# center of a $C^*$-algebra

Does there exist a $$C^*$$-algebra $$A$$ such that the center of $$A$$ is $$0$$ and $$A$$ also has a tracial state?

I know the fact that the center of $$\mathcal{K}(H)$$ is $$0$$, but $$\mathcal{K}(H)$$ has no tracial states.

• But the center of $A$ is not 0,$(0,Id)$ is the center element. – mathrookie Jan 12 at 12:38
• Oops, you're right, I'll delete that erroneous claim – Yemon Choi Jan 12 at 13:07

There are lots of AF simple nonunital C$$^*$$-algebras with (finite) traces (these have trivial centre). For example, begin with the usual $$2^{\infty}$$ UHF algebra, call it $$B$$, and take an infinite strictly increasing sequence of projections, $$(p_n)$$, and form $$\cup_n p_n B p_n$$; let $$A$$ be its closure. Then $$A$$ is simple (easy to check); it is nonunital (very easy) [and being simple, is thus centreless]; it has unique trace (given by the restriction of the unique trace on $$B$$) (easy). $$A$$ is of course a hereditary subalgebra of $$B$$, and there are hordes of similar examples.

If we set $$\alpha$$ to be the limit of the traces of the $$p_n$$, then there is a significant difference (certainly wrt classification) between the algebras we obtain when $$\alpha$$ is rational, and when $$\alpha$$ is irrational.

• Suppose $\tau$ is the trace of $B$,how to check that $\tau|_{A}$ is nonzero? – mathrookie Jan 12 at 16:30
• Since $B$ is simple, $\tau (aa*) >0$ for all $a \in B$ (since the set of $a \in B$ st $\tau (aa*) = 0$ is an ideal of $B$), hence for $a \in A$. – David Handelman Jan 12 at 19:16
• @mathrookie Of course, I meant "for all nonzero $a \in B$"; I wish we could edit comments. – David Handelman Jan 12 at 19:27
• Thank you so much!I have another qusetion:Does there exist a unital $C^*$ algebra $A$ without tracial states such that for some ideal $I$ of $A$ ,I has a tracial state and the center $Z(I)$ of $I$ is 0? – mathrookie Jan 14 at 2:09


Recall that $${\mathcal T} \subseteq \BH$$ where $$\H=\ell^2({\mathbf N}_0))$$ and $$\cT/\KH\cong C(\bT)$$. Let $$q: \cT \to C(\bT)$$ be the corresponding quotient homomorphism, let $$J=\{ f\in C(\bT) \colon f(1)=0\}$$ and let $$\cT_0= q^{-1}(J)$$. More concretely, $$\cT_0$$ is the set of closed algebra generated by all Toeplitz operators $$T_f$$ for which the symbol function $$f$$ belongs to $$J$$.

Since $$\cT_0$$ quotients onto the commutative $${\rm C}^*$$-algebra $$J$$ it has loads of tracial states (just pull back the states on $$J$$).

On the other hand, suppose $$a\in Z(\cT)$$. Then $$[a,k]=0$$ for all $$k\in \KH$$. But it is standard and not too difficult to show that if $$b\in\BH$$ satisfies $$[k,b]=0$$ for all $$k\in\KH$$, then $$b$$ is a scalar multiple of the identity. Since $$q(I_\H)\notin J$$ the only possibility is that $$a=0$$.

• Nice answer. But the statement that $\mathcal{T}_0$ consists of Toeplitz operators whose symbol vanishes at $1$; you need to throw in the compact perturbations or allow products. – Mateusz Wasilewski Jan 12 at 14:46
• Irrational rotation algebras (quantum tori) are another example. – Nik Weaver Jan 12 at 14:56
• Also, the reduced group C*-algebra of any discrete group always has a tracial state. – Nik Weaver Jan 12 at 15:03
• Or just $A = M_n$ ... – Nik Weaver Jan 12 at 15:03
• Thanks Mateusz. @NikWeaver: the OP wants $Z(A)=0$ not $\dim Z(A)=1$ – Yemon Choi Jan 12 at 15:51