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Atiyah duality is the equivalence $M/\partial M \simeq (M^{-T(M)})^\vee$, i.e. the Spanier-Whitehead dual of the space $M/\partial M$ is the Thom complex of the stable normal bundle of $M$. The theorem is proven by taking an appropriate embedding $i$ of $M$ into a sphere $S^d$, identifying the complement with the Thom complex of the normal bundle of the embedding, and constructing a duality map $M_+ \wedge M^{N(i)}\rightarrow S^d$ using the embedding.

In the case $M$ is a compact, framed manifold of dimension $n$, it is possible to describe a duality map $M_+ \wedge M_+ \rightarrow S^n$ without reference to an embedding by appealing to the fact that if we collapse everything outside a small ball around $p \in M$, we can use the framing to continuously identify it with $S^n$. This is the adjoint of a duality map $M_+ \wedge M_+ \rightarrow S^n$.

$\bf Question:$ Is it possible to create such a duality map for a framed manifold with nonempty boundary, namely one that does not invoke an embedding?

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    $\begingroup$ A construction of the pairing using the six functor formalism (and so without references to any embedding) should appear in Marco Volpe's thesis. Unfortunately I don't know what's the time frame for that. $\endgroup$ Jul 23, 2021 at 5:21
  • $\begingroup$ @DenisNardin, I am skeptical: there must be some point where the dualizing module has to be trivialized. The usual way to do to trivialize it is via a stably framed embedding. $\endgroup$
    – John Klein
    Jul 23, 2021 at 12:25
  • $\begingroup$ @JohnKlein I was taking the framing of the tangent bundle as a datum (the question is asking about framed manifolds after all), this gives you a trivialization of the stable normal bundle without any reference to an embedding, am I wrong? $\endgroup$ Jul 23, 2021 at 12:29
  • $\begingroup$ @DenisNardin that is correct. But I am not sure that suffices. At the end of the day, one must somehow obtain a "fundamental class" $\mu: S^n \to M_+$, i.e., a stable map which when followed by the diagonal gives rise to a duality pairing. The only way I know how to do this is to use an embedding. $\endgroup$
    – John Klein
    Jul 23, 2021 at 12:43
  • $\begingroup$ @DenisNardin I retract my skepticism. See my answer below for an explicit construction. $\endgroup$
    – John Klein
    Jul 23, 2021 at 13:03

2 Answers 2

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Here is another short construction which is much simpler and just takes a few lines.

  1. Let $M$ be a closed $n$-manifold. Consider the diagonal $M \to M \times M$. It is an embedding. Take its Pontryagin-Thom construction to get a map $$ M_+ \wedge M_+ \to M^\tau $$ (we have identified a tubular neighborhood of the diagonal with the total space of the tangent bundle).

  2. If $M$ is (stably) framed, then $M^\tau \simeq M_+ \wedge S^n$ (stably). Then we have the map $M_+ \wedge S^n \to S^n$ induced by smashing $M_+ \to \text{pt}_+$ with $S^n$.

  3. The composition $$ M_+ \wedge M_+ \to M^\tau \simeq M_+ \wedge S^n \to S^n $$ is what you want: it's a duality map.

This can be seen on the level of homology (but it is enough to check a map is a duality map on the level of homology).

  1. If $M$ is compact with non-empty boundary $\partial M$, then there is a map of pairs $$ (M\times M, M\times \partial M) \to (M\times M,M\times M - U) $$ where $U$ is a tubular neighborhood of the diagonal. To get this map, one might choose a collar neighborhood $C$ of $\partial M$ and thereafter identify $M$ with $M-C$. Then $M-C $ and $\partial M$ are disjoint.

The map of pairs determines a map of quotients $$ M_+ \wedge M/\partial M \to M^\tau \, , $$ and one may then proceed as in the empty boundary case, assuming that $M$ is framed, to obtain a map $$ M_+ \wedge M/\partial M \to S^n $$ which will be an $S$-duality.

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  1. Assume $M$ is a closed, smooth manifold of dimension $n$. Let $\tau^+$ be the fiberwise one point compactification of its tangent bundle. This is a fiberwise $n$-spherical fibration equipped with a preferred section (at infinity).

There is a based map $$ M_+ \to \Gamma(\tau^+) $$ where $\Gamma$ denotes the space of sections. Roughly, the map sends a point to the Pontryagin-Thom collapse of a small tubular neighborhood of that point (here we are implicitly using the exponential map to identify a tubular neighborhood of a point with its one point compactified tangent space).

