences with $\|x_i\| = \|y_i\| = 1$. It suffices to verify that the sequence defined by $$ z_n = \sum_{i=1}^n \frac{1}{2^i} x_i \otimes y_i $$ is Cauchy, but fails to converge in $X \otimes Y$ (since the infinite sum cannot be expressed as a finite sum of tensor products).

  • $\begingroup$ Actually how to argue in contradiction that if the infinite sum is a finite sum of tensor products, then what will happen? $\endgroup$ – user284331 Oct 20 '19 at 19:24
  • $\begingroup$ And the expression that $X\otimes\mathbb{C}^{2}\cong X\oplus X$ is true in algebraic, to argue that it is also topological, I switch to arguing in the universal property of projective tensor product, is there any other simpler way? $\endgroup$ – user284331 Oct 20 '19 at 19:33
  • $\begingroup$ For your second comment, it suffices to note that the map $x \oplus y \mapsto x \otimes e_1 + y \otimes e_2$ (where $\{e_1,e_2\}$ is a basis of $\Bbb C^2$) is an isomorphism $\endgroup$ – Ben Grossmann Oct 20 '19 at 19:53
  • $\begingroup$ For your first comment, actually I don't see an easy elementary approach. In the back of my head, I was using the fact that we can we can identify $X \otimes Y$ with the finite-rank operators of $B(X^*,Y)$. $\endgroup$ – Ben Grossmann Oct 20 '19 at 19:58
  • $\begingroup$ So the infinite sum is not of finite rank since $(x_{i})$ is linearly independent. Okay, nice, but I never thought it is such no elementary. Initially I was thinking about to argue in contradiction by the argument of pure linear independence, but I realize that $(x_{n})$ and $(y_{n})$ are not bases. Anyway, thanks, mate. $\endgroup$ – user284331 Oct 20 '19 at 20:02

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