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Suppose $X \subset Y$ is a pair of varieties, and $s \in H^0(N_{X/Y})$ is a section. This corresponds to a first-order deformation $X' \subset Y \times \text{Spec}(\mathbb{C}[\epsilon]/\epsilon^2)$ of $X \subset Y$.

If $\mathcal{E} \to Y$ is a vector bundle, is there a nice way to compute the cohomology of the restriction of $\mathcal{E}$ to the generic fiber of $X'$ (in terms of $X \subset Y$, $\mathcal{E}$, and $s$)?

EDIT: Oops, I meant to write "the restriction of $\mathcal{E}$ to $X'$" (not to its generic fiber, which doesn't make sense as pointed out below).

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    $\begingroup$ Dear Eric, what is the generic fiber of $X'$ ($X'$ is not reduced)? $\endgroup$
    – Piotr Achinger
    Commented Feb 23, 2013 at 21:30
  • $\begingroup$ How about tensoring the exact sequence $0\to \mathcal{O}_X \to \mathcal{O}_{X'}\to \mathcal{O}_X\to 0$ with $E$ and looking at the long exact sequence? You get a lot of information, e.g. that $H^i(X', E) = 0$ if $H^i(X, E) = 0$. $\endgroup$
    – Piotr Achinger
    Commented Feb 23, 2013 at 23:33

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On $X'$, there is an exact sequence $$ 0 \to \mathcal{O}_X \to \mathcal{O}_{X'} \to \mathcal{O}_X \to 0. $$ Note that this sequence does not carry a lot of information about the deformation, for example it often splits (e.g. if $H^1(X, T_X) = 0$) even if $X'$ is a non-trivial deformation.

Anyway, we can tensor this sequence with $E$, obtaining a short exact sequence $$ 0 \to E|_X \to E|_{X'} \to E|_X \to 0. $$ Applying cohomology, we get a long exact sequence $$ \ldots \to H^i(X, E) \to H^i(X', E) \to H^i(X, E) \to H^{i+1}(X, E) \to \ldots $$ `

In particular, if $H^i(X, E) = 0$ then $H^i(X', E) = 0$ as well, and if the initial sequence was split then you get $$ H^i(X', E) = H^i(X, E) \otimes_k k[\varepsilon]/(\varepsilon^2) = H^i(X, E)\oplus \varepsilon H^i(X, E). $$

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    $\begingroup$ Yes, this is good... But when $H^1(X,T_X) \neq 0$ --- do you know if there is an explicit formula for the boundary maps $H^i(X,E) \to H^{i+1}(X,E)$ in the above sequence? Say, in terms of the section $s \in H^0(N_{X/Y})$? (Can this map be determined from the image of $s$ in $H^1(X, T_X)$ under the boundary map for the long exact sequence $0 \to T_X \to T_Y|_X \to N_{X/Y} \to 0$?) $\endgroup$
    – Eric Larson
    Commented Feb 24, 2013 at 1:23
  • $\begingroup$ My guess is that 1) $s$ gives you an element $c$ of $H^1(X, O_X)$ (apply $Hom(-, O_X)$ to the extension $0\to O_X \to P\to N^\vee \to 0$ where $P = O_Y/I_X^2$, $N^\vee = I_X/I_X^2$, getting a connecting homomorphism $\delta: H^0(X, N)\to H^1(X, O_X)$ and let $t = \delta(s)$), 2) the boundary maps $H^i(X, E) \to H^{i+1}(X, E)$ are just cup product with $c$. It seems plausible, but I didn't check it. $\endgroup$
    – Piotr Achinger
    Commented Feb 24, 2013 at 1:30
  • $\begingroup$ In particular this would show that the boundary maps are zero if $H^1(X, O_X)=0$ e.g. $X$ is simply connected. I'm really not sure if my guess is correct, let me know if you can figure it out. $\endgroup$
    – Piotr Achinger
    Commented Feb 24, 2013 at 1:31

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