Algebras of matrix valued continuous functions.

Question

Let us consider two $C^*$ algebras $C_1$ and $C_2$.

We define $C_1$ to be the algebra of continuous functions from $f:[0,1]\to M_{2\times 2}(\mathbb{C})$ such that $f(0$) is diagonal, and define $C_2$ to be the continuous functions $f:[0,1]\to M_{2\times 2}(\mathbb{C})$ such that $f(0)$ is a scalar multiple of the identity matrix. It turns out these two algebras are not isomorphic, which seems obvious, but I can't quite see why.

We see the quotient of $C_1$ by the ideal of functions vanishing at $0$ is isomorphic to $\mathbb{C}^2$, and the quotient of $C_2$ by the same ideal is isomorphic to $\mathbb{C}$. I'm wondering if I can somehow use this observation. I'm not seeing how to though.

Any help would be much appreciated.

Answer 1

I don't see an obvious way of doing this with ideals. I would think that $C_1$ has two minimal ideals $\{f:\ f(0)_{jj}=0\}$, $j=1,2$) and $C_2$ does not, but I couldn't easily write a proof of that.

Here is another way to attack the problem (based on the above idea!). Look at the 1-dimensional representations, i.e. the characters. The kernel of a character is a maximal ideal. In $C_2$, it is easy to see that the maximal ideals are $J_t=\{f:\ f(t)=0\}$, so characters have to be point evaluations (but not all point evaluations are characters!). It follows that the only character in $C_2$ is $f\longmapsto f(0)$.

On $C_1$, on the other hand, you have two characters: $f\longmapsto f(0)_{11}$ and $f\longmapsto f(0)_{22}$. As these two characters are distinct, $C_1$ and $C_2$ cannot be isomorphic.

Answer 2

Your idea can be made complete with the usage of $K$-theory. The evaluation maps $\operatorname{ev}:C_1\to\mathbb C^2$ and $\operatorname{ev}:C_2\to\mathbb C$ induce short exact sequences $$0\to D\to C_1\to\mathbb C^2\to 0,$$ $$0\to D\to C_2\to\mathbb C\to 0,$$ Where $D=C_0((0,1],M_2(\mathbb C))$ is the cone of $M_2(\mathbb C)$. Thus $K_0(D)$ and $K_1(D)$ are trivial, and from the long exact sequences in $K$-theory, we see that $$K_0(C_1)\cong K_0(\mathbb C^2)\cong\mathbb Z^2,$$ while $$K_0(C_2)\cong K_0(\mathbb C)\cong\mathbb Z.$$ Since $K_0(C_1)\neq K_0(C_2)$, the two algebras cannot be isomorphic.