MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
Dear Eric, what is the generic fiber of $X'$ ($X'$ is not reduced)? – Piotr Achinger Feb 23 '13 at 21:30
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 $H^i(X, E) = 0$. – Piotr Achinger Feb 23 '13 at 23:33

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 $$ H^i(X', E) = H^i(X, E) \otimes_k k[\varepsilon]/(\varepsilon^2) = H^i(X, E)\oplus \varepsilon H^i(X, E). $$

share|cite|improve this answer
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 $0 \to T_X \to T_Y|_X \to N_{X/Y} \to 0$?) – Eric Larson Feb 24 '13 at 1:23
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. – Piotr Achinger Feb 24 '13 at 1:30
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. – Piotr Achinger Feb 24 '13 at 1:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.