 # ag.algebraic geometry – Subgroup of algebraic \$Ok\$-theory generated by splinter vector bundles Answer

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ag.algebraic geometry – Subgroup of algebraic \$Ok\$-theory generated by splinter vector bundles

Exercise II.6.11 and V.2.3 in Hartshorne (HH) proves that for a nonsingular round $$C$$ over an algebraically closed bailiwick $$okay$$, it follows the Grothendieck group $$operatorname{Ok}_0(C)cong operatorname{Pic}(C)oplus mathbb{Z}$$. Exercise V.2.3 proves that any rank $$r$$ regionally free sheaf $$mathcal{E}$$ on $$C$$ has a filtration
$$mathcal{E}_i$$ with $$mathcal{E}_i/mathcal{E}_{i-1}:=mathcal{L}_i$$ an invertible sheaf. Hence within the Grothendieck group $$operatorname{Ok}_0(C)$$ you get the relation

K0. $$[mathcal{E}]=sum_i [mathcal{L}_i].$$

There is a projective bundle method for the calculation of $$operatorname{Ok}_i(mathbb{P}(mathcal{E}^*))$$ for all $$igeq 0$$. I consider the next holds:

K1. $$operatorname{Ok}_i(mathbb{P}(mathcal{E}^*)) cong operatorname{Ok}_i(X)[t]/(t^{r+1})$$

the place $$mathcal{E}$$ is a regionally trifling $$mathcal{O}_X$$-module of rank $$r+1$$.

Example. In the illustration of the projective line should you let $$S:=Spec(A)$$ and $$V:=A{e_0,e_1}$$, with $$mathbb{P}^1_S:=Proj(Sym_A^*(V^*))$$ it follows

K2. $$operatorname{Ok}_i(mathbb{P}^1_S)cong operatorname{Ok}_i(S)[t]/(t^2) cong operatorname{Ok}_i(S){1,overline{t}}$$.

Hence the calculation of the upper Ok-groups of the projective line over $$S$$ reduces to the calculation of those teams for $$S$$. If $$S:=Spec(okay)$$ is the spectrum of a bailiwick mighty is thought on $$operatorname{Ok}_i(S)$$, however in common the construction of group $$operatorname{Ok}_i(S)$$ shouldn’t be identified.

Example. If $$X$$ is a spread over a bailiwick $$okay$$ the place each vector bundle/regionally free sheaf of finite rank decompose as

K3. $$mathcal{E}cong oplus_i mathcal{L}_i$$

with $$mathcal{L}_i in operatorname{Pic}(X)$$, it follows trivially that $$operatorname{Ok}_0(X)$$ is generated by lessons of invertible sheaves $$[mathcal{L}]$$. I consider there are few such varieties aside from finite disjoint unions of projective traces. In common you get a subgroup of $$operatorname{Ok}_0(X)$$ by contemplating the weather on the figure

K4. $$sum_i[ mathcal{L}_i]$$

with $$mathcal{L}_i$$ an invertible sheaf. For the identical intuition it follows this subgroup doesn’t generate the Grothendieck group in common. The identical rehearse (HH.V.2.3) proves that the cotangent bundle $$Omega$$ on the projective airplane shouldn’t be an extension of invertible sheaves. Hence projective house doesn’t answer the circumstances K0, K3 in common. There is an isomorphism

K5. $$operatorname{Ok}_0(X)_{mathbb{Q}}^{(i)} cong operatorname{CH}^i(X)_{mathbb{Q}}$$

the place $$operatorname{Ok}_0(X)_{mathbb{Q}}^{(i)}$$ is the $$i$$‘th eigenspace of algebraic Ok-theory with respect to the motion of the Adams operation, and $$operatorname{CH}^i(X)_{mathbb{Q}}$$ is the Chow group of $$X$$ with rational coefficients. The Chow group of cycles on a complicated flush projective floor
is “infinite dimensional” in common, therefore the Grothendieck group can breathe sizable.
It is a protracted standing launch drawback to compute the teams $$operatorname{CH}^i(X)$$ and $$operatorname{Ok}_i(X)$$.

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