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

   started Thom space

2. • CommentRowNumber2.
   • CommentAuthorzskoda
   • CommentTimeApr 24th 2010
   • PermaLink
   Author: zskoda
   Format: MarkdownYuli B. Rudyak, _On Thom spectra, orientability, and cobordism_, Springer 1998 [googB](http://books.google.hr/books?isbn=3540620435)

   Yuli B. Rudyak, On Thom spectra, orientability, and cobordism, Springer 1998 googB

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeApr 24th 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItexThanks. Rudyak is also the author of the corresponding [[eom]]-articles...

   Thanks. Rudyak is also the author of the corresponding eom-articles…

4. • CommentRowNumber4.
   • CommentAuthorzskoda
   • CommentTimeApr 24th 2010
   • PermaLink
   Author: zskoda
   Format: MarkdownIn noncommutative case this business of Spanier Whitehead versus Thom and Poincare duality is related to T-duality as studied by Kaminker (very spacial case, unpublished, but the role of Thom is seen better there) and then Makkai, Rosenberg, Szabo.

   In noncommutative case this business of Spanier Whitehead versus Thom and Poincare duality is related to T-duality as studied by Kaminker (very spacial case, unpublished, but the role of Thom is seen better there) and then Makkai, Rosenberg, Szabo.

5. • CommentRowNumber5.
   • CommentAuthorzskoda
   • CommentTimeApr 25th 2010
   • PermaLink
   Author: zskoda
   Format: MarkdownSomething with Whitney embedding theorem in the entry [[Thom space]] is not clear. I mean N vs n, B vs E etc. it seems there are some typoi at least there.

   Something with Whitney embedding theorem in the entry Thom space is not clear. I mean N vs n, B vs E etc. it seems there are some typoi at least there.

6. • CommentRowNumber6.
   • CommentAuthordomenico_fiorenza
   • CommentTimeApr 25th 2010
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexactually I guess the entry Zoran is referring to is [[fiber integration]]. I'm going to attemp a clean up there.

   actually I guess the entry Zoran is referring to is fiber integration. I’m going to attemp a clean up there.

7. • CommentRowNumber7.
   • CommentAuthorzskoda
   • CommentTimeApr 25th 2010
   • PermaLink
   Author: zskoda
   Format: MarkdownRight, thanks.

   Right, thanks.

8. • CommentRowNumber8.
   • CommentAuthordomenico_fiorenza
   • CommentTimeApr 25th 2010
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexcleaned a bit, please have a look. I was not sure about the meaning of $B_+$, so I left it undefined.

   cleaned a bit, please have a look. I was not sure about the meaning of $B_+$, so I left it undefined.

9. • CommentRowNumber9.
   • CommentAuthorUrs
   • CommentTimeApr 25th 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItexThanks for cleaning up my typos. > I was not sure about the meaning of $B_+$ That should be the space $B$ with a basepoint freely adjoined. But I need to better understand that collaps map...

   Thanks for cleaning up my typos.

   I was not sure about the meaning of $B_+$

   That should be the space $B$ with a basepoint freely adjoined.

   But I need to better understand that collaps map…

10. • CommentRowNumber10.
    • CommentAuthordomenico_fiorenza
    • CommentTimeApr 25th 2010
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexthe collaps is easy: you have a tubular neighborhood (which you can see as the open discs bundle in the normal bundle) and you are collapsing to a point all the space $B\times \mathbb{R}^n$ outside the tubular neighborhood. that's the same thing as having just the open disk bundle inside the closed disk bundle and collapsing to a point everythig outside the open disk bundle, i.e., the sphere bundle. this makes the thom space. from the ambient $B\times \mathbb{R}^n$ point of view, all points outside a neighborhood of the compact subspace $E$ are mapped to the point at infinity of the Thom space (the image of the sphere bundle in the quotient of the disk bundle), so the collapse map extends to a continuous map mapping the point at infinity of (the 1-point compactification of) $B\times \mathbb{R}^n$ to the point at infinity of the Thom space.

    the collaps is easy: you have a tubular neighborhood (which you can see as the open discs bundle in the normal bundle) and you are collapsing to a point all the space $B\times \mathbb{R}^n$ outside the tubular neighborhood. that’s the same thing as having just the open disk bundle inside the closed disk bundle and collapsing to a point everythig outside the open disk bundle, i.e., the sphere bundle. this makes the thom space. from the ambient $B\times \mathbb{R}^n$ point of view, all points outside a neighborhood of the compact subspace $E$ are mapped to the point at infinity of the Thom space (the image of the sphere bundle in the quotient of the disk bundle), so the collapse map extends to a continuous map mapping the point at infinity of (the 1-point compactification of) $B\times \mathbb{R}^n$ to the point at infinity of the Thom space.

