If $A$ is a nuclear $C^*$ algebra, $A^{op}$ is the opposite $C^*$ algebra, is the conclusion: "there is a unique injective $*$-homomphism $\pi \colon A\otimes A^{op}\rightarrow M(A)\otimes M(A)^{op}$" correct?

$M(A)$ is the multiplier algebra of $A$.


closed as off-topic by math112358, Saad, KReiser, max_zorn, Brahadeesh Nov 23 '18 at 6:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, KReiser, max_zorn, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.


You always have the canonical inclusion, given by $\mathrm{id} \otimes \mathrm{id}$. However, you can also consider $\alpha \otimes \beta$, where $\alpha$ and $\beta$ are automorphisms of $A$ resp. $A^{\mathrm{op}}$. In general these $*$-hom. will not be equal.

  • $\begingroup$ I have edited the post. $\endgroup$ – user42761 Oct 8 '18 at 19:51
  • $\begingroup$ No, why should it be true ? The one I wrote in my answer does not extend the identity. $\endgroup$ – user42761 Oct 11 '18 at 14:37

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