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$\newcommand{\C}{\mathbb{C}}\newcommand{\Z}{\mathbb{Z}}$ I know from Bott-periodicity that $K_0(C_0(\mathbb{C}))\simeq \Z$, is there any easy way to compute an explicit generator of $K_0(C_0(\mathbb{C}))$? By this I mean a projection $q$ in $M_k(C_0(\mathbb{C})^+)$ such that $[q]-[p_n]$ generates $K_0(C_0(\mathbb{C}))$. Moreover given an element $x\in K_0(C_0(\mathbb{C}))$ is there an easy way to compute $m\in \Z$ such that $m([q]-[p_n])=x$ (this equates to finding the inverse to the isomorphism $\Z\stackrel{\sim}{\rightarrow}K_0(C_0(\mathbb{C}))$)?

I tried digging into the isomorphism $K_0(\C)\stackrel{\sim} {\rightarrow}K_0(C_0(\mathbb{C}))$ that one obtains from Bott-periodicity but things get pretty convoluted. In the end I got a generator that doesn't facilitate any further calculations I'd like to make (like calculating the inverse of the isomorphism).

I'm trying to compute this generator to make some calculations regarding a endomorphism in $K_0(C(\C P^1))\simeq K_0(C_0(\C))\oplus K_0(\C)\simeq \Z\oplus \Z$

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The group $K_0(C_0(\mathbb{C}))$ is generated by by the class $[p_{Bott}] - [1]$ where $p_{Bott} \in M_2(C_0(\mathbb{C})^\sim)$ is the so-called "Bott projection" given by $$ p_{Bott}(z) = \frac{1}{1+|z|^2} \begin{pmatrix} |z|^2 & z \\ \overline{z} & 1 \end{pmatrix}. $$ This class comes from the tautological line bundle on $S^2 \simeq \mathbb{C}P^1$ after identifying the one-point compactification of $\mathbb{C}$ with the 2-sphere $S^2$ and performing some Serre-Swan-ifictation.

It turns out that this class is essentially what makes Bott periodicity run.

Edit: I should probably also mention that $[1]$ here is the class of the trivial line bundle on $S^2$. This appears since $p_{Bott}(\infty) = (\begin{smallmatrix} 1 & 0 \\ 0 & 0\end{smallmatrix})$.

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  • $\begingroup$ That looks much cleaner than the generator I got! Thanks a bunch :) $\endgroup$ – Julio Cáceres Jun 7 '18 at 12:59

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