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Let $M^4$ be an orientable closed 4-manifold and $c_1$ be the first Chern class of a complex line bundle on $M^4$. Let $b$ be the mod 2 reduction of $c_1$, ie $b=c_1$ mod 2.

We have a relation $w_2 b = b^2$, where $w_n$ is the $n^\text{th}$ Stiefel-Whitney class of the tangent bundle of $M^4$. This implies that if $M^4$ is spin, then the Chern number on $M^4$ must be even, ie $\int_{M^4} c_1^2 =0$ mod 2.

My question is that for any $M^4$ that is not spin, can we always find a complex line bundle on $M^4$, such that the Chern number on $M^4$ is odd, ie $\int_{M^4} c_1^2 =1$ mod 2.

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The Enriques algebraic surface has even intersection form (i.e. for any class $\beta \in H^{2}(M,\mathbb{Z})$, $\int_{M^{4}} \beta^2$ is even) but is not spin by Rochlin's theorem since the signature of the intersection form is $8$.

A simply connected $4$-manifold is spin $\iff$ the intersection form is even (which doesn't apply to the Enriques surface which has $\pi_{1} = \mathbb{Z}_{2}$).

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EDIT: This does not work, in general, as explained by Michael Albanese's comment. Thanks!

If $M$ is not spin, then $w_2(M) \neq 0$. But $w_2$ agrees with $v_2$, the second Wu class, which measures whether the intersection form of $M$ is even or odd. Thus, we can find an element $\alpha \in H^2(M;\mathbb Z)$ such that $\alpha^2$ is an odd number times the cohomological fundamental class of $M$. Now represent $\alpha$ by a map $M \to K(\mathbb Z;2) = BU(1)$, i.e., a complex line bundle $E$ on $M$, then $c_1(E) = \alpha$ is as desired.

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    $\begingroup$ If $H_1(M; \mathbb{Z})$ has no $2$-torsion, then $w_2(M) \neq 0$ implies $M$ has odd intersection form. However, as the example in Nick L's answer demonstrates, it is not true in general. The issue is that if $w_2(M) \neq 0$, then by Poincaré duality there is $x \in H^2(M; \mathbb{Z}_2)$ with $w_2(M) \cup x \neq 0$. Now $x$ lifts to an integral cohomology class if and only if $\beta(x) \in H^3(M; \mathbb{Z})$ is zero where $\beta$ denotes the Bockstein. As $\beta(x)$ is $2$-torsion, this is automatic if $H_1(M; \mathbb{Z})$ has no $2$-torsion as $H^3(M; \mathbb{Z}) \cong H_1(M; \mathbb{Z})$. $\endgroup$ – Michael Albanese Jul 2 at 14:47
  • $\begingroup$ Actually, this is the answer I want. It shows that we can find a complex line bundle on $M^4$ such that $\int_{M^4} c_1^2 = $ intersection number. Thanks! $\endgroup$ – Xiao-Gang Wen Jul 2 at 17:36

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