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I am studying homology groups and I am looking to try and develop, if possible, a little more intuition about what they actually mean. I've only been studying homology for a short while, so if possible I would prefer it if this could be kept relatively simple, but I imagine it is entirely possible there is no real answer to my query anyway.

As I said above, I want to gain a little deeper understanding of what the n-th homology group actually means: I can happily calculate away using Mayer-Vietoris but it doesn't really give me a great deal of intuition about what the n-th homology group actually means. For example, with homotopy groups, the fundamental group is in some sense a description of how loops behave on the object in question, and it is obvious to me why that is what it is for say, the torus or the circle. However, I have no idea what, if anything, I am actually saying about a triangulable object when I talk about it having 0-th homology group this or 1st homology group that.

The best I have been able to find online or in my limited book selection is the brief description "intuitively, the zeroth homology group counts how many disjoint pieces make up the shape and gives that many copies of $\Bbb Z$, while the other homology groups count different types of holes". What 'different types of holes' are there, roughly speaking? I appreciate that it may often be completely non-obvious what the low-order homology groups are for some complicated construction, but perhaps in simpler examples it might be more explicable. Are there (simple) cases where I could say, just from looking something like e.g. the torus, what its zero-th or first or second etc. homology group was based on the nature of the object? I guess in the zero-th case it is, as my source (http://teamikaria.com/hddb/wiki/Homology_groups) above says, related to the number of disjoint pieces. Can we delve deeper than this for the other homology groups?

Any book/website suggestions would be welcomed (preferably websites as I am nowhere near a library!) - I have Hatcher but not a great deal else, and I haven't gleaned as much as I wish to from that alone. Of course I know that there is a great deal we don't know about homology groups even today, so I don't expect some magical all-encompassing answer, but any thoughts you could provide would be appreciated. I hope this question is appropriate for SE Mathematics, apologies if not! -M

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    $\begingroup$ In addition to the other answers you receive that will address this more directly, you might be interested in reading about Hurewicz's Theorem, which relates homotopy and homology groups. In the case of first dimensional groups, it roughly says that the 1st dimensional homology group is what you get when you take the 1st dimensional homotopy group but don't care about the order that the paths go around a hole in; essentially forgetting that paths are oriented and just thinking about them as images of circles. Also, your intuition would probably be aided by reading about simplicial homology. $\endgroup$ – matt May 19 '11 at 20:48
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    $\begingroup$ this is no answer - but I think "cycles modulo boundaries" does provide some intuition $\endgroup$ – user8268 May 19 '11 at 21:05
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    $\begingroup$ I understand your desire to develop intuition for this, and you should pursue that to an extent, but here's another perspective. On some level, it doesn't matter what homology measures. The point is that it takes something very hard (topology) and turns it into something easy (abelian groups). If someone "hands you" two topological spaces, you basically have a useless pile of garbage. Great now you know what all the open sets are... How would you ever tell two such things apart? The answer is homology. It's easy to compute and often lets you answer a hard question with relative ease. $\endgroup$ – D Wiggles Nov 13 '16 at 6:18
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Let's restrict ourselves to orientable spaces that are homotopic to CW complexes. In low dimensions, there is a very intuitive way to think of homology groups. Basically, the rank of the $n$-th dimensional homology group is the number of $n$-dimensional “holes” the space has. As you stated in your example, for $H_0$, this is counting connected components. Moving to $H_1$, we are counting literal holes. The torus has $H_1\cong \mathbb Z \oplus \mathbb Z$ since it has two holes, one inside and one outside.

You can think of a 2-dimensional hole as an empty volume. The best analogy I’ve heard is to think of your space as an inflatable object. The rank of the second homology group is the number of different plugs you’d need to blow air into to inflate it. The torus has one empty volume, so you’d only need one plug to inflate it. If you take the wedge of two 2-spheres, you’d need two different plug to inflate it, one for each empty volume, so it has rank 2.

As is usual in topology, we now wave our hands and say “it works the same for higher dimensions.”

