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Let $A$ be a unital $C^*$-Algebra with non-trivial $K_0$ group. Define $CA = \{f\in C\big([0, 1], A\big)\,\vert\,f(0) = 0\}$. It can be shown that $CA$ is homotopic equivalent to the set $\{0\}$ and hence $K_0(CA)$ is the trivial group. Besides, define $SA = \{f\in C\big([0, 1], A\big)\,\vert\,f(0) = f(1) = 0\}$, $I_0 = \{C([0, 1])\,\vert\,f(0) = 1\}$, $I_{0, 1} = \{f\in C([0, 1])\,\vert\,f(0)=f(1)=0\}$. Then I can write $CA = I_0\bigotimes A$ and $SA = I_{0, 1}\bigotimes A$. Now, given a compact Hausdorff space $X$ that may not be contractible, and, for a point $x\in X$, I set $I_x = \{f\in C(X)\,\vert\,f(x) = 0\}$. My question is:

  1. With the given unital $C^*$-Algebra $A$ that has non-trivial $K_0$ group, is there a necessary condition, either for $X$ or $A$, so that $K_0(I_x\bigotimes A)$ is the trivial group?
  2. In general, does there exist a "minimal" closed set $F\subseteq X$ such that $K_0(I_F\bigotimes A)\neq\{0\}$ where $I_F = \{f\in C(X)\,\vert\,f\Big|_F=0\}$. "Minimal" is defined in the sense that any proper closed subset $F_0\subseteq F$ will cause $K_0(I_{F_0}\bigotimes A)=\{0\}$

This question is inspired by an Exercise 10.1 in An Introduction to $K$-theory for $C^*$-Algebra written by M. Rørdam, F. Larsen and N. Laustsen. In that exercise, a short exact sequence is given:

$$ 0\rightarrow SA\rightarrow C(\mathbb{T}, A) \overset{\underset{\large \leftarrow}{\phi}}{\underset{\rightarrow}{\pi}}A\rightarrow 0 $$

where we can embed $SA$ into $C(\mathbb{T}, A)$ since for $f\in SA$, $f(0)=f(1)$, and $\pi: f\mapsto f(1)$ and $\phi$ mapsto each $a\in A$ to a function that is constantly equal to $a$. Since $K_0$ is split exact as a functor, we then have $K_0[C(\mathbb{T}, A)]\cong K_0(SA)\bigoplus K_0(A)$. If $J$ is a maximal ideal in $C(\mathbb{T})$, then it does not seem like $K_0(J\bigotimes A) = \{0\}$ although I am not sure about this. When $X$ is contractible, then I believe, for any closed ideal (proper or not) $I$ of $C(X)$, we will have $K_0[C(X)\bigotimes A]\cong K_0(A)$

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  • $\begingroup$ I'm not sure if this is along the lines of what you're looking for in question 1: If $A$ is separable and $X$ is second countable, then the Künneth theorem (which holds when one $C^*$-algebra is abelian (or more generally, is nuclear and satisfies the UCT)) implies that $K_0(I_x \otimes A)=0$ if and only if the following four groups vanish: $K_0(I_x) \otimes K_0(A)$, $K_1(I_x)\otimes K_1(A)$, $Tor(K_0(I_x), K_1(A))$, and $Tor(K_1(I_x), K_0(A))$. I expect you can get this for arbitrary $A$ and $X$ as well by limiting over separable $C^*$-subalgebras. $\endgroup$
    – Jamie Gabe
    Commented Feb 18, 2023 at 13:37
  • $\begingroup$ Thank for your input. I am sorry to tell you I have no backgrounds related to homology and cannot understand how theorem could help. May I ask for a link to references relevant to the Künneth theorem, or its applications? $\endgroup$
    – Sanae Kochiya
    Commented Feb 20, 2023 at 19:35
  • $\begingroup$ The essence of what I meant is that if you want to check that $K_0(I_x \otimes A)$ vanishes, it is necessary and sufficient to see how $K_i(I_x)$ and $K_j(A)$ relate to each other wrt $\otimes$ and $Tor$ (as described above). You can find the Künneth theorem in [Schochet, Claude Topological methods for C∗-algebras. II. Geometric resolutions and the Künneth formula. Pacific J. Math. 98 (1982), no. 2, 443–458] $\endgroup$
    – Jamie Gabe
    Commented Feb 20, 2023 at 20:23

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