11. • CommentRowNumber11.
    • CommentAuthorUrs
    • CommentTimeApr 25th 2010
    • PermaLink
    Author: Urs
    Format: MarkdownItex> you have a tubular neighborhood (which you can see as the open discs bundle in the normal bundle) and you are collapsing to a point all the space $B \times \mathbb{R}^n$ outside the tubular neighborhood. that's the same thing as having just the open disk bundle inside the closed disk bundle and collapsing to a point everythig outside the open disk bundle, i.e., the sphere bundle. this makes the thom space. from the ambient $B \times \mathbb{R}^n$ point of view, all points outside a neighborhood of the compact subspace $E$ are mapped to the point at infinity of the Thom space (the image of the sphere bundle in the quotient of the disk bundle), Yes, that I understand. > so the collapse map extends to a continuous map mapping the point at infinity of (the 1-point compactification of) $B \times \mathbb{R}^n$ to the point at infinity of the Thom space. I still feel shaky on this bit, i.e. on how exactly the factorizing map $\tau : \Sigma^n B_+ \to Th(N_{(p,e)}(E))$ is constructed. Could you build $\tau$ for me formally? I suppose it's easy, but I don't see it yet.

    you have a tubular neighborhood (which you can see as the open discs bundle in the normal bundle) and you are collapsing to a point all the space $B \times \mathbb{R}^n$ outside the tubular neighborhood. that’s the same thing as having just the open disk bundle inside the closed disk bundle and collapsing to a point everythig outside the open disk bundle, i.e., the sphere bundle. this makes the thom space. from the ambient $B \times \mathbb{R}^n$ point of view, all points outside a neighborhood of the compact subspace $E$ are mapped to the point at infinity of the Thom space (the image of the sphere bundle in the quotient of the disk bundle),

    Yes, that I understand.

    so the collapse map extends to a continuous map mapping the point at infinity of (the 1-point compactification of) $B \times \mathbb{R}^n$ to the point at infinity of the Thom space.

    I still feel shaky on this bit, i.e. on how exactly the factorizing map $\tau : \Sigma^n B_+ \to Th(N_{(p,e)}(E))$ is constructed.

    Could you build $\tau$ for me formally? I suppose it’s easy, but I don’t see it yet.

12. • CommentRowNumber12.
    • CommentAuthordomenico_fiorenza
    • CommentTimeApr 25th 2010
    • (edited Apr 25th 2010)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexsure. as sets, $\Sigma^n B_+=B\times\mathbb{R}^n\cup \{\infty\}$ and $Th(N_{(p,e)}(E))=N_{(p,e)}\cup\{\infty\}$. let $U$ be a tubular neighborhood of $E$ in $B\times\mathbb{R}^n$ and let $\phi:U\to N_{(p,e)}$ be an isomorphism. then $\tau$ is defined as $\tau(x)=\phi(x)$ if $x\in U$; $\tau(x)=\infty$ otherwise.

    sure. as sets, $\Sigma^n B_+=B\times\mathbb{R}^n\cup \{\infty\}$ and $Th(N_{(p,e)}(E))=N_{(p,e)}\cup\{\infty\}$. let $U$ be a tubular neighborhood of $E$ in $B\times\mathbb{R}^n$ and let $\phi:U\to N_{(p,e)}$ be an isomorphism. then $\tau$ is defined as

    $\tau(x)=\phi(x)$ if $x\in U$;

    $\tau(x)=\infty$ otherwise.

13. • CommentRowNumber13.
    • CommentAuthorUrs
    • CommentTimeApr 25th 2010
    • (edited Apr 25th 2010)
    • PermaLink
    Author: Urs
    Format: MarkdownItexOkay, so we just define it by hand this way? I was hoping there'd be some universal property that gives me the factorization. Okay, then moreover I suppose we have $\Sigma^n B_+ = (B \cup \{*\}) \wedge S^n = B \times S^n/{B \times \{*\}}$ where the point of the sphere is the point at infinity in $B \times \mathbb{R}^n \cup \{\infty\}$. All right, I'll put that into the entry now. One detail that is still missing is a discussion that takes care of where we have $H$-cohomology and where we have [[reduced cohomology]]. That's related to the passage from $B$ to $B_+$, I suppose....

    Okay, so we just define it by hand this way? I was hoping there’d be some universal property that gives me the factorization.

    Okay, then moreover I suppose we have $\Sigma^n B_+ = (B \cup \{*\}) \wedge S^n = B \times S^n/{B \times \{*\}}$ where the point of the sphere is the point at infinity in $B \times \mathbb{R}^n \cup \{\infty\}$.

    All right, I’ll put that into the entry now. One detail that is still missing is a discussion that takes care of where we have $H$-cohomology and where we have reduced cohomology. That’s related to the passage from $B$ to $B_+$, I suppose….

14. • CommentRowNumber14.
    • CommentAuthordomenico_fiorenza
    • CommentTimeApr 26th 2010
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItex<blockquote><p>I was hoping there'd be some universal property that gives me the factorization.</p></blockquote> well, in a sense that's indeed the case. think of $N(E)$ inside $B\times \mathbb{R}^n$. you live on $E$ and are explring your neighborhoods. at a certain stage of your exploration you can see the boundary of $N(E)$ far away as an horizon, but have no idea of what's beyond that, so you draw a map where you have $N(E)$ in full detail and Here be dragons written all around. That's the Thom space: the normal bundle $N(E)$ plus a single <a href="http://en.wikipedia.org/wiki/Here_be_dragons">Here be dragons</a> point beyond the horizon. Then you extend your exploration, and so add to your map a lot of details, and you see that tagging thos as Here be dragons had identified a lot of distinct points. Still you can think of your exploration of not being complete, so you will still have an Here be dragons point. now if you look at the two maps you have, you see that there is a "forget the newly acquired knowledge" morphism from the more ample map to the narrower one, which maps all the points beyond the boundary of $N(E)$ in $B\times \mathbb{R}^n$ (included the Here be dragons point) to the Here be dragons point of the first map. That's Thom-Pontryagin contraction. so, more formally we can think of a topological space $X$ and of a presheaf of topological spaces over $X$ given by $U\mapsto X/(X-U)$. in the Thom-Pontryagin case, $X$ is the 1-point compactification of $B\times \mathbb{R}^n$, and the open subsets are $B\times \mathbb{R}^n$ and $N(E)$.