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    $\begingroup$ I don't completely follow your comment about the 'hose': I understand how you're describing it but I don't see quite follow your explanation of a 'void': are you saying the reason you would need two 'hoses' for the wedge of 2 2-spheres is because it has two 'components' only joined at a single point? I don't quite grasp the idea of the 2-dimensional hole I don't think, sorry. $\endgroup$ – Spyam May 19 '11 at 21:11
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    $\begingroup$ See my edits. Instead of voids, empty volumes, instead of hoses, plugs that you'd pump air into. Does that help? $\endgroup$ – Josh May 19 '11 at 21:25
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    $\begingroup$ +1 Great analogy with the plugs to blow air into -- I'm going to find that a really useful check come exam time! $\endgroup$ – Sputnik May 19 '11 at 21:33
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    $\begingroup$ @mathmo6: Everything I wrote is really just an explication of user8268's comment. A 1-hole, in the sense above, is just a 1-cycle that isn't the boundary higher dimensional object. So it generates a homology class. It turns out to be very visual for low dimensional and nice enough spaces. $\endgroup$ – Josh May 19 '11 at 23:53
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    $\begingroup$ So the homology group has one copy of $\mathbb{Z}$ for each hole. Is there a way that this explanation extends to spaces with finite homology groups? I'm specifically thinking about $H_1(\mathbb{R}P^2) = \mathbb{Z}_2$. $\endgroup$ – Mike Pierce Mar 9 '16 at 15:39
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"Cycles modulo boundaries" can get you surprisingly far. Recall that for a simplicial complex $X$, the $n^{th}$ homology $H_n(X)$ is $Z_n/B_n$, where $Z_n = \ker(d_n : C_n \to C_{n-1})$ is the group of cycles and $B_n = \text{im}(d_{n+1} : C_{n+1} \to C_n)$ is the group of boundaries. Some low-dimensional cases:

  • When $n = 0$, a cycle is a linear combination of $0$-simplices in $X$, and a boundary is a linear combination of $0$-simplices lying in the same connected component of $X$ such that the sum of their coefficients is zero. So $Z_0/B_0$ is precisely the free abelian group on the connected components of $X$.
  • When $n = 1$, a cycle is exactly what it sounds like: a linear combination of cycles in $X$ (closed paths made out of $1$-simplices). A boundary is also exactly what it sounds like: a linear combination of cycles in $X$ that bound $2$-simplices. So $Z_1/B_1$ describes the failure of $1$-cycles in $X$ to bound $2$-simplices (which is the precise sense in which it measures "$1$-dimensional holes").

For $n \ge 2$ I have trouble concisely describing what a cycle is beyond "a linear combination of $n$-simplices with zero boundary." I believe this is roughly like a linear combination of collections of $n$-simplices which together form an $n$-sphere in $X$, at least for sufficiently nice triangulations. A boundary is a linear combination of such things which bound $n+1$-simplices. So again $Z_n/B_n$ measures the failure of $n$-cycles to bound $n+1$-simplices, which is the precise sense in which it measures "$n$-dimensional holes."

$H_0$ and $H_1$ are probably easier to identify than the others in general, since connected components are intuitive and $H_1$ is just the abelianization of the fundamental group. If $X$ is a connected $n$-manifold, then $H_n \cong \mathbb{Z}$ if $X$ is orientable and $0$ otherwise, the idea being that an $n$-cycle has to involve all of the $n$-simplices in $X$ appropriately oriented so that their boundaries cancel, and such a linear combination is unique up to scalar multiplication and equivalent to providing an orientation for $X$. And if $X$ is a compact orientable $n$-manifold, then Poincaré duality indirectly relates $H_1$ to $H_{n-1}$.

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  • $\begingroup$ Very nice answer. $\endgroup$ – goblin Apr 22 '17 at 17:23
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To a certain extent, there is no great answer that works in complete generality, or at least in a way where you can have a reasonable intuition about it, enough to see at a glance what the homology of high dimensional spaces will be. Part of this is because of the difficulty of picturing higher dimensional spaces. However, we can get a glimpse of what the homology records by considering some examples.

