Let $X\subset \mathbb{P}^n$ be a smooth projective variety with ideal sheaf $I\_X$. The conormal sequence is given by


$$
0\to I\_X/I\_X^2\to \Omega\_{\mathbb{P}^n}|\_X\to \Omega\_{X}\to 0.
$$
For which varieties $X$ is the sequence above split? 


If I'm not mistaken, if $X$ a hypersurface, the sequence is split if and only if $X$ has degree 1.



When does the conormal bundle sequence split?
You are right. This is a result due to Van de Ven. 


[A. Van de Ven, A property of algebraic varieties in complex projective spaces. In: Colloque Géom. Diff. Globale (Bruxelles, 1958), 151–152, Centre Belge Rech. Math., Louvain 1959. MR0116361 (22 #7149) Zbl 0092.14004]


Even more is true. Recently [Ionescu and Repetto](http://www.informaworld.com/smpp/content~db=all~content=a794323109) proved the following generalization of Van de Ven's Theorem.



> 
> Let $X \subset \mathbb P^n$ be a smooth subvariety. If there exists a curve $C \subset X$
>  such that the restriction to $C$ of
>  the conormal sequence of $X$ splits
>  then $X$ is linear.
> 
> 
> 




---


Let me sketch a short elementary proof (of Van de Ven's result not its generalization) in the case of hypersurfaces. I will phrase it in the analytic category but once it is translated to the algebraic category, working with infinitesimal neighborhoods, I believe that what will emerge is one of the proofs in the literature. 


If the normal sequence splits then we can define a foliation $\mathcal L$ by (germs of) lines everywhere transverse to $X$ at a 
neighborhood $U$ of $X$. Since the complement of $X$ is Stein we can extend $\mathcal L$ to the whole $\mathbb P^n$. 
Therefore $\mathcal L$ is defined by a global section of $T \mathbb P^n(d-1)$ for some
$d \ge 0$. With the help of Euler's sequence, this section can be presented as a homogeneous vector field $v$
on $\mathbb C^{n+1}$ with coefficients of degree $d$. 
To compute the tangencies between $\mathcal L$ and $X$ we have just to contract
the differential $dF$ of a defining equation $F$ of $X$ with $v$. If $F$ is not linear
then the divisor on $X$ defined by the tangencies between $\mathcal L$ and $X$ (defined by $F=dF(v)=0$) will be non-empty contradicting the transversality between $X$ and $\mathcal L$. 


