# What is geometrically the Pontryagin class?

What does the Pontryagin class detects or is an obstruction to? Please avoid any answer using that it's the even Chern class of the complexified bundle or any interpretation that relies on the complexified bundle.

As related question might be the following: when one defines the obstruction classes on a rank $4$ vector bundle (and if the first three obstruction classes do vanish) then the fourth obstruction class can be decomposed as the Euler class and the first Pontryagin class (as $\pi_3(SO_4) \simeq \mathbb{Z} \oplus \mathbb{Z}$). Is there a geometric description of a system of generators in $\pi_3(SO_4)$ which is associated to these classes?

EDIT: deleted "For example, why does the first Pontryagin class distinguishes the (tangent bundles of the) exotic $4$-spheres?" as it is wrong, see Liviu's answer below.

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A related question (which I don't know the answer to): What was Pontryagin's motivation for introducing theses classes? (And if it wasn't him, who did introduce them, why?) –  Charles Rezk Dec 22 '12 at 20:01
$SO(4)$ is double covered by $SU(2)\times SU(2)$ and since $SU(2)=S^3$, $\pi_3(SO(4))=\pi_3(S^3)\times \pi_3S^3=Z\times Z$. –  Paul Dec 22 '12 at 23:19
@Charles: I would also like to hear about this if you ever get the answer. @Paul: thanks, but I was more looking to learn what is the system of generator $\langle \alpha, \beta \rangle$ of $\mathbb{Z} \times \mathbb{Z}$ so that given an element, if one writes it down as $a \alpha + b \beta$ then $a$ would be associated to the Euler class and $b$ to the Pontryagin class –  Antoine Dec 23 '12 at 10:05
Perhaps you are already aware of this, but generally one obtains interesting invariants as polynomials in the pontryagin classes rather than by looking at the pontryagin classes themselves - see the A-hat genus or the hirzebruch L-class, for instance. –  Paul Siegel Dec 23 '12 at 15:30

Pontryagin's original definition for his classes was an obstruction cycle as follows:

On the $n$ dimensional manifold $M$ take $(n-2i) +2$ vector fields in general position, and consider the points $x$ where they span a subspace (in $T_xM$) of dimension less or equal to $n-2i$. The set of such points $x$ form a cycle of codimenion $4i$ in $M.$ The dual cohomology class is $p_i(M).$

This definition might differ from the today accepted definition through Chern classes (as in the book by Milnor-Stasheff) by a second order class.

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Many thanks for your answer. Would you happen to have a precise reference for this? (regardless of the language) –  Antoine Sep 25 '13 at 15:02
I learned this from lecture course of Rokhlin, who was my first teacher in topology. This must be in one of Pontryagin's papers. If you want I can try to find out in which one. Andras –  András Szűcs Sep 25 '13 at 20:20
If you come across it some day... but it's more for bibliographical completeness (as I can't actually read Russian). Köszönom szépen. –  Antoine Sep 26 '13 at 16:35
@Antoine The reference: Понтрягин Л. С., " “Характеристические циклы дифференцируемых многообразий”, Матем. сборник, 1947, т. 21, с. 233-84. –  TT_ Sep 20 '14 at 0:52
And the link to PDF file: mathnet.ru/php/… –  TT_ Sep 20 '14 at 0:52

Some fractional Pontrjagin classes are obstructions to higher analogues of orientations/spin structures.

For example, a spin vector bundle $E \longrightarrow X$ admits a string structure if $\frac{1}{2}p_1(E) = 0$. In other words, a spin structure on $E$ determines a class $\lambda = \frac{1}{2} p_1(E) \in H^4(X; \mathbb{Z})$ such that $2\lambda = p_1(E)$, and this fractional first Pontrjagin class $\lambda$ is the obstruction to the existence of a string structure on $E$.

Similarly, if we go to the next nontrivial step on the Whitehead tower, we can try to define a so-called fivebrane structure on a string vector bundle $E \longrightarrow X$. In this case, the obstruction to the string vector bundle $E \longrightarrow X$ admitting a fivebrane structure is the fractional second Pontrjagin class $\frac{1}{6}p_2(E)$.

