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I am trying to prove something that seemed simple to me at first sight but apparently it is giving me a hard time. Here is the same question on MSE.

Let $E\subset A$ be a finite dimensional operator system with Hamel basis $\{x_1,\dots,x_n\}$ and let $B$ be an arbitrary $C^*$-algebra. I am trying to show that, if $\|\cdot\|$ is the minimal norm on $E\odot B$ (i.e. the spatial norm) and $E\otimes B:=\overline{E\odot B}^{\|\cdot\|}$, then every element of $E\otimes B$ is written uniquely as $\sum_{i=1}^nx_i\otimes b_i$. This is of course equivalent to showing that $E\odot B$ is complete with the minimal norm.

It is trivial to verify the claim for the elements of the algebraic tensor product $E\odot B$, by the following lemma:

If $X,Y$ are vector spaces and $\{a_1,\dots,a_n\}\subset X$ are linearly independent, then $$\sum_{i=1}^na_i\otimes b_i=0\implies b_1=\dots=b_n=0.$$

This lemma also takes care of uniqueness for us. But what about elements of $E\otimes B$ in general? Of course, the ideal thing would be to establish an inequality of the form $\|x_j\otimes b_j\|\leq C\|\sum_{i=1}^nx_i\otimes b_i\|$, but how?

I tried to consider $B\oplus\dots\oplus B\to (E\odot B,\|\cdot\|)$, $(b_1,\dots,b_n)\mapsto\sum_{i=1}^nx_i\otimes b_i$ and observed that this is bounded and bijective. The annoying thing: if I knew that $E\odot B$ is complete (which is what I want to show), then I could apply the open mapping theorem to deduce that the inverse is bounded, hence obtain the desired estimate.

I also tried an induction argument on $\dim(E)$ but it didn't work out.

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One way to see it is to use the dual basis of $\{x_1,\dots,x_n\}$. Let $\{\varphi_1,\dots,\varphi_n\} \subset E^{\ast}$ be functionals such that $\varphi_i(x_j)=\delta_{ij}$. Because functionals are automatically completely bounded, we get bounded maps $\Phi_i:= \varphi_i \otimes Id : E\otimes B \to B$. On the subspace $E\odot B$ we have $\Phi_i(\sum_{j=1}^{n} x_{j} \otimes b_{j}) = b_{i}$, so $\|b_{i}\| \leqslant \|\Phi_i\|\cdot \|\sum_{j=1}^{n} x_{j} \otimes b_{j}\|$. As a consequence, if a sequence $y(m):= \sum_{i=1}^{n} x_{i} \otimes b_i(m) \in E\odot B$ converges in $E \otimes B$ then for each $i\in \{1,\dots, n\}$ the sequence $b_i(m)$ converges in $B$. Let $B_i = \lim_{m\to\infty} b_i(m)$. Then $\lim_{m\to \infty} \sum_{i=1}^{n} x_i \otimes b_i(m) = \sum_{i=1}^{n} x_i \otimes B_i \in E \odot B$, hence $E\odot B$ is equal to $E\otimes B$.

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    $\begingroup$ Thank you very much, this is a very elegant argument. I guess the "non-triviality" that caused the problem lies behind the fact that the spatial tensor product of two c.b. maps is again c.b., which follows from Wittstock's theorem. Anyway, thanks again! $\endgroup$ – JustDroppedIn Apr 21 at 9:14
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Here's an approach motivated by considerations of Banach space tensor products; it owes a debt to the approach taken by Takesaki in his book, Volume 1, Chapter IV, Sections 2 and 4. Another good book is Ryan's book, "Introduction to Tensor Products of Banach Spaces"

For Banach spaces $E,F$ a norm on $E\odot F$ is a cross-norm if $\|x\otimes y\| = \|x\| \|y\|$. There are two natural cross-norms: the injective tensor norm defined by the map $E\odot F\rightarrow B(E^*,F)$, so $$ \lambda\Big(\sum_{i=1}^n x_i\otimes y_i\Big) = \sup\Big\{ \Big\|\sum_{i=1}^n f(x_i) y_i \Big\|: f\in E^*, \|f\|\leq 1 \Big\}; $$ and the projective tensor norm, $$ \pi(u) = \inf\Big\{ \sum_{i=1}^n \|x_i\| \|y_i\| : u = \sum_{i=1}^n x_i\otimes y_i \Big\}. $$ These are cross-norms. Given a cross-norm $\beta$, there is a norm $\beta^*$ on $E^*\odot F^*$ given by the natural dual-pairing between $E\odot F$ and $E^*\odot F^*$. Then $\beta^*$ is a cross-norm if and only if $\lambda \leq \beta \leq \pi$. (So such $\beta$ are in some sense well-behaved).


If $E$ is finite-dimensional, with basis $(x_i)_{i=1}^n$ then (much as Mateusz argues) we can use the dual basis $(x_i^*)$ to see that $$ \lambda\Big(\sum_{i=1}^n x_i\otimes y_i\Big) \geq K^{-1} \|y_j\| $$ for any $j$, where $K=\max_i \|x_i^*\|$. Clearly also $$ \pi(\sum_{i=1}^n x_i\otimes y_i) \leq \sum_i \|x_i\| \|y_i\| \leq L \max_i \|y_i\| $$ where $L=\sum_i \|x_i\|$. Thus both $\lambda$ and $\pi$ are equivalent to the max-norm $\max_i \|y_i\|$, and so any "nice" cross-norm on $E\odot F$ gives a norm equivalent to the direct sum of $n$ copies of $F$.


Now, it turns out that any $C^*$-tensor norm on $A\odot B$ is a "nice" cross-norm: this is shown by Takesaki on pages 207--208. As the theory of $C^*$-tensor norms is quite intricate, it's hard to say exactly why this is so. However, we then see that the dual norm on $A^*\odot B^*$ is a cross-norm (in fact, and harder to show, the dual norm on $A^*\odot B^*$ is always the same, independent of the $C^*$-tensor norm on $A\odot B$.)

The result follows.

In fact, if you are happy to use that $\|x\otimes b\| = \|x\| \|b\|$ on $E\odot B$, then already $\|\cdot\|\leq\pi$, and so it remains to show that $\lambda\leq\|\cdot\|$. We need only show that if $f\in E^*, g\in B^*$ then $f\otimes g$ induces a functional on $E\odot B$ of norm at most $\|f\| \|g\|$. Hahn-Banach $f$ to a member of $A^*$ and hence work on $A\odot B$. The result follows from a polar-decomposition argument and then a GNS argument, using the definition of the spatial $C^*$-tensor norm: this is exactly how Takesaki's book proceeds.

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