5
$\begingroup$

The unoriented bordism theory $MO$ has a map to $H\mathbb{F}_2$ which is easily described for a space $X$ by pushing forward the fundamental class of a singular manifold to $H_*(X)$. Since $MO$ and $H\mathbb{F}_2$ both factor through chain complexes, it is tempting to ask if this can be realized as a map of chain complexes. I believe this is unlikely because although all smooth manifolds admit a triangulation, this obviously cannot be done naturally. However, this makes me wonder about the following:

For a space $X$, let $D_n(X)$ denote the set of pairs $(M,f)$ where $M$ is a smooth manifold with a specified triangulation and $f$ is a map from $M$ into $X$. This is a chain complex by letting the boundary map take a manifold to its boundary. Denote the nth homology of this chain complex by $MT_n(X)$ (meant to stand for triangulated bordism). Alternatively, I believe we can describe $D_n(X)$ as tuples $(M,f,\sigma)$ where $(M,f)$ is a singular manifold and $\sigma$ is a cycle lifting the fundamental class $[M]$ subject to a few conditions.

Recall that one can describe $H\mathbb{F}_2(X)$ as the homology of the chain complex $C(X)$ where $C_n(X)$ is the set of pairs $(S,f)$ where $S$ is a simplicial complex that is the union of its n-simplices and $f$ is a map from $S$ into $X$. The boundary operator takes $S$ to the union of n-1 simplices that are incident to an odd number of n-simplices.

There is a chain map $i:D(X) \rightarrow C(X)$ given by forgetting the fact that the domain of $f$ is a manifold. As well, we may also forget the triangulation to get a map $j$ from $D(X)$ to the chain complex of singular manifolds. Since every smooth manifold has a triangulation, $j$ (hence $j_*$) is surjective. This implies $i_*$ is surjective since it factors through $j_*$ and the map $MO \rightarrow H\mathbb{F}_2$.

(1) Is $MT$ a homology theory?

(2) Supposing (1), is $MT$ a wedge of $H\mathbb{F}_2$?

(3) Supposing (1), is $MT$ a pullback of $H\mathbb{F}_2$ and $MO$?

$\endgroup$
  • 2
    $\begingroup$ Your $MT$ is either just going to be $MO_*$ or $H_*(-;\Bbb F_2)$, depending on whether or not your manifolds have corners (you never specified). The former follows from the fact that every triangulation of the boundary of a compact smooth manifold extends to a triangulation of the whole smooth manifold. The second follows from the fact a singular manifold (or singular manifold with boundary) is precisely something you can paste together from a cycle in the singular chain complex. $\endgroup$ – Mike Miller Dec 31 '19 at 11:58
  • 2
    $\begingroup$ @MikeMiller Oh, I wasn't aware of that model (I was thinking of constructing the lift to chain complexes by using the splitting of $MO$ as wedge of copies of $H\mathbb{F}_2$). I'll have to admit I don't quite understand your description though: what you describe seems to be a chain complex of monoids, not of groups (assuming the addition is given by the disjoint union). This seems a bit too simple to me to be honest, because we know that such a model cannot exist for other bordism theories (e.g. complex bordism) so why does it work in this case? $\endgroup$ – Denis Nardin Dec 31 '19 at 12:05
  • 1
    $\begingroup$ Sorry, you add the formal relation $-M = M$ so that this is a $\Bbb F_2$-vector space; in the oriented case you add the formal relations $\sqcup_n M = nM$ and $-M = \overline M$. This is a very very silly chain complex: it is set up precisely so that the cycles are closed manifolds and that the boundaries are those closed manifolds which bound compact manifolds. It works because this is the definition of bordism groups. $\endgroup$ – Mike Miller Dec 31 '19 at 12:19
  • 3
    $\begingroup$ @MikeMiller But the recipe you're giving seems to be working also for, e.g., complex cobordism and we know it cannot possibly work since $MU_*(-)$ does not factor through chain complexes... I'm probably missing something and I shouldn't be working today anyway but I'm a bit confused... $\endgroup$ – Denis Nardin Dec 31 '19 at 12:28
  • 2
    $\begingroup$ Say you have two manifolds $N_1$ and $N_2$ with diffeomorphic boundary. According to the relations Mike Miller imposed, this makes $N_1\amalg N_2$ a cycle. But it doesn't represent a well-defined cobordism class, since that depends on how exactly you glue the two boundaries together. So I disagree with this giving precisely the definition of bordism groups, and as Denis reasoned above, something like this cannot work for oriented bordism even (since $MSO$ is not an $HZ$-module) $\endgroup$ – Achim Krause Jan 1 at 9:34
6
$\begingroup$

There is an implicit assumption in your question, namely that one can define a chain complex calculating the functor $X\mapsto MO_*(X)$ which is based on maps from unoriented manifolds into $X$. As far as I know, this is not known to be the case. Though, as you point out, since $MO$ is a wedge sum of shifts of copies of $H\mathbb F_2$, it's at least remotely possible that such a complex exists. It would have to be very special, though, since most (all?) other bordism theories are not wedge sums of Eilenberg--MacLane spectra.

The following obvious thing to try definitely does not work: take $C^{MO}_*(X):=$ the free $\mathbb F_2$ vector space on all isomorphism classes of pairs $(M,f)$ where $M$ is a compact unoriented manifold with boundary and $f:M\to X$ (with the obvious boundary map $\partial(M,f):=(\partial M,f|_{\partial M})$). The homology $H^{MO}_*(X)$ is not $MO_*(X)$, even for $X=\operatorname{pt}$. Indeed, $MO_1(\operatorname{pt})=0$ (every $1$-manifold is null cobordant), whereas there is a nontrivial cycle in $C^{MO}_1(\operatorname{pt})$ given by the interval $[0,1]$ equipped with the constant map to $\operatorname{pt}$. This cycle cannot be a boundary, since the only boundaries in $C^{MO}_*$ are maps from closed manifolds.

There are, of course, a few variations one could try on the above construction, but there are also what seem to me to be a number of fundamental difficulties. Suppose we start by taking as generators pairs $(M,f:M\to X)$ with $M$ closed (these having zero boundary). To get a bordism theory as the homology, clearly we want to also add in $M$ with boundary, so that cobordant manifolds are homologous. But this creates many new cycles! Namely, we could have some formal sum $c$ of manifolds with boundary such that the terms comprising $\partial c$ all cancel (like $[0,1]$ in the example above). Moreover, there may be many different ways of realizing this cancelling geometrically by gluing together the given manifolds comprising $c$ to create a closed manifold. Note that there's no good reason that all of these resulting manifolds to be cobordant! This is somewhat unnerving, but things are about to get worse. If we forge ahead and add in isomorphism classes of pairs $(M,f:\to X)$ where $M$ is allowed to have corners, then given a cycle in this chain complex, it's no longer even clear whether it makes sense to glue up the resulting manifolds in a consistent way (note that I have written isomorphism classes here---this is very important: without these words the generators wouldn't even form a set). This issue can be remedied by fixing a set of manifolds with corners and consistent identifications of their boundary strata with other manifolds with corners on the same list. For example, this is exactly what we do to define singular homology! (take the manifolds with corners to be $\Delta^n$). I had to do something like this in section 8.7.1 of this paper, to calculate singular homology. But this seems likely to just give singular homology no matter how you do it, not any bordism theory.

$\endgroup$
  • $\begingroup$ Thank you, I had mistakenly thought that your first construction did give unoriented bordism. $\endgroup$ – Connor Malin Dec 31 '19 at 23:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.