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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.

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 To make it true for n=1, you should assume that G is not the zero bundle! – Bjorn Poonen Apr 10 2010 at 0:45
@Bjorn: you are absolutely right, thanks for catching it. – Hailong Dao Apr 10 2010 at 6:07

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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.

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@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 2010 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 2010 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 2010 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 2010 at 11:29
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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.

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@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 2010 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 2010 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 2010 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 2010 at 17:52
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you may get some indications from Atiyah's book :K-theory.

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 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 2010 at 16:55

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