Skip to main content
MathOverflow

Return to Question

added 26 characters in body
Source Link
Michael Albanese
Michael Albanese
  • 19.3k
  • 9
  • 87
  • 160

If $M$ is a smooth manifold, the total Stiefel-Whitney class of $M$ is defined to by the total Stiefel-Whitney class of the tangent bundle, i.e. $w(M) := w(TM)$.

If $M$ is a closed smooth manifold, then $w(M) = \operatorname{Sq}(\nu)$ where $\operatorname{Sq}$ is the total Steenrod square and $\nu$ is the total Wu class. One consequence of this fact is that $w(M)$ depends only on the graded algebra $H^{\bullet}(M; \mathbb{Z}_2)$ as a graded module over the Steenrod algebra. In particular, if $f : M \to N$ is a homotopy equivalence of closed smooth manifolds, then $f^*w(N) = w(M)$. So, for example, any exotic sphere has total Stiefel-Whitney class $1$ (because it is homotopy equivalent, in fact homeomorphic, to the standard smooth sphere).

Note that the expression $\operatorname{Sq}(\nu)$ does not depend on the smooth structure, so one could define the total Stiefel-Whitney class of a closed topological manifold by this expression, despite the fact such a manifold has no natural vector bundle.

Has this definition of Stiefel-Whitney classes been used to deduce any information about closed topological manifolds which admit no smooth structure?

If $M$ is a smooth manifold, the total Stiefel-Whitney class of $M$ is defined to by the total Stiefel-Whitney class of the tangent bundle, i.e. $w(M) := w(TM)$.

If $M$ is a closed smooth manifold, then $w(M) = \operatorname{Sq}(\nu)$ where $\operatorname{Sq}$ is the total Steenrod square and $\nu$ is the total Wu class. One consequence of this fact is that $w(M)$ depends only on the graded algebra $H^{\bullet}(M; \mathbb{Z}_2)$. In particular, if $f : M \to N$ is a homotopy equivalence of closed smooth manifolds, then $f^*w(N) = w(M)$. So, for example, any exotic sphere has total Stiefel-Whitney class $1$ (because it is homotopy equivalent, in fact homeomorphic, to the standard smooth sphere).

Note that the expression $\operatorname{Sq}(\nu)$ does not depend on the smooth structure, so one could define the total Stiefel-Whitney class of a closed topological manifold by this expression, despite the fact such a manifold has no natural vector bundle.

Has this definition of Stiefel-Whitney classes been used to deduce any information about closed topological manifolds which admit no smooth structure?

If $M$ is a smooth manifold, the total Stiefel-Whitney class of $M$ is defined to by the total Stiefel-Whitney class of the tangent bundle, i.e. $w(M) := w(TM)$.

If $M$ is a closed smooth manifold, then $w(M) = \operatorname{Sq}(\nu)$ where $\operatorname{Sq}$ is the total Steenrod square and $\nu$ is the total Wu class. One consequence of this fact is that $w(M)$ depends only on $H^{\bullet}(M; \mathbb{Z}_2)$ as a graded module over the Steenrod algebra. In particular, if $f : M \to N$ is a homotopy equivalence of closed smooth manifolds, then $f^*w(N) = w(M)$. So, for example, any exotic sphere has total Stiefel-Whitney class $1$ (because it is homotopy equivalent, in fact homeomorphic, to the standard smooth sphere).

Note that the expression $\operatorname{Sq}(\nu)$ does not depend on the smooth structure, so one could define the total Stiefel-Whitney class of a closed topological manifold by this expression, despite the fact such a manifold has no natural vector bundle.

Has this definition of Stiefel-Whitney classes been used to deduce any information about closed topological manifolds which admit no smooth structure?

Source Link
Michael Albanese
Michael Albanese
  • 19.3k
  • 9
  • 87
  • 160

Stiefel-Whitney classes of closed topological manifolds with no smooth structure

If $M$ is a smooth manifold, the total Stiefel-Whitney class of $M$ is defined to by the total Stiefel-Whitney class of the tangent bundle, i.e. $w(M) := w(TM)$.

If $M$ is a closed smooth manifold, then $w(M) = \operatorname{Sq}(\nu)$ where $\operatorname{Sq}$ is the total Steenrod square and $\nu$ is the total Wu class. One consequence of this fact is that $w(M)$ depends only on the graded algebra $H^{\bullet}(M; \mathbb{Z}_2)$. In particular, if $f : M \to N$ is a homotopy equivalence of closed smooth manifolds, then $f^*w(N) = w(M)$. So, for example, any exotic sphere has total Stiefel-Whitney class $1$ (because it is homotopy equivalent, in fact homeomorphic, to the standard smooth sphere).

Note that the expression $\operatorname{Sq}(\nu)$ does not depend on the smooth structure, so one could define the total Stiefel-Whitney class of a closed topological manifold by this expression, despite the fact such a manifold has no natural vector bundle.

Has this definition of Stiefel-Whitney classes been used to deduce any information about closed topological manifolds which admit no smooth structure?