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Let $A$ be a $C^*$-algebra and $A^{op}$ it's opposite $C^*$-algebra. Let $id:A\to A^{op}$ be the identity map. $id$ is positive.

The claim is: $id$ is completely positive iff $A$ is abelian.

I need this statement for further studies.

The direction $\Leftarrow$ is easier to understand (I hope that everything is correct): If $A$ is abelian, then $A^{op}$ is abelian too. One can identify $A^{op}$ with $C_0(X)$ for a locally compact Hausdorff space $X$ and consider $id$ as a map $A\to C_0(X)$. This map is positive, hence completely positive (it's a standard fact about completely positive maps). It follows that $id:A\to A^{op}$ is completely positive.

But I'm stuck to prove $\Rightarrow$. Can anybody give me some hints or does anybody know good references? It will be greatly appreciated.

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2 Answers 2

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In the unital case, it is a result of Choi that a unital 2-positive map $f:A\to B$ is a $*$-homomorphism if and only if $f(a^2) = f(a)^2$ for all self-adjoint $a\in A$. In the present case, this implies that if $\mathrm{id}:A\to A^{\mathrm{op}}$ is 2-positive, then it is a $*$-homomorphism, i.e. $\mathrm{id}(ab)=\mathrm{id}(a)\mathrm{id}(b)$, which means that $ab=ba$, so that $A$ is abelian.

By the way, the $\Leftarrow$ direction does not need Gelfand duality. $A$ being abelian means precisely that $A=A^{\mathrm{op}}$, and $\mathrm{id}:A\to A$ is trivially completely positive.

Conclusion: If $A$ is unital, then $A$ is abelian if and only if $\mathrm{id}:A\to A^{\mathrm{op}}$ is 2-positive.

I don't know whether the above argument can be generalized to cover the non-unital case as well. One way to go about this would be to show that the extension of $\mathrm{id}:A\to A^{\mathrm{op}}$ to the unitization is still 2-positive, or otherwise to go through Choi's arguments and see to what extent they actually require unitality.

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  • $\begingroup$ thanks for the input! you are right, for the $\Leftarrow$ direction it is overkill to use Gelfand duality. I will try to generalize the argument for the $\Rightarrow$ direction to the non-unital case in the next days. $\endgroup$
    – user62639
    Commented May 22, 2016 at 15:39
  • $\begingroup$ @dr.mop: even if the argument doesn't generalize to the non-unital case, one can still combine it with Nik's idea of going to the double dual. $\endgroup$
    – Tobias Fritz
    Commented May 24, 2016 at 11:35
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    $\begingroup$ But one can always extend a completely positive (contractive) map to the unitalization via $\tilde{f}(\lambda+a)=\lambda+f(a)$, and this becomes u.c.p. Hence Choi's result applies also in the non-unital case. $\endgroup$
    – Gabor Szabo
    Commented May 27, 2016 at 12:32
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I'm not sure where I saw the following argument: it might have been mentioned in a book, or a paper, or a lecture.

I seem to remember that every non-abelian ${\rm C}^*$-algebra contains a *-subalgebra isomorphic to ${\bf M}_2$. EDIT: this is not the case, as pointed out in comments: however, every non-abelian von Neumann algebra does contain a *-subalgebra isomorphic to ${\bf M}_2$. Since the double-adjoint of a c.p. map $A\to A^{\rm op}$ will be a c.p. map $A^{**} \to (A^{\rm op})^{**} = (A^{**})^{\rm op}$, one merely needs to show the identity map ${\rm j}: {\bf M}_2 \to {\bf M}_2^{\rm op}$ is not completely positive.

Well, since the transpose map is a *-isomorphism from ${\bf M}_2^{\rm op}$ onto ${\bf M}_2$, so if ${\rm j}$ were c.p. then the transpose map on ${\bf M}_2$ would also be c.p., and this is well known to be false.

There should be a more intrinsic argument, that doesn't require this "embedding of a copy of ${\bf M}_2$" result, but I can't see how to do it right now.

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    $\begingroup$ That can't be right --- it would imply that every nonabelian C*-algebra contains a nontrivial projection. $\endgroup$
    – Nik Weaver
    Commented May 21, 2016 at 23:06
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    $\begingroup$ But why not pass to the second dual? If the identity map from $A$ to $A^{op}$ is completely positive then surely this is also true of $A^{**}$ by taking the second adjoint. Then your argument works because the second dual of any nonabelian C*-algebra is a nonabelian von Neumann algebra, so it certainly contains a copy of $M_2$. $\endgroup$
    – Nik Weaver
    Commented May 21, 2016 at 23:18
  • $\begingroup$ Every non-abelian C* algebra contains a subalgebra which maps onto M$_2$? I can't remember where I saw that, either. But it's pretty easy. Take a nonzero square-zero nilpotent (these exist precisely when the C*-algebra is nonabelian); the C*algebra it generates is a quotient of the free product of two copies of the two-element group, etc. $\endgroup$
    – David Handelman
    Commented May 21, 2016 at 23:22
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    $\begingroup$ @dr.mop: $A^{**}$ is a von Neumann algebra, so it necessarily has a unit. As I said above, every nonabelian von Neumann algebra contains a copy of $M_2$. This is elementary: every von Neumann algebra is generated by its projections, so it if it nonabelian then it must contain two noncommuting projections. $\endgroup$
    – Nik Weaver
    Commented May 23, 2016 at 16:36
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    $\begingroup$ You then look at the general form of two projections in $B(H)$ and the von Neumann algebra they generate, and find a copy of $M_2$ there. For details see Takesaki vol. 1 Theorem V.1.41. $\endgroup$
    – Nik Weaver
    Commented May 23, 2016 at 16:37

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