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It is a theorem that every locally free coherent sheaf on $\mathbb{P}^1$ over an algebraically closed field is isomorphic to a unique sum of sheaves $\mathcal{O}(n)$ for various integers $n$. In particular, the K-ring of locally free coherent sheaves (or all coherent sheaves, $\mathbb{P}^1$ being nonsingular) is isomorphic to $\mathbb{Z}[t, t^{-1}]$.

The topological K-ring of vector bundles on $S^2$ is, by Bott periodicity, isomorphic to $\mathbb{Z}[H]/(H-1)^2$, where $H$ is the canonical bundle. But $S^2$ is homeomorphic to $\mathbb{P}^1_{\mathbb{C}}$.

Every locally free sheaf corresponds to a vector bundle on $S^2$. It follows that the map on the K-groups from locally free sheaves to vector bundles is surjective but not injective.

Questions:

  1. What goes wrong?

  2. Is there a version of Bott periodicity for algebraic varieties (or schemes)? (I.e., relating K-groups of $X$ and $X \times \mathbb{P}^1$.) I understand that there is one for the Picard groups.

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There are less algebraic/holomorphic sections than continuous ones, so there is nothing wrong in (1)! –  Mariano Suárez-Alvarez Oct 23 '10 at 23:46
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As for periodicity... Well, it is complicated! :) You could do much worse than read Max Karoubi's note on Bott periodicity in Topological, Algebraic and Hermitian K-theory, which you can find in his homepage at Jussieu. –  Mariano Suárez-Alvarez Oct 23 '10 at 23:50
    
@Mariano: Thanks! I agree with your first comment, but simply was used to things working out very similarly in algebraic topology as in topology (e.g. the isomorphism between sheaf cohomology and singular cohomology, the analogy between etale neighborhoods and neighborhoods in the complex topology, etc.). I take it that if you consider the Grothendieck ring of holomorphic vector bundles, it should be the same as the algebraic one, no? (This is probably in GAGA, right?). Also, thanks for the reference. –  Akhil Mathew Oct 24 '10 at 3:33
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up vote 10 down vote accepted

It is not true that the Grothendieck ring of coherent sheaves on $\mathbb{P}^1$ is isomorphic to $\mathbb{Z}[t, t^{-1}]$. Although $\mathcal{O} \oplus \mathcal{O}(2)$ is not isomorphic to $\mathcal{O}(1) \oplus \mathcal{O}(1)$, they do have the same class in $K^0$.

The definition of the Grothendieck group of coherent sheaves on a scheme $X$ is that it is generated by isomorphism classes of coherent sheaves, modulo the relation that $[A] + [C] = [B]$ whenever there is a short exact sequence $$0 \to A \to B \to C \to 0.$$ In particular, we have the short exact sequence $$0 \to \mathcal{O} \to \mathcal{O}(1)^2 \to \mathcal{O}(2) \to 0,$$ where the maps are given by $(x \ y)$ and $\binom{-y}{x}$.

This makes $K^0$ into $\mathbb{Z}[t, t^{-1}]/(t^2 - 2t +1) \cong \mathbb{Z}[u]/u^2$, just like you wanted.

When working in the categories of smooth or of topological vector bundles, all short exact sequences split, so you can get away with defining $K$-theory with direct sums. You can't do that in the coherent or the algebraic categories.

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Ah, very nice---thanks for the answer. –  Akhil Mathew Oct 26 '10 at 1:14
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