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Homology groups seem to be formally defined as quotient groups $Q$. I feel it is a bit difficult to connect this formal definition with the 'holes' intuition. Perhaps we can find a group $X$ acting on a hole ($S^1$ or the product space of $S^1$'s) so that the hole moves around and covers the manifold, and then find a homomorphism from X to a number set (e.g. $\mathbb{Z}, \mathbb{Z}\times\mathbb{Z}, {0}$), which constitutes part of a homology group.

Questions:

  1. If so, why we map the group actions to $\mathbb{Z}$ instead of $\mathbb{R}, \mathbb{N}$ (all infinite); actually it seems to me that, for example, for a group $X$ acting on a $S^1$ hole of a torus, $X$ is uncountable, and the quotient group $Q$ should be $\mathbb{R}$.
  2. for a group $X$ acting on a $S^1\times S^1$ hole of a torus, $X$ seems to contain only $I$ (identity), then why the quotient group $Q$ is still $\mathbb{Z}$ (infinite) instead of {0} (one element)?

It's said that 'The homology groups classify the cycles in a cycle group by putting together those cycles in the same class that differ by a boundary.' https://web.cse.ohio-state.edu/~dey.8/course/CTDA/homology.pdf$\quad$ This explains what the quotient group does. Then I am wondering

  1. is cycle group $Z_p(X)$ the same as the module group, or ring, $\mathbb{Z}_p$ of integers (both seem to be p-cyclic groups)? $\quad$

  2. what is the boundary group. In the link it is said that a) $B_p(X)$ is 'the image of the boundary homomorphism' and b) 'all (p − 1)-chains that can be obtained by applying the boundary operator ∂p on p-chains form a subgroup of (p − 1)-chains'. So a) a boundary operator $\partial_p$ (regarded as a homomorphism, a function) maps a $(p+1)$- chain to $p$-chains, and b) all such $p$-chains form a group, which we call a boundary group. But why $\partial_p$ operating on a chain would give a collection of, instead of just one, chain?

  3. https://jeremykun.com/2013/04/10/computing-homology/ seems to suggest $H_k(X) = \frac{Z_k}{B_k} = \frac{\mathrm{ker}\partial_k}{\mathrm{im}\partial_{k+1}}.$ Why the $k$-cyclic/boundary group equals the kernal/image of some boundary operator, respectively?

Charlie Chang
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  • Can you give an example of what you mean by 'the hole moves around and covers the manifold'? – Michael Albanese Mar 01 '22 at 08:36
  • @MichaelAlbanese like this https://mathworld.wolfram.com/CircleBundle.html. The hole could be a circle, and a fiber of circles covers the torus. – Charlie Chang Mar 01 '22 at 08:50
  • something useful: https://math.stackexchange.com/q/40149/577710 'the point is that it takes something very hard (topology) and turns it into something easy (abelian groups).' – Charlie Chang Mar 01 '22 at 09:10
  • this helps: 'the rank of the -th dimensional homology group is the number of -dimensional “holes” the space has. As you stated in your example, for 0, this is counting connected components. Moving to 1, we are counting literal holes.' $\quad$So 1) H0 counts solid pieces (not hole);$\quad$2) what matters is the number of (independent) hole types corresponding (not to the order/size of homology group but) to the rank. So Z*Z ~ 2 hole types, Z ~ 1, {0} ~ zero. It seems not to matter the number n of holes are necessary for covering the manifold. (n=size of the group X acting on the hole.) – Charlie Chang Mar 01 '22 at 09:25
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    My personal position is that the "holes" thing is completely overblown, and should only serve as a very very basic intuition, not as an actual way to understand homology. It's time to tell the truth: homology is not a thing that "counts holes". (It has the consequence that a lot of students get very attached to this "holes" thing and get confused.) – Captain Lama Mar 01 '22 at 10:26
  • @CaptainLama thanks very much. Then how to understand homology group, or calculate the quotient group? Any recommendation of articles, notes, etc.? – Charlie Chang Mar 01 '22 at 10:36
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    Homology groups can also be understood well for groups. This might be helpful for more intuition. Then there is no "excuse" with holes. Start with some easy examples, e.g., homology of cyclic groups – Dietrich Burde Mar 01 '22 at 11:42

1 Answers1

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Homology captures the idea of holes in the following way. It's not that the holes move about the space, but that the chains representing those holes are not unique. As a basic example consider the punctured plane $\mathbb R^2\setminus\{(0,0)\}$. The first homology of this space is $\mathbb Z$, but this is represented by any cycle of edges that surrounds the hole. That is why you need a quotient group so that any two cycles with the same winding number end up being homologous.

Cheerful Parsnip
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