Cobordism theory for piecewise-linear (PL) and topological manifolds

The Cobordism theory was originally developed by René Thom for smooth manifolds (i.e., differentiable), but there are now also versions for piecewise-linear and topological manifolds.

I know the Cobordism theory for smooth manifolds, for example, given two $d$-dimensional manifolds $M_1$ and $M_2$, we can ask whether they are (co)bordant via a $d+1$-dimensional manifolds $W$, such that $$\partial W= M_1 \sqcup M_2.$$ For example,

• (1) 5-dimensional Dold manifold and Wu manifold are manifolds which are cobordant to each other via 5-dimensional bordism group: $$\Omega^{SO}_5=\mathbb{Z}_2.$$ In other words, my interpretation is that the Dold and Wu manifolds are both the nontrivial generators of $\Omega^{SO}_5=\mathbb{Z}_2$.

• (2) For example, consider an oriented bordism, bordisms between $4$-manifolds. The signature $\sigma(X^{4k})$ of an oriented $4k$-manifold $X^{4k}$ is an oriented bordism invariant. Now $\sigma(S^4) = 0$ and $\sigma( C P^2) = 1.$ So $S^4$ and $C P^2$ are not oriented bordant. In fact, $\Omega_4^{{SO}} \cong \Bbb Z$ with generator $[C P^2]$.

So in the above, I give one example (Dold manifold and Wu manifold) that are cobordant via $\Omega^{SO}_5$, and another counterexample where $S^4$ and $C P^2$ are not oriented bordant $\Omega_4^{{SO}}$.

Questions: Now, I am not familiar with Cobordism theory of piecewise linear and topological manifolds.

• Can we explain how the cobordant of piecewise-linear (PL) and topological manifolds is defined, in a simple manner?

• Can we give one example and another counterexample, in dimensions 3, 4 and 5, such that two PL topological manifolds are cobordant or not cobordant, respectively?

1) The definition is no different, just replace the phrase "smooth manifold" everywhere with "PL" or "topological" manifold.

2) The book "The classifying spaces for surgery and cobordism of manifolds" by Madsen and Milgram gives a calculation of the oriented version of $\Omega^{\text{Top}}_*/\text{Tors}$ as a ring; they prove it is additionally isomorphic to $\Omega^{\text{PL}}_*/\text{Tors}$. I am not going to try to recover it from the book, though.

3) PL = Smooth in dimensions up through and including 6, so there is no difference in the corresponding bordism groups in degree up to 5. Every topological 3-manifold has a smooth structure, and every smooth 3-manifold is null-bordant. I gave a reference for an example of a topological 4-manifold that is not null-bordant in an MO question you a asked before here; in fact the calculation of the oriented version of $\Omega^{\text{Top}}_4$; the Kirby-Seibenmann invariant is an unoriented bordism invariant as well.

I don't know the answer for the group of unoriented bordism classes $\Omega_5^{\text{Top}}$. One should be able to argue in the oriented case as follows: there is an oriented bordism to a simply connected manifold, obtained by an oriented surgery on some sequence of nontrivial loops. Then the class of simply connected 5-manifolds has been classified, and in fact they are all smoothable. Hence every oriented closed topological 5-manifold is oriented-bordant to a smooth one.

• In the topological case, the unoriented bordism group $\Omega_*^{Top}$ is just a free polynomial algebra $\mathbb{Z}_2[x_2, x_4, x_5, \dots]$ over $x_n$ with $n\not = 2^k-1$, where $\deg x_n = n$. The proof is due to Dold and involves using the Pontryagin-Thom construction to reduce the problem to one about a certain classifying space, then using some algebraic machinery to compute its homotopy groups. – anomaly Sep 7 '18 at 0:04
• thanks, +1, this is a really nice discussion. – wonderich Sep 7 '18 at 1:46
• @anomaly Is the implicit claim here that $\Omega_*^{\text{Top}} \to \Omega_*$ is an isomorphism? I do not think this is true. Actually, Hsu's paper in the question I linked states that $\Omega_4^{\text{Top}} \cong (\Bbb Z/2)^3$, with the third factor given by the Kirby-Seibenmann invariant. (He writes this as $\eta_4$ and this is his Corollary 2.4(c).) – user98602 Sep 7 '18 at 17:09
• I've seen that paper. Aren't the relevant classes $x_2^2, x_4$, and $x_2^2 + x_4$? – anomaly Sep 7 '18 at 18:03
• @anomaly Those aren't linearly independent. Your ring has only two linearly independent generators in degree 4. – user98602 Sep 7 '18 at 18:09