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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 $V_n\subset \mathbb CP^{4n}$ 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?

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I wonder how you are going to prove connectedness... –  Serge Lvovski Jan 31 '13 at 18:51
    
Do you mean projective plane instead of projective line? Because all bundles on $\mathbf{CP}^n$ are linearly trivial, so then you are saying all bundles on $\mathbf{CP}^n$ split? Or did I misunderstand? –  Mahdi Majidi-Zolbanin Jan 31 '13 at 21:52
    
Mahdi, my question is about holomorphic bundles. Linearly trivial bundle is the bundle that restricts to each $\mathbb CP^1$ in $\mathbb CP^n$ as $O+O+...+O$ ($k$ copies of the structure sheaf of $\mathbb CP^1$). So for example according to my definition $O(1)$ is not linearly trivial ($T\mathbb CP^n$ neither). Serge, I agree, probably connectedness is is the most complicate thing to understand –  aglearner Jan 31 '13 at 22:38
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Serge, in fact this space will be connected. I just realised that every two points on $V_n$ can be joined by a chain of just two lines (and not $n!$ :) as I wrote previously, I changed now $n!$ for two since this works). Indeed, the dimension of the space of lines through a generic point on $V_n$ is $n!-n-1$. So the space of all paths between $x$ and $y$ can be identified with the intersection of two subvarieties of $\mathbb CP^{n!}$ of dimensions $n!-n$. All irreducible components of this intersection have dim at least $n!-2n$ hence it is connected. –  aglearner Feb 1 '13 at 1:07

3 Answers 3

up vote 2 down vote accepted

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.

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Mohan, I thought a bit more about you answer and I have some questions. 1) Could you please elaborate the phrase: "Now, by boot-strapping, since $B$ is a complete intersection on $X$, one can easily check that $E$ itself is trivial on $X$"? 2) What statement and in which book of Kollar would you like to use? 3) Note that on a quadric the union of lines through one point is not a complete intersection. 4) Is there misprint here: $H^1(E|BB(k)$? –  aglearner Feb 3 '13 at 12:03
    
I still wonder if the reasoning that I proposed in the original question can be made rigorous. In fact this reasoning can be applied even if $degX_n=n\le N/2$ a hypersurface, $X_n\subset \mathbb CP^N$, because in this case the space of length two paths between two points on $X_n$ is still connected (I think I can prove this). –  aglearner Feb 3 '13 at 12:53
    
Mohan, huge thanks for sending me the text :). Above 3) (about quadric is of course wrong...). I will try to understand your text and will write you back. Thanks again! –  aglearner Feb 3 '13 at 23:47

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.

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Sandor I consider a hypersurface of degree $n$ in $\mathbb CP^{n!}$. –  aglearner Feb 1 '13 at 0:14
    
Note that $n<< n!$. Such hypersufaces of Fano, and they have plenty of lines. So your examples are not quite related to the actual question... –  aglearner Feb 1 '13 at 0:16
    
Ah, I missed that detail. Sorry. –  Sándor Kovács Feb 1 '13 at 0:16
    
Sandor, I guess this confusion was because of $n!$... so I added a line and hope the question will not be confusing for other people who read it... –  aglearner Feb 1 '13 at 0:29
    
To be honest, I didn't pay attention to the dimension of the space. Since you said it would be a hypersurface of a given degree, I just thought the question was in an arbitrary space. The reason I haven't deleted this non-answer is to give others the chance to not make the same mistake. Cheers! –  Sándor Kovács Feb 1 '13 at 1:21

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.

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Mohan, this is interesting. I wonder if you need $n\le N+2$, so that through a generic point at least one-dimensional family of lines passes, which would imply (if full non-integrability of this distribution is established) that any two points on the hypersuface can be joined by a chain of lines? Anyway, I don't know how to make this rigorous, I and am curious to know how you would prove your claim. –  aglearner Feb 3 '13 at 0:44

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