I had been reading an article by Spencer Bloch. There is a remark in this text which states the surjectivity of a particular map between cohomology groups without explaining further. I had been trying to work it out without much success. I elaborate on this map: Let $X$ be a smooth projective hypersurface of degree $d$ in $\mathbb{P}^{2n+1}$ for some integer $n \ge 1$. Let $Z$ be a smooth subscheme in $X$ of codimension $n$. Using the natural map, say $\epsilon: \mathcal{I}_Z/\mathcal{I}_Z^2 \to \Omega^1_X \otimes \mathcal{O}_Z$, where $\mathcal{I}_Z$ is the ideal sheaf of $Z$ in $X$, one gets a map $\wedge^{n1}\epsilon:\wedge^{n1}\mathcal{I}_Z/\mathcal{I}_Z^2 \to \Omega^{n1}_X \otimes \mathcal{O}_Z$. Taking dual and tensoring with the canonical sheaf, say $K_X$, one gets a map:$\epsilon':\Omega^{n+1}_X \otimes \mathcal{O}_Z \to \wedge^{n1} \mathcal{N}_{ZX} \otimes K_X$, where $\mathcal{N}_{ZX}$ is the normal sheaf which is the dual of $\mathcal{I}_Z/\mathcal{I}_Z^2$. Bloch claims that for $d \gg 0$, the induced map $$H^{n1}(\Omega^{n+1}_X \otimes \mathcal{O}_Z) \to H^{n1}(\wedge^{n1} \mathcal{N}_{ZX} \otimes K_X)$$ is surjective. I do not understand why this is true. Any help on this question will be very much appreciated. This question is very closely related to the study of Hodge locus and deformation of algebraic cycles.
I assume that this is example 1.4 of Bloch's semiregularity paper. The problem is that you've stated this in a way that makes it harder than it is. He says that you choose $Z$ first and then you choose $X\supset Z$ with $d\gg 0$. So now it's just a matter of killing the appropriate cohomology group using Serre vanishing.
Since I really don't want to put too much effort into this, say $n=2$. Your map $\epsilon'$ is part of the sequence $$0 \to T_Z\otimes K_X\to \Omega_X^3_Z\to N_{Z/X}\otimes K_X\to 0$$ Then $H^n(T_Z\otimes K_X) = H^2(T_Z(d2(2)2))=0$ for $d\gg 0$. Which implies what you want in this case.

1$\begingroup$ Speaking for myself, thank you for clarifying the question. I was having trouble understanding how the OP's formulation could be true. $\endgroup$ – Jason Starr Jun 30 '14 at 19:16

$\begingroup$ @Arapura: Thank you very much for the answer. Sorry, for the confusion in the question. I was using a similar approach. For $n>2$, I was getting the short exact sequence, $$0 \to T_Z \wedge^{n2} T_X \otimes K_X \to \Omega^{n+1}_X \otimes \mathcal{O}_Z \to \wedge^{n1} N_{ZX} \otimes K_X \to 0$$ As you see the kernel has a term of the form $\wedge^{n2}T_X$, which is preventing me from using the Serre's vanishing theorem. May be I am doing something wrong or stupid. Could you please check. $\endgroup$ – Kali Jun 30 '14 at 19:27

$\begingroup$ userxyz: if I had to do this for myself, I would filter $\wedge^{n1}T_X$ with associated graded $\wedge^p N_{Z/X}\otimes \wedge^{n1p}T_Z$, and use the exact seq $0\to N_{Z/X}\to N_{Z/\mathbb{P}}\to O(d)\to 0$ to express everything independently of $X$. $\endgroup$ – Donu Arapura Jun 30 '14 at 19:51

$\begingroup$ @Arapura: I am learning how to deduce results in cohomology using filtration. As far as I understand I must use the related spectral sequence. Is this the method you have in mind or is there some simpler idea? $\endgroup$ – Kali Jul 1 '14 at 14:25

$\begingroup$ You can certainly use a spectral sequence, but it might be simpler to break it up into short exact sequences and use induction. $\endgroup$ – Donu Arapura Jul 1 '14 at 19:15