13
$\begingroup$

Suppose that $A,C$ are $C^*$-algebras and $\phi:A \to C$ is a completely positive, orthogonality-preserving linear map. (Orthogonality preserving means: if $a,b \in A$ satisfy $ab=0$ then $\phi(a)\phi(b) = 0$.) Then:

(i) For any $a,b,c \in A$, $$ \phi\left(ab\right)\phi\left(c\right) = \phi\left(a\right)\phi\left(bc\right) $$ (in the special case that $A$ is unital, this is equivalent to $\phi\left(a\right)\phi\left(b\right) = \phi\left(1\right)\phi\left(ab\right)$ for any $a,b \in A$);

(ii) For any $a,b \in A$, $$ \left\| \phi\left(ab\right) \right\| \leq \|a\|\cdot\left\|\phi\left(b\right)\right\|; $$

(iii) If $A$ is unital and simple, then for any $a \in A$, $$ \left\| \phi\left(a\right) \right\| = \|a\|\cdot\left\|\phi\left(1\right)\right\|. $$

In fact, there is a rich structure theorem about completely positive, orthogonality-preserving maps (in the literature, they are called "order zero" instead of "orthogonality-preserving" ), Theorem 2.3 of Winter, Zacharias, "Completely positive maps of order zero," Münster J. Math, 2009 (see also Corollary 3.1); and I can prove these statements easily using the structure theorem. But, my question is: can we prove any of the facts above directly (without appealing to this structure theorem)?

(I am intentionally not restating the structure theorem here because my question is about not using it.)

$\endgroup$
8
  • 4
    $\begingroup$ Don't you mean $\phi(a)\phi(b)=0$ in the first paragraph? Also note that the definition of "orthogonal" in the cited paper is more complicated: $a \perp b$ iff $ab=ba=a^* b = a b^* = 0$. $\endgroup$ – Martin Brandenburg Mar 20 '12 at 10:04
  • 1
    $\begingroup$ @Martin: yes, and it's now fixed. I recognize that there are other definitions of orthogonality. The one I gave is rather weak (of course they coincide for self-adjoint elements). These two notions of orthogonality then give two notions of orthogonality-preserving, which turn out to be equivalent. (Weak-orthogonality)-preserving implies (strong-orthogonality)-preserving is immediate, and the other direction follows from the structure theorem I mentioned (eg. in the unital case, if $ab=0$ then by (i), $\phi(a)\phi(b)=\phi(1)\phi(ab) = 0$). $\endgroup$ – Aaron Tikuisis Mar 20 '12 at 15:13
  • 1
    $\begingroup$ I'd be happy to see a direct proof that (strong-orthogonality)-preserving implies (i),(ii), or (iii), or even a direct proof that it implies (weak-orthogonality)-preserving. $\endgroup$ – Aaron Tikuisis Mar 20 '12 at 15:14
  • 3
    $\begingroup$ This doesn't qualify as a direct proof, but (i) can be deduced as a special case of a different general result for Banach algebras, and the hypothesis of complete positivity can be dropped. See the abstract of Maps preserving zero products, J. Alaminos, M. Brešar, J. Extremera, A. R. Villena. Studia Math. 193 (2009), 131-159. dx.doi.org/10.4064/sm193-2-3 (To apply the result mentioned there, note that any unital C*algebra is linearly spanned by the unitaries). IIRC, their proof rests on the fact that the diagonal subgroup of ${\mathbb T}\times{\mathbb T}$ is a set of synthesis $\endgroup$ – Yemon Choi Mar 20 '12 at 21:52
  • 2
    $\begingroup$ I should thank Leonel Robert for helping me to realize that the transpose map is only (strong-orthogonality)-preserving, and not (weak-orthogonality)-preserving. For example, if $$ A= \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} $$ and $$ B= \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} $$ then $AB=0$ but $A^TB^T\neq 0$. We can also see that (i) doesn't hold for the transpose map. Thus, the result of Alaminos, Brešar, Extremera, and Villena only applies to (weak-orthogonality)-preserving maps and not (strong-orthogonality)-preserving ones. $\endgroup$ – Aaron Tikuisis Mar 23 '12 at 15:35
1
$\begingroup$

The general form of a (bounded) orthogonality preserving linear mapping between C*-algebras is obtained here: http://www.sciencedirect.com/science/article/pii/S0022247X08007245

$\endgroup$
0
$\begingroup$

For a proof of the fact (i) "ϕ(ab)ϕ(c)=ϕ(a)ϕ(bc) and ϕ(a)ϕ(b)=ϕ(1)ϕ(ab)", see M.A. Chebotar, W.-F. Ke, P.-H. Lee, N.-C. Wong Mappings preserving zero products, Studia Math., 155 (1) (2003), pp. 77-94 MR1961162 (2003m:47066)

It is really surprising (at least to me) that few operator algebra people knowing of the notion of zero product preserving maps, on which a few dozen papers published since 1980s.

$\endgroup$
2
  • $\begingroup$ This proof only works with the extra assumption that the subalgebra generated by the idempotents is dense, right? $\endgroup$ – Ulrich Pennig Mar 15 '18 at 16:05
  • $\begingroup$ There are many cases. The one about algebras generated by idempotents, e.g., W*-algebras, is easier than the others. But for general C*-algebras, one can make use of open projections (and functional calculus) to get similar results for bounded zero preserving linear maps. Look at the last section of the mentioned Studia paper, or e.g., N.-C. Wong, “Zero product preservers of C*-algebras”, Contemporary Math., 435 (2007), 377-380. $\endgroup$ – NC Wong Mar 16 '18 at 17:20

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.