3

The Leray–Hirsch theorem says that given a fiber bundle $F \to E \to B$ such that $H^*(F)$ is free (as a module over whatever coefficient ring $k$) and, for each $n \geq 0$ there is a set of classes $H^n(E)$ whose restrictions form a basis for the cohomology of each fiber, we have an $H^*(B)$-module isomorphism

$$H^*(E) \cong H^*(B) \otimes H^*(F).$$

It is further the case that if $k$ is a field of characteristic zero (I think any field will do) and $E \to B$ is a principal $G$-bundle for some compact, connected Lie group $G$, then the isomorphism

$$H^*(E) \cong H^*(B) \otimes H^*(G)$$

can in fact be taken to be a ring isomorphism. This can be seen as a Künneth theorem, even though the bundle is not a product.

The proof that I know involves connections and is scattered through a few chapters of lemmas in a fairly lengthy book. I feel that (okay, would like it to be the case that) the action $E \times G \to G$, restricting on each fiber to the multiplication of $G$, and the fact that $H^*(G)$ is naturally a Hopf algebra should together lead to some sort of diagram-theoretic proof of this Künneth theorem.

So, is there some obvious reason for this ring isomorphism?

jdc
  • 4,934
  • I do not think there will be. If the proof is not easy for vector bundles, it would be more difficult for principal bundles in general. – Bombyx mori Jan 09 '15 at 01:56
  • The proof is easy for vector bundles; in that case, the projection $p\colon E \to B$ is a homotopy equivalence. – jdc Jan 09 '15 at 16:32
  • No, I do not think so. The proofs I read are not trivial. You can check Hatcher's book for more detail. – Bombyx mori Jan 09 '15 at 16:33
  • I'm starting to become worried we're conducting two different conversations here. You don't think it's clear that a vector bundle (fiber $\mathbb{R}^n$ contractible) deformation retracts onto its zero section? – jdc Jan 09 '15 at 16:39
  • You need the base space to be paracompact. And even with paracompactness I think the proof is not trivial. To prove it rigorously takes almost half an hour at least. – Bombyx mori Jan 09 '15 at 16:42
  • I am absolutely fine assuming paracompactness, but even without paracompactness, I'm not sure I understand the difficulty, for vector bundles. Without loss of generality, assume the transition functions between trivializations are norm-preserving; we can do that because $GL(n,\mathbb R)$ deformation retracts onto $O(n)$. Then radial deformation retractions $(v,b) \mapsto ((1-t)v,b)\colon \mathbb{R}^n \times U \to \mathbb{R}^n \times U$ on trivializations piece together coherently, no? I thought paracompactness was used for showing bundles over contractible bases were trivial. – jdc Jan 09 '15 at 20:19
  • I think there is the issue of infinitely many (possibly uncountable) different gluing maps over the base space. So if you want $GL(n,\mathbb{R})\rightarrow O(n)$ for each one of them in the same time it would not be entirely trivial, and the proof would take quite a few lines. If you work over a Riemannian manifold the situation is better, but even then you have to construct radial deformations globally. Over a really nasty space where the fibers over different $U_{i}$s intersect, I think the desired global map is not trivial to construct. – Bombyx mori Jan 09 '15 at 23:02
  • I would think the deformation retraction $GL(n,\mathbb{R}) \to O(n)$ exists "on its own" outside the bundle and then applies uniformly. Let's try it. The bundle can be seen as the disjoint $\Big[\coprod_\alpha \big({\alpha} \times U_\alpha \times \mathbb{R}^n)\Big]/\big[(\beta,x,v)\sim(\alpha,x,g_{\alpha\beta\alpha}(x)v)\big]$, where $\alpha$ and $\beta$ range over an index set for your trivializing cover and $g_{\alpha\beta}\colon U_\alpha \cap U_\beta \to GL(n,\mathbb{R})$ are functions. – jdc Jan 11 '15 at 02:16
  • I see what you mean, as the retraction by Gram-Schmidt can be made global. But I still think there is the issue of intersecting fibers over disjoint $U_{i}$s. You could have $E$ and $B$ homotopic, but there is no deformation retraction possible. – Bombyx mori Jan 11 '15 at 02:36
