8
$\begingroup$

Let $M$ be an $n$ dimensional smooth manifold and let $j: M \to \mathbb{R}^{m}$ be an embedding. Associated to this embedding we can form the "collapse map" which is a pointed map from a sphere to the Thom space of the normal bundle $S^{m}=(\mathbb{R}^m)^{+} \to Th(N_j)$ (which depends on the choice of tubular neighborhood). The homotopy class of the collpase map is however well defined as a function of the isotopy class of the embedding (notice the homotopy type of the Thom space depends only on the isotopy class of the embedding). The "collapse map" is, in laconic terms, nothing more than a well defined function of sets:

$$\{{ \text{embeddings } M \hookrightarrow \mathbb{R}^m \}/\sim_{\text{isotopy}}} =: E \longrightarrow \coprod_{[j] \in E} [S^m,Th(N_j)]$$

This might be fine for the most part but it seems desirable (even from the POV of applications) to have a version of this construction which works for families.

Question: Is there a "natural" (in the sense that it does what you expect it to do for families of embeddings) map (more accurately a homotopy class of maps) of spaces:

$$Emb(M,\mathbb{R}^m) \to \coprod_{[j] \in \pi_0(Emb(M,\mathbb{R}^m))} Map_*(S^m,Th(N_j))$$

Which on connected components induces the collapse map construction above?

$\endgroup$
  • 1
    $\begingroup$ The boxed equation does not make sense to me: You take the coproduct over equivalence classes, and then take a representative for each one. How do you choose the representative? To me this sound like an evil thing to do: You want to pick a preferred representative out of an equivalence class. $\endgroup$ – Thomas Rot May 22 '18 at 11:53
  • $\begingroup$ @ThomasRot Perhaps I don't understand your complaint but If you declare that space means "homotopy type" then I don't see a problem. As I said the homotopy type of the Thom space only depends on the isotopy class of the embedding. $\endgroup$ – Saal Hardali May 22 '18 at 11:55
  • $\begingroup$ don't take it as a complaint, but as a lack of my understanding. If you declare space to be a homotopy type, a map should be a homotopy class right? Isn't your proposed map then the composition $Emb(M,\mathbb{R}^m)\rightarrow E\rightarrow \coprod_{[j]}[S^m,Th(N_j)]$? $\endgroup$ – Thomas Rot May 22 '18 at 13:24
  • 2
    $\begingroup$ @SaalHardali: There is something strange about this set-up. Often in the P-T construction one obtains a map to the Thom space of the tautological bundle over a Grassmannian, by composing the map you discussed with the classifying map for the normal bundle. It is only after doing this that it really makes sense to compare the P-T maps of two different embeddings: otherwise isotopic embeddings give classes in the homotopy of different spaces $Th(N_j)$ and $Th(N_{j'})$, and while these are homotopy equivalent they are not canonically so. $\endgroup$ – Oscar Randal-Williams May 22 '18 at 14:45
  • 4
    $\begingroup$ @SaalHardali Let me be more direct: I don't believe the function you have in mind is well-defined. Let $j' \in [j]$. It has a collapse map in $\pi_m(Th(N_{j'}))$, and you want to associate to it an element of $\pi_m(Th(N_{j}))$. You want to do so by saying that there exists a homotopy equivalence from $Th(N_{j'})$ to $Th(N_{j})$, which there does: in fact, there exist many; which do you choose? $\endgroup$ – Oscar Randal-Williams May 22 '18 at 16:09
9
$\begingroup$

The right framework of definitions is as follows.

