# Bordism for oriented triangulable manifolds without smooth differentiable structures

We know the bordism group for oriented smooth differentiable structures such as $$\Omega_d^{SO}$$ that requires the special orthogonal group structure on the tangent bundle $$TM$$ of manifold $$M$$.

$$\Omega_0^{SO}=\mathbb{Z}$$ $$\Omega_1^{SO}=0$$ $$\Omega_2^{SO}=0$$ $$\Omega_3^{SO}=0$$ $$\Omega_4^{SO}=\mathbb{Z}$$ $$\Omega_5^{SO}=\mathbb{Z}/2$$ $$\Omega_6^{SO}=0$$ $$\Omega_7^{SO}=0$$ $$\Omega_8^{SO}=\mathbb{Z} \oplus \mathbb{Z}$$ $$\Omega_9^{SO}=\mathbb{Z}/2 \oplus \mathbb{Z}/2$$ $$\Omega_{10}^{SO}=\mathbb{Z}/2$$ $$\Omega_{11}^{SO}=\mathbb{Z}/2$$

Some of the cobordism invariants for these bordism groups only require the Stiefel-Whitney class $$w_j(TM)$$. But $$w_j(TM)$$ can be defined on oriented triangulable manifolds without smooth differentiable structures. So those cobordism invariants may still survive when we loosen the condition to study the oriented triangulable manifolds without smooth differentiable structures.

Do we know any bordism theory for oriented triangulable manifolds without smooth differentiable structures that produces the similar bordism groups as above, but only maintain those cobordism invariants that only require oriented triangulable manifolds (such as those only require Stiefel-Whitney class $$w_j(TM)$$)? That is, could we remove the $$SO$$ structure above, and only keep the triangulation, say $$trg$$, and study its bordism group? I expect something like:

$$\Omega_5^{triag}=\mathbb{Z}/2, \text{: cobordism invariant: w_2w_3}$$

$$\Omega_9^{triag}=\mathbb{Z}/2 \oplus \dots \text{: cobordism invariant: w_4w_5}$$

$$\Omega_{13}^{triag}=\mathbb{Z}/2 \oplus \dots \text{: cobordism invariant: w_6w_7}$$

$$\Omega_{4k+1}^{triag}=\mathbb{Z}/2 \oplus \dots \text{: cobordism invariant: w_{2k}w_{2k+1}, for k\geq1}$$

I suppose the Piecewise Linear PL is similar to what I want. I do not think I am considering the topological manifold Top. But I do not require the smooth structure, nor the $$SO$$ or $$O$$ tangential structure. But I am not sure that the Stiefel-Whitney class $$w_j(TM)$$ requires the PL structure. The Stiefel-Whitney class $$w_j(TM)$$ may require only the triangulation and branching rule on manifold $$M$$.

• I don't understand what category of manifold you are interested in. Are you talking about the image of the forgetful map $PL \to TOP$? Dec 8, 2021 at 17:10
• $w_j(TM)$ does not require a triangulation (a $PL$ structure). The tangent microbundle of a topological manifold has Stiefel-Whitney classes. In fact, even a (stable) spherical fibration on a space X has SW classes; $w_j$ can be defined as the class that corresponds via the Thom isomorphism to $Sq^j$ of the Thom class. Dec 8, 2021 at 17:34
• The paper "Classification of simplicial triangulations of topological manifolds" of Galewski and Stern relates triangulations to lifts of the microtangent bundle to a space $\operatorname{BTri}$. If there is a positive answer to your question, presumably the homology theory is represented by some $\operatorname{MTri}$. Dec 8, 2021 at 17:58 I am not sure whether this answers your question in full, but I can suggest an idea. It is based on triangulated, oriented, labelled manifolds. You can drop the orientation and then move from coefficients in $$\bf Z$$ to $$\bf Z/2Z$$.

Define $$\Omega_0^{tol}$$ to be the group of cobordism classes of two-coloured points. The superscript $$tol$$ stands for triangulated, oriented, labelled. Using the notation from your question, $$\Omega_0^{tol} = \Omega_0^{SO} = \bf Z$$.

