On defining homology groups

Question

I have been trying to understand what homology groups are "talking about," and now I am wondering if the following works as a definition of homology.

But first, some illustration of what it is inspired by: if $D\subset\Bbb C$ is a domain it seems elements of $H_1(D)$ can be interpreted as contours to integrate over, considered with orientation (sign) and multiplicity. If $\oint_\gamma f(z)dz=0$ for all of the functions $f\in{\cal O}(D)$ (i.e. holomorphic $f:D\to\Bbb C$) then we treat $\gamma$ as $0$ in $H_1(D)$. Thus, loops are trivial if they can be contracted to a point within the domain $D$. On the other hand, if a loop goes around a punctured point in $D$, there will be holomorphic functions in ${\cal O}(D)$ with a simple pole at the point and integrating such a function over the contour will yield a nonzero value.

There are some more relations satisfied by these contours-with-multiplicity. If $-\gamma$ is the same contour as $\gamma$ but traversed in the opposite direction then $\oint_{-\gamma} f(z)dz=-\oint_\gamma f(z)dz$ and hence $\oint_{\gamma+(-\gamma)}f(z)dz=0$ for all $f$, so we may conclude $\gamma+(-\gamma)=0$. And if $\gamma_1$ and $\gamma_2$ go around two distinct punctures respectively (and no others) with the same orientation then $\gamma_1+\gamma_2$ is equivalent to any contour that loops around both punctures (but no others) in the same direction. Without loss of generality I think we can consider $H_1(D)$ to be generated by simple loops.

To generalize to $n$ dimensions, we need vector calculus and differential forms to replace complex analysis. From what I understand, Stokes' and de Rham's theorems say that elements of homology are oriented domains of integration for differential forms counted with multiplicity. (I have not delved more deeply than this intuitive understanding, though.) In that spirit, I want to define my chain groups $C_n({\cal M})$ as $\Bbb Z$-linear combinations of closed, compact, oriented $n$-dimensional submanifolds, subject to the following two relations:

• If $\cal A, B$ intersect only on their boundary and $\cal A\cup B$ has a consistent orientation then we identify $\cal A\cup B$ with $A+B$. I believe an equivalent condition is that if $\cal A\cap B\ne\varnothing$ and have the same restricted orientation on $\cal A\cap B$ then $\cal A+B=(A\cup B)+(A\cap B)$.
• If $\cal C,C'$ are the same submanifold but with opposite orientations, then $\cal C'=-C$.

One may take the usual topological boundary (in which orientation is inherited) and extend linearly to obtain the boundary maps $\partial_n:C_n({\cal M})\to C_{n-1}({\cal M})$. Then $\partial^2=0$ and we may define the homology group $H_n({\cal M})$ as generated by the boundaryless chains modulo boundaries in the usual fashion ($H_n=\ker(\partial_n)/{\rm img}(\partial_{n+1})$). Does this work as a viable definition?

Just as with contours in complex analysis, it seems we can homotope any submanifold and still represent the same element of homology. (For instance, consider a sphere $\cal A$ around the origin in $\Bbb R^3-0$ with orientation chosen to make outward-pointing normals by the right-hand-rule. Say we homotope this to a smaller sphere $\cal B$ within $\Bbb R^3-0$. Then $\cal A-B$ will be the oriented boundary of some oriented annular region and hence $\cal A-B\equiv{\rm 0}~\Rightarrow A=B$ in $H_2(\Bbb R^3-0)$.

If we put a nice CW complex structure on $\cal M$, it feels like any submanifold (a generator) in the chain group $C_n({\cal M})$ may be homotoped to one comprised entirely of oriented cells from the complex. Thus, using CW complexes should represent a "discretization" of the definition I'm submitting that allows for finite computations to be humanly carried out. Moreever, every compact manifold can be triangulated, so it seems we can also relate it to singular homology.

So, anyway, to repeat my question: does this work as a viable definition of homology? Does this idea have a name? Does it seem more natural than other definitions? Are there technical obstacles to relating it to usual definitions of homology? If this doesn't work, why not?

One idea offered in chat is that I may not be able to homotope (or isotope, whatever may be necessary to make things work out) two knots in a way that can be broken apart into annuli (if annuli bound the before/after submanifolds, that would make them all equal in $H_1$). The moral of that lesson is that we might need to introduce a third relation in $C_n({\cal M})$, that isotopic submanifolds are equal.

Answer 1

You run into two technical problems with this attempt to describe homology, but they're not "obvious" problems. (In fact, your description is at least partially what motivated Poincare to define homology in the first place.)

The first is the restriction to submanifolds. There are too many technical details about embedding submanifolds: how many submanifolds there are, cutting submanifolds into pieces, etc etc. Homology instead often talks about manifolds with a map to your given manifold instead, which you can think of as a parametrized object that you can do an integral over. (These are technical details that are also part of multivariate integration, dealing with the definition of an integral over a submanifold without choosing a parametrization.) More specifically is this problem: determining when a manifold is a boundary of another. It may be a boundary, but can you guarantee that the thing it's a boundary of will be nicely embedded?

