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Let's assume we have a sheaf of spectra on some scheme. As an example I will assume that we are working with the $K$-theory sheaf. There are certain local to global spectral sequences, like descent spectral sequence. If one knows the algebraic $K$-groups on some open cover and the intersections of the opens, then it is possible to use the descent spectral sequence to recover the algebraic $K$-group globally. I was wondering whether it is possible to recover the algebraic $K$-groups by knowing its values on local rings?

Another way of looking at the problem is the following: Let's assume for two schemes $X$ and $Y$ we have a morphism $f:X\rightarrow Y$. Assume we have two open coverings of $X$ and $Y$ denoted by $U_{\alpha}^X$ and $U_{\alpha}^Y$ in a way that $f$ induces a map from $\bigcap_{\alpha\in I}U_{\alpha}^X$ to $\bigcap_{\alpha\in I}U_{\alpha}^Y$ ($I$ is a finite collection of indices) which that map induces isomorphism of algebraic $K$-groups then by descent spectral sequence this implies that $f$ induces isomorphisms $K_n(X)\cong K_n(Y)$.

Now instead of working with an open cover, assume $f$ induces isomorphism of algebraic $K$-groups on the local rings of $X$ and $Y$, is it necessarily true that $K_n(X)\cong K_n(Y)$.

Another version: Is it enough these isomorphisms to happen for the local rings of closed points?

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  • $\begingroup$ Closed points only, I believe, are ruled out: for a smooth scheme over a field, say, they have isomorphic neighborhoods, right? $\endgroup$ Commented Mar 3, 2021 at 6:16
  • $\begingroup$ I see. For example blow-ups of projective space locally have neighbors isomorphic to an open in $\mathbb{A}^n$. But blow-up does not induce isomorphism of algebraic $K$-groups. I wonder whether this could be fixed under some extra conditions. $\endgroup$
    – user127776
    Commented Mar 3, 2021 at 6:23
  • $\begingroup$ can you clarify your hypotheses? bijection between which opens? $\endgroup$
    – user175135
    Commented Mar 3, 2021 at 6:54
  • $\begingroup$ Isn't $K(O_{X,x})$ the stalk at $x$ of the sheaf $U\mapsto K(U)$ on $X$ ? Let me call that sheaf $K_X$ and $K_Y$ for $Y$, then you get a map $f^{-1}K_Y\to K_X$. Your hypothesis seems to be exactly that this map is an equivalence on stalks i.e. that it's an equivalence; you can then take the global sections of that. Going from there to $K(Y)$ seems more complicated $\endgroup$ Commented Mar 3, 2021 at 9:08
  • $\begingroup$ Right, if $X$ has a very small image in $Y$, the condition on local rings tells you basically nothing about $Y$, so there's essentially no way of getting $K(Y)$ from that. Just take $Y = X\sqcup Z$ and the canonical inclusion of $Z$ $\endgroup$ Commented Mar 3, 2021 at 9:11

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