Atiyah duality without reference to an embedding

Question

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?

Answer 1

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).

4. 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.

Answer 2

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).

2. 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.)