Vtyjkyuk

Given $X=0 cup 1/n$. We can show that it is a compact Hausdorff space.
Now take $M_n$ to be the matrix algebra. I am confusing on showing that both $C(X, M_2)$ and $B=f in C(X, M_2)$ where $f(0)$ is diagonal$$ are $C^*$-algebras and finding the two-sided closed ideals with respect to each other.

I am trying to define the norm $||f||:=sup||f(x)||$ for all $xin X$ in which the right-hand side norm is just the operator norm defined for matrix algebra, for $f(x)$ is some matrices in $M_2$. So when calculating the operator norm, does it just mean to find the largest eigenvalue? Also take the $f^*(x)=barf(x)$, could anybody tell me how to move on for showing these two spaces are $C^*$-algebra? I understand we need to check the conditions for *-algebra, but what is the difference between these two examples?

I'll be assuming that your set is $X=0cup1/n: nin mathbb N$ (or did you really wanted $X$ with two points?)

The operator norm of a matrix is the largest eigenvalue only if the matrix is positive (or some special selfadjoint cases). In general, the operator norm of a matrix is the largest singular value.

Completeness: if $f_k$ is a Cauchy sequence, it follows that $f_k(x)$ is Cauchy in $M_2$ for each $x$. So convergent (because $M_2$ is complete) and there exists $f(x)=lim f_k(x)$. A typical $varepsilon$ argument shows that $f$ is continuous. For $B$ the argument is the same, and if $f_k(0)$ is diagonal for all $k$, then $f(0)$ is diagonal.

As $Bsubset C(X,M_2)$ and closed, all you need to do is show that it is a $*$-algebra.

The norm you defined is a norm, just using that the operator norm is a norm. Indeed, you have
$$
|f+g|=supf(x)+g(x).
$$
As $|f(x)+g(x)|leq |f(x)|+|g(x)|leq|f|+|g|$, the triangle inequality is proven. If $|f|=0$, then $f=0$ by definition of function. And $|lambda f|=|lambda|,|f|$ again follows from the fact that the operator norm is a norm.

The C$^*$-identity: I assume that the adjoint you want is $f^*(x)=f(x)^*$. Then
$$
|f^*f|=sup=sup^2: xin X=|f|^2.
$$
As for the ideals, since $M_2$ is simple, it is not hard to show in general that the (closed) ideals of $C(X,M_2)$ are of the form $$tag*I_Y=_Y=0,$$ where $Ysubset X$ is closed. The closed subsets of $X$ are those that

• are finite; or,




• are infinite and contain $0$

So any such $Y$ gives rise to an ideal of $C(X,M_2)$ via $(*)$.

As for ideals of $B$, note that if $Jsubset B$ is an ideal, then for each $xin X$, $$J_x=f(x): fin J$$ is an ideal of $M_2$. As the latter is simple, either $J_x=0$ or $J_x=M_2$. And $J_0$ is an ideal of $M_2$ that consists of diagonal matrices, so $J_0=0$. If $Y=x: J_x=0$, then $Y$ is closed (using the continuity of the functions), $0in Y$, and
$$tag**
J=fin B: f.
$$
Indeed, the argument above showed that the inclusion $subset$ holds. For the reverse inclusion, see the argument here.