    I was hoping there’d be some universal property that gives me the factorization.

    well, in a sense that’s indeed the case. think of $N(E)$ inside $B\times \mathbb{R}^n$. you live on $E$ and are explring your neighborhoods. at a certain stage of your exploration you can see the boundary of $N(E)$ far away as an horizon, but have no idea of what’s beyond that, so you draw a map where you have $N(E)$ in full detail and Here be dragons written all around. That’s the Thom space: the normal bundle $N(E)$ plus a single Here be dragons point beyond the horizon. Then you extend your exploration, and so add to your map a lot of details, and you see that tagging thos as Here be dragons had identified a lot of distinct points. Still you can think of your exploration of not being complete, so you will still have an Here be dragons point. now if you look at the two maps you have, you see that there is a “forget the newly acquired knowledge” morphism from the more ample map to the narrower one, which maps all the points beyond the boundary of $N(E)$ in $B\times \mathbb{R}^n$ (included the Here be dragons point) to the Here be dragons point of the first map. That’s Thom-Pontryagin contraction.

    so, more formally we can think of a topological space $X$ and of a presheaf of topological spaces over $X$ given by $U\mapsto X/(X-U)$. in the Thom-Pontryagin case, $X$ is the 1-point compactification of $B\times \mathbb{R}^n$, and the open subsets are $B\times \mathbb{R}^n$ and $N(E)$.

15. • CommentRowNumber15.
    • CommentAuthorUrs
    • CommentTimeApr 26th 2010
    • PermaLink
    Author: Urs
    Format: MarkdownItexOkay, thanks, I see what you are saying. I would still hope that there is a formal way to say it, but for the moment I am happy. The key point here is that the collaps map in a way achieves to turn the n-dimensional fiber of $E$ into an $n$-sphere, i.e. giving us a map from the trivial n-sphere bundle $B \times S^n$ to (the Thom space of) our original bundle $E \to B$. This is the crucial step for the fiber integration, as it allows to reduce it to what we called "integration without integration" over the n-sphere, which is just n-fold looping. The reason why I keep asking for a more abstract way to say this is that we want to make sense of this in contexts like DW-theory, where the bundles over which we want to fiber-integrate are far from being like finite.dimensional manifolds. If we had a more formal way to say it for manifolds, that more formal way might generalize to these more general setups, one would hope.

    Okay, thanks, I see what you are saying. I would still hope that there is a formal way to say it, but for the moment I am happy.

    The key point here is that the collaps map in a way achieves to turn the n-dimensional fiber of $E$ into an $n$-sphere, i.e. giving us a map from the trivial n-sphere bundle $B \times S^n$ to (the Thom space of) our original bundle $E \to B$. This is the crucial step for the fiber integration, as it allows to reduce it to what we called “integration without integration” over the n-sphere, which is just n-fold looping.

    The reason why I keep asking for a more abstract way to say this is that we want to make sense of this in contexts like DW-theory, where the bundles over which we want to fiber-integrate are far from being like finite.dimensional manifolds. If we had a more formal way to say it for manifolds, that more formal way might generalize to these more general setups, one would hope.

16. • CommentRowNumber16.
    • CommentAuthordomenico_fiorenza
    • CommentTimeApr 26th 2010
    • (edited Apr 26th 2010)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexyes, that's exacty how it should work. indeed, a natural morphism $H^n(E,A)\to H^{n-k}(B,A)$ should be read as $\pi_0\mathbf{H}(E,\mathbf{B}^n A)\to \pi_0\mathbf{H}(B,\Omega^k\mathbf{B}^n A)=\pi_0\mathbf{H}(\Sigma^n B,\mathbf{B}^n A)$. so what the Thom-Pontryagin construction seems to be doing is trading an arbitrary bundle with $k$-dimensiona fibre for a trivial sphere bundle. and this is what we need to understand. I've added an example at [[fiber integration]]

    yes, that’s exacty how it should work. indeed, a natural morphism $H^n(E,A)\to H^{n-k}(B,A)$ should be read as $\pi_0\mathbf{H}(E,\mathbf{B}^n A)\to \pi_0\mathbf{H}(B,\Omega^k\mathbf{B}^n A)=\pi_0\mathbf{H}(\Sigma^n B,\mathbf{B}^n A)$. so what the Thom-Pontryagin construction seems to be doing is trading an arbitrary bundle with $k$-dimensiona fibre for a trivial sphere bundle. and this is what we need to understand.

    I’ve added an example at fiber integration

17. • CommentRowNumber17.
    • CommentAuthorUrs
    • CommentTimeApr 26th 2010
    • PermaLink
    Author: Urs
    Format: MarkdownItex> I've added an example at fiber integration Maybe you could say in more detail what you think "integration against the fundamental class" means for a general cohomology theory. It seems to me that by itself this is just a synonym for "fiber integration" for a bundle over the point, so to some extent what you write seems to beg the question. No? Maybe I am mixed up.

