9
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Andras' answer to this question might give another geometric interpretation. They're talking specifically about the Pontryagin class of the tangent bundle, but I imagine a similar interpretation works for other vector bundles. $\endgroup$ – Kevin Casto Apr 13 '17 at 3:42
  • $\begingroup$ @KevinCasto I was aware of that question when I asked. I was partially wondering whether anyone could confirm that this still held, but also I feel it doesn't say enough. What's the difference, say, between vector bundles where $p_1$ is -1, 0, and 1? Is there something tangible about the geometry of the bundle one can point to that tracks the difference? $\endgroup$ – Brian Klatt Apr 13 '17 at 4:38
  • $\begingroup$ @KevinCasto Also the dimension of the manifold is the same as the rank of the tangent bundle but what if the dimension and rank differ? I'm not very sure of how that would affect the interpretation given in that other answer. $\endgroup$ – Brian Klatt Apr 13 '17 at 14:13
6
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Perhaps it's a matter of taste, but I don't feel particularly enlightened by this though I appreciate this perspective, which is challenging for me as a (budding) Riemannian geometer. (A technical gripe: the bundle I'm most interested in over my 4-manifold has rank 3 and you start your response talking about a rank 4 bundle. Not sure how substantially this modifies the discussion that follows.) $\endgroup$ – Brian Klatt Apr 14 '17 at 0:45
  • $\begingroup$ @Brian: I can see what you mean; I guess how you value things really depends on what part of math you focus on... (I am an algebraic topologist). I think in case you are interested 3-dimensional bundles, the situation is actually getting easier: $BSO(3)$ is a quite tractable space, in particular it is rationally equivalent to $\mathbb HP^{\infty}$. Not sure how helpful this will be to you, though... $\endgroup$ – Jens Reinhold Apr 14 '17 at 1:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.