4
$\begingroup$

This is the setting we are working in:

$M$ is a closed, smooth $n$-manifold embedded in $\mathbb{R}^{n+k}$ with a chosen embedding $e\colon M^n\to \mathbb{R}^{n+k}$. It is $E$-oriented, for $E$ a connected ring spectrum, i.e. there exists a fundamental class $$z \in E_n(M)$$ which is mapped, for every $m\in M$ to the generator of $E_n(M,M\setminus \{m\})$ via the inclusion $i_m\colon M \to M,M\setminus \{m\}$.

What I want to do: I want to follow the proof of Kochman (sadly the interesting page is not available for preview, it is Prop $4.3.5$ page 136-137 $b)$ implies $a)$.)

Little recap of proof: We will build local Thom classes for the restriction of the normal bundle and then proceed to glue them together using M-V. Let $\{N_i\}_{i=1}^s$ be a family of nbhds diffeo to open disks which over the manifold and let $z_i\in E_n(N_i,N_i\setminus m_i)$ be the image via excision and inclusion of the fundamental class.

Notice that we have non degenerate pairings $$E_n(N_i,N_i\setminus m_i)\otimes E^n(N_i,N_i\setminus m_i)\to \pi_0E$$ given suspension isomorphism and multiplication. These permit us to define the classes $z^i \in E^n(N_i,N_i\setminus m_i)$ as dual of the $z_i$'s. Then identifying $\nu^{-1}(N_i)/(\nu^{-1}(N_i)\setminus N_i)$ with the Thom space of the restriction of the normal bundle we have the following chain of identifications $$\widetilde{E}^k(\text{Th}(\nu_{|N_i}))\cong \widetilde{E}^k(D_n^+\wedge S^k)\cong \widetilde{E}^0(D^+_n)\cong \widetilde{E}^0(S^0)\cong \widetilde{E}^n(S^n)\cong E^n(N_i,N_i\setminus m_i) $$ which permits us to consider the class $t^i \in \widetilde{E}^k(\text{Th}(\nu_{|N_i}))$ define as the image through this chain (backwards) of the $z^i$'s.

Here is the problem: Kochman claims that after a standard application of M-V we can inductively build our global Thom class via a gluing of these $t^i$'s. What I can't prove is how to prove that these classes coincides on the intersection of the $N_i$'s.

My problems are depicted in the following diagram (I'm not claiming it's commutative!): we can assume to have a good cover and let $N_3=N_1 \cap N_2$ to make things a little bit easier. One has to check that the restriction of the two lateral Thom classes induce the same element in the upper central group. What we have is the lower part of the diagram and the fact that the elements in the second row come from a fixed element in the third row.

enter image description here

I tried translating my compatibility condition in some commutativity property of this diagram, but speaking of commutativity here is a little bit awkward. Recall that the vertical maps $\phi$'s are suspension isos AND some chosen diffeomorphisms from the quotients to the spheres AND the duality. Far from canonical I'd say. But since this is what I have I fear that one must conclude something from this.

There is a similar argument on Switzer's book, page $319$ prop $14.18$. But there he uses micro-bundles and orientation as a Thom class of the tangent micro-bundle. I tried adapt the proof to no avail. (And it doesn't seem a trivial application of M-V).

So can someone give me some hints about this?

(I've posted this question on M.Se but receive little attention there. I decided to post it here, hoping to draw attention of people working in this field and maybe used to this kind of reasoning.)

$\endgroup$
  • 1
    $\begingroup$ I'm not sure of how to adapt this proof, but you can recycle Switzer's proof by identifying $E_n(TM,TM\smallsetminus M)$ with $E_n(M\times M,M\times M\smallsetminus M)$ via the Pontryagin-Thom construction. $\endgroup$ – Denis Nardin Jun 5 '16 at 18:01
  • $\begingroup$ @Denis Nardin Mmh not sure what you are suggesting actually. $\endgroup$ – Riccardo Jun 5 '16 at 18:59
  • 1
    $\begingroup$ I'm merely saying that, since the inclusion of a tubular neighborhood of the diagonal $(TM,TM\smallsetminus M)\subseteq (M\times M,M\times M\smallsetminus M)$ induces an equivalence in relative homology then the vector bundle $TM$ is orientable if and only if the tangent microbundle is. There are many ways to prove the result you cited, but I'm not sure of how to fix the one you are trying to understand. $\endgroup$ – Denis Nardin Jun 5 '16 at 19:20
  • $\begingroup$ Yes, I'd be interested in fixing my argument. Thanks for the suggestion though, I'll keep it in mind $\endgroup$ – Riccardo Jun 6 '16 at 13:03
  • 3
    $\begingroup$ Reference to Switzer is wrongheaded. Here is a quote from Stong's Bulletin review of that book: The most crucial mistake is in 14.8 on p. 311, where the author says that $E^0(B) = 0$ if $E$ is a connected spectrum when in fact one needs $\pi_q(E)=0$ for $q>0$. This error is compounded in the following discussions of orientation. In particular, 14.9 is false except for ordinary cohomology and the proof of 14.18 is only valid then. $\endgroup$ – Peter May Jun 6 '16 at 15:11

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.