# Elements of the minimal tensor product of a finite dimensional operator system and a $C^*$-algebra

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.

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 $$E\otimes B$$.

• 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! – JustDroppedIn Apr 21 at 9:14

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.