  1. You have a space $E=\text{Emb}(M,\mathbb{R}^n)$ of smooth embeddings, topologised in a way that respects all derivatives. In more detail, we give $C^\infty(M)$ the smallest topology such that the inclusion in $C(M)$ is continuous, as is the map $C^\infty(M)\to C^\infty(M)$ given by any smooth vector field. We then topologise $E$ as a subspace of $C^\infty(M)^n$. One needs to check that for sufficiently small open sets $U\subseteq E$, there is a continuous choice of diffeomorphisms $r_{e,e'}\colon\mathbb{R}^n\to\mathbb{R}^n$ for $(e,e')\in U^2$ such that $r_{e,e'}\circ e'=e$.
  2. Over $E\times M$ there is a vector bundle $\mu$, whose fibre at $(e,m)$ is $\mathbb{R}^n\ominus e_*(T_mM)$. Thus, $\mu|_{\{e\}\times M}$ can be identified with the normal bundle $\nu_e$. One needs to check that for sufficiently small open sets $U\subseteq E$, the restriction $\mu|_{U\times M}$ is isomorphic to a bundle pulled back from $M$, so $\nu_e$ is essentially independent of $e$ for $e\in U$.
  3. There is also a space $F$ of thickened embeddings. A point of $F$ is a pair $(e,f)$ where $e\in E$, and $f$ is an embedding of the total space of $\nu_e$ in $\mathbb{R}^n$, with $f(m,u)\simeq e(m)+u$ to first order in $\|u\|$. There is an evident way to topologise $F$ so that it has the same kind of $E$-local triviality as $\mu$.
  4. One then needs to check that the projection $q\colon F\to E$ is a weak equivalence (or even a homotopy equivalence). This can be done by elaborating the standard proof of uniqueness of tubular neighbourhoods, combined with some paracompactness technology if you want an actual homotopy equivalence.
  5. There is an evident fibrewise Pontrjagin-Thom construction $p\colon F_+\wedge S^n\to (F\times M)^{q^*\mu}$, and a projection $q'\colon (F\times M)^{q^*\mu}\to (E\times M)^\mu$. If we regard the homotopy category as a category of fractions in which weak equivalences are inverted, then we now have a morphism $p'=q'\circ p\circ(q\wedge 1)^{-1}\colon E_+\wedge S^n\to (E\times M)^\mu$. This is the most natural incarnation of the Pontrjagin-Thom construction.
$\endgroup$
  • $\begingroup$ There is a typo near the end: the eventual target should be a Thom space over $E \times M$, not $E$. $\endgroup$ – Oscar Randal-Williams May 22 '18 at 17:37
  • $\begingroup$ @OscarRandal-Williams: you are right, I have corrected it. $\endgroup$ – Neil Strickland May 22 '18 at 17:54
4
$\begingroup$

The situation is somewhat easier to describe if one replaces the embeddings of the closed codimension $(m-n)$manifold $M$ with the embeddings of the total space of a disk bundle of a rank $(m-n)$-vector bundle over $M$.

If $\xi$ is a smooth vector bundle over $M$ of rank $(m-n)$, with disk bundle $D(\xi)$, then there is a restriction map of embedding spaces $$ E(D(\xi),\Bbb R^m) \to E(M,\Bbb R^m)\, , $$ There is an evident Pontryagin-Thom map $$ E(D(\xi),\Bbb R^m) \to \Omega^mM^\xi $$ where the target it the $m$-fold loop space of the Thom space of $\xi$. The map is given by sending an embedding to the map $S^m \to M^\xi$ given by collapsing the complement of the image of the embedding to a point.

The above restriction map sits in a homotopy fiber sequence, $$ E(D(\xi),\Bbb R^m) \to E(M,\Bbb R^m) \to \text{maps}(M,BO_{m-n}) $$ where the base space is the space of maps from $M$ to the Grasmannian of $(m-n)$-planes in $\Bbb R^\infty$. The displayed fiber is the one taken at the basepoint defined by $\xi$.

There is another homotopy fiber sequence $$ \Omega^m M^\xi \to D_m(M) \to \text{maps}(M,BO_{m-n}) $$ where $M^\xi$ is the Thom space of $\xi$ and $D_m(M)$ is the space consisting of pairs $(\xi,g)$ in which $\xi$ is as above and $g: S^m \to M^\xi$ is a based map.

The first fiber sequence maps to the second one, after thickening up $E(M,\Bbb R^m)$ in the way that Neil describes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.