The second is that homology only corresponds to manifolds that are built out of simple pieces ("triangulable" manifolds), and only regards two as equivalent if they're the common boundary of something triangulable, and the triangulations are compatible, and... Basically everything having to do with triangulations turned out to be extremely messy.

Mathematicians do have a definition of something like homology defined in terms of manifolds. It's called bordism, and there is a natural map from bordism to homology that's not an isomorphism. In particular, there are spaces with homology elements that don't come from any manifold at all, due to the negative solution to the Steenrod problem.

Answer 2

Let me try to take your proposal seriously and see what happens. Your setting is the one of a smooth (everything will be smooth from now on) manifold $M$, where the group of $k$-cycles is the free abelian group with the basis given by oriented $k$-dimensional closed submanifolds $c\subset M$ and the subgroup of boundaries is (essentially) generated by boundaries of compact $k+1$-dimensional submanifolds in $M$.

Note. You are taking a further quotient given by identification $-c=\bar{c}$, where $\bar{c}$ is an oriented manifold equipped with the opposite orientation. I will keep this in mind.

The result is an abelian group which I denote $h_k(M)$, but the question what is it good for and can you compute it?

Everything is clear if $dim(M)=1$. After a bit of work (you will need to prove, among other things, smooth Jordan curve theorem in $R^2$), you prove that for oriented surfaces $M$, $h_k(M)\cong H_k(M)$, where $H_k$ is the singular homology group. Then you try $M=RP^2$ and you realize (to your great distress) that $h_k(M)$ has infinite rank of cardinality continuum! Independent elements of $h_1(M)$ are represented by projective lines on $RP^2$. That's not good, since maybe you are hoping for an analogue of Euler's formula (Euler characteristic equals to the alternating sum of dimensions of homology group). You might declare at this point that you are interested only in oriented manifolds. Fine. Let's move to dimension 3. You prove (with some difficulty) that $h_k(S^3)$ is isomorphic to $H_k(S^3)$, so nothing surprising here. However, in dimension 4 you look at $S^4$ and realize that you have no idea what to do. Does $h_2(S^4)$ have finite rank or not? (I do not know either but I did not spend much time thinking about it.)

Edit. Mike Miller noted in a comment that every surface in $S^4$ will bound a submanifold, so $h_2(S^4)=0$. Moreover, whenever $H_{n-2}(M)=0$, $h_{n-2}(M)=0$ as well, for every oriented $n$-dimensional manifold $M$.

Then you look at $CP^2$ and, again discover that $rank(h_2(CP^2))$ has cardinality of continuum! (Now because of complex projective lines.) At this point you might declare that you are only interested in homology of domains in $R^n$. Fine: You then find that $R^8$ contains a copy of $CP^2$ which is nontrivial as an element of $h_4(R^8)$. (Since $CP^2$ does not bound any compact 5-manifold.)

At this point you probably are ready to give up on using embedded manifolds as your cycles and boundaries. After all, in order to integrate differential forms, embeddedness is irrelevant. You use smooth maps of oriented manifolds (with boundary) into $M$ in order to define cycles and boundaries. This leads you to rediscover the bordism groups $O_k(M)$ of $M$. These are interesting groups which are discussed in other answers. However, even the manifold consisting of a single point has nontrivial bordism groups (already $O_4(\{p\})\ne 0$), again because of $CP^2$. You may not like it either and decide to allow "smooth manifolds with mild singularities" for your chains. After all, you do not need smoothness in order to integrate. What constitutes "mild singularities" is debatable. The classical singular homology theory provides one answer. Another answer appears in the work of Mattias Kreck "Differential Algebraic Topology", which might be what you are searching for. In the end, it yields groups isomorphic to the singular homology groups, see this paper by Kreck and Singhof. Sullivan's "Infinitesimal Computations in Topology" is also relevant here.

Lastly, as for the request for

"an answer that addresses the technical problems in this proposal and makes it either a equivalent to a well-known existing approach, or else superior to other approaches",

it is akin to a request to show that a turtle either has equal speed or is faster than a hare.

Answer 3

This is a supplement to Tyler's answer (it should be a comment, but it's too long).

As Tyler mentions, the Hauptvertmutung (any two triangulations have a common refinement) is false for topological spaces with dimension greater than 2. This breaks the sentence "simplicial (co)homology is an approximation of oriented cobordism (co)homology theory in all dimensions." However, you are onto something, so I figured that it's worth mentioning a few related things.

