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I've took the opportunity to join the community, because I didn't find a satisfying explanation to the following fact.

Let $S^2$ the 2-sphere, let $H$ the tautological line bundle. Assume that $$ K(S^2) \approx \mathbb{Z}[x]/(x-1)^2$$ (thanks to the map $x \mapsto H$)

Then for every source I've found over the net, it's obvious that $\tilde{K}(X)$ is generated as an abelian group from $H-1$.

I can't prove this fact, according to me, if $K(X)$ has as additive basis {1,H} (because i'm killing every factor of degree $\geq 2$ right?), then thanks to the relation $$\tilde{K}(X) \approx K(X)/\mathbb{Z}$$ I'm tempted to conclude that I'm killing the $1$ of the additive basis and so I have only $H$ as a generator. But it seems that it is not the case. Thanks for every advice!

(I've tagged the questions as K-theory, if more tags are needed feel free to edit)

Luigi M
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1 Answers1

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The equation you write $\tilde{K}(X)\cong K(X)/\mathbb{Z}$ is confusing, it should be

$$K(X)=\mathbb{Z}\oplus\tilde{K}(X)$$ its a small thing but I think its the source of your difficulty. $\tilde{K}(X)$ is the kernel of the degree map

$$\deg:K(X)\rightarrow \mathbb{Z}$$ now $H$ has dimension $1$ so its not in the kernel, but $H-1$ has dimension zero so it is in the kernel so in fact the decomposition reads,

$$K(S^2)=\mathbb{Z}\oplus \langle H-1\rangle$$

Rene Schipperus
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  • thanks for the answer! By degree map you mean the map $E-E' \mapsto deg(E){x_0}-deg(E'){x_0}$ which in other words subtract the dimension of the fiber over a point of the first bundle from the dimension of the fiber over the same point of another? – Luigi M Jul 07 '14 at 15:06
  • Yes that is it, its called degree ? or is there another name ? – Rene Schipperus Jul 07 '14 at 15:08
  • oh, I don't know, maybe it's its natural name/definition, (i'm following hatcher, and he doesn't mention it) – Luigi M Jul 07 '14 at 15:13
  • @ReneSchipperus very nice answer! just to be sure I'm following your reasoning, the main fact is that I have as a basis {1,H}, so every combination of $H$ AND trivial line bundles (n) which are mapped to 0 by the degree must be a multiple of H-1 right? – Riccardo Jul 07 '14 at 15:19
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    @Riccardo Yes you could write a general element as $nH-m$ the degree will be zero only for $n=m$ so you have $n(H-1)$ as a general element of the kernel. – Rene Schipperus Jul 07 '14 at 15:28