# 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

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

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 $\operatorname{Th}(\xi)$ and $X^{\xi}$.

###### Remark 1

If $\xi=X\times{\mathbb{R}}^{d}$ is a trivial vector bundle, then its Thom space is homeomorphic to $\Sigma^{d}X_{+}$, where $X_{+}$ stands for $X$ with an added disjoint basepoint, and $\Sigma^{d}$ 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 ThomSpace 2013-03-22 15:40:46 2013-03-22 15:40:46 antonio (1116) antonio (1116) 5 antonio (1116) Definition msc 57R22 msc 55R25 Thom space disk bundle sphere bundle