---
tag: 5152FTFE13
kind: definition
name: Thom space
msc: [55R25, 57R20]
defines: 5MNYT3AFPH
uses: [3VQ2V5HH3R, 0BR3FY5Q8M, 0NB0AWGJBW, 0GWN3CMRK8]
sources:
  - url: https://en.wikipedia.org/wiki/Thom_space
---

## Definition

Let $\pi \colon E \to B$ be a real vector bundle of rank $n$ with a continuous [[inner product|0NB0AWGJBW]] on its fibers.
Let $D(E) = \set{v \in E : \abs{v} \le 1}$ be the disc bundle and $S(E) = \set{v \in E : \abs{v} = 1}$ the unit [[sphere bundle|3VQ2V5HH3R]].
The *Thom space* of $E$ is the [[quotient space|0BR3FY5Q8M]]
$$\operatorname{Th}(E) = D(E) / S(E),$$
in which $S(E)$ is collapsed to one point, the base point.

## Remarks

Up to homeomorphism, $\operatorname{Th}(E)$ does not depend on the inner product.
If $B$ is compact Hausdorff, $\operatorname{Th}(E)$ is homeomorphic to the [[one-point compactification|0GWN3CMRK8]] of $E$.
The notations $T(E)$, $B^E$ and $M(E)$ are also used.

Example: if $B$ is a point, then $E = \R^n$, $D(E) = D^n$ and $S(E) = S^{n-1}$, so $\operatorname{Th}(E) = D^n / S^{n-1} \cong S^n$.
More generally, for the trivial bundle $B \times \R^n$ over a compact space $B$, the Thom space is $(B \times D^n)/(B \times S^{n-1})$, the $n$-fold reduced suspension of $B$ with a disjoint base point added.
The Thom space of the tautological line bundle over $\R P^m$ is $\R P^{m+1}$.

Non-example: the disc bundle $D(E)$ without the collapse is not the Thom space; it deformation retracts onto the zero section, so it is homotopy equivalent to $B$.
For $B$ a point and $n \ge 1$ this gives a disc, which is contractible, while $\operatorname{Th}(E) = S^n$ is not.