Pontryagin class $p_1$ on (non-)spin manifolds

Question

Pontryagin classes are cohomology groups with degree a multiple of four, which are defined for real vector bundles.

Let us consider $p_1$ of $SO(N)$ bundle, the 4th Pontryagin class of $SO(N)$, for $N=3$.

1. How can we prove that $p_1$ can be only an even integer on the spin manifold?
2. How can we prove that $p_1$ can be an integer on the non-spin manifold?

(What are explicit examples for $p_1 \in \mathbb{Z}$, or $p_1=1$ on which non-spin manifold? e.g. $\mathbb{CP}^2$ or $\mathbb{RP}^4$ and what elses?)

Two References I find may be helpful to answer this (somewhat technical) are here:

1. Chern-Simons invariants, SO(3) instantons, and Z/2-homology cobordism, by Matthew Hedden, Paul Kirk

2. On the Pontryagin Classes of Certain SO(n)-Bundles Over Manifolds, by Michel A. Kervaire

---

Note added:

Given any $SO(3)$ connection $A$ on a bundle $E$ over a 4-manifold $M$, let $F(A)$ denote its curvature 2-form. Define the Pontryagin charge of $A$ to be the real number $$ p_1(A)=-\frac{1}{8\pi^2}\int_{M} \operatorname{Tr}(F(A)\wedge F(A)) , $$ provided this integral converges. When $M$ is closed, $p_1(A)=\langle p_1(E),[M]\rangle\in \mathbb{Z}$ at the least, but it could also be $2\mathbb{Z}$ or $4\mathbb{Z}$. It looks that we need to use some properties of instanton solution whose curvature form satisfies the equation $F(A)=-\star F(A)$ to prove the above two statements I made. However, it is not obvious to me that how the spin or non-spin manifold enter to affect the instanton solution. I suppose we need to use $w_1(M)$ and $w_2(M)$ as zeros or non-zeros neatly in order to show the (1) and (2).

Answer 1

Restrict to $SO(3)$-bundles (you can take the direct sum with a trivial rank $n-3$ bundle to get an $SO(n)$-bundle with the same characteristic classes).

There is a relation on characteristic classes: $w_2(E)^2 = p_1(E) \pmod 4$, where we use the Pontryagin product $H^2(X;\Bbb Z/2) \otimes H^2(X;\Bbb Z/2) \to H^4(X;\Bbb Z/4)$. $SO(3)$-bundles over a 4-complex are completely classified by Dold-Whitney; there is a bundle with characteristic classes $(w_2, p_1) \in H^2(X;\Bbb Z/2) \times H^4(X;\Bbb Z)$ iff $w_2^2 = p_1 \pmod 4$.

So the point is to determine the values $w_2^2$ can take. The statement that $X$ has even intersection form literally means that the map $H^2(X;\Bbb Z/2) \to H^4(X;\Bbb Z/2)$, $x \mapsto x^2$, is zero. So $p_1$ must be even, as it reduces to $0$ mod $2$. This is the only relation you have: by the classification of even symmetric bilinear forms you can always find a vector that squares to $2$ (in $E8$ pick anything on the diagonal, in $H$ pick $x+y$). So you can realize every even integer as $p_1(E)$ for some bundle $E$.

Every spin manifold has even intersection form. A simply connected non-spin manifold does not have even intersection form, but this doesn't mean you can choose $p_1$ arbitrarily: for $SO(3)$-bundles over $\Bbb{CP}^2$, $p_1(E)$ must be 0 or 1 mod 4.

---

If $E = TM$ (an $SO(4)$-bundle), then the simplest form of the Hirzebruch signature theorem says that $\sigma(M) = p_1(TM)/3$, where $\sigma$ is the signature. The signature of a smooth spin 4-manifold is divisible by 16 by Rokhlin's theorem, so $p_1(TM)$ is divisible by $48$.