    I’ve added an example at fiber integration

    Maybe you could say in more detail what you think “integration against the fundamental class” means for a general cohomology theory. It seems to me that by itself this is just a synonym for “fiber integration” for a bundle over the point, so to some extent what you write seems to beg the question. No? Maybe I am mixed up.

18. • CommentRowNumber18.
    • CommentAuthorUrs
    • CommentTimeApr 26th 2010
    • PermaLink
    Author: Urs
    Format: MarkdownItex> so what the Thom-Pontryagin construction seems to be doing is trading an arbitrary bundle with k-dimensiona fibre with a trivial sphere bundle. and this is what we need to understand. Yes. And I expect we need to think a bit more about geometric versus categorical spheres again, in this context. Generally one would hope to make sense of fiber integration in cohomology in any loally contractible oo-topos: we use the geometric paths and spheres defined intrinsically by the oo-topos in place of $\mathbb{R}$ and $S^1$. Hmm. On the other hand, to get the expected degree shift, we might want to evaluate against categorical spehers after all. Hm. Given a bundle $E \to B$ in an arbitrary (locally contractible) oo-topos, what chances do we have to get canonical morphisms $ B \times LConst S^n \to somewhere$, where "somewhere" is something like the Thom object of $E$?

    so what the Thom-Pontryagin construction seems to be doing is trading an arbitrary bundle with k-dimensiona fibre with a trivial sphere bundle. and this is what we need to understand.

    Yes. And I expect we need to think a bit more about geometric versus categorical spheres again, in this context.

    Generally one would hope to make sense of fiber integration in cohomology in any loally contractible oo-topos: we use the geometric paths and spheres defined intrinsically by the oo-topos in place of $\mathbb{R}$ and $S^1$.

    Hmm. On the other hand, to get the expected degree shift, we might want to evaluate against categorical spehers after all.

    Hm. Given a bundle $E \to B$ in an arbitrary (locally contractible) oo-topos, what chances do we have to get canonical morphisms $B \times LConst S^n \to somewhere$, where “somewhere” is something like the Thom object of $E$?

19. • CommentRowNumber19.
    • CommentAuthordomenico_fiorenza
    • CommentTimeApr 26th 2010
    • (edited Apr 26th 2010)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexUrs, at #17: yes, that is what I intended: once one has a notion of integration over a fiber, one automatically gets a notion of integration against a fundamental class. so the first lines in the example should have served as a definition of $\int_E$. I see it is written in a confusing way :( at #18 yes, and there's also the orientability issue to take into account. on general abstract nonsense grounds, I expect to have an obstruction to our trading business :) (lunch time... I'l be back in a while)

    Urs,

    at #17: yes, that is what I intended: once one has a notion of integration over a fiber, one automatically gets a notion of integration against a fundamental class. so the first lines in the example should have served as a definition of $\int_E$. I see it is written in a confusing way :(

    at #18 yes, and there’s also the orientability issue to take into account. on general abstract nonsense grounds, I expect to have an obstruction to our trading business :)

    (lunch time… I’l be back in a while)

20. • CommentRowNumber20.
    • CommentAuthordomenico_fiorenza
    • CommentTimeApr 26th 2010
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexso.. I was thinking we could stress to the maximum the point of view in Cohen-Klein: the only Thom spaces we are interested in are Thom spaces of normal bundles $N(E)$ for high codimension embeddings. these organize themselves into a spectrum: the Thom spectrum $E^{-T_E}$ of the virtual negative tangent bundle $-T_E$. having at disposal a notion an a construction of a Thom space for an arbitrary vector bundle is useful, but in the end we should consider it only as an auxiliary tool, here. the only object we are really interested is the spectrum $E^{-T_E}$. and this we know from Cohen-Klein is homotopy equivalent to the mapping spectrum $Maps(E,S^0)$. so what we need is to rephrase fiber integration in terms of morphisms of mapping spectra.

    so.. I was thinking we could stress to the maximum the point of view in Cohen-Klein: the only Thom spaces we are interested in are Thom spaces of normal bundles $N(E)$ for high codimension embeddings. these organize themselves into a spectrum: the Thom spectrum $E^{-T_E}$ of the virtual negative tangent bundle $-T_E$. having at disposal a notion an a construction of a Thom space for an arbitrary vector bundle is useful, but in the end we should consider it only as an auxiliary tool, here. the only object we are really interested is the spectrum $E^{-T_E}$. and this we know from Cohen-Klein is homotopy equivalent to the mapping spectrum $Maps(E,S^0)$. so what we need is to rephrase fiber integration in terms of morphisms of mapping spectra.

21. • CommentRowNumber21.
    • CommentAuthorUrs
    • CommentTimeApr 26th 2010
    • PermaLink
    Author: Urs
    Format: MarkdownItexYes, that sounds good. That's why I thought Cohen-Klein made a progress towards a more abstract formulation of fiber integration. But then I still need to understand this better: we now probably want to say that given any bundle $E \to B$ we first form $Maps(B,S^0) \to Maps(E,S^0)$ and then take yet another spectrum $A$ and form $$ \pi_0 Maps(Maps(E,S^0), A) \to \pi_0 Maps(Maps(B,S^0), A) $$ and regard this as fiber integration in ??-A-??-cohomology? No, this can't be right. So what now?

    Yes, that sounds good. That’s why I thought Cohen-Klein made a progress towards a more abstract formulation of fiber integration.