First, understanding the finer details of homology is hard, so we will look at a simpler invariant, the betti numbers $b_i=\dim_{\mathbb{Q}} H_i(X;\mathbb{Q})$. This disregards all sorts of subtle information which is safe to ignore for starting to build your intuition.

$b_0$ is the number of connected components. $H_1(X)$ is the abelianization of $\pi_1(X)$, and so $b_1$ is the number of "independent loops" in $X$, or the number of holes . The best example of this looking at surfaces. Take the $3$-dimensional sphere, and attach $g$ handles to it. Now, take the surface which is the boundary of this three dimensional object. For each handle, we have a loop that goes around the handle (in the way you might pick up a coffee mug), and a loop that goes from the body, into the handle and back again. For this, imagine that inside the $3$-dimensional space, you took a rubber band that went through the handle, and you shrunk the band until it was taut against the surface.

So we have $b_0=1$, $b_1=2g$, and $b_2=1$ (In general, with a compact orientable $n$-manifold, $b_n=1$). However, instead of viewing the surface as coming from attaching handles to the $3$-sphere, we can view it as punching $g$ holes through the sphere. The fact that $b_1=2g$ and not $g$ might be a little disconcerting, as we have $g$ holes. This is remedied somewhat by the fact that the handle-body (the object before we took the surface) is homotopy equivalent to a wedge of $g$ circles, and so it does have $b_1=g$. However, this shows how subtle it is to say that $b_1$ is "the number of holes", as it makes it difficult to say what exactly a hole is.

To get a grasp about what's going on in the other directly, let's consider $\mathbb{R}^n\setminus \{0\}$. This is putting a hole in the space. Let's try to detect the hole by using spheres. We can't encase the hole inside of a $S^k$ with $k<n-1$, because there is enough ambient space that we can move the sphere past the hole. However, we can encase it inside of an $n-1$ sphere. You can build some more intuition about detecting holes with spheres by looking at $\mathbb{R}^n\setminus \mathbb{R}^m$, and thinking about when we can encase the "hole" inside a sphere.

Unfortunately, this has a lot more to do with higher homotopy groups. There are relations between homology and homotopy groups, but the connection is subtle. For example, while the Hurewicz isomorphism theorem gives a close connection in one dimension (for any particular space), there is a Hurewicz map in all dimensions, but it is not in general easy to understand what it does.

One good way of understanding homology of CW complexes is with cellular homology. In general, this is still hard to understand, but if you have a buffer between the dimensions of the spheres used to construct the space (e.g. complex projective spaces, which are built up out of even-dimensional spheres), homology is counting the spheres used to construct the space.

Beyond these things, I can only recommend that you let your intuition be built up from examples. Sadly, there is no magic bullet that will give you a great understanding.

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Begin with dimension $1$, as we can use the motivation behind homotopy groups. So, we have $1$-dimensional loops, and we count the number of 'distinct' loops which wind round the holes, if any. The circle has one loop going round itself $n$ times, so the homology group of dimension one is $\Bbb Z$. With the disc, all loops can be shrunk to a point, so the homology group is the trivial group. Same with the sphere. A torus however has $2$ 'distinct' (neither can be continuously transformed into the other) loop types, one going along, and one across it, so we have $\Bbb Z \times \Bbb Z$.

Moving on to $2$ dimensions; what corresponds to a $1$-dimensional loop, i.e. a $2$-dimensional loop, is a surface! (Rotate a circle around its diameter).So, just like the circle had $\Bbb Z$ as its $1$-dimensional homology group, the sphere has $\Bbb Z$ for its $2$-dimensional homology group, because you can wrap the surface of a sphere with a surface (cling film?) $n$ times, which may not be shrunk down to a point without straying off the surface of the sphere. $H_2$ takes care of surfaces in $3$ dimensions, and that is the furthest one can go in our puny human brains. Beyond that, we have to let the algebra take over. Let loose all the generators on to the surface, and see what information they bring back!

As for the $0$-th dimension, $H_0$ just reflects the connectedness of the surface. For a simply-connected one it is $\Bbb Z$. Don't ask me why, I can't picture the $0$-th dimension either.

If there is a twist involved, like with the Möbius strip, or the Klein bottle, we need $Z_2$ to generate the twist, so that appears in the sequence of groups.

