Applied topology 6: Homology

[@davidhand9721]

Can someone please explain _how_ homology groups are abelian groups? In homotopy, the group operation was very well defined; if you had homotopy < A, B > then AB was a path that went around B then went around A. It was a function that mapped a t parameter to a point in the space. It all made perfect sense at every step. In homology, we just sort of assert that the "group" allows us to put things together in any order, but I struggle to imagine an object or function that is both the same regardless of order _and_ allows any number of path crossings _and_ has an inverse. What _is_ A + B (being 1-simplices, for example) if they don't form a full cycle? What is the difference between that and 2A - C? Maybe I'm just having a math-foreigner freakout. Maybe I'm looking at it as a scientist and I want objects to _mean_ something, and I'm offended that we're just sort of inventing rules. Maybe I'm looking at it like a coder, thinking that if two things are different, they must contain or refer to something different from one another. Either way, it would be _very_ helpful if someone could give me some sense of what we are doing by adding and (scalar) multiplying these little path fragments together. Even just an analogy would make a world of difference to me.

[@aatrn1]

Let me start with an analogy, and then discuss homology with "Z/2Z coefficients" (whatever that means), and then briefly mention homology with Z coefficients at the end (which is what you are asking about most directly). Okay, here's the analogy. If A and B are sets, then (A union B) is the same as (B union A). So the operation of set union is commutative. There's a related operation called "exclusive union", where (A exclusive union B) contains all of the elements of (A union B) that are not in both A and B --- see www.oreilly.com/library/view/sql-and-relational/9781449319724/ch07s01.html. It turns out that we still have (A exclusive union B) = (B exclusive union A). In a very real sense, homology with Z/2Z coefficients forms an abelian group simply because exclusive union is commutative. Now let me discuss homology with Z/2Z coefficients. Intuitively, Z/2Z coefficients means we can ignore signs and orientations, since 1+1=0 in Z/2Z, which means 1 is its own inverse in Z/2Z. Okay, in this context, a cycle is simply a set of simplices with empty boundary. So, a 1-cycle is a set of edges with empty boundary. These 1-cycles generate the 1-dimensional homology group. And, the group operation is exclusive union. So the sum or the concatenation of two 1-cycles is simply the exclusive union of both sets of edges. It doesn't matter in which order you take the exclusive union --- you still get the same exclusive union set --- which again has no boundary. This is why homology is abelian. In your post, you talk about orientations, and you include negative signs. So I can tell you are reading about homology with Z coefficients, instead of homology with Z/2Z coefficients. But, it turns out that even when you include orientations and multiplicities, homology is still abelian for the same reason. When you have Z coefficients, a cycle is a collection of oriented simplices, perhaps counted with multiplicity, that has no boundary. But again, when you take two such cycles A and B and add them, you still get the same multiplicities regardless of whether you do this addition as A+B or as B+A.

[@InfiniteQuest86]

This might be a stupid question, but wouldn't the homology groups only have a vector space structure if you operate over a field? If you operate over integers, wouldn't you be reduced to a Z-module, hence why you have to use rank rather than dimension?

[@HenryAdamsMath]

Great question - and you're 100% right! I'm being very intuitive in this video, and I should have specified that if you take homology with coefficients in a field then you get a vector space, but if you take homology with coefficients in an abelian group, then (as you say) the homology groups are abelian groups but not necessarily vector spaces. And even when you take coefficients in a field, it's only loosely true that the rank of the i-dimensional homology vector space is the "number of i-dimensional holes in that space", since your answer can change with the field of coefficients (as it does for the Klein bottle)!

[@user-tp5ux6mu7n]

Thank you for the great video! Following the linear combination example, is it safe to say that we only count the basis of holes instead of the actual number of holes? For example a tetrahedron skeleton will only have 3 1-d holes instead of 4? If that's the case, then shouldn't the 0-d homology has the rank of n-1 instead of n (if my understanding about the so-called 'connected components' is correct)?

[@HenryAdamsMath]

Thanks for the great question. That's right, the "number of holes" (i.e. the Betti number) is the dimension of the homology group (which is a vector space when coefficients are taken in a field). So as you say, you only count the number of basis elements needed to generate all of the holes. Thus, the 1-skeleton of a tetrahedron only has three 1-dimensional holes. And the 1-skeleton of a cube only has five 1-dimensional holes, as the "sixth" is actually the sum of the other five. Regarding 0-dimensional homology, people actually do it both ways! With "homology", the rank of 0-dimensional homology is the number of connected components. But in many settings it is much more convenient to instead work with "reduced homology", in which case the rank of 0-dimensional reduced homology is one fewer than the number of connected components. See for example en.wikipedia.org/wiki/Reduced_homology and math.stackexchange.com/questions/51142/use-of-reduced-homology and (in particular) Proposition 3.1 of ncatlab.org/nlab/show/reduced+homology. That Proposition 3.1 explains that the rank of 0-dimensional reduced homology is always one fewer than the rank of standard 0-dimensional homology. So, in summary, people definitely use both conventions!

[@TheTessatje123]

Thanks! This is very intuitive 🙂So the homology of a shape will give you the number of holes in different dimensions. But what is homology groups? What does the set and group operation consist of?

[@aatrn1]

Homology groups take a bit of work to define. See en.wikipedia.org/wiki/Homology_(mathematics) for example. I also recommend the book "Basic Topology" by Armstrong. But once you've put in the work to define homology groups, they can be efficiently computed using linear algebra.

[@gumalak4804]

Great video!

[@aatrn1]

Thanks so much!