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6 Replaced n! by 4n and made the idea of proof more affirmative

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 $V_n\subset \mathbb CP^{n!}$ CP^{4n}$is trivial. (note that$n!>>n$this is why I chose this dimension, but I guess$10n$or$10n^3$will do as well) 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 of lines that join from$x$with to$y$should be is a connected projective variety, while all trivialization of$E$over$y$is an affine variety. Question 2. Does this reasoning sound plausible? 5 added 122 characters in body 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$V_n\subset \mathbb CP^{n!}$is trivial. (note that$n!>>n$this is why I chose this numberdimension, but I guess$10n$or$10n^3$will do as well) Idea of the proof. Trivialize One can easily see that on$V_n$any two points can be joined by a chain of two projective lines. So let us trivialize the bundle at one point$x\in V_n$. Then extend this trivialization along each connected chain of$n!$2$ 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 of lines that join $x$ with $y$ should be a connected projective variety, while all trivialization of $E$ over $y$ is an affine variety. (I use $n!$ just as a big number).

Question 2. Does this reasoning sound plausible?

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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 $V_n\subset \mathbb CP^{n!}$ is trivial. (note that $n!>>n$ this is why I chose this number)

Idea of the proof. Trivialize the bundle at one point $x\in V_n$. Then extend this trivialization along each connected chain of $n!$ 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 of lines that join $x$ with $y$ should be a connected projective variety, while all trivialization of $E$ over $y$ is an affine variety. (I use $n!$ just as a big number).

Question 2. Does this reasoning sound plausible?

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