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What's the K-group of a surface? I also want to know how to calculate such group and if there is a explicit characterization of the generators.

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If you have a "fundamental domain" for your surface, then you can use basic obstruction theory to compute this explicitly (or at least reduce it to pure group theory). Recall that we have the natural isomorphism $K(X) \cong [X,BU\times \mathbb{Z}]$, and $\pi_1(BU)\cong \pi_0(U)=0$. So, the 1-skeleton of any CW-complex maps essentially trivially into $BU$, and hence $[\Sigma,BU] \cong \pi_2(BU)$ for any such surface $\Sigma$ with a fundamental domain. Then, we can compute that $\pi_2(BU) = \mathbb{Z}$ by Bott periodicity (or you can easily compute by hand that $\pi_1(U)=\mathbb{Z}$). So, $K(\Sigma)\cong \mathbb{Z} \times \mathbb{Z}$.

So, this brings us to computing $K(S^2)$: $$K(S^2) \cong [S^2,BU \times \mathbb{Z} ] \cong \pi_2(BU) \times \mathbb{Z} \cong \mathbb{Z} \times \mathbb{Z}.$$ The nontrivial copy of $\mathbb{Z}$ is generated by the (vdim-0) Hopf bundle, namely the canonical bundle over $\mathbb{CP}^1$ minus $\underline{\mathbb{C}}$. Moreover, If $\Sigma$ is a surface with a fundamental domain (i.e. any surface of positive genus), then by the description I've given above, you can obtain a generator for the first copy of $\mathbb{Z}$ (i.e., the copy not picking off virtual dimension) via pullback along the map $\Sigma \rightarrow S^2$ which collapses the 1-skeleton to a point.

Of course, you could have just done this with arbitrary CW decompositions all along. But most surfaces admit one with a single 2-cell, which simplifies the computation considerably.

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With the caveat that I haven't done this computation myself, I suggest having a look at the Atiyah-Hirzebruch spectral sequence, which computes generalised cohomology of a space in terms of the ordinary cohomology and the coefficients of the generalised theory.

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I agree - this and the Chern character (which can get you the non-rational part) –  Juan S Sep 17 '12 at 11:36
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Doesn't the Chern character only give you the rational part? –  Aaron Mazel-Gee Sep 21 '12 at 11:49
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