This is a slightly revamped version of a question I asked on the stackexchange forum. That question was asking if the Pontryagin-Thom constructon respects the suspension operation, alluding to stable homotopy. I now would like to dig deeper on this allusion.

For a compact simply-connected oriented manifold $M$, view the Pontryagin-Thom construction as the bijective correspondence between $[M,S^r]$ and the set of (appropriate equivalence classes of) framed submanifolds of codimension $r$ in $M$. A quick subquestion is: Is this in some way functorial?
For applying suspension $\Sigma^1$, I would ideally like to keep our general $M$, but for now I would be happy understanding what occurs for $M\approx S^n$. I am actually unsure if this will work for general $M$ because suspension doesn't typically produce a manifold right? -- Maybe we can tweak it (smooth the corners, or homotope it) to produce a manifold.
In perhaps a non-rigorous sense, does $\Sigma^1$ represent a natural transformation for the above construction? In particular, under $\Sigma^1$ we pass from $[M,S^r]$ to $[\Sigma^1M,S^{r+1}]$. So I would expect to get a correspondence $\Sigma^1\lbrace\text{framed }(n-r)\text{-submanifolds in }M\rbrace\simeq\lbrace\text{framed }(n-r)\text{-submanifolds in }\Sigma^1M\rbrace$.

Now if this all works out, what happens on repeated iterations $\Sigma^n$? Does the Pontryagin-Thom construction blossom into anything under the infinite-suspension $\Sigma^\infty$?

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    $\begingroup$ Perhaps this should be tagged at.algebraic topology? $\endgroup$ – Mark Grant Nov 22 '12 at 10:10

Johannes has answered your first two questions, let me have a stab at the final question.

The answer is yes, it blossoms into bordism classes of framed immersions of codimension $r$. That is to say, the pointed stable homotopy group $$\lim_{n\to\infty}[\Sigma^n M_+, S^{n+r}]\cong [M_+,\Omega^\infty \Sigma^\infty S^r] $$ is in one-to-one correspondence with the bordism classes of framed immersions $N\looparrowright M$ of codimension $r$, at least when $r\geq 1$. (Here the $+$ denotes one-point compactification, which means adding a disjoint base-point since $M$ is already compact.) This is sometimes called the Pontrjagin-Thom-Wells theorem, the gereralisation being due to Robert Wells (Cobordism Groups of Immersions, Topology 5 (1966)). The same is also true for connected Thom spaces $T\xi$ other than $S^r$, giving bordism of immersions with other structures on their normal bundles, eg orientations.

The idea of the correspondence is as follows. Given a map $$(M\times \mathbb{R}^n)_ + \simeq \Sigma^n M_{+} \to \Sigma^n T\xi = T(\xi\oplus \varepsilon^n)$$ we can make it transverse to the zero section in $\xi\oplus\varepsilon^n$, and therefore obtain a codimension $r+n$ embedding in $M\times\mathbb{R}^n$ with $n$ independent normal vector fields. By performing an isotopy we can make these normal vector fields point parallel to the $\mathbb{R}^n$ directions, and thus obtain an embedding which projects down to a codimension $r$ immersion in $M$ (this is the content of the Compression Theorem of Rourke and Sanderson). To go the other way, start with an immersion $N\looparrowright M$ whose normal bundle is classified by a map $\nu\to\xi$ and compose with the embedding $M\times\{0\}\hookrightarrow M\times\mathbb{R}^n$. If $n$ is sufficiently large then this will be regularly homotopic to an embedding whose normal bundle is classified by $\nu\oplus\varepsilon^n\to \xi\oplus\varepsilon^n$. Now take the collapse map of this.

  • $\begingroup$ This is great! My knowledge of stable homotopy is empty, but what is going on here that in the limit we end up with immersions instead of embeddings? And I would definitely appreciate details on your sketch of the correspondence -- it seems exciting but I can't piece it all together. $\endgroup$ – Chris Gerig Nov 22 '12 at 5:37
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    $\begingroup$ The main geometric idea is that immersions in $M$ with $\xi$-structure on their normal bundle (which just means there is a bundle map $\nu\to\xi$) can be "decompressed" to embeddings in $M\times\mathbb{R}^n$ with $\xi\oplus\varepsilon^n$ structure on their normal bundle, for $n$ sufficiently large. Then the Compression Theorem allows you to go the other way. A few more details and references can be found in section 2 of arxiv.org/pdf/math/0504152v1.pdf. $\endgroup$ – Mark Grant Nov 22 '12 at 7:26
  • $\begingroup$ And perhaps I should add that every immersion $M\looparrowright N$ is regularly homotopic to an embedding as soon as the dimension of $N$ is more than twice that of $M$. $\endgroup$ – Mark Grant Nov 22 '12 at 12:04

Answer to the quick subquestion: yes, this is functorial, in the following sense. The set of bordism classes of framed submanifolds of $M$ of codimension $n$ is functorial. Let $f:N \to M$ be a map, make it transverse to a given framed submanifold and take the preimage. The Pontrjagin-Thom argument proves that this is well-defined on the level of bordism classes and homotopy classes.

Answer to the second question: it is not clear in which sense you consider the suspension of a manifold as a manifold. However, the set of pointed homotopy classes $[\Sigma M_+; S^{n+1} ]$ is in bijection with the set of bordism classes of all framed compact codimension $n+1$ submanifolds of $\mathbb{R} \times M$. The suspension $[M;S^n] \to [\Sigma M_+; S^{n+1} ]$ takes a submanifold of $M$ and considers it as a submanifold of $\mathbb{R} \times M$, with the product framing and of one codimension higher.

EDIT: in connection with Pontrjagin-Thom constructions, you should always talk about pointed maps. If the basepoint lies in $M$, then $[M;S^n]$ corresponds to submanifolds of $M$ avoiding the basepoint. If you want to talk about all submanifolds or free homotopy classes, you need $[M_+;S^n]$ (extra basepoint added).

  • $\begingroup$ Good answer. I'm fairly sure you need a disjoint base-point somewhere, since $M$ is compact and so the Thom space of $M\times\mathbb{R}$ is $\Sigma M_+$. $\endgroup$ – Mark Grant Nov 21 '12 at 21:58
  • $\begingroup$ I like this concise response! One slight confusion: When $M=S^m$ (i.e. when we definitely have a manifold under suspension) then our submanifold of $\mathbb{R}\times S^m$ is really a submanifold of the defined quotient $S^{m+1}$, so what is the "product" framing here? $\endgroup$ – Chris Gerig Nov 22 '12 at 5:26
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    $\begingroup$ The Pontrjagin-Thom collapse map for a compact submanifold in $M$ gives a map to a Thom space which sends everything "at infinity" to the base-point of that Thom space. Thus it should really be thought of as a map from the one-point compactification of $M$. In particular, the homotopy group $\pi_n(S^r)$ classifies codimension $r$ framed embeddings in $\mathbb{R}^n$, rather than $S^n$. $\endgroup$ – Mark Grant Nov 22 '12 at 10:15

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