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 $\xi =X\times {ℝ}^d$ 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.