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I am having some difficulties understanding the difference between simplicial and singular homology. I am aware of the fact that they are isomorphic, i.e. the homology groups are in fact the same (and maybe this doesnt't help my intuition), but I am having trouble seeing where in the setup they differ.

To my understanding, the singular chain complex on a space $X$ consists of the free abelian groups generated by the sets of $n$-simplices in X, where an $n$-simplex in this context is a continuous map $\sigma : \Delta^n \to X$ from the standard geometric $n$-simplex $\Delta^n$ to $X$, with boundary map $\partial_n = \sum_{i=0}^n (-1)^i d_i$ where $d_i: C_n (X) \to C_{n-1} (X)$ is the $i$th face map ("deleting" the $i$th vertex). The singular homology groups are then the homology groups of this complex (ie. $H_n(X)=\ker(\partial_n)/\text{im}(\partial_{n+1})$).

Now for the simplicial homology, we have a simplicial complex $S$, which is a set of (abstract?) ordered simplices, such that a face of any simplex in $S$ is itself a simplex in $S$. Then we form the simplicial chain complex where $C_n(S) = \mathbb{Z}[S_n]$, where $S_n \subset S$ is the set of $n$-simplices in $S$, i.e. the free abelian group generated by $S_n$. This complex has boundary operator $\partial_n = \sum_{i=0}^n (-1)^i d_i$, where $d_i$ is the $i$th face map. The homology groups of this is $H^\Delta_n(S) = \ker(\partial_n) / \text{im}(\partial_{n+1})$. Now for this to make any sense in a topological framework, we have the realization of $S$, $|S| = \coprod (S_n \times \Delta^n) / (d_i \sigma, y) \sim (\sigma, d^iy)$ for all $(\sigma, y) \in S_n \times \Delta^{n-1}$ and $d^i$ is the coface map. (As I understand it, $S$ is a blueprint of how to "assemble" the geometric $n$-simplices to form a space). And then of course, if you want to talk about a specific space $X$, you need to find a simplicial complex $S$, whose realization is homeomorphic to $X$.

I can see very well that these two are two very different ways of building up the framework, but what I don't understand is where in practice it differs. Don't they both require that you find a way to divide $X$ into $n$-simplices? The only difference I see, is whether you map from $\Delta^n$ into $X$ before or after you form your homology groups, but there must be something I'm missing...

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Singular homology does NOT require you to divide $X$ into $n$-simplices. In particular, note that in singular homology. The $\sigma: \Delta^n \to X$ needs not be a homeomorphism of any sort - it is only required to be continuous and can be wildly singular. (I think this is the reason for the name "singular" homology). This gives singular homology much advantage over simplicial homology: you don't need to have a simplicial structuer to start with, and you don't need to show that two simplicial structure on the same space gives you same homology. –  user27126 Nov 4 '12 at 18:19
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Here is an observation that may also help clarify things. Note that for any space $X$ you can form the singular simplices $S_*(X)$, which is one of these abstract simplicial complex things that you're referring to. The set $S_n(X)$ is exactly the set of continuous maps $\Delta^n \rightarrow X$. Now, there is an evident map $|S_*(X)| \rightarrow X$, and this is a weak homotopy equivalence; in particular, the simplicial homology of $S_*(X)$ is nothing more or less than the singular homology of $X$. –  Aaron Mazel-Gee Nov 4 '12 at 19:28
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It is also nice to notice that in simplicial homology, your chain groups will be relatively small. In all your favorite examples, each will have finite rank because you only have a finite number of simplices in each dimension. In singular homology, the rank however is the number of maps of a simplex into your space. Because you can send the simplex to any single point you choose, the rank will be at least the number of points in your space, that is to say almost always uncountable! So the singular chain groups are much much larger, but it homology of course it all cancels out. –  Oliver Nov 4 '12 at 19:54
    
Ah, I think I understand. Just to clarify one thing; when in the framework of singular homology do you then (have to) consider all possible maps $\Delta^n \to X$, or do you just choose some covering of $X$? –  user48168 Nov 4 '12 at 21:07
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You consider all possible maps. –  user27126 Nov 5 '12 at 3:19
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1 Answer 1

Just to sum up, mostly for my own reference, but I thought others might find it useful. (I am new to the site, so please excuse me if this shouldn't be an answer...)