    But then I still need to understand this better: we now probably want to say that given any bundle $E \to B$ we first form $Maps(B,S^0) \to Maps(E,S^0)$ and then take yet another spectrum $A$ and form

    $$\pi_0 Maps(Maps(E,S^0), A) \to \pi_0 Maps(Maps(B,S^0), A)$$

    and regard this as fiber integration in ??-A-??-cohomology? No, this can’t be right. So what now?

22. • CommentRowNumber22.
    • CommentAuthordomenico_fiorenza
    • CommentTimeApr 26th 2010
    • (edited Apr 26th 2010)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexI'll randomly use homology and reduced homology here, since I'm not confident of which duality involves which. some abstract nonsense picture seems to be emerging: mixing page 6 in Cohen-Klein with page 20 in Adam's Infinite loop spaces, the Thom isomorphism with $\mathbb{R}$ coefficients seems to be something like $\tilde{H}_{q+dim M}(M,\mathbb{R})=\lim_{n\to\infty}\pi_{n+q}(M^{-T_M}\wedge h\mathbb{R}_n)=\pi_q^{stab}(M^{-T_M}\wedge h \mathbb{R})$, where, as in Cohen-Klein, $h\mathbb{R}$ is the Eilenberg-MacLane spectrum of $\mathbb{R}$. if so, using the mapping spectrum $maps(M,S^0)$ as a definition of the Thom spectrum $M^{-T_M}$, we could state Thom isomorphism as $$ \tilde{H}_{q+dim M}(M,\mathbb{R})=\pi_q^{stab}(maps(M,S^0)\wedge h\mathbb{R}) $$ but the, Cohen-Klein again tells us that $maps(M,S^0)\wedge h\mathbb{R}\cong maps(M,S^0)$ (this will probably be obvious as soon as I'll manage smash products of spectra..), so we arrive to the following formulation of the Thom isomorphism: $$ \tilde{H}_{\bullet+dim M}(M,\mathbb{R})=\pi_\bullet^{stab}(\mathbf{Sp}(M,h\mathbb{R})) $$ where $\mathbf{Sp}$ is the category of spectra. then, a fibration of compact manifolds $F\to E\to B$ induces by pullback a map $$ \pi_\bullet^{stab}(\mathbf{Sp}(B,h\mathbb{R}))\to\pi_\bullet^{stab}(\mathbf{Sp}(E,h\mathbb{R})) $$ i.e., $$ \tilde{H}_{\bullet+dim M}(B,\mathbb{R})\to \tilde{H}_{\bullet+dim P}(E,\mathbb{R}) $$ or, shifting the degrees by $- dim P$, $$ \tilde{H}_{\bullet-dim F}(B,\mathbb{R})\to \tilde{H}_{\bullet}(E,\mathbb{R}) $$ which dually becomes $$ H^\bullet(E,\mathbb{R})\to H^{\bullet-dim F}(B,\mathbb{R}) $$ in perfect agreement with nPOV: <i>cohomology is the wrong name for cohomotopy</i>. moreover in this form statements and constructions seem to admit a generalization to an arbitrary spectrum, not only $h\mathbb{R}$.

    I’ll randomly use homology and reduced homology here, since I’m not confident of which duality involves which. some abstract nonsense picture seems to be emerging:

    mixing page 6 in Cohen-Klein with page 20 in Adam’s Infinite loop spaces, the Thom isomorphism with $\mathbb{R}$ coefficients seems to be something like $\tilde{H}_{q+dim M}(M,\mathbb{R})=\lim_{n\to\infty}\pi_{n+q}(M^{-T_M}\wedge h\mathbb{R}_n)=\pi_q^{stab}(M^{-T_M}\wedge h \mathbb{R})$, where, as in Cohen-Klein, $h\mathbb{R}$ is the Eilenberg-MacLane spectrum of $\mathbb{R}$. if so, using the mapping spectrum $maps(M,S^0)$ as a definition of the Thom spectrum $M^{-T_M}$, we could state Thom isomorphism as

    $$\tilde{H}_{q+dim M}(M,\mathbb{R})=\pi_q^{stab}(maps(M,S^0)\wedge h\mathbb{R})$$

    but the, Cohen-Klein again tells us that $maps(M,S^0)\wedge h\mathbb{R}\cong maps(M,S^0)$ (this will probably be obvious as soon as I’ll manage smash products of spectra..), so we arrive to the following formulation of the Thom isomorphism:

    $$\tilde{H}_{\bullet+dim M}(M,\mathbb{R})=\pi_\bullet^{stab}(\mathbf{Sp}(M,h\mathbb{R}))$$

    where $\mathbf{Sp}$ is the category of spectra. then, a fibration of compact manifolds $F\to E\to B$ induces by pullback a map

    $$\pi_\bullet^{stab}(\mathbf{Sp}(B,h\mathbb{R}))\to\pi_\bullet^{stab}(\mathbf{Sp}(E,h\mathbb{R}))$$

    i.e.,

    $$\tilde{H}_{\bullet+dim M}(B,\mathbb{R})\to \tilde{H}_{\bullet+dim P}(E,\mathbb{R})$$

    or, shifting the degrees by $- dim P$,

    $$\tilde{H}_{\bullet-dim F}(B,\mathbb{R})\to \tilde{H}_{\bullet}(E,\mathbb{R})$$

    which dually becomes

    $$H^\bullet(E,\mathbb{R})\to H^{\bullet-dim F}(B,\mathbb{R})$$

    in perfect agreement with nPOV: cohomology is the wrong name for cohomotopy. moreover in this form statements and constructions seem to admit a generalization to an arbitrary spectrum, not only $h\mathbb{R}$.

