orthogonal projections in $C^*$ algebra

Question

Suppose $A$ is an arbitrary $C^*$ algebra,can $A$ be linear spanned by all orthogonal projections of in it ?

If not,is there a relationship between a $C^*$ algebra and all orthogonal projections in it?

Answer 1

No relation in general.

There are many C$^*$-algebras that only have one projection, namely $0$. For instance $C_0(T)$ for any locally compact, non-compact, connected $T$. Say, $C_0(\mathbb R^n)$ for any $n$.

Among unital C$^*$-algebras, you always have $0$ and $1$ as projections. But often they are the only ones. Among the most famous cases are $C[0,1]$ and $C_r^*(\mathbb F_n)$, $n\in1+\mathbb N$.

At the other end of the spectrum, some C$^*$-algebras are indeed spanned by their projections (no closure needeed; this is a very non-trivial result). Among them are $B(H)$ for any dimension of $H$, $UHF(2^\infty)$, Irrational Rotation Algebras, Bunce-Dedens algebras, and any simple unital purely infinite algebra, like the Cuntz algebras $\mathcal O_n$ (and more generally the Cuntz-Krieger algebras).

The algebra of compact operators $K(H)$, when $H$ is infinite-dimensional, gives an example of a C$^*$-algebra that is the closed span of its projections, but it is not equal to the span of its projections.

Answer 2

No. The Jiang-Su algebra is infinite dimensional and it's only projections are $0$ and $1$.

Answer 3

For a von-Neumann algebra $A$, the linear span of all orthogonal projections of $A$ is dense in $A$ respect to the norm. When $A$ is finite-dimensional, it is linear spanned by all orthogonal projections in it.