# tracial state on a unital infinite dimensional simple $C^*$ algebra

If $$A$$ is a finite dimensional simple $$C^*$$ algebra,then it has the form of $$M_n(\mathbb{C})$$,which has unique tracial state.

My question is:If $$A$$ is an unital infinite dimensional simple $$C^*$$ algebra,what is the structure of $$A$$?Does it have a relationship with matrix algebras?I cannot think of a concrete example (unital infinite dimensional simple $$C^*$$ algebra whose tracial state exists).

For a "matrix-like" example, you can take UHF$$(2^\infty)$$. It's simple and it has a unique tracial state.

But there are lots of examples of different kinds. If you take any infinite discrete group $$G$$, you can form $$C_r^*(G)$$, the reduced C$$^*$$-algebra (i.e., the C$$^*$$-algebra generated by the left-regular representation). For any $$G$$, this always has a faithful trace, namely $$\tau(x)=\langle xe,e\rangle,$$ where $$e$$ is the vector induced by the unit of $$G$$. And, for many groups, $$C_r^*(G)$$ is simple. The free groups, for instance So $$C_r^*(\mathbb F_n)$$, $$n\in\mathbb N$$ are examples of infinite-dimensional, simple C$$^*$$-algebras with a faithful trace.

Among many, $$G$$ above could be any of

• $$\mathbb F_n$$, $$n\in\mathbb N$$
• PSL$$_n(\mathbb Z)$$, $$n\geq2$$
• non-trivial free products
• non-solvable subgroups of PSL$$_2(\mathbb R)$$
• torsion-free non-elementary Gromov hyperbolic groups
• any Zariski-dense subgroup with with centre reduced to $$\{1\}$$ in a connected semi-simple real Lie group without compact factor
• centerless mapping class groups and outer automorphism groups of free groups
• irreducible Coxeter groups which are neither finite nor affine
• Tarski monster groups
• Not necessarily, but it is in many cases. – Martin Argerami Nov 2 '18 at 15:05
• In $\ell^\infty$, there are uncountably many pairwise orthogonal infinite projections. These pass to the quotient. Any C$^*$-algebra with uncountable many pairwise orthogonal projections cannot have a faithful state. You would have, for any finite set $F$, $$1=\varphi(1)\geq\varphi(\sum_{ j\in F}p_j)=\sum_{j\in F}\varphi(p_j).$$ It follows that $\sum_j\varphi(p_j)<\infty$, but this requires only countably many to be nonzero. – Martin Argerami Nov 2 '18 at 16:38