  • I guess "intersecting fibers" may not be an appropriate way to describe it. What I mean is something similar to this construction at here: http://math.stackexchange.com/questions/126342/question-about-an-exercise-in-hatchers-book-algebraic-topology – Bombyx mori Jan 11 '15 at 02:42
  • I wasn't done, but am on a poor connection. The main reason I am so convinced that the difficulties are illusory is this old conversation: http://math.stackexchange.com/questions/710400/what-are-other-examples-of-characteristic-numbers#comment1790305_867747 – jdc Jan 11 '15 at 02:55
  • "We can consider the disjoint union $g = \coprod g_{\alpha\beta}\colon \coprod (U_\alpha \cap U_\beta) \to GL(n,\mathbb R)$ and then follow with the deformation retraction to $O(n)$, no?" My worry about this was only whether the resulting collection was still a cocycle, and could be arranged to remain so throughout the deformation retraction. – jdc Jan 11 '15 at 02:56
  • No, I do not think that (the comments made by Qiaochu) would solve the problem. You may be able to get a $\textit{weak deformation retract}$, but not a deformation retract. – Bombyx mori Jan 11 '15 at 02:57
  • My wonder is whether this is now a Cech cohomology problem, in this reformulation. I think the comments work, because, first, a weak homotopy equivalence should be all that's necessary to ensure equivalence of categories of bundles, and second, because the Milnor versions of $BGL(n,\mathbb{R})$ and $BO(n)$ are indeed genuinely homotopy equivalent. – jdc Jan 11 '15 at 02:59
  • I think cocycle condition is something serious I missed - and it is not clear to me how to generalize this for non-locally finite cases. I do not know much about Cech cohomology, so I cannot comment on it. What I am worried is you can have vector bundles living in some ambient space (for convenience let it be $\mathbb{R}^{n}$) and the topology in the ambient space interfere the deformation argument you needed. Maybe I am kind of stupid, but I think you can get a weak homotopy equivalence via the long exact sequence of homotopy groups right away. – Bombyx mori Jan 11 '15 at 03:07
  • Sorry for the late response on this; for some reason I just saw it. I think the argument in the link works, and it seems to me that the fact it works indicates that there is some way to make my argument above work, but I am not sure immediately how the one translates to the other. I suspect it has to do with the Serre spectral sequence in Cech cohomology of the fibration $GL(n,\mathbb R) \to EGL(n,\mathbb R) \to BGL(n,\mathbb R)$. – jdc Jan 16 '15 at 03:22
  • Sorry I cannot comment on it, as my level is too low (I do not know much about Serre spectral sequence). I think Qiaochu or Ryan Budney might be able to say more on this. Anyways it would be nice if you can show Leray-Hirsh for principal bundles with some simple argument. I think the situation is analogous but more difficult. – Bombyx mori Jan 16 '15 at 03:26
  • It's okay. It's been nice of you to continue the discussion so long! – jdc Jan 16 '15 at 03:33

1 Answers1

1

This is true for the dumbest of possible reasons. If $G$ is a compact, connected Lie group, then after inverting finitely many primes in the coefficient ring, $H^*(G) = \Lambda P$ is an exterior algebra.

Over characteristics other than 2, exterior algebras are free commutative graded algebras (CGAs), so any surjection $H^*(E) \to H^*(G)$ splits. One simply lifts a basis for $P$ up to $H^*(E)$ and notes that these elements span an exterior subalgebra of $H^*(E)$ because it is a CGA. This isomorphic subalgebra then lets one define a CGA map $H^*(B) \otimes H^*(G) \to H^*(E)$ which commutes with the maps from $H^*(B)$ and to/from $H^*(G)$. At least if each $H^j(B)$ is of finite rank, this map is bijective simply by rank considerations.

jdc
  • 4,934