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Let me recall the definition of the Bott periodicity isomorphism in the context of $C^*$-algebras. We take a (class of) projection $p \in M_n(A^+)$ and map it to the class of $M_n(A)$ valued loop $f_p$ which is defined by $$f_p(z)=1_n+(z-1)p$$ (here $A^+$ is a unitization of $A$ (even if $A$ already has a unit) and $1_n$ is the identity matrix in $M_n(A)$). Then the Bott isomorphism is defined as $$\beta: K_0(A) \ni [p]-[q] \mapsto [f_pf_q^*] \in K_1(SA).$$ In the classical (meaning: commutative) context Bott isomorphism is defined as multiplication by $[\gamma]-1$ where $\gamma$ is the tautological (complex) line bundle over compex projective space (which is the two sphere $S^2$).

When the (unital) $C^*$-algebra is commutative then it is of the form $C(X)$ for some compact $X$. Then there is a natural isomorphism between $K^0(X)$ and $K_0(C(X))$. Is this isomorphism compatible with the Bott periodicity maps? If the answer is yes, why it is so?

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    $\begingroup$ The answer is yes. In short, the Bott isomorphism as you defined it is the same thing as K-theory product with the $K_1$ class of the unitary valued function $z \mapsto \overline{z}$ on $S^1$. There are isomorphisms $K_1(C(S^1)) \cong K_0(C(S^2)) \cong K^0(S^2)$ which respect K-theory products, and under this identification $z \mapsto \overline{z}$ corresponds to the traditional Bott class in topological K-theory. $\endgroup$ Commented Aug 20, 2016 at 21:44
  • $\begingroup$ Thank you, could you please provide more details? This question bothers me for a long time and I havent found the answer anywhere in the literature. $\endgroup$
    – truebaran
    Commented Aug 20, 2016 at 21:52
  • $\begingroup$ Dear Paul, as you can see, I offered the bounty for this question, if you could turn your comment into answer (providing some more details) I will be happy to accept it (forgive me putting such a comment but I;m not quite sure whether the person which wrote a comment under my post will get the information that bounty was offered for the question). $\endgroup$
    – truebaran
    Commented Oct 2, 2016 at 17:16

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