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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 $V \to X$ 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?

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 $E$-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?

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 $E$-Thom isomorphism via the AHSS in a good bit of detail.