# Lawson and Michelsohn's proof of the splitting principle for oriented vector bundles

I'm trying to understand the proof of the following result in Lawson & Michelsohn's Spin Geometry:

Proposition 11.2. Let $$E$$ be an oriented real vector bundle of dimension $$2n$$ over a manifold $$X$$. Then there is a smooth proper fibration $$\pi\colon \mathscr S_E\to X$$ such that $$\pi^*\colon H^*(X)\to H^*(\mathscr S_E)$$ is injective and the bundle $$\pi^*(E\otimes \mathbb C)$$ splits into complex line bundles: $$\pi^*(E\otimes \mathbb C) \cong \ell_1\oplus \overline {\ell_1} \oplus \dots \oplus \ell_n \oplus \overline {\ell_n}$$ where $$\overline{\ell_j}$$ denotes the "inverse" or "complex conjugate" bundle to $$\ell_j$$. In fact, there is a splitting $$\pi^*(E) \cong E_1\oplus \dots \oplus E_n$$ into oriented real $$2$$-plane bundles such that $$E_k\otimes \mathbb C = \ell_k \oplus \overline{\ell_k}$$ for each $$k$$.

The proof is based on constructing the bundle $$p:G(E)\to X$$ whose fiber at a point $$x\in X$$ is the set of all oriented $$2$$-dimensional subspaces of $$E_x$$, so the pullback bundle $$p^*E$$ splits into the tautological oriented $$2$$-plane bundle $$E_1\to G(E)$$ plus its orthogonal complement, and then proceeding inductively. This part I understand.

What is perplexing me is their argument that $$p$$ induces an injection on cohomology. It goes like this: "The argument given for Theorem C.14 adapts immediately to prove that the homomorphism $$p^*: H^*(X;\mathbb Z) \to H^*(G(E);\mathbb Z)$$ is injective." The theorem referred to is this:

Theorem C.14. Suppose $$E\to X$$ is a complex vector bundle of rank $$k$$. Then $$H^*(\mathbb P(E);\mathbb Z)$$ is a free $$H^*(X;\mathbb Z)$$-module with basis $$1,u,u^2,\dots,u^{k-1}$$ [where $$\mathbb P(E)$$ is the projectivization of $$E$$, $$\ell$$ is the tautological complex line bundle over $$\mathbb P(E)$$, and $$u = c_1(\ell)$$].

The proof of Theorem C.14 is a fairly standard application of the Leray-Hirsch theorem, using the fact that the restriction of $$u$$ to each fiber generates the integral cohomology ring of the fiber, which is a copy of $$\mathbb C\mathbb P^{k-1}$$.

But in the case at hand, the fiber of $$G(E)\to X$$ is a copy of the oriented Grassmannian $$\widetilde G_2(\mathbb R^{2n})$$, and I'm not aware of any argument that there exist global cohomology classes on $$G(E)$$ whose restrictions to each fiber freely generate the integral cohomology groups of the fiber (as an abelian group), which is what's needed to apply Leray-Hirsch. What am I missing?

• Example 5 of Peter May's note on the splitting principle may be helpful. Commented Oct 13, 2021 at 0:23
• @MichaelAlbanese: Thanks. I've seen that argument. It can be used to prove Lawson & Michelsohn's Proposition 11.2; but it doesn't seem to justify the claim that $H^*(X;\mathbb Z) \to H^*(G(E);\mathbb Z)$ is injective, which I'd still like to understand. Commented Oct 13, 2021 at 0:29
• @JackLee: I am not even beginner on VB but one can find similar theorem by a glimpse into Hatcher VBK with this simple argument: the fact that $1$ is among the basis elements implies that $p^*$ is injective. Commented Oct 13, 2021 at 11:58
• The notes linked by @MichaelAlbanese show that the map is injective for coefficients in a Ring where 2 is invertible. I don't know how one deals with the two torsion. I believe the most import example to consider whould be the universal bundle over $X=BSO(2n)$, I believe then one has $G(E) = B(SO(2)\times SO(2n-2))$ and expect (didn't think too much) that $G(E) \to X$ is then a part of the fibration sequence $SO(2) \times SO(2n-2) \to SO(2n) \to \tilde{G}_2(\mathbb{R}^{2n}) \to G(E) \to BSO(2n)$. But of course this doesn't fix the issues of the proof in the general case... Commented Oct 13, 2021 at 13:15
• @C.F.G: Thanks for pointing out the typo. Fixed now. As for Hatcher's argument, the statement you quoted is predicated on the facct that there are cohomology classes on $\mathbb P(E)$ that restrict to a basis for the cohomology of each fiber. (In this case he's talking about K-theory classes, but the argument is the same.) That's what I don't see how to prove for $G(E)$. Commented Oct 13, 2021 at 17:47

For odd vector bundles with non-trivial euler class $$p^*$$ cannot be injective, and it seems, one can transport this problem to the even case. Please let know if I made any mistakes...

Consider the long exact Bockstein sequence $$\ldots \to H^*(-;\mathbb{Z}) \overset{2\cdot}{\to} H^*(-;\mathbb{Z}) \overset{\rho_2}{\to} H^*(-;\mathbb{Z}/2) \overset{\beta_2}{\to} H^{*+1}(-;\mathbb{Z}) \to \ldots$$

Let $$V \to X$$ be an oriented vector bundle of rank $$2n-1=3$$ with

1. $$w_{3}(V) \neq 0 \in H^{3}(X;\mathbb{Z}/2)$$
2. $$\beta_2 w_2(V) = e(V) \in H^{3}(X;\mathbb{Z})$$

(I believe 2. is a general fact, but I couldn't find a reliable source..)

Add a trivial line bundle $$E = V \oplus 1$$. We have $$w_{3}(E) = w_{3}(V) \neq 0$$.

Now For $$p: G(E) \to X$$ as described above, we have $$p^*(E) \overset{\sim}{=} P_1 \oplus P_2$$ where $$P_1$$ and $$P_2$$ are oriented 2-plane bundles.

It is a general fact that the $$\rho_2$$ maps the euler class of an oriented vector bundle to its top Stiefel Whitney class, which means that the top SW class is in the kernel of $$\beta_2$$, for example $$\beta_2 w_2(P_i) = 0$$.

Since $$w_1(P_i) = 0$$, we have, that $$p^*w_{2}(E) = w_{2}(P_1) + w_2(P_2) \in \ker\beta_2.$$

Using that the Bockstein is natural we have $$p^*e(V) = p^*\beta_2w_2(V) = \beta_2p^*w_2(E) = 0.$$ Since $$e(V) \neq 0$$ that means $$p^*$$ is not injective.

In this answer Qiaochu Yuan constructed such a vector bundle $$V \to \mathbb{RP}^2\times \mathbb{RP}^2$$ with the $$w_3(V) \neq 0$$ property. Using $$Sq^1w_2(V) = \rho_2 \beta_2(w_2(V)) = w_3(V) = \rho_2(e(V))$$ one gets $$\beta_2 w_2(V) = e(V)$$, since $$H^3(\mathbb{RP}^2\times \mathbb{RP}^2;\mathbb{Z})$$ only consists of $$2$$- torsion.