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Let $\xi_n$ be an orientable $n$-dimensional vector bundle over a pointed space $B_n$. We can consider the relative Serre Spectral Sequence $$ H_p(B_n; h_q(D(\xi_n|\ast),S(\xi_n|\ast))\Rightarrow h_{p+q}(D(\xi_n),S(\xi_n)) $$ which can be rewritten as $$ H_p(B_n; \tilde{h}_q(S^n))\Rightarrow \tilde{h}_{p+q}(M\xi_n) $$ where $M\xi_n$ is the Thom Space of the vector bundle $\xi_n$. Via the suspension isomorphism we have equivalently $$ H_p(B_n; \tilde{h}_{q-n}(S^0))\Rightarrow \tilde{h}_{p+q}(M\xi_n) $$ using the Thom-Isomorphism $H_p(M\xi_n)\cong H_{p-n}(B_n)$ we get $$ H_{p+n}(M\xi_n; \tilde{h}_{q-n}(S^0))\Rightarrow \tilde{h}_{p+q}(M\xi_n)$$ which seems the Atiyah-Hirzebruch Spectral sequence for $M\xi_n$ in the sense that I'm unable to prove that under these transformations the differentials coincide with the one that comes from the AHSS, in other words that what I got is really the AHSS. How can I show that this is indeed the case?

Since these transformations are not given by a geometrical map, I'm unable to boil everything down to a simply naturality argument.

I'm interested in this fact because I'm reading this paper about the construction of the James Spectral Sequence where this fact is crucial for the identification of the second differential of this spectral sequence (Prop 1) and seems to be trivial.

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NOTE For simplicity of notation I'm going to work only with ordinary cohomology, although it doesn't really matter (any cohomology theory will do).

The best way to see if two spectral sequences are the same is to compare their exact couples. If they originate from the same exact couple, they are the same spectral sequences, differential and all. Fix a CW structure on the base $B$. Then the exact couple originating the Serre spectral sequence is

$$ \require{AMScd} \begin{CD}\cdots @>>> H^*(p^{-1}B^{m})@>>> H^*(p^{-1}B^{m-1})@>>> \cdots\\ {} @VVV {} @VVV {}\\ {} @.H^*(p^{-1}B^{m+1},p^{-1}B^m) @. H^*(p^{-1}B^m,p^{-1}B^{m-1)}) \end{CD}$$ (sorry I can't do diagonal arrows, but they should go one step up and to the left).

To write down the exact couple for the AHSS you need to choose a CW structure on $M\xi$. If we could choose one such that the $m$-skeleton is $M(\xi|_{B^{m-n}})$, then the corresponding exact couple would be

$$ \require{AMScd} \begin{CD}\cdots @>>> \tilde H^*(M(\xi_{B^{m-n}}))@>>> \tilde H^*(M(\xi_{B^{m-n-1}}))@>>> \cdots\\ {} @VVV {} @VVV {}\\ {} @.H^*(M(\xi_{B^{m-n+1}}),M(\xi_{B^{m-n}})) @. H^*(M(\xi_{B^{m-n}}),M(\xi_{B^{m-n-1}})) \end{CD}$$

which by the Thom isomorphism is isomorphic to the previous one, thus proving our thesis.

So the problem boils down to construct a suitable CW structure on $M\xi$, or more precisely to prove that $M\xi_{B^{m+1}}$ can be constructed by $M\xi_{B^m}$ by attaching $(m+n+1)$-cells. This is very easy, since you can work one cell at a time on $B^{m+1}$, and it is left as an exercise for the reader. For more detail on this see also this question.

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  • $\begingroup$ Dear Denis, thanks a lot. I wasn't thinking in terms of exact couples, but everything is much easier if done as you did. Thanks again! $\endgroup$ – Riccardo Sep 12 '16 at 7:47

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