Talk:K-groups of oriented surface groups (Ex)

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First we look at the Atiyah-Hirzebruch spectral sequence. Recall that a closed orientable surface is either S^2 or aspherical. Hence if G is not the trivial group, then the surface is itself a model for BG. Let us fix the genus g of the surface. Then the integral homology looks like

\displaystyle  H_*(BG;\Z) \cong \begin{cases} \Z & \text{ if } * \in \{0,2\}, \\ \Z^{2g} & \text{ if } * = 1, \\ 0 & \text{ else. }\end{cases}

Moreover since the coefficients of KU are only \Z or 0 we see that all E^2-terms of the spectral sequence are free abelian groups. By purely formal reasons there cannot be any non-trivial differential (the only possible ones have target the y-axis of the spectral sequence, but since the coefficients in the AHSS always split off, these maps must have trivial image). Hence also the E^\infty-terms of the spectral sequence are torsion free. In particular there are no extension problems to solve and we can read off the abutment to be

\displaystyle  K_0(BG) \cong \Z^2 \text{ and } K_1(BG) \cong \Z^{2g} .

We want to remark that also for other homology theories there cannot occur non-trivial extensions when we want to compute the abutment from the E_\infty-page. This follows from the following observation. The surface BG is given by a pushout

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where the attaching map is of the form x_1 y_1 x_1^{-1} y_1^{-1} \dots x_g y_g x_g^{-1} y_g^{-1} where the 1-cells are indexed over \{x_i,y_i\}. In particular we see that suspending this map once, it becomes null-homotopic (higher homotopy groups are abelian). Hence, stably, the top cell of BG splits off and we can use the Mayer-Vietoris sequence to compute the homology of BG for any homology theory in terms of the coefficients.

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