How do homology groups work? Looking at the wikipedia article, it lists, for example, $H_k(S^1) = \mathbb Z$ for $k = 0,1$ and ${0}$ otherwise. It also says that $H_k(X)$ is the k-dimensional holes in $X$. Thus, there is a 0-d hole and a 1-d hole. I see a 2-d hole, but neither of 1-d nor 0-d. This trend continues for the other example listed. What is it that I am completely missing? Thank you.

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    $\begingroup$ The problem is that unless you have a sensible definition of what a «$k$-dimensional hole» is, you are trying to make sense of a statement which does not have one. One way to define the number of $k$-dimensional holes in a space $X$ i s by declaring that it is the rank of $H_k(X)$... $\endgroup$ – Mariano Suárez-Álvarez Jul 5 '14 at 5:02
  • $\begingroup$ I'm sorry. It really bothers me when people do something careless like that. $\endgroup$ – Thoth19 Jul 5 '14 at 5:23
  • $\begingroup$ Well, what bother you is math, then, because this is how much of math works. The day you see the definition of the category of coherent sheaves on a non-commutative projective variety, you are going to have a fit! In any case, there is absolutely nothing careless about it. $\endgroup$ – Mariano Suárez-Álvarez Jul 5 '14 at 5:41
  • $\begingroup$ I meant talk about something when it matters how it is defined in particular to that problem. Like oh so many questions about proving something about the nature of $e$. $\endgroup$ – Thoth19 Jul 5 '14 at 5:55
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    $\begingroup$ Well, how something is defined is of the utmost importance, and even more important is to have a definition. In the questioon, you are asking how homology counts/measures holes, but to have an answer you need a definition of what a hole is. Now that is a rather difficult problem! Homology and homotopy were invented, in fact, precisely to provide a precise definition. $\endgroup$ – Mariano Suárez-Álvarez Jul 5 '14 at 5:58

Actually, homology groups of a k-dimensional space do measure holes, but for dimensions $1,2,..,k-1$. For dimension zero, they measure the number of connected components , and for the top homology, they measure whether the space is orientable or not. A hole is informally defined as an obstruction to shrinking an n-dimensional object within the space, so that a 1-d hole in a topological space of dimension larger than one an obstruction to shrinking a curve within the space into a point, though ( see comments) you may consider a non-zero top -dimensional homology as a hole .

Notice that, algebraically, we define a hole to be a cycle that does not bound, i.e., we say that the homology is non-trivial , or that there is an n-hole if the quotient $Z_n/B_n \neq {id}$. If you look, e.g., at the case of a 2-torus $T^2:= S^1 \times S^1$ , you will see that, e.g., a meridian is a cycle that does not bound, because its removal will not disconnect the space. Similarly for any strictly latitudinal curve. These two cycles (simple-closed curves in the space) generate the homology of the torus.

An additional property of homology groups is that EDIT: this applies to most spaces you will run into unless you do specialized work ( see Mariano's comment below) , for a space $X$ of topological dimension: http://en.wikipedia.org/wiki/Topological_dimension greater than $k$, we have that $H_k(X) =0$. Notice this is not true for some of the groups $\pi_n(X)$.

  • $\begingroup$ Your last statement only works for appropriate definitions of dimension. There are examples of subsets of the plane with non-zero homology groups in infinitely many dimensions, for example, due to Barratt and Milnor (see ams.org/journals/proc/1962-013-02/S0002-9939-1962-0137110-9) $\endgroup$ – Mariano Suárez-Álvarez Jul 5 '14 at 5:05
  • $\begingroup$ @MarianoSuárez-Alvarez: I just edited. Would that be topological dimension? $\endgroup$ – user99680 Jul 5 '14 at 5:07
  • $\begingroup$ In the top dimension, homology also measures the nymber of holes, really. A closed connected $n$-manifold, say, has one $n$-hole if it is orientable and zero if not, by definition of hole if you want (and this is just as much cheating as defining holes for lower $k$ in the way you did!) $\endgroup$ – Mariano Suárez-Álvarez Jul 5 '14 at 5:09
  • $\begingroup$ So, to clarify, the point(or at least one of the points) of homologies is to talk about holes in a rigorous way. So in the case of a Torus ($S^1 x S^1$) we have Z for 0-d because it is connected, and we have ZxZ for 1-d because there are circles that we can kinda "shove in there" that would prevent us from squeezing the shape in. Where does the 2-d hole come from (as wikipedia mentions)? Is it the "munchkin," part of the torus? $\endgroup$ – Thoth19 Jul 5 '14 at 5:26
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    $\begingroup$ One can never tell what nature is going to throw at you! One better be at least prepared :-) $\endgroup$ – Mariano Suárez-Álvarez Jul 5 '14 at 6:00

The $k$-th homology group measures how many $k$-dimensional subthings in $X$ are there which are not the boundary of $(k+1)$-dimensional subthings of $X$. I am not being precise in what a thing is in this, because that is difficult and, in a sense, the precise definition of a thing is what the actual definition of homology encodes.

One ideal realization of this would be the following: if $X$ is a space, we could define $H_k$ te the set of maps $f:M\to X$ from a $k$-dimensional manifold $M$ to $X$ which are not the restriction of a map $g:N\to X$ from a $(k+1)$-dimensional manifold with boundary $N$ to $X$ such that $\partial N=M$ and $g|_M=f$. Now this definition relfects the idea in the first paragraph but does not have the technical properties one needs to do anything sensible. So one has to adapt.

This idea is good enough to see why a torus $T$ has a $2$-whole: the identity map $id:T\to T$ does not extend to any $3$-manifold $N$ with boundary $\partial N=T$. Of course, how do you prove this? Well, the great minds of the people that developed homology theory devised the actual definition so as to be able to prove this!

Later. The above idea does not work. One way to improve it is the following. Let $Z_k(X)$ be the free abelian group with basis the set of all maps $f:M\to X$ defined on a $k$-dimensional orientable and oriented manifold $M$ (we call such maps $k$-cycles in $X$) and now declare two such $k$-cycles $f_1:M_1\to X$ and $f_2:M_2\to X$ to be equivalent (homologous is the usual term) if there is a map $g:N\to X$ from a $(k+1)$-dimensional manifold $N$ with boundary, such that the boundary $\partial N$ is the disjoint union of $M_1$ and $M_2$, and such that the restriction of $g$ to $\partial N$ coincides with $f_1$ and $f_2$ (I am not giving all details here...) Let $H_k(X)$ now be the quotint of $Z_k(X)$ by that relation. This is a much better approximation to the actual homology (it is even equal to it in some cases, iirc)

  • $\begingroup$ This idea is of course similar one of the ways homotopy groups can be defined: the $k$th homotopy group of a space $X$ is essentially the set of maps $S^k\to X$ defined on the $k$-dimensional sphere which do not extend to a map $B^{k+1}\to X$ on the whole $(k+1)$-ball (which contains $S^k$ as its boundary) $\endgroup$ – Mariano Suárez-Álvarez Jul 5 '14 at 5:53
  • $\begingroup$ So you see it a as a matter of detecting cobordisms within a space? $\endgroup$ – user99680 Jul 8 '14 at 7:32
  • $\begingroup$ Well, that is precisely what the equivalence relation of two cycles being homologous means! (for an appropriate interpretation of what «a cobordism within a space» means) $\endgroup$ – Mariano Suárez-Álvarez Jul 8 '14 at 7:44
  • $\begingroup$ No, I agree, I just never formally saw it that way. $\endgroup$ – user99680 Jul 8 '14 at 7:47

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