Restricting split short exact sequence of quasi-coherent sheaves to their coherent sub-sheaves.

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.)

• 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. Dec 30, 2020 at 4:47
• $X\setminus U$ is not affine, $U=X\setminus Z$ is affine. Yes $U$ is the open affine complement. Dec 30, 2020 at 4:48
• Let me clarify and edit the question then. Dec 30, 2020 at 4:50
• 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. Dec 30, 2020 at 4:57
• 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. Dec 30, 2020 at 5:01

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).
• 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. Dec 30, 2020 at 5:41
• @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 :) Dec 30, 2020 at 19:21
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 $$M, M(1), \ldots , M(n), \ldots$$