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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeApr 24th 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItexstarted [[Thom isomorphism]]

   started Thom isomorphism

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeJun 2nd 2016
   • (edited Jun 2nd 2016)
   • PermaLink
   Author: Urs
   Format: MarkdownItexThe standard derivation of the Thom iso via a "relative" Serre spectral sequence for a "relative fibration" feels a little inefficient. There should be a way to use just a plain Serre spectral sequence for a plain Serre fibration. Let me see: Given a vector bundle $V \to X$ of rank $n$, let me write $$ E \coloneqq D(V)/_{B} S(V) $$ for the _fiberwise_ quotient, an $n$-sphere bundle over $X$ $$ \array{ S^{n-1} &\hookrightarrow& D^n &\longrightarrow& S^n \\ \downarrow && \downarrow && \downarrow \\ S(V) &\hookrightarrow& D(V) &\longrightarrow& E \\ \downarrow && \downarrow && \downarrow \\ B &=& B &=& B } $$ This way the reduced cohomology of the Thom space is the relative cohomology fo $E$ relative $B$: $$ \tilde H^\bullet(Th(V)) \simeq H^\bullet(E,B) \,. $$ Moreover, this $E$ has a section. Therefore combining the exact sequence in unreduced relative cohomology (horizontally) with the Thom-Gysin sequence (vertically) gives $$ \array{ && H^\bullet(B) \\ && \downarrow & \searrow^{\mathrlap{\simeq}} \\ H^\bullet(E,B) &\longrightarrow& H^\bullet(E) &\longrightarrow& H^\bullet(B) \\ && \downarrow \\ && H^{\bullet-n}(B) } \,. $$ By the iso in the top right (from the section of $E$) there is a splitting of the horizontal sequence $$ \begin{aligned} H^\bullet(E) & \simeq H^\bullet(E,B) \oplus H^\bullet(B) \\ & \simeq \tilde H^\bullet(Th(V)) \oplus H^\bullet(B) \end{aligned} $$ but this then implies from exactness of the vertical sequence an iso $$ \tilde H^\bullet(Th(V)) \overset{\simeq}{\longrightarrow} H^{\bullet-n}(B) \,. $$ That should be the Thom isomorphism, no?

   The standard derivation of the Thom iso via a “relative” Serre spectral sequence for a “relative fibration” feels a little inefficient. There should be a way to use just a plain Serre spectral sequence for a plain Serre fibration.

   Let me see:

   Given a vector bundle of rank , let me write

   for the fiberwise quotient, an -sphere bundle over

   This way the reduced cohomology of the Thom space is the relative cohomology fo relative :

   Moreover, this has a section.

   Therefore combining the exact sequence in unreduced relative cohomology (horizontally) with the Thom-Gysin sequence (vertically) gives

   By the iso in the top right (from the section of ) there is a splitting of the horizontal sequence

   but this then implies from exactness of the vertical sequence an iso

   That should be the Thom isomorphism, no?

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeJun 3rd 2016
   • PermaLink
   Author: Urs
   Format: MarkdownItex> That should be the Thom isomorphism, no? Yes, I think one may show this. I have spelled out the proof this way [here](https://ncatlab.org/nlab/show/Thom+isomorphism#ProofOfThomIsomorphismViaFiberwiseThomSpaceFibration).

   That should be the Thom isomorphism, no?

   Yes, I think one may show this. I have spelled out the proof this way here.

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeJun 7th 2016
   • PermaLink
   Author: Urs
   Format: MarkdownItexThe argument in [Kochmann 96, prop. 4.3.6](https://ncatlab.org/nlab/show/Thom%20isomorphism#Kochmann96) for the $E$-Thom isomorphism in generalized $E$-cohomology has an elegant strategy: argue that the operation of pullback followed by cup product with the $E$-Thom class induces on the second page of the respective relative $E$-Atiyah-Hirzebruch spectral sequences the Thom isomorphism with an induced orientation class in ordinary cohomology with coefficients in $\pi_0(E)$. But I feel dubious about Kochmann's argument as to why this holds. Even if his argument holds water, it relies, he says, on the assumption that $\pi_0(E)$ is cyclic, which is an assumption that shouldn't be there. I am thinking an argument should go as follows: since the Thom space of a rank $n$ vector bundles has vanishing ordinary cohomology in degrees $\lt n$, the relative Atiyah-Hirzebruch spectral sequence gives an isomorphism $$ H^n(D(V),S(V); \pi_0(E)) \simeq E^n(D(V), S(V)) $$ and under that iso orientations should go to orientations. Is there any textbook (or similar) that would use this approach to the $E$-Thom isomorphism?

   The argument in Kochmann 96, prop. 4.3.6 for the -Thom isomorphism in generalized -cohomology has an elegant strategy: argue that the operation of pullback followed by cup product with the -Thom class induces on the second page of the respective relative -Atiyah-Hirzebruch spectral sequences the Thom isomorphism with an induced orientation class in ordinary cohomology with coefficients in .

   But I feel dubious about Kochmann’s argument as to why this holds. Even if his argument holds water, it relies, he says, on the assumption that is cyclic, which is an assumption that shouldn’t be there.

   I am thinking an argument should go as follows: since the Thom space of a rank vector bundles has vanishing ordinary cohomology in degrees , the relative Atiyah-Hirzebruch spectral sequence gives an isomorphism

   and under that iso orientations should go to orientations.

   Is there any textbook (or similar) that would use this approach to the -Thom isomorphism?

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeJun 17th 2016
   • PermaLink
   Author: Urs
   Format: MarkdownItexI have added pointer to the note * Riccardo Pedrotti, _Complex oriented cohomology -- Orientation in generalized cohomology_, 2016 ([pdf](https://ncatlab.org/nlab/files/PedrotticECohomology2016.pdf)) which spells out the proof of the $E$-Thom isomorphism via the AHSS in a good bit of detail.

   I have added pointer to the note

   • Riccardo Pedrotti, Complex oriented cohomology – Orientation in generalized cohomology, 2016 (pdf)

   which spells out the proof of the -Thom isomorphism via the AHSS in a good bit of detail.