# On the role of orientation in Homology

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.

• Without an orientation, what is the difference between an element $a \in \Delta_n(X)$ and $-a \in \Delta_n(X)$? Jun 11 '20 at 0:20
• Thanks for your reply. But why is that difference important? Isn't Δ𝑛(𝑋) defined as a formal sum of the chains. So since it's an abstract group, -a and a would be different right? Jun 11 '20 at 0:22
• You're right, I was trying to give intuition about how orientations sort of naturally reflect the inverse of a simplex (and by extension a chain). Orientations give this group nice properties. Let me try and illustrate with a good example. Jun 11 '20 at 0:30
• Okay, ignoring an example, you need this orientation so that $\partial^2 = 0$, if you look at Lemma 2.1 in Hatcher, you will see this there. Your question could obviously now be why do we want $\partial^2 = 0$, but this statement really encapsulates all of the homological algebraic machinery going on. Jun 11 '20 at 0:35
• You can define a homology without regard to orientations. One way is homology with $\mathbb{Z}/2$ coefficients. This often encodes way less information, and so we lose a lot of insight. Jun 11 '20 at 0:36

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 $$\{v_0,\ldots,v_n\}$$.