What does positivity of the first Pontryagin number of a vector bundle tell us?

Question

Some context:

In the theory of compact, oriented Riemannian Einstein 4-manifolds, there is a a fundamental topological constraint that is implied by the Einstein equations. To wit, if $\chi$ and $\tau$ denote the Euler characteristic and signature of a fixed 4-manifold $M$, then $$2\chi\pm3\tau=\textstyle\frac{1}{8\pi^2}\displaystyle\int_M\,\big(\displaystyle\frac{R^2}{12}-\mathring{|Rc|}^2+|W^{\pm}|^2 \big)\,dV$$ for $any$ Riemannian metric on $M$. As the traceless Ricci tensor $\mathring{Rc}$ vanishes identically for an Einstein metric, these 2 (well, 1 if $\tau=0$) topological invariants are non-negative if an Einstein metric exists on $M$. This is known as the Hitchin-Thorpe inequality.

One can calculate that in fact $2\chi \pm 3\tau=p_1(\Lambda^2_{\pm})$ by using the connection which is induced on $\Lambda^2_{\pm}$ by the Levi-Civita connection, so the presence of an Einstein metric on $M$ means that the first Pontryagin number of the bundles $\Lambda^2_{\pm}$ is non-negative.

My question is plainly this: Sticking with a 4-manifold $M$, what does it mean in a tangible, geometric sense, to say that a vector bundle over $M$ has non-negative/positive first Pontryagin number? Or perhaps more generally, what does the quantity $p_1$ in fact quantify? Does it say something about generic sections of the bundle (I vaguely recall that the Steifel-Whitney classes quantify something like this), or something else palpable?

I'd be perfectly happy if one were to restrict discussion to the bundles $\Lambda^2_{\pm}$ above or maybe general rank-3 bundles if either simplifies the discussion.

Answer 1

Suppose $M^4$ is a simply-connected 4-manifold, and $V$ is a 4-dimensional real vector bundle over $M$. Then $V$ is classified by a map $M \to BSO(4)$. Rationally, the cohomology of this space has the following form:

$$H^{\ast}(BSO(4);\mathbb Q) = \mathbb Q[p_1,e].$$

Here $p_1$ is the first Pontrygion class, and $e$ is the Euler class (the square of this class is equal to the second Pontryagin class $p_2$). The obstruction for $V$ to admit a non-vanishing section that you mention above is given by $e$, not by $p_1$.

In fact we can change $V$ in such a way that $e$ becomes $0$, but $p_1$ does not change. If we do this, we can split off a line bundle, which corresponds to lifting the classifying map to a map $M \to BSO(3)$. Now there is (since $M$ is simply connected) no obstruction to split off another line bundle, we thus finally get that $V$ is classified by a map $V \to BSO(2) = \mathbb CP^{\infty}$, so in fact $V$ comes from a complex line bundle $\xi$. This complex line bundle has a Chern class $c_1(\xi) \in H^2(M),$ which is the obstruction to trivializing it. By definition, $p_1(V) = -c_1(\xi)^2 \in H^4(M)$. I hope this satisfies your request for understanding $p_1$ geometrically...

In fact, the difference between $e$ and $p_1$ is on e of the things that Milnor exploits in his famous paper "On manifolds homeomorphic to the 7-sphere" (I highly recommend reading this paper!), which was what he got the Fields medal for.