# Thom space

Let $\xi \to X$ be a vector bundle^{} over a topological space^{} $X$. Assume that $\xi $ has a Riemannian metric. We can form its associated disk bundle $D(\xi )$ and its associated sphere bundle $S(\xi )$, by letting

$$D(\xi )=\{v\in \xi :\parallel v\parallel \le 1\},S(\xi )=\{v\in \xi :\parallel v\parallel =1\}.$$ |

The *Thom space* of $\xi $ is defined to be the quotient space^{} $D(\xi )/S(\xi )$, obtained by taking the disk bundle and collapsing the sphere bundle to a point. Notice that this makes the Thom space naturally into a based topological space^{}.

Two common forms of notation for the Thom space are $\mathrm{Th}(\xi )$ and ${X}^{\xi}$.

###### Remark 1

If $\xi =X\times {\mathbb{R}}^{d}$ is a trivial vector bundle, then its Thom space is homeomorphic^{} to ${\mathrm{\Sigma}}^{d}{X}_{+}$, where ${X}_{+}$ stands for $X$ with an added disjoint basepoint, and ${\mathrm{\Sigma}}^{d}$ stands for the based suspension iterated $d$ times. Thus, we may think of ${X}^{\xi}$ as a “twisted suspension” of ${X}_{+}$.

###### Remark 2

If $X$ is compact^{}, then ${X}^{\xi}$ is homeomorphic as a based space to the one-point compactification of $\xi $. Even if $X$ is not compact, ${X}^{\xi}$ can be obtained by doing a one-point compactification on each fiber and then collapsing the resulting section^{} of points at infinity to a point.

###### Remark 3

The choice of Riemannian metric on $\xi $ does not change the homeomorphism^{} type of ${X}^{\xi}$, and, by the previous remark, the Thom space can be described without reference to associated disk and sphere bundles.

Title | Thom space |
---|---|

Canonical name | ThomSpace |

Date of creation | 2013-03-22 15:40:46 |

Last modified on | 2013-03-22 15:40:46 |

Owner | antonio (1116) |

Last modified by | antonio (1116) |

Numerical id | 5 |

Author | antonio (1116) |

Entry type | Definition |

Classification | msc 57R22 |

Classification | msc 55R25 |

Defines | Thom space |

Defines | disk bundle |

Defines | sphere bundle |