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Let $X$ be a smooth projective surface and $C$ a Cartier divisor on $X$. Denote by $\mathcal{H}^1_C(\mathcal{O}_X)$ the sheaf associated to the presheaf $U \mapsto H^1_{C \cap U}(\mathcal{O}_X|_U)$. Let $j:X\backslash C \to X$ be the natural immersion. Using the local cohomology sequence (see Hartshorne Ex. III.$2.3$) and the fact $\mathcal{H}^0_C(\mathcal{O}_X)=0$, we have a short exact sequence, $$0 \to \mathcal{O}_X \to j_*\mathcal{O}_{X\backslash C} \xrightarrow{\delta_i} \mathcal{H}^1_C(\mathcal{O}_X) \to 0.$$

My question is about two conflicting descrptions of $\mathrm{Tor}^1_X(\mathcal{O}_C,\mathcal{H}^1_C(\mathcal{O}_X))$. I elaborate on them.

Since $C$ is a Cartier divisor, for any small enough open affine subset $U$ of $X$, $C \cap U$ is defined by exactly one equation, say $f_U$, on $U$. Then, $\mathrm{Tor}^1_U(\mathcal{O}_{C \cap U},\mathcal{H}^1_C(\mathcal{O}_X)|_U)$ consists of all elements of the form $\delta_i|_U(g/f)$, where $g \in \mathcal{O}_X(U)$ and $\delta_i|_U:\Gamma(U,j_*\mathcal{O}_{X\backslash C}) \to \Gamma(U,\mathcal{H}^1_C(\mathcal{O}_X))$ described above. Indeed, this follows directly from the above short exact sequence and the fact that $\mathrm{Tor}^1_U(\mathcal{O}_{C \cap U},\mathcal{H}^1_C(\mathcal{O}_X)|_U)$ consists of elements of the form $\delta_i|_U(g/f_U^r)$ such that $f.\delta_i|_U(g/f_U^r)=0$ (use that $\mathcal{O}_C(U)=\mathcal{O}_X(U)/(f_U)$). It seems to me that there is a natural morphism from $\mathcal{O}_X(C)$ to $\mathcal{T}or^1_X(\mathcal{O}_C,\mathcal{H}^1_C(\mathcal{O}_X))$ induced by $\delta_i$, the kernel of which is $\mathcal{O}_X$. To summarize, I suspect that there exists a short exact sequence of the form $$0 \to \mathcal{O}_X\ \to \mathcal{O}_X(C)\to \mathcal{T}or^1_X(\mathcal{O}_C,\mathcal{H}^1_C(\mathcal{O}_X)) \to 0$$ This is my first interpretation.

My second interpretation is as follows. By checking on open affine sets, we observe that $j_*\mathcal{O}_{X\backslash C} \otimes_{\mathcal{O}_X} \mathcal{O}_C=0$ as well as $\mathcal{T}or^1_X(\mathcal{O}_C,j_*\mathcal{O}_{X\backslash C})=0$. This would imply by the above short exact sequence, after applying $- \otimes_{\mathcal{O}_X} \mathcal{O}_C$, that $\mathcal{T}or^1_X(\mathcal{O}_C,\mathcal{H}^1_C(\mathcal{O}_X)) \cong \mathcal{O}_C$.

But both the interpretations cannot be correct due to the short exact sequence $$0 \to \mathcal{O}_X \to \mathcal{O}_X(C) \to \mathcal{O}_C \otimes _{\mathcal{O}_X} \mathcal{O}_X(C) \to 0 $$

What am I doing wrong? Any idea or reference on this question will be most welcome.

