*Recall a definition*. Let $V\subset \mathbb CP^n$ be a projective variety
and $E$ be a holomorphic vector bundle on it. We call $E$ *linearly trivial* if the restriction of $E$ to any projective line in $V$ is trivial.


It is well known that any linearly trivial bundle on $\mathbb CP^n$ itself is trivial (see Okonek, Schneider, Spindler). 


**Question 1.** I think that I have an idea of a generalization of this statement and would like
to ask you if this generalization is known?


**Generalized statement**. For any integer $n>0$ 
any linearly trivial bundle on any smooth degree $n$ hypersuface $V\_n\subset \mathbb CP^{4n}$ is trivial. 


*Idea of the proof*. One can easily see that on $V\_n$ any two points can be joined by a chain of two projective lines. Moreover for two points $x,y$ the set of such two-lines paths from $x$ to $y$ is a connected projective variety.
So let us trivialize the bundle at one point $x\in V\_n$. Then extend this trivialization along each connected chain of $2$ lines on $V\_n$ starting at $x$.
I think that the extension will be independent of the choice of a chain since 
the space of all chains from $x$ to $y$ is a connected projective variety, 
while all trivialization of $E$ over $y$ is an affine variety. 


**Question 2.** Does this reasoning sound plausible?



Linearly trivial bundles on hypersufaces in $\mathbb CP^n$
I didn't want to comment, since it might take longer. I need more than the fact that through a general point of $X\subset \mathbb{P}^N$ of degree $n\leq N-2$, there is at least $N-n-1$ dimensional family of lines, but exactly of that dimension. Once we have that (and it is proved in Kollar's book, and this is where I need generality of the hypersurface), the subvariety $B\subset X$ of the union lines passing through this point is a complete intersection on $X$. So, the vector bundle $E$ on $X$ restricted to $B$ will be trivial mimicking the proof in Okonek et. al. $\dim B\geq 2$ by our assumption on $N,n$. So, $H^1(E|BB(k))=0$ for all $k$ since $E|B$ is trivial and $B$ is a complete intersection. Now, by boot strapping, since $B$ is a complete intersection on $X$, one can easily check that $E$ itself is trivial on $X$. If you need more details, please let me know.