Moving on to 1-dimensional closed manifolds. Here, these are oriented, 3-coloured circles. The key ingredient to determine $$\Omega_1^{tol}$$ is the generalized Lemma of Sperner in combination with a degree function on the labelling.

It says: If you triangulate a 2-disk and label it with 3 colours, such that the labelling or colouring of the boundary circle has degree $$k\in\bf Z$$, then $$\sum \Delta^+ - \sum\Delta^- = k$$ where $$\Delta^+$$ are 3-coloured triangles with one orientation and $$\Delta^-$$ are 3-coloured triangles with the opposite orientation. In his original 1928 paper, Sperner treats the case $$k=1$$ (for $$d$$-dimensional simplices).

My illustration at the top shows an example for $$k = 2$$ where both triangles have the same orientation. You can see the details about degree and orientation in the reference below. It basically works like this: the degree of the labelling or colouring is defined through counting the colour changes e.g. from blue to yellow on the boundary circle $$M$$. I have put $$+1$$ and $$+1$$ in the diagram to show that count.

Now here is the key idea. Translate that into cobordism. See the 3-coloured triangles as holes, and see all other triangles as 2-dimensional, filled triangles. Saying it differently, regard the 3-coloured triangles as circles $$S^1$$ while the other triangles form a triangulated 2-manifold $$X$$. To illustrate this idea, the 3-coloured triangles $$N_1$$ and $$N_2$$ in my picture are not filled while the interiors of all other triangles are shaded (with the meaning they are subsets of a 2-manifold; their union defines a 2-manifold).

So, translated to cobordism, the generalized Lemma of Sperner asserts: for each oriented, 3-coloured circle $$M$$ with labelling degree $$k$$ you can find a 2-manifold $$X$$ (triangulated, oriented, 3-coloured disk with holes) such that $$\partial X = M \sqcup -N_1 \sqcup -N_2 \sqcup \ldots \sqcup -N_k$$

This means $$M \cong N_1 \sqcup N_2 \sqcup \ldots \sqcup N_k$$ and you have a cobordism equivalence relation. Here, $$-N_i$$ is the inverse of $$N_i$$ which just means two vertex colours of the triangle are flipped (i.e. orientation is reversed).

Importantly, we need to be careful with the orientations here. If you start with an orientatin for $$M$$, the orienation of $$X$$ and $$N_i$$ have to be compatible (such that the orientations of $$M$$ and $$N_i$$ can be seen as induced by the orientation of $$X$$).

The degree $$k$$ of the labelling can take any value $$k\in \bf Z$$, and indeed $$\Omega_1^{tol}$$ is generated by a 3-coloured triangle with degree $$+1$$. In summary $$\Omega_1^{tol}={\bf Z} \neq \Omega_1^{SO}$$ so this really defines a cobordism theory that is different from the oriented smooth one.

The generalized Lemma of Sperner also holds for higher dimensions with the same formula for simplices (and of course $$d+1$$ colours for dimension $$d$$). Hence, if you restrict yourself to spheres $$S^d$$, you will get $$\Omega_d^{tol}=\bf Z$$ (but I am not sure this restriction is meaningful). If you can show that the generalized Lemma of Sperner (or variants of it) also holds for the torus, the n-holed torus etc, you can determine $$\Omega_2^{tol}$$ for surfaces and move on to higher dimensional manifolds.

Regarding degree of labelling and generalized Lemma of Sperner, here is the reference: Michael Prüfer and Hans Willi Siegberg (1981): Complementary Pivoting and the Hopf Degree Theorem. Journal of Mathematical Analysis and Applications, Volume 84, Issue 1, November 1981, Pages 133-149.

• To see the fully colored simplex as a hole in a manifold - this is a pretty revolutionary idea! Sperner's lemma is almost 100 years old but this is the first time I see such an idea. You should publish that. Jun 17 at 16:25