3
$\begingroup$

Let $X$ be a projective variety, $Z$ a hypersurface section and $U \overset{def}= X \setminus Z$ its complement, an open affine subscheme of $X$. Let $i:U \hookrightarrow X$ be the corresponding open embedding. Given a coherent sheaf $M$ on $U$ we can consider the quasi-coherent sheaf $i_*M$. This seems to have a filtration of coherent sheaves in the following manner: Let's assume we are working on an affine chart and $W=\text{Spec}(R)$ is an affine in this chart and the equation of $Z$ corresponds to the element $f$ in $R$. The quasi-coherent sheaf $i_*M|_{W\cap U}$ on $W$ has this increasing filtration that $M_0=M|_W$, $M_1=\frac 1fM|_W$, $M_2=\frac 1{f^2}M|_W, \ldots$. This filtration seems to glue and give a filtration by coherent sheaves $(i_*M)_0\subset (i_*M)_1 \subset \ldots$ of $i_*M$. (Please correct me if I am wrong). Note that for each $j$, $i^*(i_*M)_j\cong i^*i_*M\cong M$.

Now let's consider a short exact sequence of vector bundles on $X$ like $0\rightarrow E_1 \rightarrow E_2 \rightarrow E_3 \rightarrow 0$. This leads to a short exact sequence of quasi-coherent sheaves $0\rightarrow i_*i^*E_1\rightarrow i_*i^*E_2\rightarrow i_*i^*E_3\rightarrow 0$ which restricts back to the original short exact sequence on $U$. Note that this exact sequence of quasi-coherent sheaves is split. This is because short exact sequence of vector bundles split on $U$. If the contents of the first paragraph is correct, my question is, does this split short exact sequence of quasi-coherent sheaves restrict to a split short exact sequence of coherent sheaves of the form $0\rightarrow (i_*i^*E_1)_j\rightarrow (i_*i^*E_2)_j\rightarrow (i_*i^*E_3)_j\rightarrow 0$ for each $j\geq 0$? (Here $j$ is the same indexing of the filtration defined in the first paragraph.)

$\endgroup$
5
  • $\begingroup$ I think we need some clarification. Is $U$ an open subscheme of $X$ here? Because the assumption $X \setminus U$ is affine is one I know if it is either an open affine or a closed affine; open affine would imply $U$ is closed (where the choice of the letter $U$ makes little sense IMO) and closed affine would mean $U$ open but would also imply that $X \setminus U$ is a closed affine subset of the projective variety $X$, which would mean it is a point. In any case, I don't know exactly what's the setting you're considering. $\endgroup$ Dec 30, 2020 at 4:47
  • $\begingroup$ $X\setminus U$ is not affine, $U=X\setminus Z$ is affine. Yes $U$ is the open affine complement. $\endgroup$
    – user127776
    Dec 30, 2020 at 4:48
  • $\begingroup$ Let me clarify and edit the question then. $\endgroup$ Dec 30, 2020 at 4:50
  • 1
    $\begingroup$ Your statement in the first paragraph is correct. It's basically equivalent to the following statement in module theory that if $R$ is a ring and $f$ is a non-zero divisor, then the $R$-submodule $\frac 1{f^n} R$ of $R_f$ satisfies $R_f \otimes_R \frac 1{f^n} R_f \simeq R_f$, which is sort of clear because $\frac 1{f^n} R_f$ is already a subset of $R_f$, so tensoring with $R_f$ over $R$ just computes its $R_f$-span within $R_f$, which is clearly $R_f$ since it contains $1$. This isomorphism is natural in $R$ and $f$, so you can glue it over all affine charts and get your result for sheaves. $\endgroup$ Dec 30, 2020 at 4:57
  • $\begingroup$ I modified your question to use the notation $\frac 1{f^n}$ instead of $f^{-1}$, because $f^{-1}$ is used a lot when $f$ is a morphism and $f^{-1}$ the inverse image functor. I wanted to avoid confusion. $\endgroup$ Dec 30, 2020 at 5:01

2 Answers 2

2
$\begingroup$

What you're doing when you compute $(i_*M)_j$ in the affine setting is that you have a module $M$ and you tensor up with $\frac 1f R$ over $R$, which is an $R$-submodule of $R_f \otimes_R M$. So in other words, $(i_*M)_j = \frac 1{f^j}R \otimes_R M$. I think your question can be converted into a module-theoretic question, i.e. is the $R$-module $\frac 1{f^j}R$ flat. I think your intuition for this statement comes from the fact that $\frac 1{f^j}R$ is an invertible $R$-module, since $$ f^j R \otimes_R \frac 1{f^j} R \simeq R $$ via the multiplication map, so that $\frac 1{f^j} R$ is an invertible $R$-module; this would be the module-theoretic equivalent of "twisting" when dealing with sheaves, in some sense I can't put my finger on since it's been a while. I'm not sure, but I think you may need the assumption that $f$ is a non-zero divisor so that the map $R \to R_f$ is injective and the computations in the above statement (i.e. the 'multiplication map') can happen within $R_f$.

Also because it's been a while, I don't feel super confident in my intuition that tells me that the above implies that $f^j R$ is faithfully flat for $j \in \mathbb Z$, but it seems to make sense when $X$ is integral, because then $R$ is integral, and saying that $f^jR$ is flat is basically saying that $f^jR_{\mathfrak p}$ is a free $R_{\mathfrak p}$-module of rank $1$ for any $\mathfrak p \in \mathrm{Spec}(R)$, and the isomorphism $R_{\mathfrak p} \to f^j R_{\mathfrak p}$ can clearly be given by multiplication by $f^j$ when both modules are seen as submodules of the quotient field $Q(R)$ (which is possible when $R$ is integral).

Hope that helps,

$\endgroup$
2
  • 1
    $\begingroup$ The assumption that $f$ is a non-zero divisor is valid. Because in my original problem I'm assuming everything is smooth and $Z$ is a smooth hyperplane section given by Bertini's theorem which implies $f$ is a non-zero divisor. Thanks for your intuition. $\endgroup$
    – user127776
    Dec 30, 2020 at 5:41
  • $\begingroup$ @user127776 You're welcome! I think my idea of "twisting" comes from ideas related to the Rees algebra and the associated graded ring, because you can consider the ideal generated by $f$ and the associated algebra of $f^jR / f^{j+1}R$ for $j \ge 0$, and in that sense the twist was achieved by multiplying by $f$, not by literally twisting as in scheme theory. I spent a lot of time dealing with $\mathbb Z$-graded algebras and graded ideals so in those contexts I was literally twisting but in this case I think it just looks similar but not related. But I think you can work with what I produced :) $\endgroup$ Dec 30, 2020 at 19:21
0
$\begingroup$

The answer to the second paragraph is negative. For example the zeroth filtration of the short exact sequence involving $i_*i^*E_k$'s just recovers the original short exact sequence. Every coherent sheaf $M$ such that $M|_U$ is isomorphic to a fixed coherent sheaf $E$, injects into the quasi-coherent sheaf $i_*E$. Applying the filtration corresponding to $M$, just reproduces the coherent sheaves $M, M(1), \ldots , M(n), \ldots$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .