# Kirby–Siebenmann class and the 4th Stiefel-Whitney class: $\operatorname {ks} (M)$ v.s. $w_4(M)$

Kirby–Siebenmann class $\operatorname {ks} (M)$ is an element of the fourth cohomology group $$\operatorname {ks} (M)\in H^{4}(M;\mathbb {Z} /2)}$$ which must vanish if a topological manifold $M$ is to have a piecewise linear structure. It is named for Robion Kirby and Larry Siebenmann.

However, the 4th Stiefel-Whitney class $w_4(M)$ also is an element of the fourth cohomology group $$w_4(M) \in H^{4}(M;\mathbb {Z} /2).$$

Questions:

(a) How are Kirby–Siebenmann class $\operatorname {ks} (M)$ and the 4th Stiefel-Whitney class $w_4(M)$ related?

(b) We say that the obstruction of the spin structure is the non-vanishing of the 1st or 2nd Stiefel-Whitney class $w_1(M)\neq 0$ or $w_2(M)\neq 0$. Is the obstruction of having piecewise linear (PL) structure has something to do with

• the non-vanishing of the Kirby–Siebenmann class $\operatorname {ks} (M)$?

• the non-vanishing of the 4th Stiefel-Whitney class $w_4(M)$?

Initially, Stiefel-Whitney classes are only defined for smooth manifolds (if the manifold is not smooth, it doesn't have a tangent bundle). However, Wu's theorem states that $w = \operatorname{Sq}(\nu)$ and we can use this as a definition of Stiefel-Whitney classes for topological manifolds; in particular, $w_4(M)$ is defined even if $M$ does not admit a smooth structure or even a PL structure.

Freedman's Theorem states every non-degenerate bilinear form $b$ arises as the intersection of a closed, simply connected four-manifold. Moreover, if $b$ is even, there is a unique such manifold up to homeomorphism, and if $b$ is odd, there are two up to homeomorphism and they are distinguished by the Kirby-Siebenmann invariant.

As $E_8$ is an even form, there is a unique (up to homeomorphism) closed, simply connected four-manifold with intersection form $E_8$; we denote it by $M_{E_8}$.

As $(1)$ is an odd form, there are (up to homeomorphism) two closed, simply connected four-manifolds with intersection form $(1)$. One is $\mathbb{CP}^2$ which has zero Kirby-Siebenmann invariant (it admits a smooth structure and hence a PL structure). We denote the other manifold by $*\mathbb{CP}^2$ and note that it has non-trivial Kirby-Siebenmann invariant.

Recall that for a closed smooth four-manifold $M$, the Stiefel-Whitney number $\langle w_4(M), [M]\rangle$ is the mod $2$ reduction of the Euler characteristic $\chi(M)$; this is Corollary $11.12$ of Milnor and Stasheff's Characteristic Classes. The same is true for topological manifolds (using the definition of Stiefel-Whitney classes outlined above), see this question.

With all of this in mind, we can now see that $\operatorname{ks}(M)$ and $w_4(M)$ are unrelated as the following table demonstrates.

$$\begin{array}{c|cc} M & \operatorname{ks}(M) & w_4(M)\\ \hline S^4 & 0 & 0 \\ \mathbb{CP}^2 & 0 & \neq 0 \\ M_{E_8} & \neq 0 & 0 \\ *\mathbb{CP}^2 & \neq 0 & \neq 0 \end{array}$$

If the four-manifold $M$ admits a PL structure, then $\operatorname{ks}(M) = 0$, but the converse is not true. For example, $M_{E_8}\# M_{E_8}$ has trivial Kirby-Siebenmann invariant but it does not admit a PL structure (every PL manifold of dimension less than or equal to $7$ is smoothable, but $M$ is not smoothable by Donaldson's theorem). The class $w_4(M)$ has nothing to do with the obstructions to admitting a PL structure (both $S^4$ and $\mathbb{CP}^2$ admit PL structures).

• thanks +1, I appreciate this quick answer! – wonderich Sep 7 '18 at 20:50
• This is an interesting fact which I didn't know that $w_4(M_{E_8})$ is zero. :) – Anubhav Mukherjee Sep 9 '18 at 0:50