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It's maybe worth noting that that pattern doesn't continue step-by-step. Moving from $B\mathrm{Fivebrane}$ to the next stage in the Whitehead tower is not described by Pontryagin classes. This isn't a very exciting thing to point out, though; for almost identical reasons, lifting unoriented real vector bundles to oriented ones and then to spin ones is also not controlled by Pontryagin classes. –  Eric Peterson Dec 22 '12 at 22:06

There is a paper by Paul Bressler:

The ﬁrst Pontryagin class

http://arxiv.org/abs/math/0509563

According to him:

We give a natural obstruction theoretic interpretation to the ﬁrst Pontryagin class in terms of Courant algebroids. ........

Thus, (A,h , i) admits a (globally deﬁned) Courant extension if and only if the the Pontryagin class of (A,h , i) vanishes.

More generally, the ﬁrst Pontryagin class with values as above may be associated to a transitive Lie algebroid (see A.1), say, A, together with an invariant symmetric pairing h , i on the kernel of the anchor map and will be denoted Π(A,h , i). 1 The Pontryagin class of a principal bundle is deﬁned as the Pontryagin class of the Atiyah algebra of the bundle.

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thanks, I'll try to look through this... –  Antoine Dec 23 '12 at 9:50

I don't think that $p_1$ distinguishes the tangent bundles of exotic $4$-spheres (if any). On an oriented smooth $4$-manifold $M$ Hirzebruch signature formula states that

$${\rm sign}(M)=\frac{1}{3}\int_M p_1(TM).$$

The signature of any homology $4$-sphere is zero since there is no homology in the $4$-th dimension.

1. There is one stupid way in which $p_1$ describes an obstruction, because $p_1$ is the $2$-nd Chern class of the complexification, and Chern classes have obstruction-theoretic descriptions.

2. The first Pontryagin class of a $4$-manifold $M$ appears in a nice integral formula of MacPherson and it involves the singularities of generic maps $M\to \mathbb{R}^4$. (I do not remember the reference at this moment.)

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My mistake about the 4-spheres: I thought it was in a paper of Briekorn (which is about 7-spheres and does not mention Pontryagin class so I was way off)... –  Antoine Dec 23 '12 at 9:59

Worth mentioning separately I believe: in "A combinatorial formula for the Pontrjagin classes", Gelfand and Macpherson construct something like analogs of Segre classes for Pontryagin classes: for a triangulation of a manifold $X$ they invoke oriented matroids to produce explicit rational simplicial cycles on its barycentric subdivision which are Poincaré duals of inverses $\bar p_i(X)$ of the Pontryagin classes of $X$.

They also describe (on half a page!) a version of the Chern-Weil theory for Pontryagin classes of a vector bundle $E$ with connection on a manifold $M$ which shows relationship between their approach and the "standard" one. It is so concise and enlightening that I decided just to reproduce it here. They consider the Grassmanian bundle $\pi:\mathscr Y\to M$ of codimension 2 planes in $E$, together with the principal bundle $\rho:\mathscr Z\to\mathscr Y$ corresponding to the tautological quotient 2-plane bundle over $\mathscr Y$. The connection on $E$ gives them a 1-form $\Theta$ on $\mathscr Z$ with coefficients in the orientation sheaf of $\mathscr Z$ and a curvature form $\Omega$ on $\mathscr Y$ determined by $\rho^*\Omega=d\Theta$. Their formula then is$$\bar p_i(E)=(-1)^i\pi_*\Omega^{\dim(E)-2(i-1)}.$$

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I think the positive answer to your question follows from what I wrote above. Here is a hint: Let $\xi \to M^m$ be a bundle, and let $\nu^q$ be a normal bundle of $M$ (in $R^{m+q}$). Take t generic sections of $\xi$ and consider the points where they generate a subspace of rank $t-2$. Apply the original, above mentioned method of computing the Pontryagin class to the manifold N, which is the total space of the bundle $\nu \oplus \xi$. Since the total space $P^{q+m}$ of $\nu$ is parallelizable, it has $q+m$ everywhere independent vector fields. One can take the $t$ sections of $\xi$, extend them to vectorfields on $N$ trivially, and the $q+m$ vectorfields on $P$, extend them also trivially to vector fields on the whole manifold $N$. If we did the "trivial extensions" appropriately, then the claim follows.