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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 '10 at 0:45
@Bjorn: you are absolutely right, thanks for catching it. – Hailong Dao Apr 10 '10 at 6:07

4 Answers 4

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 '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

According to this preprint, over a connected proper algebraic variety $X$ there is a universal reductive group $G$ such that isomorphism classes of vector bundles of rank $n$ are in bijection with isomorphism classes of $n$-dimensional representations of $G$. Furthermore the component group is $\pi_1(X)$.

So the cancellation property you seek holds for any simply-connected proper algebraic variety, because it holds for representations of any connected reductive group. We can see this by characters - a representation is uniquely determined by its character, and because the group is connected, there are no zero-divisors in the ring of characters.

This certainly includes $\mathbb P^n$.

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Very nice application of O'Sullivan's (rather abstract) result. You should mention it to him. – abx Jul 13 at 7:23

The answer to your question is yes, if $F$ is a direct sum of line bundles. So, let us assume that and its rank is $m$. And as you observed, we will assume that we are working over $\mathbb{P}^2$. Let $0\to F_0\to F_1\to E\to 0$ be the minimal resolution of $E$, where $F_i$ are direct sum of line bundles with rank of $F_0=n$ so that rank of $F_1=n+m$. Tensoring with $G$, we get $0\to G\otimes F_0\to G\otimes F_1\to G\otimes F\to 0$, the last by the assumption. Taking cohomologies, letting $H_*^0(G)=M,H_*^1(G)=N$, we get an exact sequence $N^n\to N^{n+m}\to N^m$, which for length considerations (length of $N$ is finite) can easily seen to be exact on the left (and right). In particular, we have surjectivity of global sections and thus an exact sequence $0\to M^n\to M^{n+m}\to M^m\to 0$. This splits (I have forgotten whose theorem it is, but I think Ihave read it recently in some comment by Graham Leuschke) which is impossible since the first map has all entries in the maximal ideal unless $n=0$.

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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 '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

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