Linearly trivial bundles on hypersufaces in $\mathbb CP^n$ - MathOverflow

Linearly trivial bundles on hypersufaces in $\mathbb CP^n$

2013-01-31T10:42:51Z

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?

Answer by Sándor Kovács for Linearly trivial bundles on hypersufaces in $\mathbb CP^n$

2013-01-31T23:48:54Z

Apparently I didn't read the question correctly and so this is not an answer....

This is not true as stated.

Example 1 Let $V$ be a hypersurface that does not contain a line. For instance, every general surface of degree at least $4$ in $\mathbb P^3$ is such, because they have Picard number $1$ and hence cannot have any non-trivial curves on them. Any line bundle on such a $V$ is linearly trivial, because the condition is satisfied vacuously.

OK, so let's assume that $V$ contains lines.

Example 2 Let $V$ be an irrational scroll and consider the globally generated but not ample line bundle that induces the morphism that collapses the lines. This is linearly trivial because the only lines are the ones that the morphism collapses.

OK, so let's assume that $V$ is connected by (chains of) lines. However, that probably implies that $V$ is linear, so you don't gain anything.

In any case, you need a new formulation for this.

Answer by Mohan for Linearly trivial bundles on hypersufaces in $\mathbb CP^n$

2013-02-02T20:49:39Z

I do not think the answer to your question is known. After your question, I thought about it and I think I can see how to prove it for general hypersurfaces of degree $n$ in $\mathbb{P}^N$ with $N\geq n+2$. I do not yet see how to do this for all smooth hypersurfaces. If you wish (on the other hand, it is your idea and you should see what you see fit) I can try to explain my idea.

Answer by Mohan for Linearly trivial bundles on hypersufaces in $\mathbb CP^n$

2013-02-03T01:45:35Z

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.