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I am reading a paper on Chiral Differential Operators

http://arxiv.org/pdf/hep-th/0604179v3.pdf

and it says on page 23 that a line bundle over a manifold C can be characterized completely by its restriction to a non-trivial two-cycle, such as a two-sphere in C.

The paper refers a book to check this fact: "Differential Forms in Algebraic Topology" by Raoul Bott & Loring W. Tu.

http://www.amazon.com/Differential-Algebraic-Topology-Graduate-Mathematics/dp/0387906134

I could not find any help from the book.

Edit: I have posted the same question on Math Stack exchange. https://math.stackexchange.com/questions/628329/restriction-of-a-line-bundle-to-a-a-two-cycle

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  • $\begingroup$ This is a basic consequence of the classifying-space formalism for describing vector bundles over a space. The classifying space for complex line bundles is an Eilenberg-MacLane space $\mathbb CP^\infty \simeq K(\mathbb Z,2)$, which corresponds to 2-dimensional cohomology. Bott and Tu, as well as a basic algebraic topology text like Hatcher or May's is a good place to start. Bredon's text would be fairly self-contained for this, as the underlying geometry for this construction is fairly direct. $\endgroup$
    – Ryan Budney
    Commented Jan 6, 2014 at 21:03
  • $\begingroup$ There's basically three main observations. 1) "vector bundles embed", or more precisely, vector-bundles are pull-backs of canonical bundles over grassmannians. 2) The homotopy long exact sequence of a fibration. 3) The obstruction theory observation that cohomology is maps into Eilenberg-Maclane spaces. $\endgroup$
    – Ryan Budney
    Commented Jan 6, 2014 at 21:08
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    $\begingroup$ It seems to me that the given statement is only true if the 2-dimensional cohomology of the manifold $C$ is cyclic. If $C=S^2\times S^2$, for instance, then its easy to construct non-isomorphic line bundles whose restrictions to the first sphere factor are isomorphic. $\endgroup$
    – Mark Grant
    Commented Jan 6, 2014 at 21:27
  • $\begingroup$ Please add a Link here to the question at math.se and backwards. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 7, 2014 at 3:53

2 Answers 2

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Alex is on the right track, but I don't think this is correct in general. It is true that a bundle over a manifold $C$ is characterized by its first Chern class $c_1 \in H^2(C;{\bf Z})$. It is not true that a cohomology class in $H^2$ is determined by its evaluation on all cycles; this would be tantamount to saying that the natural map $H^2(C;{\bf Z}) \to Hom(H_2(C);{\bf Z})$ is always an injection. But it is not: by the universal coefficient theorem, torsion in $H_1(C)$ gives rise to torsion in $H^2(C;{\bf Z})$ that is not detected by this map. A correct (but not particularly useful) assertion is that $c_1$ (and hence the bundle) is determined by its evaluation on all $2$-chains.

On the other hand, in the linked paper, the authors (if I understand the physics-y terminology correctly) are only interested in the case of a line bundle (the determinant bundle of some family of operators) on $\Sigma \times B$ where $\Sigma$ is a complex curve and $B$ is a $2$-sphere. In that setting, there is no torsion in the cohomology group, so $c_1$ is indeed determined by its values on a basis for $H_2(\Sigma \times B); {\bf Z})$.

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  • $\begingroup$ The authors are interested in a line bundle L on C. C is the configuration space of a manifold M. They say that instead of studying the properties of L on C, we can study its properties on a two-sphere B which is a subset of C. $\endgroup$
    – Ali Shehper
    Commented Jan 7, 2014 at 11:26
  • $\begingroup$ And you said that $c_1$ is determined by its evalutation on all 2-chains. How is that true? $\endgroup$
    – Ali Shehper
    Commented Jan 7, 2014 at 11:27
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    $\begingroup$ @AliShehper: This is true by the definition of the integer cohomology group. $\endgroup$
    – Misha
    Commented Jan 7, 2014 at 13:28
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The statement is not true: the bundle is characterized by its restriction to all $2$-cycles, i.e., by its $2$-cohomology class (Chern class $c_1$). This is a simple fact in algebraic topology. One (the easiest, but not the most elementary) way to prove it is the exponential exact sequence $0\to\mathbb{Z}\to\mathbb{C}\to\mathbb{C}^*\to0$ of sheaves and the associated cohomology sequence (in either topological or smooth category; note that this does not work in the analytic category, though).

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