23. • CommentRowNumber23.
    • CommentAuthordomenico_fiorenza
    • CommentTimeApr 26th 2010
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexcleaning up a bit, a Thom isomorphism should be a duality between $H^\bullet(M,A)$ and $\pi_{\bullet-dim M}(M,A)$. but I admit I don't know what <i>duality</i> should mean in this generality.

    cleaning up a bit, a Thom isomorphism should be a duality between $H^\bullet(M,A)$ and $\pi_{\bullet-dim M}(M,A)$. but I admit I don’t know what duality should mean in this generality.

24. • CommentRowNumber24.
    • CommentAuthorUrs
    • CommentTimeApr 26th 2010
    • PermaLink
    Author: Urs
    Format: MarkdownItex> cleaning up a bit, a Thom isomorphism should be a duality between $H^\bullet(M,A)$ and $\pi_{\bullet-dim M}(M,A)$. but I admit I don't know what <i>duality</i> should mean in this generality. If it really reduces to that point, then we are lucky, because this I understand (and we talked about it just recently): we have $$ H^{-n}(M,A) \simeq H^0(M, \Omega^n A) \simeq \pi_0 Maps(M, \Omega^n A) \simeq \pi_n Maps(M , A) $$ by (very) general abstract nonsense. But how is that enough now?? :-)

    cleaning up a bit, a Thom isomorphism should be a duality between $H^\bullet(M,A)$ and $\pi_{\bullet-dim M}(M,A)$. but I admit I don’t know what duality should mean in this generality.

    If it really reduces to that point, then we are lucky, because this I understand (and we talked about it just recently):

    we have

    $$H^{-n}(M,A) \simeq H^0(M, \Omega^n A) \simeq \pi_0 Maps(M, \Omega^n A) \simeq \pi_n Maps(M , A)$$

    by (very) general abstract nonsense. But how is that enough now?? :-)

25. • CommentRowNumber25.
    • CommentAuthorzskoda
    • CommentTimeApr 26th 2010
    • (edited Apr 26th 2010)
    • PermaLink
    Author: zskoda
    Format: Markdown> in perfect agreement with nPOV: cohomology is the wrong name for cohomotopy This is old Eckmann-Hilton (which is not a perfect duality) hence long before nPOV... On the other hand, the Thom isomorphism + Poincare duality = Spanier-Whitehead duality.

    in perfect agreement with nPOV: cohomology is the wrong name for cohomotopy

    This is old Eckmann-Hilton (which is not a perfect duality) hence long before nPOV...

    On the other hand, the Thom isomorphism + Poincare duality = Spanier-Whitehead duality.

26. • CommentRowNumber26.
    • CommentAuthordomenico_fiorenza
    • CommentTimeApr 26th 2010
    • (edited Apr 26th 2010)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexsorry, I forgot a "stab": a Thom isomorphism should be a duality between $H^\bullet(M,A)$ and $\pi_{\bullet-dim M}^{stab}(M,A)$. apart from this detail, which I hope won't destroy the general abstract nonsense above which we very recently talked about, what we miss is the shift in the degrees and the duality: a morphims $E\to B$ would induce by functoriality a morphism $\pi^{stab}_{n-dim E} \mathbf{Sp}(B,A)\to \pi^{stab}_{n-dim E} \mathbf{Sp}(E,A)$; in turn, this should induce a morphism $H^{n}(E,A)\to H^{n-dim E+dim B}(B,A)$. it is this last step the duality I'm not able to see. but since I know that from nPOV cohomolgy is dual to homotopy, I'm confident there's a natural way to state and see this duality.

    sorry, I forgot a “stab”: a Thom isomorphism should be a duality between $H^\bullet(M,A)$ and $\pi_{\bullet-dim M}^{stab}(M,A)$. apart from this detail, which I hope won’t destroy the general abstract nonsense above which we very recently talked about, what we miss is the shift in the degrees and the duality: a morphims $E\to B$ would induce by functoriality a morphism $\pi^{stab}_{n-dim E} \mathbf{Sp}(B,A)\to \pi^{stab}_{n-dim E} \mathbf{Sp}(E,A)$; in turn, this should induce a morphism $H^{n}(E,A)\to H^{n-dim E+dim B}(B,A)$. it is this last step the duality I’m not able to see. but since I know that from nPOV cohomolgy is dual to homotopy, I’m confident there’s a natural way to state and see this duality.

27. • CommentRowNumber27.
    • CommentAuthorzskoda
    • CommentTimeApr 26th 2010
    • PermaLink
    Author: zskoda
    Format: MarkdownHow could nForum ruin 20 minutes of my work...I just pressed wrong button and can not get back to my edit...

    How could nForum ruin 20 minutes of my work...I just pressed wrong button and can not get back to my edit...