This is a geometric interpretation of the multiplication of singular cohomology classes (from the Princeton Companion):

Let $S$ and $T$ be closed oriented submanifolds of X, of codimension i and j respectively. By moving $S$ slightly (which doesn't change its class in $H^i(X)$) we can assume that $S$ and $T$ intersect transversely, which implies that the intersection of $S$ and $T$ is a smooth submanifold of codimension $i+j$ in $X$. Then the product of the cohomology classes $[S]$ and $[T]$ is simply the cohomology class of their intersection $[S \cap T] \in H^{i+j}(X)$.

This fails when we can't assume that $S$ and $T$ intersect transversely, in which case we can use Intersection (co)homology which allows for "perversity" (i.e., how far cycles are allowed to deviate from transversality).

However, the operations of most cohomology theories come from richer and different geometric (or algebraic) structures. There are even cohomology theories for which there is no known notion of cocycles (e.g., elliptic cohomology theories, tmf)!

If geometric interpretations of (co)homology are interesting to you, I highly recommend that you start by getting to know and love characteristic classes via Milnor and Stasheff.

The relations you wish to impose on your combinations of closed compact oriented n-dimensional submanifolds also remind me of the algebra of chains (where the union of n-simplexes is an n-chain). The sum $C_1 + C_2$ is defined as the k-chain made up of the k-simplexes in $C_1$ or $C_2$ but not both, and this sum is commutative, associative, and for every k-chain $C$ there is a unique k-chain $D$ such that $C + D = \emptyset$.

As a parting note, it looks like you might also be moving toward accidentally re-deriving the Poincare index theorem, which certainly has a Stokes-flavor to it (for it connects the behaviour of a vector field inside of a cell with its behavior on the boundary). First, choose a circle $\gamma$ about the critical point $P$ so that within and on $\gamma$ the vector field $V$ vanishes except at $P$. The index of $V$ at $P$, denoted $I(P)$, is defined as the winding number $W(\gamma)$ of $V$ on $\gamma$.

Let $V$ be a continuous vector field. Let D be a cell and $\gamma$ its boundary. Supposing that $V$ is not zero on $\gamma$, then:

$$W(\gamma) = I(P_1) + ... + I(P_n)$$

where $P_1, P_2, ..., P_n$ are the critical points of $V$ inside $D$.

Answer 4

The questioner is right to identify one of the inspirations for homology as coming from integration theory; this is emphasised in the article by S. Lefschetz in the book "History of Topology", edited I.M. James. It seems also that early articles on Betti nubers and torsion coefficients wanted to take "cycles modulo boundaries" but were not so clear about the meaning of those. It was Poincaré who introduced the notion of formal sums of oriented simplices which led to the equation $\partial \partial=0$ which we all know and love. The idea of formal sums surely came from integration theory where it was convenient to write $$\int_C fdz + \int _D f dz = \int_{C+D} f dz. $$

This formal sum contrasted with the explicit (partial) composition of paths which led to Poincaré introducing the fundamental group of a pointed space. Algebraic topologists in the early 20th century were hoping for higher dimensional nonabelian version of the fundamental group, but these were dashed by the acceptance of the abelian higher homotopy groups of a pointed space. This abelian property can be explained by saying that a group object in the category of groups is an abelian group.

That situation is changed if one considers groupoids: groupoid objects in the category of groups are more complicated than groups, and are equivalent to crossed modules.

This has led to the theory given an exposition in the book Nonabelian Algebraic Topology: filtered spaces, crossed complexes, cubical homotopy groupoids (EMS Tracts in Math vol 15, 2011). The question refers to a CW-, i.e. cell-, decomposition of a manifold: this gives a special case of a filtered space, which could also come from a handlebody decomposition. For a filtered space $X_*$ one can give a homotopical definition of a crossed complex $\Pi X_*$, a kind of chain complex with operators which is nonabelian in dimensions $\leqslant 2$ and includes information on the fundamental groupoid. A key to the main results of the book is using compositions of higher dimensional cubes to prove a theorem of the Seifert-van Kampen type which allow one to compute $\Pi X_*$ in terms of colimits and so to prove that $\Pi X_*$ for a CW-filtration $X_*$ is "free on the cells of $X_*$". To go back to your situation, if all the submanifolds $M$ are allowed a "connected" filtration $M_*$, e.g. from a cell-decomposition, then the same SvK theorem could determine $\Pi X_*$ in terms of the $\Pi M_*$. Interpretation of this, and investigating the necessary conditions, would seem to be a research programme, but has possibilities!

The aim of putting this material together into a book was to enable it more easily to be evaluated, for example as to whether it helped to understand "what homology is talking about", and also to give an exposition of the relation with basic homotopical results, such as the Relative Hurewicz Theorem, without invoking the "formal sums", i.e. $\mathbb Z$-linear combinations, of the standard homology theory. Key insights came from J.H.C. Whitehead's 1949 paper "Combinatorial Homotopy II", particularly a result there on free crossed modules.

The cubical methods are useful for subdivision arguments, and also for discussing homotopies and products.