On the Completeness of Projective Tensor Products

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.

• 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 $dim(Y) = m$. – 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).
• 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? – user284331 Oct 20 '19 at 19:33
• 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 – Ben Grossmann Oct 20 '19 at 19:53
• 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 $B(X^*,Y)$. – Ben Grossmann Oct 20 '19 at 19:58
• 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. – user284331 Oct 20 '19 at 20:02