28. • CommentRowNumber28.
    • CommentAuthorzskoda
    • CommentTimeApr 26th 2010
    • (edited Apr 26th 2010)
    • PermaLink
    Author: zskoda
    Format: MarkdownItexExcerpt from a lecture of Kaminker which I was typing in 2003 1. Alexander duality. Let $X$ be a finite CW complex and $X \subseteq S^{n+1}$. Then there is a map $$ X \times (S^{n+1}\backslash X) \to S^n,\,\,\,\,\,\, (x,y) \mapsto \frac{x-y}{\|x - y\|}.$$ This element determines an element $$\delta \in H^n( X \times (S^{n+1}\backslash X)).$$ In topology there is a slant product operation, sort of ``dividing'', and in this case one can do slant product with $\delta$. This way one obtains a map $$ \delta_\slash : H_{i}(X)\to H^{n-i}(S^{n+1}\backslash{X}). $$ (\delta_\slash command used, can not see it on my computer. ON the other hand, my previous try == WHY frac{}{} does not work and give slash but gives empty space!!! now \slash gives bad character but maybe is my firefox (btw I instaleld stix fonts!) and finally if I do \delta_{\slash} I get itex error!) This map is an isomorphism and it is called the Alexander-Čech duality. It can be considered for infinite complexes as well, but in that case one has to change the flavour of (co)homology theories involved. $H_i$ is then the **Steenrod-Sitnikov homology** and $H^{n-i}$ has to be cohomology (?) 2. Spanier-Whitehead duality. This is a generalization of the Alexander duality where $S^{n+1}\backslash X$ is replaced by any space $D_n(X)$, together with an element $\delta$ such that the corresponding map $$\delta_\slash : H_i(X) \to H^{n-i}(\mathcal{D}_n X)$$ is an isomorphism. It follows that one may replace $\mathcal{D}_n X$ by its suspension and so on, hence the stable homotopy theory is a natural setup for this duality. 3. Poincare duality. For Poincare duality, the input is an $n$-dimensional *oriented* manifold $M^n$. For our purposes, it is instructive to explain it in terms of Spanier-Whitehead duality. To this aim, embed $M$ in a high dimensional $S^p$, consider its normal bundle $\nu M$ in $S^p$, which is also naturally oriented, and its one-point compactification, namely the Thom space $(\nu M)^+$. Take $\mathcal{D}_p X := (\nu M)^+$. Unlike the Spanier-Whitehead duality, the Thom isomorphism $H^i(M) \to H^{p-n+i}((\nu M)^+)$ utilizes the orientation on both $M$ and $(\nu M)^+$. In this setup, the Poincare duality is simply the composition of the Thom isomorphism and the Spanier-Whitehead duality, i.e. the diagram -- I will display this later, my diagram is xypic -- commutes. 3' The same picture holds for **K-theory** instead of (co)homology, but as an appropriate replacement of the orientation, we should assume a $Spin^{\mathbb{C}}$-structure on $M^n$.

    Excerpt from a lecture of Kaminker which I was typing in 2003

    1. Alexander duality. Let $X$ be a finite CW complex and $X \subseteq S^{n+1}$. Then there is a map

    $$X \times (S^{n+1}\backslash X) \to S^n,\,\,\,\,\,\, (x,y) \mapsto \frac{x-y}{\|x - y\|}.$$

    This element determines an element

    $$\delta \in H^n( X \times (S^{n+1}\backslash X)).$$

    In topology there is a slant product operation, sort of “dividing”, and in this case one can do slant product with $\delta$. This way one obtains a map

    $$\delta_\slash : H_{i}(X)\to H^{n-i}(S^{n+1}\backslash{X}).$$

    (\delta_\slash command used, can not see it on my computer. ON the other hand, my previous try == WHY frac{}{} does not work and give slash but gives empty space!!! now \slash gives bad character but maybe is my firefox (btw I instaleld stix fonts!) and finally if I do \delta_{\slash} I get itex error!)

    This map is an isomorphism and it is called the Alexander-Čech duality. It can be considered for infinite complexes as well, but in that case one has to change the flavour of (co)homology theories involved. $H_i$ is then the Steenrod-Sitnikov homology and $H^{n-i}$ has to be cohomology (?)

    1. Spanier-Whitehead duality. This is a generalization of the Alexander duality where $S^{n+1}\backslash X$ is replaced by any space $D_n(X)$, together with an element $\delta$ such that the corresponding map

    $$\delta_\slash : H_i(X) \to H^{n-i}(\mathcal{D}_n X)$$

    is an isomorphism. It follows that one may replace $\mathcal{D}_n X$ by its suspension and so on, hence the stable homotopy theory is a natural setup for this duality.

    1. Poincare duality. For Poincare duality, the input is an $n$-dimensional oriented manifold $M^n$. For our purposes, it is instructive to explain it in terms of Spanier-Whitehead duality. To this aim, embed $M$ in a high dimensional $S^p$, consider its normal bundle $\nu M$ in $S^p$, which is also naturally oriented, and its one-point compactification, namely the Thom space $(\nu M)^+$. Take $\mathcal{D}_p X := (\nu M)^+$. Unlike the Spanier-Whitehead duality, the Thom isomorphism $H^i(M) \to H^{p-n+i}((\nu M)^+)$ utilizes the orientation on both $M$ and $(\nu M)^+$. In this setup, the Poincare duality is simply the composition of the Thom isomorphism and the Spanier-Whitehead duality, i.e. the diagram

    – I will display this later, my diagram is xypic –

    commutes.