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  • $\begingroup$ I do not understand your first construction. Perhaps this is the source of the confusion: for an ideal sheaf $\mathcal{I}$ in $\mathcal{O}_X$, for an $\mathcal{O}_X$-module $\mathcal{F}$, there is a canonical isomorphism between $\text{Tor}^{\mathcal{O}_X}_1(\mathcal{O}_X/\mathcal{I},\mathcal{F})$ and the kernel of the natural map $\mathcal{I}\otimes_{\mathcal{O}_X} \mathcal{F} \to \mathcal{F}$. If $\mathcal{I}$ is an invertible sheaf, the domain of that morphism locally is isomorphic to $\mathcal{F}$. However, globally, the Tor sheaf maps to $\mathcal{I}\otimes_{\mathcal{O}_X}\mathcal{F}$. $\endgroup$
    – Jason Starr
    Commented Oct 8, 2015 at 13:08
  • $\begingroup$ @JasonStarr I elaborate on the first construction a bit. We know that for any ring $R$ and $r \in R$ a non-zero divisor, $Tor^1_R(R/(r),M)$ consists of those elements $m \in M$ such that $r.m=0$. Using this description, we see that (under the above notations) $f.\delta_i(g/f^r)=\delta_i(g/f^{r-1})$ is zero if and only if $r-1=0$. Does this help? I do not think I use the local identification of $\mathcal{F}$ that you mention, but will check once again. I also wanted to clarify, is the second observation correct or it might also be wrong? $\endgroup$
    – user46578
    Commented Oct 8, 2015 at 13:25
  • $\begingroup$ Consider the computation of $\text{Tor}^R_1(R/I,M)$ where $I$ is not a principal ideal. This should clear up your confusion. Invertible sheaves (e.g., fractional ideals) are "locally" isomorphic to $R$, but the canonical isomorphism does not use $R$, but $I$. $\endgroup$
    – Jason Starr
    Commented Oct 8, 2015 at 13:36

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I believe I understand now your first construction. You are comparing the following two short exact sequences. $$\begin{array}{ccccccccc} 0 & \rightarrow & \mathcal{O}_X & \rightarrow & \mathcal{O}_X(C) & \rightarrow & \mathcal{O}_X(C)/\mathcal{O}_X & \rightarrow& 0 \\ & & =\downarrow & & \downarrow & &\downarrow \\ 0 & \rightarrow & \mathcal{O}_X & \rightarrow & j_*\mathcal{O}_{X\setminus C} & \rightarrow & \mathcal{H}^1_C(\mathcal{O}_X) & \rightarrow & 0\end{array}$$ This induces a morphism of Tor sheaves, $$ \textit{Tor}^{\mathcal{O}_X}_1(\mathcal{O}_X/\mathcal{O}_X(-C),\mathcal{O}_X(C)/\mathcal{O}_X) \to \textit{Tor}^{\mathcal{O}_X}_1(\mathcal{O}_X/\mathcal{O}_X(-C),\mathcal{H}^1_C(\mathcal{O}_X)),$$ and this turns out to be an isomorphism. Moreover, from my comment above, there is an isomorphism of $\textit{Tor}^{\mathcal{O}_X}_1(\mathcal{O}_X/\mathcal{O}_X(-C),\mathcal{O}_X(C)/\mathcal{O}_X)$ with the kernel of the morphism $$r:\mathcal{O}_X(-C)\otimes_{\mathcal{O}_X}(\mathcal{O}_X(C)/\mathcal{O}_X) \to (\mathcal{O}_X(C)/\mathcal{O}_X).$$ Since $r$ is a zero morphism, this finally gives an isomorphism, $$\textit{Tor}^{\mathcal{O}_X}_1(\mathcal{O}_X/\mathcal{O}_X(-C),\mathcal{H}^1_C(\mathcal{O}_X)) \cong \mathcal{O}_X(-C)\otimes_{\mathcal{O}_X}(\mathcal{O}_X(C)/\mathcal{O}_X).$$ Of course the second sheaf is isomorphic to $\mathcal{O}_X/\mathcal{O}_X(-C)$.

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  • $\begingroup$ This makes a lot of sense. This clears up my mistake. Thank you very much. $\endgroup$
    – user46578
    Commented Oct 8, 2015 at 15:20

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