# Pontrjagin number label of bordism class

I am attempting Exercise 12.15 in Dan Freed's notes. It says the following:

suppose $M$ is a closed oriented $4k$-manifold whose rational bordism class is the sum $c_{i_1 \cdots i_r} y^{i_1} \cdots y^{i_r}$ under the isomorphism $\mathbb{Q}[y^1,y^2,y^3,\dots] \stackrel{\cong}{\to} \Omega^{SO} \otimes \mathbb{Q}$, then $c_{i_1 \cdots i_r}$ is the Pontrjagin number $p_{i_1 \cdots i_r}$ of the stable normal bundle over $M$.

I am stuck with the identification of $y^I$ and also his hint to use Pontrjagin-Thom collapse which should bring in the stable normal bundle.

Could somebody help me? If the proof is lengthy then could somebody points me to a reference?

I'm going to outline an argument, but it's sketchy at parts and I leave it to you to fill in the gaps. For convenience, I'm going to abbreviate $x^I := x^{i_1} \cdots x^{i_r}$ and $p_I = p_{i_1} \cdots p_{i_r}$.

Idea 1: If you squint at Theorems 12.4 and 12.9, you should believe that the class $x^I \in \pi_{4k} MSO$ is "dual" to the product of the Pontryagin classes $p_I$.

Idea 2: Recall the Pontryagin-Thom construction. In particular, recall how to go from an element $\pi_{4k} MSO$ to a (cobordism class of) a $4k$-manifold. Roughly an element $x \in \pi_{4k} MSO$ corresponds to a map $S^{4k+l} \to MSO(l)$ for $l \gg 0$. Since $S^{4k+l}$ is compact, this map factors through the Thom space of the tautological oriented $l$-plane bundle $\gamma_l^+$ over the oriented Grassmannian $\operatorname{Gr}^+(m,l)$ for $m \gg 0$. The preimage of the zero section $\operatorname{Gr}^+(m,l) \subset \operatorname{Th}(\gamma_l^+)$ in $S^{4k+l}$ is the wanted $4k$-manifold $M$. In other words, we have \begin{array}{ccccc} M & \rightarrow & S^{4k+l} \\ \downarrow & & \downarrow & \overset{x}{\searrow} \\ \operatorname{Gr}^+(m,l) & \xrightarrow[0]{} & \operatorname{Th}(\gamma_l^+) & \hookrightarrow & MSO(l) \end{array} where the left square is a pullback. By abuse of notation, I'm also going to call all the vertical maps $x$.

We want to compute the Pontryagin number of the stable normal bundle $\nu_M$. Since we've taken $l \gg 0$, I'm going to suppress mentioning "stable".

Idea 3: Identify this normal bundle using naturality. By construction, $\nu_M = (x^I)^* \nu_{\operatorname{Gr}}$ is the pullback of the normal bundle of the Grassmannian $\operatorname{Gr}^+(m,l)$ in $\operatorname{Th}(\gamma_l^+)$. This normal bundle is just $\gamma_l^+$.

The Pontryagin number we want to compute is $$\langle p_I(\nu_M), [M] \rangle = \langle (x^I)^* p_I(\gamma_l^+), [M] \rangle.$$ By definition of the Pontryagin classes using the universal bundle, $p_I(\gamma_l^+)$ is the cohomology class $p_I \in H^* \operatorname{Gr}^+(m,l)$.

But the conclusion of idea 1 is that $x^I$ is "dual" to $p_I$, which more precisely means in cohomology $(x^I)^* p_J$ is $[M]^\vee$ (dual to the fundamental class $[M] \in H_{4k} M$) if $I = J$, and $0$ otherwise.

So we have shown that $$\langle p_I(\nu_M), [M] \rangle = \langle \delta_{I = J} [M]^\vee, [M] \rangle = \delta_{I = J}.$$

The rest of the argument is straightforward.

• I see! Many thanks for the sketch of proof! I think the key fact is that $\text{Gr}^+(m,l)$ as a zero section of $\text{Th}(\gamma_l^+) \to \text{Gr}^+(m,l)$, because from it we can (1) construct the stable normal bundle over $M$ by pullback and (2) introduce the homotopy type (equivalently the bordism class) of $x$ which defines $M$, am I right? (I am trying to summarize your proof into one sentence which I always try to do in learning math : ) – PhysicsMath Aug 7 '16 at 21:36
• One minor question: if you look at 12.4 and 12.9 the dual'' is actually dual up to cup product with Thom space (from the Thom isomorphism). I actually was not very comfortable about what Thom space we use for the cup product and why we can safely ignore that. – PhysicsMath Aug 7 '16 at 21:42
• Also is it a typo that you use $S^{4k}$ instead of $S^{4k+l}$ in the commutative diagram? – PhysicsMath Aug 7 '16 at 21:43
• Thanks, fixed the "$+l$" typo. As for the other comment, the Thom class is essentially uniquely determined, which is why I am sloppy with the Thom isomorphism. I think this is the same "sloppiness" when I abuse notation and call all the vertical maps in the diagram $x$. In any case, it's probably a good exercise for you to verify that the "duality" between $x^I$ and $p_I$ that you get from all the isomorphisms (Thom, etc.) gives me what I use, which is the formula $(x^I)^* p_I = [M]^\vee$ (for suitably interpreted $x^I$). – JHF Aug 7 '16 at 22:07