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    $\begingroup$ I like this answer -- I think it would help to mention though that for dimension two it has to be a closed surface (i.e. a boundary-less surface, boundary in the sense of geometry and not homology, although the two uses coincide here I think). At least I think that is true, it is suggested in Massey, although being a novice to the subject I am not entirely certain. $\endgroup$ – Chill2Macht Nov 12 '16 at 18:32
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The questions of what should be homology theory are quite ancient, with apparently the older papers on Betti numbers and torsion coefficients quite vague about what were cycles and boundaries. See Dieudonné's book on the history of algebraic topology. Poincaré made a new start with chains as formal sums of oriented simplices and this has developed ever since particularly with singular homology. For a discussion of the history of ordered and oriented homology, see this paper. However students naturally feel difficulty with thinking about what really is a cycle or boundary.

My paper Modelling and Computing Homotopy Types: I starts by discussing anomalies in algebraic topology at the border between homology and homotopy. It goes on to discuss the approach of the book Nonabelian Algebraic Topology: EMS 2011, which gives an approach based on homotopy theory, without any simplicial approximation, or assuming singular homology theory. See the previous link for reviews and a pdf. The story is quite complicated, but Part I discusses history, intuitions, restricts to results in dimensions 1 and 2 and includes nonabelian calculations of second relative homotopy groups, not available otherwise

This direct and homotopical approach to a generalisation of cellular homology requires new ideas and proofs, involving strict higher homotopy groupoids defined in terms of homotopy classes of certain mappings of cubes into filtered space.

Just to detail some key ingredients: In this theory, an $n$-chain on a filtered space $X_*$ is for $n=1$ an element of $\pi_1(X_1,X_0)$ and of $\pi_n(X_n, X_{n-1},x)$ for some $x \in X_0$ if $n >1$ (and a $0$-chain is a point of $X_0$). So for $n>2$ a chain is represented by a map $(E^n, S^{n-1}, 1) \to (X_n,X_{n-1}, X_0)$ and this chain is a cycle if the restriction to $(S^{n-1}, 1) \to (X_{n-1}, X_0)$ is nullhomotopic. These are some of the ingredients of $\Pi(X_*)$, the fundamental crossed complex of the filtered space $X_*$. So $\Pi$ is homotopically defined.

To get to explicit calculations, we need a Seifert-van Kampen Theorem for $\Pi$, which is proved using cubical groupoid methods; these allow for "algebraic inverses to subdivision" in all dimensions. As one application, the Relative Hurewicz Theorem is deduced from this theorem.

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  • $\begingroup$ The third link does not seem to be working. :( $\endgroup$ – byk7 Apr 28 at 20:46
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    $\begingroup$ @byk7 Thaanks. Link now fixed. $\endgroup$ – Ronnie Brown May 6 at 14:24
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We can read some geometric intuition about homology groups directly from the definition of the singular homology groups. We may also have a look at cellular homology to see what we can extract.

Singular homology. Let $X$ be a space, and let $S:=S_\cdot X$ be its singular chain complex with differential (or boundary map) $d_\cdot$. We call $Z_n(S):=\text{ker}d_n$ the $n$-cycles of $S$, and $B_n(S):=\text{im}d_{n+1}$ the $n$-boundaries of $S$. A pair $f,g$ of $n$-chains of $S$ are homologous if their difference $f-g$ is a boundary. In other words, if there exists an $n+1$-chain $h$ such that $dh=f-g$; we really do mean they are a boundary!

This is analogous to homotopy. A pair $f,g:X\to Y$ are homotopic if there exists a map $h:X\times I\to Y$ such that

enter image description here

commutes (where $i_0$ and $i_1$ are the inclusion maps at $0$ and $1$ respectively). In other words; two maps $f,g:X\to Y$ are homotopic if they are restrictions to the boundary of a full map $h:X\to Y$.

Intuitively, homology finds $n$-holes, and indeed different types of $n$-holes. A way to think about it is that homology finds holes by keeping track of which $n$-chains are the boundary of an $(n+1)$-chain. Indeed, if you cannot fill in your $n$-chain there must be something in your space which is missing.

