I had a question regarding the importance of orienting simplices for the construction of the simplicial homology groups. In Hatcher, $\Delta_n(X)$ is defined as a group of n-chains, constructed using the constituent n-simplices of the $\Delta$– complex as basis elements. For this, he orients each of the n-simplices and then defines the corresponding boundary homomorphisms. Can't this be done without orienting the $\Delta$-complex? What role is played by this process of orienting it? Thanks for your answer.
On the role of orientation in Homology
algebraic-topologyhomology-cohomology
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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..
It sounds as if you're confused about the difference between singular/simplicial homology and cellular homology. If you look at the homology chapter of Hatcher's book again, you'll see that these subjects are covered in some detail.
The relevant details are as follows:
The homology of a space is a deep property of that space that is hidden behind a frosted glass wall. That is to say, it should be considered as an extremely fundamental topoogical invariant, but not one which is readily accessible to us. This means that it is difficult to talk about the meaning of the homology groups - there is a nice interpretation in terms of the $n$-dimensional 'holes' in the surface, but that breaks down once we start encountering spaces whose homology groups exhibit non-trivial torsion. This is why we often have to do a lot of messy work in order to get anywhere with homology. Simplicial homology is not a nice piece of mathematics, and it's not obvious why we care about it at all and why we are prepared to invest so much effort studying it. The reason is that it gives us a way to reach through the glass and grasp the homology of the space.
The nice thing is, once we have got a foothold on the homology groups of a few spaces, we can apply various rules again and again to compute many more homology groups, and the whole subject opens up to us.
Pretty soon, we can identify exactly which tools we are using again and again. This leads into the idea of defining a 'homology theory' axiomatically: we just specify that it is a rule that takes a topological space and spits out a sequence of groups in such a way that all our useful tools work. These tools - homotopy invariance, excision, the dimension rule, additivity and the LES of a pair - are known as the Eilenberg-Steenrod axioms.
We then see a reason for computing simplicial homology in the first place - it gives us a model of the Eilenberg-Steenrod axioms. Simplicial homology satisfies the Eilenberg-Steenrod axioms, and so we are guaranteed that A) It is a homotopy invariant and B) we can compute it in a nice way using the other tools. Homotopy invariants that are easy to compute are extremely useful to algebraic topologists!
Where does cellular homology come into this? Well, the other cool thing about homology theories is that they tend to agree. Remember what I said about homology being a deep but somehow 'hidden' feature of a space? Well, simplicial homology is one way to reach into the heart of a space and extract this important invariant. But there are a number of other ways, and it turns out that cellular homology (computing homology from the chain complex of $n$-cells) is one such way. Moreover, cellular homology is much easier to compute than simplicial homology.
The problem is, in order to prove anything about cellular homology, we need to have the theory of homology set up already. So we need some homology theory already set up that satisfies the Eilenberg-Steenrod axioms. That's where simplicial homology (or, more commonly, its cousin singular homology) comes into play. We need to define it first, but once we've used that to set up the theory, we can use it to prove that cellular homology A) satisfies the Eilenberg-Steenrod axioms and B) agrees with our previous definition of homology, and then we can use the (easier) cellular homology to compute homology groups for cell complexes from then on.
Important point (suggested by Mike Miller in the comments): The definition of cellular homology involves picking a choice of cell decomposition for our space. A cell complex may admit many different cell decompositions. How do we know that the cellular homology we get is independent of our choice of decomposition? Why, because we know it is isomorphic to singular homology, which is a topological(/homotopy/weak homotopy) invariant.
Best Answer
Although you do not mention it in your question, it it obvious that you are interested in the simplicial homology of $\Delta$-complexes.
As an $n$-simplex Hatcher understands an ordered $n$-simplex which is an $(n+1)$-tuple $[v_0,\ldots,v_n]$ of vertices $v_i$. This means that an $n$-simplex contains more information than the set $\{v_0,\ldots,v_n\}$ of its vertices - in fact, if we take different orderings of the set of vertices, then this yields different $n$-simplices. Do not confuse this with the concept of an oriented simplex which is usually defined as an equivalence class of ordered simplices, two ordered simplices being equivalent if they originate from each other by an even permutation of their vertrices (i.e. we have $[v_0,\ldots,v_n] \sim [v_{\pi(0)},\ldots,v_{\pi(n)}]$ for each even permutation $\pi$).
The boundary homomorphism $\partial_n : \Delta_n(X) \to \Delta_{n-1}(X)$ is defined on the generators $\sigma^n : [v_0,\ldots,v_n] \to X$ by $$\partial_n(\sigma^n) = \sum_{i=0}^n (-1)^n \sigma^n \mid [v_0,\ldots,\hat{v}_i,\ldots,v_n] .$$ The ordered $(n-1)$-simplices $[v_0,\ldots,\hat{v}_i,\ldots,v_n]$ are the faces of $[v_0,\ldots,v_n]$. More precisely, $[v_0,\ldots,\hat{v}_i,\ldots,v_n]$ is the $i$-th face of $[v_0,\ldots,v_n]$. In the above formula it is essential that we associate the sign $(-1)^i$ to the $i$-th face $[v_0,\ldots,\hat{v}_i,\ldots,v_n]$. Only these signs allow to show that $\partial_{n-1}\partial_n = 0$.
If you work with unordered $n$-simplices, i.e. with the sets $\{v_0,\ldots,v_n\}$, then we obtain of course a set of $n+1$ unordered $(n-1)$-simplices $\{v_0,\ldots,\hat{v}_i,\ldots,v_n\}$ which we may call the faces of $\{v_0,\ldots,v_n\}$, but we do not have any chance to reasonably define the notion of an $i$-th face of the set $\{v_0,\ldots,v_n\}$.
The geometric meaning of orderings of vertices if explained on p. 105.