Let $A$ and $B$ be $C^*$-algebras and let $\alpha$ be any $C^*$-norm on the algebraic tensor product $A⊙B$. Why is $A⊗_{\alpha}B$ is the subalgeba of $A\otimes_{max} B$?

In many reference books,they mentioned that there was a natural surjective * homomorphism from $A\otimes_{max} B$ to $A⊗_{\alpha}B$ ,which reveals that $A⊗_{\alpha}B$ is the subalgeba of $A\otimes_{max} B$.

According to the definition,$A\otimes_{max} B$ denote the $||.||_{max}$, $A⊗_{\alpha}B$ is the completion of $A⊙B$ with respect to $\alpha$,$||.||_{max}$ is the maximal $C^*$-norm,I prove that $A\otimes_{max} B$ is the subalgebra of $A⊗_{\alpha}B$(refer to Completion with respect to stronger norm is no subset?) .
Can anyone point out my mistake?Thanks in advance!


It isn't. The existence of a $*$-epimorphism $\phi:A\to B$ does not tell you that $B$ can be seen as a subalgebra of $A$; only as quotient, which is most often not a subalgebra.

To see an easy example, let $A=C_0(\mathbb R)$, $B=\mathbb C$. Then you have for instance the $*$-epimorphism $\phi:A\to B$ given by $\phi(f)=f(0)$, but no subalgebra of $C_0(\mathbb R)$ is isomorphic to $\mathbb C$.

  • $\begingroup$ There was a natural surjective * homomorphism from $A\otimes_{max} B$ to $A⊗_{\alpha}B$,can we deduce that $A\otimes_{max} B$ is larger than $A⊗_{\alpha}B$? $\endgroup$ – math112358 May 13 '18 at 9:19
  • $\begingroup$ You would have to tell me what "larger" means. $\endgroup$ – Martin Argerami May 13 '18 at 13:56
  • $\begingroup$ I mean $ A\otimes_\gamma B$ is the subset of $A\otimes_{max} B$? I prove that $A\otimes_{max} B$ is the subset of $ A\otimes_\gamma B$ above,Can you point out my mistake?Thanks! $\endgroup$ – math112358 May 13 '18 at 14:17
  • $\begingroup$ I cannot point a mistake because I see no proof. But it is just not true, as the answer to the question you quoted shows. A Cauchy sequence for $\|\cdot\|_\alpha$ needs not be Cauchy for $\|\cdot\|_\max$. $\endgroup$ – Martin Argerami May 13 '18 at 14:25
  • 1
    $\begingroup$ In an algebra where both $\|\cdot\|_\alpha$ and $\|\cdot\|_\infty$ are defined, yes. $\endgroup$ – Martin Argerami May 17 '18 at 12:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.