Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $E,F,G$ be algebraic vector bundles over $\mathbb P_{\mathbb C}^n$. My general question is:

Assume $E\otimes G \cong F\otimes G$, under what conditions can one conclude that $E\cong F$?

Some easy answers (if I am not mistaken): one can when $n=1$ or when $G$ is a line bundle. At this point I am mostly interested in the case when $E$ is a direct sum of line bundles, but any comments/reference/solutions/analogues about other cases would be appreciated.

share|improve this question
2  
To make it true for n=1, you should assume that G is not the zero bundle! –  Bjorn Poonen Apr 10 '10 at 0:45
    
@Bjorn: you are absolutely right, thanks for catching it. –  Hailong Dao Apr 10 '10 at 6:07
add comment

3 Answers 3

It seems, that in the case in which E and F are direct sums of line bundles (and G is non-zero!), you can reconstruct E and F knowing that $E \otimes G \simeq F \otimes G$: this simply imitates the proof of the fact that vector bundles on $\mathbb{P}^1$ are sums of line bundles. Indeed, since the reconstruction is fine in the case in which G is a line bundle, we may replace E and F by E(e) and F(e) for any integer e: hence, exchanging if necessary E and F, we may suppose that E has a section and E(-1),F(-1) do not have sections. Let g be the integer such that G(g) has a section and G(g-1) does not. Thus we have $E \otimes G(g) \simeq F \otimes G(g)$; by considering global sections, we deduce that the multiplicity of the number of trivial direct summands in E is the same as the multiplicity in F. Remove the copies of $\mathcal{O}$ from both E and F and repeat.

share|improve this answer
    
@damiano: That's right! I think the tricky part is: given E splits, showing F has to split too. I think one can do it, but it will not be very easy. –  Hailong Dao Apr 14 '10 at 17:48
    
A quick comment: to show that F splits, one can restrict to some hyperplane and reduce to the situation $n=2$. –  Hailong Dao Apr 14 '10 at 17:58
    
You might be able to use some kind of Harder-Narasimhan filtration argument to show that if one of the two bundles splits, then the other one must split as well. I do not know how to do this, it just seems a natural generalization of the above argument. –  damiano Apr 14 '10 at 18:19
    
To handle the case that $E$ and $F$ are direct sums of line bundles, it is easier to restrict to a line; sums of line bundles are determined by their restrictions to a line. Also, the case that $E$ is a direct sum of copies of the same line bundle follows by the same trick from Horrock's theorem that a vector bundle with splitting type $(d, \dots, d)$ on all lines splits. –  Angelo Apr 15 '10 at 11:29
add comment

Well, I am by no means an algebraic geometer and maybe this is not even helpful, but anyway:

If you have a bundle $G^\perp$, such that $G \oplus G^\perp \simeq \underline{\mathbb{C}^n}$ is trivial and $E \otimes G^\perp$ is isomorphic to $F \otimes G^\perp$, then you at least get $E^n \simeq F^n$, which is kind of analogous to your remark concerning line bundles. But I better stop mumbling trivialities now.

share|improve this answer
    
@Ulrich: Do you mean to say $G\otimes G^{\perp}$ is trivial? I am fairly certain this forces $G$ to split, and so have to be direct sum of $\mathcal O(n)$ for same $n$. –  Hailong Dao Apr 13 '10 at 20:04
    
I'm sure he meant direct sum, as evidenced by his orthogonal complement notation, where such examples come from. –  Adam Gal Apr 13 '10 at 20:24
    
Sorry, I may have caused some confusion by my comparison with the case of line bundles: I did mean the direct sum and the isomorphism $E^n \simeq E \otimes \underline{\mathbb{C}^n} \simeq E \otimes (G \oplus G^\perp) \simeq (E \otimes G) \oplus (E \otimes G^\perp) \simeq (F \otimes G) \oplus (F \otimes G^\perp) \simeq F^n$. –  Ulrich Pennig Apr 14 '10 at 9:22
    
@Ulrich: I was indeed confused by the line bundle analogue, because there you have to tensor with $G^*$. But here you will need to know that $E\otimes G^{\perp} \cong F\otimes G^{\perp}$ as well! Is that easy to see? –  Hailong Dao Apr 14 '10 at 17:52
add comment

you may get some indications from Atiyah's book :K-theory.

share|improve this answer
8  
This is not useful at all. You could probably give a more precise reference/solution since the question is very specific. –  Somnath Basu Apr 10 '10 at 16:55
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.