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The category of $C^{*}$ algebras is denoted by $\mathcal{A}$.

Is there a functor $\mathcal{F}$ on $\mathcal{A}$ which send each object $A\in \mathcal{A}$ to its center $Z(A)$. In the other words, can we extend the maping $A\mapsto Z(A)$ on objects to a functor on this category?

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    $\begingroup$ Algebra homomorphisms don't always send centers to centers, so the obvious idea of defining $Z(\alpha\colon A \to B)$ to be the restriction of $\alpha$ to $A(Z)$ doesn't work. $\endgroup$ Sep 30, 2015 at 15:55
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    $\begingroup$ Just as general background information: it is a standard exercise in various introductions to category theory, to show that there is no functor on the category of groups which sends each group to its centre. If one knows this, then one would guess that the answer to your question is negative, as indeed seems to be the case $\endgroup$
    – Yemon Choi
    Sep 30, 2015 at 21:25
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    $\begingroup$ Centers tend fit into category theory in a different way, as being the natural automorphisms of identity functors. (e.g. for a ring $R$, view it as a preadditive category with one object. Or view it as $R$-Mod) $\endgroup$
    – user13113
    Oct 1, 2015 at 0:49

2 Answers 2

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Here is another attempt at proving no such functor exists — I apologize to Chris and to Manny if something like this is already in the papers which they cite.$\newcommand{\Mat}{{\bf M}}\newcommand{\Cplx}{{\bf C}}\newcommand{\Cst}{{\rm C}^*}$

Let $\Mat_2$ denote the algebra of $2\times 2$ complex-valued matrices, and let $D_2$ be the algebra $\Cplx\oplus\Cplx$ regarded as the subalgebra of $\Mat_2$ consisting of diagonal matrices. Let $$ A = \{ f\in C([0,1], \Mat_2) \mid f(1)\in D_2\}. $$ Since $Z(\Mat_2)=\Cplx$, one shows (with a small appeal to continuity) that $$ Z(A)= \{ f\cdot I_2 \mid f\in C([0,1])\}\cong C([0,1]).$$

Let $\phi: D_2 \to A$ be the $*$-homomorphic embedding $\phi(d)(t) = d$ for all $t\in [0,1]$ and all $d\in D_2$. Let ${\rm ev}_1 : A\to D_2$ be the homomorphism "evaluate at $1$".

Now suppose $\mathcal F$ is a functor on the category of $\Cst$-algebras and $*$-homomorphisms, such that ${\mathcal F}(B)=Z(B)$ for every $\Cst$-algebra $B$. Then ${\mathcal F}({\rm ev}_1) \circ {\mathcal F}(\phi)$ must be the identity homomorphism on ${\mathcal F}(D_2)=D_2$, and thus the identity homomorphism on $D_2$ must factor somehow through the algebra $Z(A)\cong C([0,1])$. But this is impossible, since there is no injective algebra homomorphism from $D_2$ into $C([0,1])$. One way to see this last claim is to note that $D_2$ contains two non-trivial projections $e_1$ and $e_2$ which sum to $1$, while the only projections in $C([0,1])$ are $0$ and $1$.


Update 7th Oct. 2015

In response to a comment: we can find examples of a retract $D\to B \to D$ (that is, the composition of these two maps is the identity map on $D$) where $D$ is commutative and infinite-dimensional and $B$ has trivial centre.

Here is one way. Let $A$ be an infinite, countable group, and $H$ a countable amenable group, equipped with an action $\alpha: A\to {\rm Aut}(H)$. Form the semidirect product $G=H\rtimes_\alpha A$, which can be defined as the set $H\times A$ equipped with the product $$ (h,a)(k,b) = (h\cdot\alpha_a(k), ab). $$ The subset $\{e_H\} \times A$ is a subgroup of $G$; there is a quotient map of $G$ onto $A$, defined by "projection in the second variable"; and doing the inclusion $A \to G$ followed by the quotient $G\to A$ gives the identity map on $A$. (In other words, we have a retract.)