First some preliminary notions:

For a topological space $X$, an $n$-simplex in $X$ is a continuous map $\Delta^n \to X$ from the standard geometric $n$-simplex $\Delta^n$ into $X$. The maps $d^i: \Delta^{n-1} \to \Delta^{n}$, sends $\Delta^{n-1}$ to the face of $\Delta^n$ sitting opposite the $i$th vertex of $\Delta^n$.

An ordered $n$-simplex is a partially ordered set $n_+ = \{ 0 < 1 < \cdots < n \}$. The $n+1$ elements of $n_+$ is called the vertices of $\sigma$. The subsets of $n_+$ are called the faces of $\sigma$. There are morphisms of simplices $d^i: (n-1)_+ \to n_+$ called coface maps, given by $d^i((n-1)_+) = \{ 0 < 1 < \dots < î < \cdots < n \}$ omitting the $i$th vertex of $n_+$.

Then for the two homologies:

The singular (unreduced) chain complex on a space $X$, is the chain complex $$\cdots \xrightarrow{\partial_{n+1}} C_n(X) \xrightarrow{\partial_n} C_{n-1}(X) \xrightarrow{\partial_{n-1}} \cdots C_1(X) \xrightarrow{\partial_1} C_0(X) \to 0$$ where $C_n(X)$ is the free abelian group $\mathbb{Z}[S_n(X)]$ generated by the set $S_n(X) = \{ \sigma : \Delta^n \to X \}$ of all $n$-simplices in $X$ (i.e. the set of all continuous maps $\Delta^n \to X$). The boundary maps $\partial_n : C_n(X) \to C_{n-1}(X)$ is given by $\partial_n (\sigma) = \sum_{i=0}^{n}(-1)^i \sigma d^i : \Delta^{n-1} \to \Delta^n \to X$.

The $n$th homology group $H_n(X) = \ker(\partial_n) / \text{im}(\partial_{n+1})$ of this complex is the $n$th singular homology group of $X$.

A simplicial complex $S$ is a set $S = \bigcup_{n=0}^{\infty} S_n$ where $S_n = S(n_+)$ being a set of ordered $n$-simplices, such that a face of any simplex in $S$ is itself a simplex in $S$. The simplicial chain complex $$\cdots \xrightarrow{\partial_{n+1}} C_n(S) \xrightarrow{\partial_n} C_{n-1}(S) \xrightarrow{\partial_{n-1}} \cdots C_1(S) \xrightarrow{\partial_1} C_0(S) \to 0$$ consists of the free abelian groups $C_n(S) = \mathbb{Z}[S_n]$ generated by the $n$-simplices. The boundary map $\partial_n : C_n(S) \to C_{n-1}(S)$ is given by $\partial_n(\sigma) = \sum_{i=0}^n (-1)^i d_i \sigma$ where $d_i = S(d^i) : S_n \to S_{n-1}$ is the face maps $d_i(\sigma) = \sigma \circ d^i$.

The $n$th homology groups of this complex $H^\Delta_n(S) = \ker(\partial_n) / \text{im}(\partial_{n+1})$ is the $n$th simplicial homology group of $S$.

Lastly we have the realization of $S$, $|S| = \coprod (S_n \times \Delta^n) / \left((d_i \sigma, y) \sim (\sigma, d^iy) \right)$ for all $(\sigma, y) \in S_n \times \Delta^{n-1}$, where $d_i \sigma \times \Delta^{n-1}$ is identified with the $i$'th face of $\sigma \times \Delta^n$.

Then if you want to say something about a specific space $X$, you need to find a simplicial complex $S$, whose realization is homeomorphic to $X$ (i.e. you triangulate $X$ and find the homology groups of the resulting simplicial complex).

NOTE: Feel free to edit any mistakes and clarify where you find it necessary. I'm still not 100% comfortable with it yet..

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