# Singular Homology Groups from Compact Oriented Submanifolds?

In the mathematics literature, for example in Lee's Introduction to Smooth Manifolds, the singular homology groups of a smooth manifold $M$ are defined using singular $p$-chains, which are formal superpositions of singular $p$-simplices. For the purposes of this question I will define singular $p$-simplices as smooth maps from the standard $p$-simplices to $M$. By contrast, in the physics literature, for example "String Theory and M-theory", by Becker/Becker/Schwarz, chains are instead defined as formal superpositions of oriented compact submanifolds of $M$. Do these define the same singular homology (and cohomology) groups for all smooth manifolds $M$? I believe this reduces to the question of whether or not any oriented compact submanifold $N$ in $M$ can be tiled by the image of a set of singular $p$-simplices whose orientations are consistent with the orientation of $N$. Is this correct? If not, why not?

• Please google "the steenrod realization problem". This is a foundational problem in algebraic topology. It has a nice answer -- Glen Bredon's book "Topology and Geometry" covers it pretty well. – Ryan Budney Feb 27 '17 at 19:37
• Thanks! As a physicist it is sometimes hard to find these things. I'm curious to see where the proof of the de Rham theorem fails if one uses submanifolds. – DanielHarlow Feb 27 '17 at 19:51
• @RyanBudney, why not post your comment as an answer? – HJRW Feb 27 '17 at 19:55

Expanded version of my comment:

Please google "the steenrod realization problem". This is a foundational problem in algebraic topology. It has a nice answer -- Glen Bredon's book "Topology and Geometry" covers it pretty well.

Bredon's book is not the full story but it gets you quite close to it.

For example if you use $\mathbb Z_2$ coefficients then every homology class is realizable as a manifold. Similarly, if you use real coefficients (or rational) then a multiple of every homology class is realizable.

The difficulty comes with odd torsion classes, when you use integral homology. Some of these classes are not realizable. Bredon's book gives you a good idea of how the machine works.

Similarly, for integral homology it's known that in certain dimension ranges (both low and high dimensional) Steenrod realization holds. It's the middle-dimensional homology classes (or close to it) in odd torsion that are problematic.

Although I have not read Becker/Becker/Schwarz, my guess is they are using homology with real coefficients. This would explain why they are comfortable with singular manifolds and not the more conventional homology formalism with formal linear combinations of singular simplices, etc.

edit: I suppose it should be noted that the more you look at this problem the more "layers" it has. If you demand your manifolds be genuine sub-manifolds and not singular maps from manifolds to a given manifold, then the problem is more subtle. Many of the results I mention above still hold. But generally speaking less is known at this level of specificity. If this is what you are interested in, the Bredon book gets you started on it as well. Serre gave a rather beautiful approach to this problem, which is what Bredon describes.

• Non-torsion classes can be non-realizable also. I learned this from a paper by Kotschick et al. that I can't find right now. Consider the compact Lie group $Sp(2)$, which is a simply connected 10-manifold. Its integral cohomology is an exterior algebra on generators $x_3$ and $x_7$ of degrees 3 and 7, and $x_3$ is not dual to a submanifold. The reason is that there's a mod 3 cohomology operation $P$ taking $\bar{x}_3$ to $\bar{x}_7$ (by calculations of Borel-Serre), so $\bar{x}_3 \cup P\bar{x}_3\neq 0$, yet the relevant universal Thom class $u$ satisfies $u\cup Pu =0$. – Tim Perutz Feb 27 '17 at 21:14
• I'm now wondering if the question really is about Steenrod realization. The question seems to ask whether homology classes are realized by submanifolds, while Steenrod realization seems to concern realization by singular submanifolds. – HJRW Feb 27 '17 at 21:17
• @HJRW: there's two versions of steenrod realization. The singular version, and the submanifold version. The answers to the two are different but in many ways very similar. The embedded version has less complete answers, yes. I tried to keep my answers vague enough so that they apply fairly broadly. But maybe I should clarify a little. – Ryan Budney Feb 27 '17 at 21:19
• @HJRW: well, "formal superpositions" is fairly general -- it looks to me like this could very well coincide with singular maps. But I have not read the book. – Ryan Budney Feb 27 '17 at 21:24
• Thanks for all the comments! I am interested in both the singular version and the submanifold version. About the coefficient group, I am predominantly interested in the case where the coefficients are real, as indeed is what is discussed by Becker/Becker/Schwarz, but I am also interested in the case with integer coefficients. I'm also happy to learn more general statements: this seems like a very basic question in algebraic topology, and I'm surprised it is not discussed in the introductory treatments of the subject that I've found so far. – DanielHarlow Feb 27 '17 at 22:39

Kreck proved that homology classes can always be represented by maps from certain singular manifolds (stratifolds)to a topological space, see his book. Apparently, singularities of codimension $\ge 2$ are necessary to produce correct homology groups. Two embedded stratifolds are cohomologous if there exists a stratifold-bordism between them.

Indeed, if you consider a singular $k$-cycle in $M$ (or any topological space $X$), then "closed" implies that the $k$-simplices involved can be glued along their $(k-1)$-faces. But there is no control on what happens along $\ell$-simplices for $\ell\le k-2$.

• There's quite a lot of work on this in the literature, mostly in the 70's and 80's. Buoncristiano has a book on it, I believe. Fred Cohen has the beautiful result that mod-2 homology is "singular bordism with a braid group structure on the stable normal bundle" (or something like that, it's been a while since I've been thinking about this) – Ryan Budney Mar 1 '17 at 0:57