Note, for later reference, that $$(h,a)^{-1} = ({\alpha_a}^{-1}(h^{-1}), a^{-1} )$$ and so, using the fact that $A$ is abelian, a short calculation gives $$ (h,a) (k,b) (h,a)^{-1} = ( h\alpha_a(k)\alpha_b(h^{-1}), b). $$

I now want to impose conditions on $\alpha$ which will ensure that $G$ is an ICC group, i.e. all conjugacy classes except $\{e\}$ are infinite. It is well known that if $G$ is countable and ICC then its group von Neumann algebra ${\rm VN}(G)$ has trivial centre: since the centre of $\Cst_r(G)$ is contained in the centre of ${\rm VN}(G)$, we deduce that $\Cst_r(G)$ has trivial centre.

To find such conditions: note that $(0,a)(k,b)(0,a)^{-1} = (\alpha_a(k), b)$ and $(h,0)(e,b)(h,0)^{-1} = (h\alpha_b(h^{-1}), b)$.

So if we assume that $\alpha$ satisfies the following conditions

1) for each $k \in H \setminus\{e_H\}$, the orbit $\{\alpha_a(k) : a\in A\}$ is infinite

2) for each $b\in A\setminus\{e_A\}$, the set $\{h \alpha_b(h^{-1}): h\in H\}$ is infinite

then $G$ will be ICC, as desired, and so $\Cst_r(G)$ will have trivial centre. $\newcommand{\Cst}{{\rm C}^*}$

Recall that on the category ${\sf Grp}$ of (discrete) groups and homomorphisms, the full group $\Cst$-algebra defines a functor ${\sf Grp}\to {\sf Cstar}$ is functorial. Since $A \to G \to A$ is a retract in ${\sf Grp}$, we get a retract $\Cst(A)\to \Cst(G)\to\Cst(A)$. Clearly $\Cst(A)$ is commutative and infinite-dimensional. Now since $A$ and $H$ are amenable, so is $G$; hence $\Cst(G)=\Cst_r(G)$, which has trivial centre by the remarks above.

To get an actual, concrete example: let $A=\{2^n : n \in {\mathbb Z}\}$ be the multiplicative group formed by all integer powers of $2$, and let $H$ be the set of dyadic rationals, regarded as an additive group. The action of $A$ on $H$ is just by scaling, i.e. for $a\in A$ and $h\in H$ we have $\alpha_a(h)=ah$. It is easily checked that both conditions 1) and 2) are satisfied; note that $e_H=0$, $e_A=1$, and $h\alpha_b(h^{-1})$ is just $h-bh$.

[I learned of the existence of solvable countable ICC groups in a seminar many years ago, which is why I knew there should be some way to manufacture examples that did what you want, once one has the idea of trying to build an example in ${\sf Grp}$ and transport it functorially to ${\sf Cstar}$.]

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  • $\begingroup$ thanks again for the answer. Can one construct an alternative example : a commutative infinite dim A and a B with trivial center with morphisms $\alpha:A\to B$ and $\beta:B \to A$ such that $\beta \circ \alpha$ is an automorphism? $\endgroup$ Oct 6, 2015 at 15:01
  • $\begingroup$ @AliTaghavi I think we can do this, and I have updated my answer to give a reasonably complete sketch. $\endgroup$
    – Yemon Choi
    Oct 7, 2015 at 18:36
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This paper proves that there is no functor from the category of C* algebra (and morphisms) to the category of locales/topological spaces/a lot of other things that extend the gelfand duality and send matrix algebra for $n >2$ to some non-empty space. Composing your eventual "center functor" with gelfand spectrum would give such a functor.

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    $\begingroup$ In fact, this sort of thing fails already in case $n = 2$, using more elementary arguments. See Proposition 4.5 of the following, noting that the morphisms in question are $*$-homomorphisms in case the field is $\mathbb{C}$: arxiv.org/abs/1101.2239 $\endgroup$ Sep 30, 2015 at 18:35
  • $\begingroup$ @Simon thank you very much for your very interesting answer and helpful link. $\endgroup$ Oct 4, 2015 at 11:27
  • $\begingroup$ @Simon Thanks again for your answer. according to the previous version of your answer, are you sure that one can omit the action on morphisms? $\endgroup$ Oct 6, 2015 at 14:52
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    $\begingroup$ @AliTaghavi : I removed that part of my answer because Yemon Choi answer below gives a more convincing prof of this fact. $\endgroup$ Oct 6, 2015 at 19:52
  • $\begingroup$ @SimonHenry thank you very much for your comment. $\endgroup$ Oct 6, 2015 at 19:54

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