TITLE: Chern number on non-spin manifold
QUESTION [7 upvotes]: 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.

REPLY [11 votes]: 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 Rokhlin'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 $\pi_{1} = \mathbb{Z}_{2}$).

REPLY [2 votes]: 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.