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Calculate the Chern classes $$ c_n \in H^{2n}(S^{2n})$$ for the generator of the group $$ K(S^{2n})$$ where $S^{2n} $ - sphere of dimension $ 2n $, $ K(S^{2n})$ - group from K-theory.

I found the following fact about the K-group $$ K(S^0) = K(S^{2k})= \mathbb{Z} \times \mathbb{Z},\quad K(S^1) = K(S^{2k+1}) =\mathbb{Z} $$ How to calculate the Chern classes for sphere $S^{2n}$ if we know that $\eta$ is generator for $K(S^{2n})$?

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Note that $K(S^{2n}) = \mathbb{Z}\times\mathbb{Z}$ has many different sets of generators and these can have different Chern classes, so the question doesn't have a well-defined answer. For one particular choice of generators, the first $\mathbb{Z}$ is generated by the trivial complex line bundle $\varepsilon^1_{\mathbb{C}}$ which has $\operatorname{ch}(\varepsilon^1_{\mathbb{C}}) = 1$. The interesting case is the generator of the other copy of $\mathbb{Z}$ (the one that survives in reduced $K$ theory).

Recall that the Chern character is a map $\operatorname{ch} : K(X) \to H^*(X; \mathbb{Q})$. For $X = S^{2n}$, it follows from Bott Periodicity for K theory that the image of the Chern character is actually contained in $H^*(S^{2n}; \mathbb{Z})$ (see Proposition 6.1. of this nice paper by Konstantis and Parton). Note that on $S^{2n}$ the Chern character takes the form $\operatorname{ch}(E) = \operatorname{rank}(E) + \operatorname{ch}_n(E)$. Using identities betweeen symmetric polynomials, one can show that

$$\operatorname{ch}_n(E) = \frac{(-1)^{n+1}c_n(E)}{(n-1)!}.$$

As $\operatorname{ch}_n(E) \in H^{2n}(S^{2n}; \mathbb{Z})$ we see that $c_n(E)$ must be divisible by $(n - 1)!$.

It turns out that there is a complex vector bundle $E \to S^{2n}$ with $c_n(E) = (n-1)!\,\alpha$ where $\alpha$ is a generator of $H^{2n}(S^{2n}; \mathbb{Z}) \cong \mathbb{Z}$. Blaine Lawson once told me that you can take $E$ to be the positive complex spinor bundle.

One can use these observations to show that the only spheres which admit almost complex structures are $S^2$ and $S^6$. For more details, see the aforementioned paper.

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  • $\begingroup$ So, because the image of the Chern character is the integral cohomology, $c_n(E)/(n-1)!$ is an integer for any $E$. This raises the question: what is this integer that is intrinsically associated to the vector bundle E? In fact there is an answer: it is the index of the signature operator coupled to E. $\endgroup$
    – Oliver Nash
    Commented Jun 6, 2018 at 13:14
  • $\begingroup$ @OliverNash: I believe that index of the signature operator coupled to $E$ is $2^n$ times the integer you mention, see Theorem III.13.9 of Lawson and Michelsohn's Spin Geometry. On the other hand, the index of the Dirac operator coupled to $E$ has this integer as its index, see Theorem III.13.10. $\endgroup$
    – Michael Albanese
    Commented Jun 6, 2018 at 13:59
  • $\begingroup$ Many apologies if, as it seems, I am spreading misinformation. Glancing quickly at L&M, I can't quite line up what you're saying though. Furthermore, is not Theorem IV 1.4 support for my remark? [I'll check carefully later --- for now I have to get back to the day job!] $\endgroup$
    – Oliver Nash
    Commented Jun 6, 2018 at 14:09
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    $\begingroup$ There is an error in the proof of Theorem IV 1.4, it should be $\operatorname{ch}_2E$ not $\operatorname{ch}E$. Making this adjustment, the conclusion is $\operatorname{ind}(D^+) = \frac{2^n}{(n-1)!}c_n(E)$ which is not strong enough to prove the theorem. However, using the Dirac operator coupled to $E$ gives the stated index and the correct conclusion. $\endgroup$
    – Michael Albanese
    Commented Jun 7, 2018 at 18:11

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