A thought experiment that helped me: We may try consider various spaces, and try to construct a chain or two which are not the boundary of a higher chain. What can we for instance say about a chain realizing the circle centered at $0$, in the plane punctured at $0$? What happens of we have two such chains? Are they homologous and how?

Cellular homology. We can probably read even stronger geometric intuition from cellular homology. Indeed if $X$ is a CW complex, then it has a filtration $X_0\subset \cdots \subset X$, and it is natural to think that the homology of the $X_i$ together with the data of the inclusion maps (how $X$ is glued together) should determine the homology of $X$.

This is indeed the case, and in fact in a very elegant fashion: The space $X_i/X_{i-1}$ is a wedge of $i$-spheres and as such it has free abelian homology concentrated at $i$ (precisely, if $k$ is the number of $i$-spheres of $X_i/X_{i-1}$, then $\tilde H_j(X_i/X_{i-1})=\mathbb{Z}^k$ for $i=j$ and $0$ otherwise). Using this and the data of the filtration of $X$ to its skeleton, we may tie the groups together in a long exact sequence which produces homology data based on the degrees of gluing maps (in other words, data about how $X$ is glued together from spheres and disks).

This reminds me that I should read more about cellular homology; I'll come back to this answer!

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  • $\begingroup$ Do you mean a full map $h: X \times I \to Y$? Anyway I like this answer, although I don't see yet the claimed analogy to homotopy from what is written (I am also a novice to the subject). $\endgroup$ – Chill2Macht Nov 12 '16 at 18:26
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The best explanation I know is in Hatcher's "Algebraic Topology", pages 108-109.

One starts (as it happened historically) with separating vs. nonseparating compact hypersurfaces in compact connected manifolds. The nonseparating hypersurfaces are homologically nontrivial while separating ones are trivial. Then one realizes that a separating hypersurface is the one which bounds a codimension 0 submanifold. Then, one can attempt to define $k$-cycles in a topological space $X$ as $k$-dimensional embedded (oriented) submanifolds $C\subset X$ where a cycle is trivial if it bounds in $X$ an embedded oriented $k+1$-dimensional submanifold $W\subset X$: $C=\partial W$ and the orientation on $C$ is induced from the one on $W$. This concept is quite geometric and intuitive.

Sadly, dealing with embedded objects in $X$ is unsatisfactory for various reasons which are discussed in great detail here, although in many interesting situations, they suffice. (And they suffice for the intuition of homology.)

The next attempt then is to consider (continuous) maps to $X$ from compact manifolds and compact manifolds with boundary which are not necessarily embeddings. This leads to an interesting and fruitful concept of the (oriented) "bordism groups" of $X$. Here one looks at continuous maps $f: C\to X$ (with $C$ a closed oriented, possibly disconnected, $k$-dimensional manifold). Tentatively, call such map a $k$-cycle in $X$. Tentatively, call such a cycle trivial if there exists a compact oriented $k+1$-dimensional manifold $W$ with $\partial W=C$ and $f$ extending to a continuous map $f: W\to X$. More generally, two $k$-cycles are equivalent, $f_1: C_1\to X$ is equivalent to $f_2: C_2\to X$ if there exists a compact oriented $k+1$-diimensional manifold $W$ with $\partial W= C_1\cup C_2$ (the orientation of $W$ should induce the orientation of $C_1$ and the opposite of the orientation of $C_2$) such that $f_1\sqcup f_2$ extends to a map $f: W\to X$. One can convert this to a group $O_k(X)$ by taking as the sum the disjoint union and $-(f: C\to X)=f: (-C)\to X$, where $-C$ is $C$ with orientation reversed.

This is all fine and well and geometric, but is not the (ordinary) homology theory since $O_4(point)\ne 0$, while $H_4(point)=0$.

