In Wegge-Olsen’s book K-Theory and C$ ^{*} $-Algebras, there is an outline of a proof of Bott Periodicity (the proof is due to George Elliott, Toshikazu Natsume and Ryszard Nest). The first step of the proof outline goes as follows:

Suppose that there are natural transformations

  • $ \Phi^{0} $ from the $ K_{0} S $-functor to the $ K_{1} $-functor and
  • $ \Phi^{1} $ from the $ K_{1} S $-functor to the $ K_{0} $-functor,

where $ S $ denotes the suspension functor. In other words, for every C$ ^{*} $-algebra $ \mathscr{A} $ (not necessarily unital), there are abelian-group homomorphisms $$ \Phi^{0}_{\mathscr{A}}: {K_{0}}(S(\mathscr{A})) \to {K_{1}}(\mathscr{A}) \quad \text{and} \quad \Phi^{1}_{\mathscr{A}}: {K_{1}}(S(\mathscr{A})) \to {K_{0}}(\mathscr{A}) $$ such that for every C$ ^{*} $-algebraic homomorphism $ \alpha: \mathscr{A} \to \mathscr{B} $, the following diagrams commute: $$ \require{AMScd} \begin{CD} {K_{0}}(S(\mathscr{A})) @>{{K_{0} S}(\alpha)}>> {K_{0}}(S(\mathscr{B})) \\ @V{\Phi^{0}_{\mathscr{A}}}VV @VV{\Phi^{0}_{\mathscr{B}}}V \\ {K_{1}}(\mathscr{A}) @>>{{K_{1}}(\alpha)}> {K_{1}}(\mathscr{B}) \end{CD} \quad \quad \quad \begin{CD} {K_{1}}(S(\mathscr{A})) @>{{K_{1} S}(\alpha)}>> {K_{1}}(S(\mathscr{B})) \\ @V{\Phi^{1}_{\mathscr{A}}}VV @VV{\Phi^{1}_{\mathscr{B}}}V \\ {K_{0}}(\mathscr{A}) @>>{{K_{0}}(\alpha)}> {K_{0}}(\mathscr{B}) \end{CD} $$ Suppose further that $ \Phi^{0}_{\mathbb{C}} $ and $ \Phi^{1}_{\mathbb{C}} $ are isomorphisms. Then prove that $ \Phi^{0}_{\mathscr{A}} $ and $ \Phi^{1}_{\mathscr{A}} $ are isomorphisms for every C$ ^{*} $-algebra $ \mathscr{A} $.

Wegge-Olsen’s hint is to consider the commutative diagrams above for the homomorphism \begin{align*} \alpha: \mathbb{C} & \to \mathscr{A} \otimes \mathbb{K}; \\ 1 & \mapsto p \otimes e_{11}, \end{align*} where $ p $ is a fixed choice of a projection in $ \mathscr{A} $ and $ e_{11} $ is a rank-$ 1 $ projection in $ \mathbb{K} \stackrel{\text{def}}{=} {M_{\infty}}(\mathbb{C}) $.

My difficulty:

I do not see how \begin{align*} {K_{0}}(\alpha): & {K_{0}}(\mathbb{C}) \to {K_{0}}(\mathscr{A} \otimes \mathbb{K}), \\ {K_{1}}(\alpha): & {K_{1}}(\mathbb{C}) \to {K_{1}}(\mathscr{A} \otimes \mathbb{K}), \\ {K_{0} S}(\alpha): & {K_{0}}(S(\mathbb{C})) \to {K_{0}}(S(\mathscr{A} \otimes \mathbb{K})) \quad \text{and} \\ {K_{1} S}(\alpha): & {K_{1}}(S(\mathbb{C})) \to {K_{1}}(S(\mathscr{A} \otimes \mathbb{K})) \end{align*} are isomorphisms. If this hurdle can be overcome, then $ \Phi^{0}_{\mathscr{A}} $ and $ \Phi^{1}_{\mathscr{A}} $ are clearly isomorphisms.

Thank you for any assistance!

  • 4
    $\begingroup$ Obviously, $K_0(\alpha)$ and the rest of the mappings are not isomorphisms. Because otherwise it would imply that all $C^\ast$-algebras have isomorphic $K_0$-groups (and $K_1$-groups). My guess is that he has shown that the suspension functor is surjective. Then using the above technique he tries to show that the suspension is injective as well. $\endgroup$
    – user23860
    Jan 29, 2014 at 11:45

1 Answer 1


A vague guess:

Consider something by putting your two Diagramms together (for the map $\alpha:\mathbf{C} \rightarrow A$) so that they fuse to one diagram with three lines. In the first diagram you replace $A$ by $S(A)$ so that double Suspension $SS(A)$ is involved.

The first column is then an isomorphism $K_0(SS(\mathbf{C}) \cong K_0(\mathbf{C})$.

Chase in the Diagramm, to see that the second column is also something (a isomorphism (??!)) for the subgroup generated by $[p]$. Then vary over all $p$.

Maybe something like this.. or a starting Point to think further.

  • $\begingroup$ I was thinking about this problem too. Following the suggestion in this answer, I think I can get the surjectivity of $\Phi^0_A$ and $\Phi^1_A$ for all $A$, but I couldn't figure out the injectivity. $\endgroup$
    – cyc
    Jan 15, 2018 at 22:26

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