  1. The above map induces a stable map, which on zeroth spaces is a scanning map $$ Q(M_+) \to \Gamma^{\text{st}}(\tau^+) $$ where $\Gamma^{\text{st}}$ is the corresponding space of stable sections (i.e., sections of the corresponding stable spherical fibration) and $Q = \Omega^\infty\Sigma^\infty$ is the stable homotopy functor.

Now the key fact is this: the scanning map is always a homotopy equivalence (I do not have a reference; maybe it's due to Graeme Segal). This is a version of "Poincare duality by scanning."

  1. To get your duality map, we stably trivialize $\tau$ (using the fact that that $M$ is stably framed). Then the scanning map is adjoint to stable map $$ M_+ \wedge M_+ \to S^n $$ which will then be a duality map.

  2. Alternatively, take the $n$-fold loops of the scanning map, to obtain a homotopy equivalence $$ \Omega^n Q(M_+) \to \Omega^n \text{maps}(M_+,Q(S^n)) = \text{maps}(M_+,Q(S^0))\, . $$ The right side has a preferred point given by the unit map $M_+ \to S^0$, so the left side gives a preferred stable homotopy class $$ \mu: S^n \to M_+ \, , $$ which is a fundamental class for $M$ in stable homotopy, in the sense that the composition $$ S^n\overset{\mu}\to M_+ \overset{\text{diagonal}}\longrightarrow M_+\wedge M_+ $$ is a duality map.

  3. Remark: Having a duality map is almost the same thing as having a Euclidean "embedding" if by the latter we mean Poincare embedding in some sphere.

Here's why:

(a). A choice of duality map $$ S^d \to M_+ \wedge M^\nu $$ gives us a (stable) map $\mu: S^d \to M^\nu$ (here $M^\nu$ is the Thom space of the stable normal bundle). The map $\mu$ is $S$-dual to the unit map $M_+\to S^0$.

(b). Let $A$ be the fiberwise one point compactification of the stable normal bundle with section $M\to A$. Then $M^\nu = A/M$. Represent $\mu$ as an unstable map $S^d \to M^\nu$ at the expense of stabilizing $\nu$ and $d$.

Then the data give a homotopy pushout diagram $\require{AMScd}$ \begin{CD} A @>>> M^\nu \cup_\mu D^{d+1} \\ @VVV @VVV\\ M @>>> S^{d+1} \end{CD} and the square is a gluing diagram for a Poincare embedding of $M$ in $S^{d+1}$ with complement $M^\nu \cup_\mu D^{d+1}$.

(This trick is implicitly in a paper by Browder from the 1966 Proceedings of the ICM in Moscow.)

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  • $\begingroup$ Thank you for the answer; I was wondering if your map $M_+ \rightarrow \Gamma(\tau^+)$ can be adapted to a map $M/\partial M \rightarrow \Gamma(\tau^+)$ if $M$ has a boundary. I am worried about continuity issues if one just naively extends by sending everything to the point at infinity. $\endgroup$ Jul 23, 2021 at 15:35
  • $\begingroup$ No. I think not--that won't work. I think it's better to think of it as a map $M_+ \to \Gamma_{\text{cs}}(\tau_+)$, where the target is sections with compact support. This works for any finite dimensional manifold without boundary, not necessarily compact. In fact, you need this version of the map to prove the scanning result by induction, since locally any manifold looks like Euclidean space. $\endgroup$
    – John Klein
    Jul 23, 2021 at 15:39
  • $\begingroup$ Well we can always pass to the interior of our manifold with boundary in that case. Would you happen to have a reference that treats the noncompact case? $\endgroup$ Jul 23, 2021 at 15:52
  • $\begingroup$ like I wrote already, I don't have a reference, but I think the scanning result is due to Segal. I have my own approach to these problems which I wrote about 20 years ago. But my approach is different. $\endgroup$
    – John Klein
    Jul 23, 2021 at 16:00
  • $\begingroup$ @ConnorMalin yes, If M is compact, framed and with boundary, then stable sections with compact support coincides with the function space of stable maps $M/\partial M \to S%n$. Taking the adjoint the construction yields a stable duality map $M_+ \wedge M/\partial M \to S^n$. $\endgroup$
    – John Klein
    Jul 25, 2021 at 16:48

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