# Chern number on non-spin manifold

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.

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 $$\pi_{1} = \mathbb{Z}_{2}$$).
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.
• 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 $H^3(M; \mathbb{Z}) \cong H_1(M; \mathbb{Z})$. – Michael Albanese Jul 2 at 14:47
• 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! – Xiao-Gang Wen Jul 2 at 17:36