    3’ The same picture holds for K-theory instead of (co)homology, but as an appropriate replacement of the orientation, we should assume a $Spin^{\mathbb{C}}$-structure on $M^n$.

29. • CommentRowNumber29.
    • CommentAuthorzskoda
    • CommentTimeApr 26th 2010
    • (edited Apr 26th 2010)
    • PermaLink
    Author: zskoda
    Format: MarkdownItexAgain the nForum replaces my numbering 1 2 3 by his own 1 1 3...How to surpress numbering ? I hate MS Word-defaults entering the arena of LaTeX control... The loss of data can not happen in nlab when using firefox. If made a mistake just take the back in the firefox. Back in nForum is creating a refresh and erases my old data, while back button in nlab is back button and even if I made some changes between a save option and getting back the signal from the server, after I press back button the changes in between the loads are still there in my edit window, what I like a lot, when being on slow connection. I just created [[Jerome Kaminker]] page.

    Again the nForum replaces my numbering 1 2 3 by his own 1 1 3…How to surpress numbering ? I hate MS Word-defaults entering the arena of LaTeX control…

    The loss of data can not happen in nlab when using firefox. If made a mistake just take the back in the firefox. Back in nForum is creating a refresh and erases my old data, while back button in nlab is back button and even if I made some changes between a save option and getting back the signal from the server, after I press back button the changes in between the loads are still there in my edit window, what I like a lot, when being on slow connection.

    I just created Jerome Kaminker page.

30. • CommentRowNumber30.
    • CommentAuthordomenico_fiorenza
    • CommentTimeApr 26th 2010
    • (edited Apr 26th 2010)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexmmm... so, for (nice) topological spaces, the [[slant product]] is a map $$ H_q(X,A)\otimes H^n(X\times Y,A')\to H^{n-q}(Y,A\otimes A') $$ so, what we need is a pairing between homology and cohomology, and this is what I'm missing in the most abstract formulation. edit: created a stub for [[slant product]].

    mmm… so, for (nice) topological spaces, the slant product is a map

    $$H_q(X,A)\otimes H^n(X\times Y,A')\to H^{n-q}(Y,A\otimes A')$$

    so, what we need is a pairing between homology and cohomology, and this is what I’m missing in the most abstract formulation.

    edit: created a stub for slant product.

31. • CommentRowNumber31.
    • CommentAuthorUrs
    • CommentTimeMay 31st 2011
    • PermaLink
    Author: Urs
    Format: MarkdownItexadded to [[Thom space]] the crucial suspension property $$ Th(\mathbb{R}^n \oplus V) \simeq S^n \wedge Th(V) $$ that gives [[Thom spectra]]

    added to Thom space the crucial suspension property

    $$Th(\mathbb{R}^n \oplus V) \simeq S^n \wedge Th(V)$$

    that gives Thom spectra

32. • CommentRowNumber32.
    • CommentAuthorUrs
    • CommentTimeMay 18th 2016
    • (edited May 18th 2016)
    • PermaLink
    Author: Urs
    Format: MarkdownItexI gave _[[Thom space]]_ a somewhat pedantic definition section ([here](https://ncatlab.org/nlab/show/Thom+space#Definition)) and then added to the properties section [here](https://ncatlab.org/nlab/show/Thom+space#Properties) somewhat pedantic details on the statement that the Thom space of the Whitney sum with a trivial bundle is reduced suspension of the original Thom space.

    I gave Thom space a somewhat pedantic definition section (here) and then added to the properties section here somewhat pedantic details on the statement that the Thom space of the Whitney sum with a trivial bundle is reduced suspension of the original Thom space.

33. • CommentRowNumber33.
    • CommentAuthorUrs
    • CommentTimeJun 2nd 2016
    • PermaLink
    Author: Urs
    Format: MarkdownItexI have added ([here](https://ncatlab.org/nlab/show/Thom+space#HomotopyTheoreticNature)) a brief discussion of the homotopy-theoretic nature of the Thom-space construction, and its equivalence, when over a CW-complex, with, in particular, the mapping cone of the inclusion of the complement of the 0-section.

    I have added (here) a brief discussion of the homotopy-theoretic nature of the Thom-space construction, and its equivalence, when over a CW-complex, with, in particular, the mapping cone of the inclusion of the complement of the 0-section.

34. • CommentRowNumber34.
    • CommentAuthorUrs
    • CommentTimeJun 7th 2016
    • PermaLink
    Author: Urs
    Format: MarkdownItexadded the statement that the reduced cohomology of the Thom space of a vector bundle of rank $n$ vanishes in degrees $\lt n$. ([here](https://ncatlab.org/nlab/show/Thom+space#OrdinaryCohomologyOfThomSpaceInLowDegree))

    added the statement that the reduced cohomology of the Thom space of a vector bundle of rank $n$ vanishes in degrees $\lt n$. (here)

35. • CommentRowNumber35.
    • CommentAuthornLab edit announcer
    • CommentTimeDec 7th 2022
    • PermaLink
    Author: nLab edit announcer
    Format: MarkdownItextypo fix: morhism -> morphism Arun Debray <a href="https://ncatlab.org/nlab/revision/diff/Thom+space/36">diff</a>, <a href="https://ncatlab.org/nlab/revision/Thom+space/36">v36</a>, <a href="https://ncatlab.org/nlab/show/Thom+space">current</a>

    typo fix: morhism -> morphism

    Arun Debray

    diff, v36, current