nLab Pontrjagin-Thom collapse map

Context

Manifolds and cobordisms

manifolds and cobordisms

Contents

Idea

Given an embedding of manifolds $i:X↪Y$, the Thom collapse map is a useful approximation to its would-be left inverse.

Definition

All topological spaces in the following are taken to be compact.

Let $X$ and $Y$ be two manifolds and let

$i:X↪Y$i : X \hookrightarrow Y

be an embedding. Write ${N}_{i}X$ for the normal bundle ${i}^{*}TY/TX$ of the immersion $i$ of $X$ and let $f:{N}_{i}X\to Y$ be any tubular neighbourhood of $i$. Finally write $\mathrm{Th}\left({N}_{i}X\right)$ for the Thom space of the normal bundle.

Definition

The collapse map (or the Pontrjagin-Thom construction) associated to $i$ and the choice of tubular neighbourhood $f$ is

${c}_{i}:Y\to Y/\left(Y-f\left({N}_{i}X\right)\right)\stackrel{\simeq }{\to }\mathrm{Th}\left({N}_{i}X\right)\phantom{\rule{thinmathspace}{0ex}},$c_i : Y \to Y/(Y - f(N_i X)) \stackrel{\simeq}{\to} Th(N_i X) \,,

where the first morphism is the projection onto the quotient topological space and the second is the canonical homeomorphism to the Thom space of the normal bundle.

Since every point of ${N}_{i}X$ is associated to a particular point of $X$, this map can be refined to a map

$Y\to {X}_{+}\wedge \mathrm{Th}\left({N}_{i}X\right)$Y \to X_+ \wedge Th(N_i X)

If $Y={S}^{n}$ for some $n\in ℕ$, then this refined Thom collapse map induces a stable map $S\to {\Sigma }_{+}^{\infty }X\wedge {\Sigma }^{-n}\mathrm{Th}\left({N}_{i}X\right)$, where $S$ denotes the sphere spectrum. This stable map is the unit which exhibits the suspension spectrum ${\Sigma }_{+}^{\infty }X$ as a dualizable object in the stable homotopy category. See n-duality and fixed point index?.

Equivalently, one may proceed as follows. For a framed manifold i.e. a manifold ${M}^{n}$ with a chosen trivialization of the normal bundle ${N}_{i}\left({M}^{n}\right)$ in some ${R}^{n+r}$ one has $T{N}_{i}\left({M}^{n}\right)\cong {\Sigma }^{r}\left({M}_{+}^{n}\right)$ where ${M}_{+}^{n}$ is the union of ${M}^{n}$ with a disjoint base point. Identify a sphere ${S}^{n+r}$ with a one-point compactification ${R}^{n+r}\cup \left\{\infty \right\}$. Then the Pontrjagin-Thom construction is the map ${S}^{n+r}\to \mathrm{Th}\left({N}_{i}X\right)$ obtained by collapsing the complement of the interior of the unit disc bundle $D\left({N}_{i}{M}^{n}\right)$ to the point corresponding to $S\left({N}_{i}{M}^{n}\right)$ and by mapping each point of $D\left({N}_{i}{M}^{n}\right)$ to itself. Thus to a framed manifold ${M}^{n}$ one associates the composition

${S}^{n+r}\to \mathrm{Th}\left({N}_{i}X\right)\cong {\Sigma }^{r}{M}_{+}^{n}\to {S}^{r}$S^{n+r}\to Th(N_i X)\cong \Sigma^r M^n_+\to S^r

and its homotopy class defines an element in ${\pi }_{n+r}\left({S}^{r}\right)$.

Properties

Proposition

For given $i$ all collapse maps for different choices of tubular neighbourhood $f$ are homotopic.

Proof

By the fact that the space of tubular neighbourhoods (see there for details) is contractible.

The following terms all refer to essentially the same concept:

References

An illustration is given on slide 15 of

More details are in

Revised on June 17, 2013 17:43:57 by Urs Schreiber (82.113.99.37)