In order to recover the ordinary homology, one needs to relax the notion of manifolds (manifolds with boundary). Instead of manifolds one settles for pseudo-manifolds. One can think of pseudomanifolds as manifolds with singularities where singularities occur at a codimension 2 subset. To be more precise, take a finite $k$-dimensional simplicial complex which has the property that every simplex of dimension $k-1$ is the common boundary of at most two and at least one $k$-dimensional simplices. The result is a $k$-dimensional compact pseudomanifold with boundary. Its boundary is the union of all $k-1$-dimensional simplices each of which is the boundary of exactly one $k$-dimensional simplex. As an example, think of a compact triangulated surface where two distinct vertices are glued together. One then defines an oriented pseudomanifold by requiring all $k$-dimensional simplices to be oriented so that if $F$ is a common codimension 1 face of two distinct $k$-simplices then these simplices induce opposite orientations on $F$.

Thus, the complement to the $k-2$-dimensional skeleton in an oriented pseudomanifold is an oriented manifold (possibly with boundary). All the non-manifold points belong to the $k-2$-dimensional skeleton.

With these substitutes of manifolds, one can now describe homology geometrically:

$k$-cycles in $X$ are continuous maps $f: C\to X$ from oriented compact $k$-dimensional pseudomanifolds without boundary. The negative of a $k$-cycle is obtained by reversing the orientation of $C$ (reverse orientations of all its top-dimensional simplices). The sum of two cycles can be taken to be the disjoint union. For instance, in this setting, $$ 2(f: C\to X)= (f: C\to X) \sqcup (f: C\to X). $$ The zero $k$-cycle is understood to be the map of an empty set. Two cycles are homologous: $$ [f_1: C_1\to X]= [f_2: C_2\to X] $$ if there exists an oriented $k+1$-dimensional pseudomanifold $W$ with $\partial W= C_1\sqcup (-C_2)$ and an extension $f: W\to X$ of the map $f_1\sqcup f_2$. A cycle $f_1: C_1\to X$ is a boundary if it is homologous to the zero cycle, i.e. if there exists $W$ as above with $\partial W=C_1$.

To see how to identify this with the usual singular homology, read Hatcher. For instance, every compact connected oriented $k$-dimensional pseudomanifold without boundary $M$ has its "fundamental class", i.e. the sum of its top-dimensional oriented simplices $$ \sum_i \Delta_i^k, $$ understood as an element of $C_k^{sim}(M)$ (the simplicial chain complex). Then for every continuous map $f: M\to X$, we obtain the associated element of $Z^{sing}_k(X)$ (a singular cycle) by taking $$ \sum_i (f: \Delta_i\to X). $$

See also M.Kreck, Differential Algebraic Topology: From Stratifolds to Exotic Spheres for a development of the homology theory from this point of view.

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    $\begingroup$ (In case it helps other readers, $O_i(point) = 0$ for $i\leq 3$ because all orientable $i$-manifolds bound an $i+1$ manifold for $i\leq 3$. On the other hand, there are $4$-manifolds which do not bound any $5$ manifold, e.g. $\mathbb{C}P^2$.) $\endgroup$ – Jason DeVito Jan 10 at 21:07
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People make this so complicated!

Consider a smooth surface $S$ and a closed differential $1$-form $\omega = f \, dz$.

If you have a curve $C \subset M$ you can take the integral $\int_C \omega$ and get a number. Stokes's theorem says that if you deform the curve $C$, the value of this integral doesn't change. So instead of thinking of the integral over a curve, you can think of taking the integral over an equivalence class of curves.

Moreover, the same holds for the differential form: instead of integrating a particular form you can integrate an equivalence class of forms over an equivalence class of curves, i.e. you can talk about the number $\int_{[C]} [\omega]$.

Of course this works in arbitrary dimensions. de Rham's theorem establishes that the pairing you get is in fact perfect, a fact known in topology as Poincare duality. (Poincare, I'm told, had no idea what was going on with Poincare duality -- he just observed that something was going on with Betti numbers and came up with some incomprehensible "proof" of it.)

Of course this is only good in the differentiable setting. But the definition of homology goes through in the setting of algebraic topology, and then you can simply define cohomology using the duality properties you expect it to have, which turns out to work pretty well.

(What's the point of doing that, rather than just sticking with homology? Well, it gives you a generalization of the wedge product on differential forms, and there's no nice version of that for homology.)

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