It is a wellknown result of Griffiths that the pieces of Hodge filtration of a smooth hypersurface $X:= (f=0)$ of degree $d$ in $\mathbb{P}^{n}$ are isomorphic to graded pieces of the Jacobian ring associated to $X$ i.e the ring $R= \mathbb{C}[x_{1},...x_{n}]/J$ where $J$ is the ideal generated by the partial derivatives of $f$ . The isomorphism is given by means of the socalled Poincare residue. There is another isomorphism $R^{d} \cong H^{1}(X,T_{X})$ (both are isomorphic to the space of infinitesimal deformaions of $X$) . I don't know how this isomorphism is explicitly given. In other words, I am wondering that in a concrete example how one can explicitly find the image of an element of $H^{1}(X,T_{X})$ in $R^{d}$under this last isomorphism. More exactly if I have a 1Cech cocycle of the bundle $T_{X}$, how can I explicitly associate a homogenous polynomial of degree $d$ in $R^{d}$ to it? For simplicity you can consider the family $y^{4}= x(x1)(x+1)(x\lambda)$ of plane curves of degree $4$.

1$\begingroup$ Just to clarify, the theorem of Griffiths is only valid for the "primitive" Hodge structure of the hypersurface, not the "full" Hodge structure. $\endgroup$– Jason StarrJul 17 '12 at 12:23

$\begingroup$ Thanks Jason! I know this, but since the question is not about the Hodge structure, I just did not emphasize on it. $\endgroup$– JackJul 17 '12 at 13:09
For $d\geq 5$, for a smooth plane curve of degree $d$, it is simply not true that the natural map $R_d \to H^1(X,T_X)$ is surjective: there are plenty of abstract deformations which are not plane curves. I am including below the proof that the natural map is surjective when $n\geq 4$, when $n=3$ and $d\leq 3$, and when $n=2$ and $d\leq 4$.
The basic idea is to look at the short exact sequence of sheaves $$ 0 \to T_X \to T_{\mathbb{P}^n}_X \to \mathcal{O}_{\mathbb{P}^n}(d)_X \to 0.$$ Using the long exact sequence of cohomology, once you prove that $h^1(X,T_{\mathbb{P}^n}_X)$ equals $0$, then it follows that the connecting map $$H^0(X,\mathcal{O}_{\mathbb{P}^n}(d)_X) \to H^1(X,T_X)$$ is surjective. Of course the image of $H^0(X,\mathcal{O}_{\mathbb{P}^n}(d)_X)$ is $R_d$. Finally, using the long exact sequence of cohomology associated to the Euler sequence, $$ 0 \to \mathcal{O}_{\mathbb{P}^n}_X \to \mathcal{O}_{\mathbb{P}^n}(1)^{\oplus n+1}_X \to T_{\mathbb{P}^n}_X \to 0,$$ to prove that $h^1(X,T_{\mathbb{P}^n}_X)$ equals $0$, it suffices to prove that $h^1(X,\mathcal{O}_{\mathbb{P}^n}(1)_X)$ and $h^2(X,\mathcal{O}_X)$ both equal $0$. These cohomologies can be computed from the usual computation of $H^q(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n}(r))$ using the long exact sequence of cohomology associated to the short exact sequence, $$ 0 \to \mathcal{O}_{\mathbb{P}^n}(rd) \to \mathcal{O}_{\mathbb{P}^n}(r) \to \mathcal{O}_{\mathbb{P}^n}(r)_X \to 0 $$ For $n\geq 3$, always $h^1(X,\mathcal{O}_{\mathbb{P}^n}(1)_X)$ equals $0$, and the second vanishing holds so long as $\text{dim}(X) \geq 3$, i.e., $n\geq 4$. It is not hard to prove that the second vanishing also holds if $n=3$ and $d\leq 3$. It definitely fails if $n=3$ and $d=4$, i.e., it fails for quartic K3 surface (although, in fact, the natural map is surjective again for $d\geq 5$).
This brings us to the case that $n=2$, i.e., $X$ is a plane curve. By adjunction and Serre duality on $X$, $h^1(X,\mathcal{O}_X(d))$ equals $0$. So the natural map is surjective if and only if $h^1(X,T_{\mathbb{P}^n}_X)$ equals $0$. By the Euler sequence again, this holds if and only if the map $$ H^1(X,\mathcal{O}_X) \to H^1(X,\mathcal{O}_X(1))^{\oplus 3} $$ is surjective. If $d$ equals $1$, $2$ or $3$, the second group is already zero. Finally, if $d$ equals $4$, your case, then by Serre duality the transpose map is $$ H^0(X,\mathcal{O}_X)^{\oplus 3} \to H^0(X,\mathcal{O}_X(1)) $$ So you are reduced to the question of whether the restriction map '$$H^0(\mathbb{P}^2,\mathcal{O}_{\mathbb{P}^2}(1)) \to H^0(X,\mathcal{O}_X(1)) $$' is surjective, i.e., whether the next term $H^1(\mathbb{P}^2,\mathcal{O}(3))$ equals $0$, which it does.
Once again, for $d\geq 5$, the natural map is not surjective.
Regarding your original question, how to "transform" a cocycle into an element of $R_d$: on the basic open affine covers of $\mathbb{P}^2$, write the corresponding Cech 1cocycle for $T_{\mathbb{P}^2}_X$ as the coboundary of Cech 0cochain, which you can do since the cohomology group vanishes. Then take the image 0cochain in $\mathcal{O}(d)_X$. The coboundary of this 0cochain will be the image of your original 0cochain in $\mathcal{O}_X(d)$, which is zero since the composite map $T_X \to \mathcal{O}_X(d)$ is zero. So your 0cochain is a 0cocycle, i.e., a global section in $R_d$.

$\begingroup$ Dear Jason, Thanks alot for your complete answer. I have a question about what you said: "it is not true that the natural map $R_{d}$ to $H^{1}(X,T_{X})$ is surjective: there are plenty of deformations which are not plane curves". So if $n=2$ and we have a family of plane curves,it must be true that at least the image of the Kodairaspencer map of the family in $H^{1}(X,T_{X})$ must be contained in $R_{d}$. Because the KS map parametrizes the infinitesimal deformations of $X$ in the family which are all plane curves and based on your argument the image of KS must be in $R_{d}$. Is this true? $\endgroup$– JackJul 17 '12 at 18:19

$\begingroup$ Of course I mean a general family of smooth plane curves of degree $d$, which $d$ can be greater than 6. $\endgroup$– JackJul 17 '12 at 18:30

1$\begingroup$ Dear Jack  Precisely $R_d$ parameterizes first order deformations of $X$ as an abstract projective manifold which come from first order deformations of $X$ as a closed submanifold of $\mathbb{P}^n$. This part can be seen directly: given a base space / parameter space $B$ for deformations of $X$ with local coordinates $t_1,\dots,t_d$, and given a deformation of $X$ with defining equation $F(t_1,...,t_n;x)$, then the image in $R_d$ of the generator $\partial/\partial t_i$ of $T_0 B$ is just $\partial F/\partial t_i$. $\endgroup$ Jul 17 '12 at 18:32

$\begingroup$ Dear Jason I again thank you very much for your illuminating answers. $\endgroup$– JackJul 18 '12 at 7:52