Let $A$ be a $C^*$-algebra. I want to prove that $$ \phi : C_0(X) \otimes_\alpha A \to C_0(X,A): f \otimes a \mapsto a \cdot f $$ is an isomorphism, where $\alpha$ is a norm making the tensor product into a $C^*$-algebra. Note that $C_0(X)$ is commutative hence nuclear. Since $\phi(C_0(X) \otimes A)$ is dense (with $\otimes$ the algebraic tensor product) one is done, if the restriction of $\phi $ to $C_0(X) \otimes A$ is an isometry. Then $\phi$ is clearly surjective and also injective. (On the algebraic tensorproduct $\phi$ is injective but I think this does not necessarily extend to the completion.)

So my question would be what one can take for $\alpha$ to prove in an efficient way that $\phi$ is an isometry.

  • $\begingroup$ The map is always injective, right? Why not then take $\|x\|_\alpha:= \|\phi(x)\|$? $\endgroup$ – s.harp Sep 7 '16 at 19:59

By construction, $\phi$ is a $*$-epimorphism, if it is well-defined. The argument for well-definedness, as usually happens, is tied with the proof that $\phi$ is one-to-one.

The key fact is the following:

Lemma. The following statements are equivalent:

  1. $\sum a_j\otimes f_j=0$

  2. There exists $T\in M_n(\mathbb C)$ such that $\sum_kT_{kj}a_k=0$, and $\sum_jT_{kj}f_j=f_k$.

Now, if $\sum a_j\otimes f_j=0$, we take $T$ from the Lemma, and then $$ \sum_ka_kf_k=\sum_k\sum_jT_{kj}a_kf_j=\sum_j\left(\sum_kT_{kj}a_k\right)f_j=0. $$ So $\phi$ is well-defined.

Conversely, if $\sum_ja_jf_j=0$, by re-arranging the sum we may assume that the $a_j$ are linearly independent. But then, for any any $t\in X$, $\sum f_j(t)a_j=0$, which implies $f_j(t)=0$ by the linear independence. Thus $f_j=0$ and so $\sum a_j\otimes f_j=0$ (in the re-arranged version, but this is implies the original version is also zero).

So $\phi$ is one-to-one. We can use this to define $$\|x\|_\alpha:=\|\phi(x)\|.$$ Since $\alpha$ is a norm and $C_0(X)$ is nuclear, we get that $\|x\|_\alpha=\|x\|$, i.e. that $\phi$ is isometric.

An isomorphism is isometric, so as expected $\|x\|_\alpha=\|\phi(x)\|$ for any $x\in C_0(X)\otimes_\alpha A$ and any tensor norm you could define on $C_0(X)\otimes A$.

  • $\begingroup$ I have a question concerning the injectivity. You proved: $\psi (\sum a_j\otimes f_j )=0$ then $\sum a_j\otimes f_j=0$. I'm wondering if this is enough? Since Injectivity on $C_0(X)\otimes A$ doesn't imply injectivity on the closure in general (however, I know if this map is isometric on a dense subset, then it extends to an isometry on the closure). I overlook something.. Can you explain why $\phi$ is injective on the closure again, please? I'd really appreciate that. $\endgroup$ – Sabrina G. Nov 16 '16 at 21:42
  • $\begingroup$ Definitely a good point. I need to address that. $\endgroup$ – Martin Argerami Nov 17 '16 at 3:05
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    $\begingroup$ @toto: I have modified the last part of the answer, using s.harp's idea, to show that $\phi$ is isometric. $\endgroup$ – Martin Argerami Nov 17 '16 at 4:29
  • $\begingroup$ thank you very much! Now I understand the argument! $\endgroup$ – Sabrina G. Nov 17 '16 at 7:56
  • $\begingroup$ How to show that $\phi$ is surjective? $\endgroup$ – math112358 Aug 3 at 10:46

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