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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?

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  • $\begingroup$ Example 5 of Peter May's note on the splitting principle may be helpful. $\endgroup$ Oct 13, 2021 at 0:23
  • $\begingroup$ @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. $\endgroup$
    – Jack Lee
    Oct 13, 2021 at 0:29
  • $\begingroup$ @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. $\endgroup$
    – C.F.G
    Oct 13, 2021 at 11:58
  • $\begingroup$ 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... $\endgroup$
    – Jonas
    Oct 13, 2021 at 13:15
  • $\begingroup$ @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 $G(E)$. $\endgroup$
    – Jack Lee
    Oct 13, 2021 at 17:47

1 Answer 1

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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.

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