# Number of connected components of a $C^{*}$ algebra

Inspired by the concept in the following post What are these compact sets called? We introduce the following concept:

Let $A$ be a unital $C^{*}$ algebra. We consider the unitary equivalent relation on the set of all projections in $A$. We denote by $n(A)$, the number of equivalents classes of projections in $A$. The number of connected components of $A$ is denoted by $ncc(A)$ and is defined by $ncc(A)=log_{2} n(A)$. For a non unital $C^{*}$ algebra $A$, we define $ncc(A)=ncc(M(A))$ where $M(A)$ is the multiplier algebra of $A$. This is a natural definition because for a locally compact Hausdorff space $X$ we have $ncc(C_{0}(X))=$ The number of connected components of $X$.

The concept in the above post is a motivation to ask:

Question 1: Assume that We have an extension $0\to A\to B\to C\to 0$ of $C^{*}$ algebras with $ncc(A)< \infty$ and $ncc(C)< \infty$. Does this implies that $ncc(B)$ is finite, too? How can $ncc(B)$ be controlled by $ncc(A)$ and $ncc(C)$?

Question 2: How can $ncc(A\otimes_{min} B)$ be controlled in term of $ncc(A)$ and $ncc(B)$? When $A$ is commutative, is it true to say that $ncc(A\otimes B) \leq ncc(A).ncc(B)$?

Note: The commutative case gives us enough motivation for such noncommutative questions.

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As a sign of good faith, why don't you tell us what you've been able to work out when A is a noncommutative finite-dimensional Cstar algebra? – Yemon Choi May 27 '14 at 4:11
@YemonChoi For the matrix algebra n(A) is n+1.So when A,C are finite dimensional algebra, the first question is obvious, the second is not a so hard(for large value n, m) – Ali Taghavi Feb 6 at 10:55
Thanks. So according to your definition, $ncc(M_n({\bf C}))= \log_2(n+1)$? This seems slightly odd, and it's not clear to me why this definition is better than looking at equivalence classes of projections in stabilized versions of the algebra (i.e. $K_0)$. – Yemon Choi Feb 6 at 15:30
Also, why does an answer for the matrix algebra case tell you what is going on for the finite-dimensional case? It seems perfectly possible that if $A\subset M_n({\bf C})$ then two projections in $A$ could be unitarily equivalent in $M_n({\bf C})$ without being unitarily equivalent in $A$... – Yemon Choi Feb 6 at 15:32
@YemonChoi Thank s for your comments.yes it is possible. for example $A=M_{2}(\mathbb{C})\oplus M_{2}(\mathbb{C})\subset M_{4}(\mathbb{C})$. Let $e\in M_{2}(\mathbb{C})$ be the projection on $x$-axis. Then $e\oplus I_{2}$ and $I_{2} \oplus e$ are similar in the whole algebra but not similar in $A$. after all, I think the main question in the post is obvious for finite dim algebras since any such algebra is a direct sum of matrix algebras. – Ali Taghavi Feb 6 at 16:49