4
$\begingroup$

In this question which I flag it as a community wiki, I search for a big list of $C^{*}$ algebras(and a big list of criterions) which do not admit a non trivial idempotent $C^{*}-$morphism.

I know 2 cases: Simple $C^{*}$ algebras, $B(H)$, where $H$ is a separable Hilbert space.

$\endgroup$
8
  • $\begingroup$ @DuchampGérardH.E. Thank you for your comment. can I ask you to more explain? By i dempotent morphism I mean a C* morphism $T:A \to A$ with $T^{2}=T$ $\endgroup$ Commented Aug 12, 2015 at 12:16
  • $\begingroup$ I replied too quickly, sorry ! I withdraw my past comment and will think of a better one. Thank you for insisting on the adjective "idempotent". $\endgroup$ Commented Aug 12, 2015 at 18:09
  • $\begingroup$ @DuchampGérardH.E. You are well come and thanks for your attention to my question. Regarding B(H), the argument which was in my mind is the following: every C* morphism on B(H) is injective, otherwise we would obtain a C* embedding of the Calkin algebra B(H)/K(H) into B(H). But it is well known that the calkin algebra can not be embedded in B(H) $\endgroup$ Commented Aug 12, 2015 at 18:30
  • $\begingroup$ Do you use the fact that, if $H$ is separable and of infinite dimension, the Calkin algebra $B(H)/K(H)$ is simple ? $\endgroup$ Commented Aug 12, 2015 at 18:36
  • $\begingroup$ @DuchampGérardH.E. as i wrote in previous comment, I learned from various reference (book or oral communication) that the Calkin algebra can not be embedded in B(H). I confess that I did not nter in the proof of this fact. $\endgroup$ Commented Aug 12, 2015 at 18:42

2 Answers 2

9
+25
$\begingroup$

You are looking for C*-algebras $\mathcal{A}$ which lack a nontrivial $*$-homomorphism $\phi: \mathcal{A} \to \mathcal{A}$ satisfying $\phi^2 = \phi$. That is equivalent to having a surjective $*$-homomorphism $\phi: \mathcal{A} \to \mathcal{B}$ together with a $*$-homomorphism $\psi: \mathcal{B} \to \mathcal{A}$ satisfying $\phi\circ\psi = {\rm id}_\mathcal{B}$. That is, your problem is equivalent to finding C*-algebras which cannot be the middle term of a nontrivial split exact sequence $$0 \to \mathcal{I} \to \mathcal{A} \to \mathcal{B} \to 0$$ (where "nontrivial" means neither $\mathcal{I}$ nor $\mathcal{B}$ is zero).

Well, being the middle term of a nontrivial split exact sequence is a very unusual property. "Most" C*-algebras in nature won't have this property. Any C*-algebra which is a direct sum of two nontrivial C*-algebras does have this property, as does any C*-algebra which is the unitization of another C*-algebra. (By $0 \to \mathcal{A} \to \tilde{\mathcal{A}} \to \mathbb{C} \to 0$.) Beyond that it gets kind of hard to come up with examples that have the property you are trying to avoid.

$\endgroup$
1
  • $\begingroup$ Thank you very much for your interesting answer.However in NC case the situation is unusual, but in commutative case we have an abundance of such situation. the property is equivalent to "Retract" in general topology. I realized this operator language for retract of general topology from page 3 of the Book of Allen Hatcher algebraic topology. any way it would be interesting to think whether this property is invariant by tensor product or is invariant from A to M_{n}(A). $\endgroup$ Commented Aug 12, 2015 at 18:39
1
$\begingroup$

This is not a direct answer, rather a small contribution of what happens in the opposite case as well as an answer to a comment of the OP. Hope that it can help.

Of course, if your $C^*$-algebra $\mathcal{B}$ admits such an idempotent endomorphism (call it $\alpha$), you have a (straightforward) structure theorem $$ \mathcal{B}=ker(\alpha)\oplus im(\alpha)=\mathcal{I}\oplus \mathcal{B}_1 \qquad\qquad (*) $$ $\mathcal{I}$ is a two-sided ideal and $\mathcal{B}_1$ a sub-$C^*$-algebra. Even if this direct sum is not trivial, the decomposition of $1_\mathcal{B}$ can be trivial.

Before giving examples, I elaborate a bit on the vein of retracts indicated by Ali and underlined by Nik. I provide it in the framework of operator-valued algebras in order to answer a comment of the OP, as $\mathcal{C}(K)\widehat{\otimes} M_n(\mathbb{C})\simeq \mathcal{C}(K,M_n(\mathbb{C}))$.

Let $\mathcal{A}$ be a $C^*$-algebra and $K_1\subset K_2$ be two compact Hausdorff sets, we suppose that it exists a continuous retraction $r : K_2\rightarrow K_1$. Let $$ res_{1,2} : \mathcal{C}(K_2,\mathcal{A})\rightarrow \mathcal{C}(K_1,\mathcal{A})\mbox{ and } \delta_r : \mathcal{C}(K_1,\mathcal{A})\rightarrow \mathcal{C}(K_2,\mathcal{A}) $$
be respectively the natural restriction and $\delta_r(f)=f\circ r$. One checks immediately that\ $res_{1,2}\circ \delta_r=Id_{\mathcal{C}(K_1,\mathcal{A})}$ and that $\alpha=\delta_r\circ res_{1,2}$ is an idempotent $*$-endomorphism of $\mathcal{C}(K_2,\mathcal{A})$.

We come back to the notation around (*).

Below two examples : in the first, $\alpha$ is due to the multiplication by an idempotent (which serves as unit for $\mathcal{B}_1=Im(\alpha)$) and the second one, transformed in its noncommutative incarnation, where this is not the case (one can take $\mathcal{A}=M_n(\mathbb{C})$ with $n\geq 2$, but it is true for all $\mathcal{A}$).

Example 1 Direct sum of two (non trivial) $C^*$-algebras : in this case $1_{\mathcal{B}_1}\not= 1_\mathcal{B}$

Example 2 Where $1_{\mathcal{B}_1}=1_\mathcal{B}$ and therefore the projection of the unity on $\mathcal{I}$ is zero. Take the following real intervals $$ K_1=[0,1];\ K_2=[0,2], \mathcal{B}=\mathcal{C}(K_2,\mathcal{A}),\ \mathcal{B}_1=\mathcal{C}(K_1,\mathcal{A})\ . $$ The retraction $r : K_2\rightarrow K_1$ being given by the identity on $[0,1]$ and $r(t)=r(1)=1$ on $[1,2]$. It becomes clear that, for $f\in \mathcal{B}$, $\alpha(f)=g$ where $g(x)=f(x)$ on $K_1=[0,1]$ and $g(x)=f(1)$ on $[1,2]$ then the image of $\alpha$ is the sub-$C^*$-algebra of (continuous) functions which are constant on $[1,2]$, it contains the identity and the supplement $\mathcal{I}$ is the space of functions vanishing at $1$. We are in the second case.

$\endgroup$
3
  • $\begingroup$ Your example 2 is just a special case of the general phenomenon of topological retracts (noted by Ali Taghavi in his comment to my answer). $\endgroup$
    – Nik Weaver
    Commented Aug 13, 2015 at 14:18
  • 1
    $\begingroup$ Could anyone explain this downvote ? $\endgroup$ Commented Aug 14, 2015 at 16:27
  • $\begingroup$ @DuchampGérardH.E. Thank you very much for your interesting answer, as I gave an upvote to it. $\endgroup$ Commented Aug 19, 2015 at 6:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .