Thom space

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 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.