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arXiv 2402.15329math.AG
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簇的朴素 $\mathbb A^1$-连通分量函子的迭代

Iterations of the functor of naive $\mathbb A^1$-connected components of varieties

Nidhi Gupta

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15 pages, comments are welcome, v3: final version before page proofs, accepted for publication in NYJM
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AI总结：

本文证明构造概形的万有A¹-不变量商时，朴素A¹-连通分量函子的无限次迭代是必要的，并构造簇X_n使S^n(X_n)≠S^{n+1}(X_n)且S^{n+2}(X_n)=*。

摘要：

对于 $Sm/k$ 上的任意集合层 $\mathcal F$，众所周知，$\mathcal F$ 的万有 $\mathbb A^1$-不变量商由层 $\mathcal S^n(\mathcal F)$ 的余极限给出，其中 $\mathcal S(F)$ 是 $\mathcal F$ 的朴素 $\mathbb A^1$-连通分量层。我们证明，在构造概形的万有 $\mathbb A^1$-不变量商时，这些朴素 $\mathbb A^1$-连通分量的无限次迭代确实是必需的。对于每个 $n$，我们构造一个 $\mathbb A^1$-连通簇 $X_n$，使得 $\mathcal S^n(X_n)\neq \mathcal S^{n+1}(X_n)$ 且 $\mathcal S^{n+2}(X_n)=*$。

英文摘要：

For any sheaf of sets $\mathcal F$ on $Sm/k$, it is well known that the universal $\mathbb A^1$-invariant quotient of $\mathcal F$ is given as the colimit of sheaves $\mathcal S^n(\mathcal F)$ where $\mathcal S(F)$ is the sheaf of naive $\mathbb A^1$-connected components of $\mathcal F$. We show that these infinite iterations of naive $\mathbb A^1$-connected components in the construction of universal $\mathbb A^1$-invariant quotient for a scheme are certainly required. For every $n$, we construct an $\mathbb A^1$-connected variety $X_n$ such that $\mathcal S^n(X_n)\neq \mathcal S^{n+1}(X_n)$ and $\mathcal S^{n+2}(X_n)=*$.

机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。

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