# 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! 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.