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

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:

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).

• Too many questions in one post, (too broad); too little work from you, far too little forms of other context. (off topic, missing context). – Namaste Jul 23 '18 at 20:11
• wonderich I'm merely a user committed to keeping this site as healthy as it can be. Do not assume I voted to close; you cannot know now whether I did, or no. I merely brought to your attention issues that could, foreseeably be issues that yield close votes. Primarily, I merely want users with over 1K in rep, who should know better than to write PSQ's that are too long, showing no work at all, to set better examples that don't defy the guidelines for writing a good question, like you have here. On behalf of all new users. – Namaste Jul 23 '18 at 20:20
• I would add some of what you just told me, into your post: Tell us that your know of the following two references [references], say a little something about the instanton number (defined in the paper), explain that you were not able to prove/or answer your questions, and, ideally, add a little work you've done, even if you get stuck, or it's wrong. – Namaste Jul 23 '18 at 20:29
• Very nice post! Thanks! – Namaste Jul 23 '18 at 20:46
• Your notation is very confusing. If one writes $p_k(M)$ where $M$ is an oriented manifold, this means "the $k$th Pontryagin class of $TM$", which is an element of $H^{4k}(M;\Bbb Z)$. When $M$ is a Lie group its tangent bundle is trivial and these all vanish. There are other ways I might try to interpret your first question but none really make sense. Further, I suspect you mean $p_1$, not $p_4$. The question in your note makes sense (where $E$ is an $SO(3)$-bundle over a 4-manifold). But what this invariant is has nothing to do with whether or not $M$ is spin, because $E$ is arbitrary. – user98602 Jul 24 '18 at 7:43

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 $48$.