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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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