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I come across a statement like

Let $L$ be a complex line bundle on a manifold. $c_1(L)=0$ mod $2$ if and only if there exists a line bundle $K$ such that $L\cong K^{\otimes 2}$.

How can one prove this statement? Does one need some further assumption?

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Complex line bundles are classified by the first Chern classes. If the class of a bundle is twice the class of another, then the former bundle is the square of the latter. –  Mariano Suárez-Alvarez Dec 27 '12 at 5:30
    
I think that complex line bundles are classified by elements in $H^1(M,\mathcal{O}^{\times})$ and the first Chern class is the image of the boundary map of the exponential exact sequence $0\rightarrow \mathbb{Z}\rightarrow \mathcal{O} \rightarrow \mathcal{O}^{\times} \rightarrow 0$. I am not sure if this map is injective and first Chern classes really classifies complex line bundles. –  M. K. Dec 27 '12 at 6:38
    
In the smooth categry, $\mathcal O$ is a fine sheaf, so it is acyclic for cohomology. –  Mariano Suárez-Alvarez Dec 27 '12 at 6:42
    
You are right. I usually work in algebraic or holomorphic category and am being confused. Thanks. –  M. K. Dec 27 '12 at 6:46

1 Answer 1

up vote 2 down vote accepted

Mariano's comment has essentially answered the question, but I'll go ahead and flesh it out.

On any manifold $X$, there is an isomorphism of groups $$\mathrm{Pic}(X) \xrightarrow{~\cong~} H^2(X; \mathbb{Z}),$$ $$L \mapsto c_1(L).$$ Now if $$c_1(L) \equiv 0 \pmod 2,$$ then there is some element $a \in H^2(X; \mathbb{Z})$ such that $$c_1(L) = 2a.$$ The above isomorphism tells us that there exists a complex line bundle $K \in \mathrm{Pic}(X)$ such that $c_1(K) = a$ and $$K \otimes K \mapsto 2a = c_1(L).$$ Then $$K \otimes K \cong L,$$ so that $K$ is a square root of $L$.

For a proof of the above isomorphism, see for example Proposition 3.10 in Allen Hatcher's unfinished book Vector Bundles and $K$-Theory.

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Do you mean $K\otimes K \mapsto 2a$? –  Michael Albanese Dec 27 '12 at 6:53
    
@MichaelAlbanese Yes, thanks for noticing the typo. –  Henry T. Horton Dec 27 '12 at 16:25

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