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I'm an absolute beginner to the concepts of algebraic topology and we're only now covering the basics of singular homology in class. I'm horribly confused by a few things, particularly by this example where we compute the homology of a single point set, $X = \{x_0\}$. I'd be incredibly grateful if I could try explain how I see the problem and someone could explain my misapprehensions.

We begin by saying that $C_n(X) = \mathbb{Z}$; this is a freely generated abelian group whose generators are all the possible singular n-simplices that could exist in the space $X$, and these singular n-simplices are themselves are maps that embed the regular simplices in $X$.

Then since only one singular simplex exists per dimension because they're mapping into a singular point we can define: $\sigma_n : \Delta^n \rightarrow \{x_0\}$

Then computing the boundary of each of these:

$\partial_0(\sigma_0) = 0$... Does this holds for all 0-d simplices?

$ \partial_1(\sigma_1) = \sigma_0 - \sigma_0 = 0$ Is this because there is only one permitted 0-simplex in $X$?

$ \partial_2(\sigma_2) = \sigma_1 - \sigma_1 + \sigma_1 = \sigma_1$ Is this because there is only one permitted 1-simplex in $X$? Also, is it possible for $\sigma_2$ to be orientated such that this gives the same result but with a negative sign?

So the chain complex looks like:

$C_2 = \mathbb{Z} \xrightarrow{\partial_2} C_1 = \mathbb{Z} \xrightarrow{\partial_1} C_0 = \mathbb{Z} \xrightarrow{\partial_0} 0$ because there is exactly one generator per $C_n$?

And so, $H_0(X) = \dfrac{\ker(\partial_0)}{Im({\partial_1})} = \mathbb{Z} / \{0\} = \mathbb{Z}$

$H_1(X) = \dfrac{\ker(\partial_1)}{Im({\partial_2})} = \mathbb{Z} / \mathbb{Z} = \{0\} $

Is this right?

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Pretty much everything you've said is correct. Because the boundary of an n-simplex is defined to be a sum of the restrictions of the map to various "faces" of the $n$-simplex, each of which is therefore an $n$-simplex...then if there's only one $0$-simplex, the boundary of the 1-simplex has to be $\sigma_0 \pm \sigma_0$. (And the sign is determined by the definition.)

Regarding $\sigma_2$ and orientation: No, it's not possible to orient it so that it gives a negative sign. The simplex $\sigma_2$ is a particular map from $\Delta^2$ (which I like to think of as $\{(x,y,z) \mid 0 \le x,y,z \le 1, x + y + z = 1 \}$) to the single point (call it $P$) defined by $(x, y, z) \mapsto P$. That's all it is. There's no "orientation" of the map.

It's possible to choose a different generator for $C_2$, namely $-\sigma_2$. But that doesn't change $\sigma_2$ itself.

Otherwise...you've got it exactly right.

Singular homology appears, at first, to be just weird. If you do simplicial homology, everything is nice and finite and manageable, so why go to the set of all possible continuous maps? Well you might ask. One answer is that after you prove a few more things, you'll find that the two techniques yield exactly the same results for simplicial complexes, so it doesn't matter. But that doesn't address "Why make it so complicated???" A good answer to that is "because not everything whose homology you wish to compute is actually as nice as a simplicial complex," so you either have to go through the (complicated, to me at least) simplicial approximation theorem, or you have to just say that this singular homology isn't looking so bad after all. And after you've done a few examples like this one, and the homology of a circle, and a few others, you rapidly learn how to do stuff where you never need to think about the singular simplices themselves, just as when you learn that an ordered pair is defined by $$(a, b) := \{ \{a\}, \{a, b\}\}, $$ it seems really awkward and messy. But then you prove that $(a, b) = (c, d)$ if and only if $a = c$ and $b = d$, and thereafter you use only that lemma, and never look at the formal set-theory definition again.

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  • $\begingroup$ This is a nice answer. +1. A comment: while simplicial approximation in simplicial homology is a hard proof, so is excision in singular homology, so it seems that each version has its pros and cons. $\endgroup$ – Matematleta Oct 12 '19 at 14:46
  • $\begingroup$ You're completely right, and I was perhaps too glib -- excision IS a pain. ;) $\endgroup$ – John Hughes Oct 12 '19 at 17:14
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    $\begingroup$ Regarding the question "Why make it so complicated?", there is a good answer even if the only topological spaces you care about are simplicial complexes. Namely, singular homology is (almost) obviously a topological invariant. But the question of whether simplicial homology is a topological invariant is not at all obvious... until you prove that simplicial homology is isomorphic to singular homology. $\endgroup$ – Lee Mosher Oct 13 '19 at 0:47
  • $\begingroup$ Nice point, Lee. Simplicial homology is (fairly) clearly invariant under simplicial "homeomorphisms" (I assume, being unable to recall the definitions/proof), but at the core of topology are continuous maps, not simplicial ones. $\endgroup$ – John Hughes Oct 13 '19 at 4:09
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It looks to me like you understand this very well. It is no harder to compute $\text{all}$ homology groups of a one-point space $X$, for $n\ge 1:\ H_n(X)=0.$

as you point out, $C_n(X)=\langle \sigma_n\rangle$ the cyclic group generated by any $\sigma_n:\Delta_n\to X.$ Then, with the face maps $\epsilon^n_i:\Delta_{n-1}\to \Delta_n,$ we have $\partial \sigma_n=\sum^n_{i=0}(-1)^n\sigma_n\epsilon^n_i$. But $\sigma_n\epsilon^n_i=\sigma_{n-1}$ (because there is only one simplex in each $C_n(X)$) so we get $\partial \sigma_n=\left(\sum^n_{i=0}(-1)^n\right)\sigma_{n-1}.$ This sum is zero if $n$ is odd and $\sigma_{n-1}$ if $n$ is even. The upshot of this is that $\partial=0$ if $n$ is odd and $\partial$ is an isomorphism if $n$ is even. Conclude by inspecting the sequence $$C_{n+1}(X)\overset{\partial}\to C_{n+1}(X)\overset{\partial}\to C_n(X)$$

and repeating your analysis for $n$ odd and $n$ even.

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