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In the reduced K-theory we have the exact sequence of pair $$ \tilde{K}(S(X/A)) \to \tilde{K}(SX) \to \tilde{K}(SA) \to \tilde{K}(X/A) \to \tilde{K}(X) \to \tilde{K}(A) $$ If choose $X$ to be a surface $\Sigma$ and $A$ to be its 1-skeleton, we end up with the sequence $$ \tilde{K}(S^3) \to \tilde{K}(S\Sigma) \to \tilde{K}\left(\bigvee_{1\leq i \leq n} S^2\right) \to \tilde{K}(S^2) \to \tilde{K}(\Sigma) \to \tilde{K}\left(\bigvee_{1\leq i \leq n} S^1\right),$$ where $n$ is the number of closed curves in the 1-skeleton. Notice that I am using a cell decomposition with one 0-cell and one 2-cell and that the suspension of a bouquet of circles is a bouquet of spheres.

My problem is: since the first and the last term of that sequence are trivial, we only have to know what is the map in the middle to determine $K^{-1}(\Sigma)$ and $K(\Sigma)$, but how do I calculate this map?

I would prefer a more explicit answer.

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4  
The map $X \to X/A$ splits stably (in fact, after one suspension), so the middle map is zero. –  Oscar Randal-Williams Jun 30 '13 at 20:21
    
@OscarRandal-Williams, you mean it splits in $K$-theory or for surfaces, right? –  Sean Tilson Jun 30 '13 at 22:43
1  
Are you asking about closed, orientable surfaces only? In that case, the suspension is just a wedge of spheres (the suspension of the attaching map of the 2-cell, which is a product of commutators, becomes nullhomotopic after suspension - this is essentially due to the fact that $\pi_2$ is abelian). So then, as Oscar says, the middle map splits. In the non-orientable (closed) surface case, the attaching map involves a square, which becomes multiplication by 2 after suspension. So the middle map is zero on n-1 factors and multiplication by 2 on the last factor. –  Dan Ramras Jul 1 '13 at 4:12
    
I think Oscar explanation is extremely intuitive. –  Fernando Muro Jul 1 '13 at 7:26
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