Do we distinguish two singular simplices if they have different vertex orders?

We define a $$\textbf{singular n-simplex}$$ in $$X$$ to be a continuous map $$\sigma:\Delta^n\to X$$ where $$\Delta^n$$ is the standard $$n$$-simplex. Now, as an example, Let $$X$$ be a singleton $$\{p\}$$. Then is the number of $$1$$-simplices in $$X$$ one or two? I think it could be one because there is only one continuous map on $$\Delta^1$$ into $$X$$. But since $$\Delta^1$$ has two orders $$[v_0,v_1]$$ and $$[v_1,v_0]$$, I think the answer could be two (namely, $$[v_0,v_1]\to X$$ and $$[v_1,v_0]\to X$$).

• The definition means exactly what it says, so you have already answered your question yourself. Dec 6 '21 at 4:44
• @Eric So is the answer one? I don't know if the information about the vertex order is included in the definition of standard simplices. So I don't know whether I should say that there is just one $\Delta^1$ or two $\Delta^1$s.
– zxcv
Dec 6 '21 at 4:46

The answer is one. There is only one continuous map here, so the answer is one. Actually, both the maps that you are mentioning are the same map.

This is an interesting question.

You got two answers conforming that there is only one continuous map on $$\Delta^1$$ into $$X$$, and of course this is correct. The standard $$n$$-simplex $$\Delta^n$$ is a topological space and nothing else, therefore the continuous maps living on $$\Delta^n$$ do not depend on a vertex ordering.

But if we consider the complete singular complex $$C_*(X) = (C_n(X), \partial_n )$$ of $$X$$, we see that the boundaries $$\partial_n : C_n(X) \to C_{n-1}(X)$$ depend on a particular vertex ordering of $$\Delta^n$$. In fact, the standard $$n$$-simplex is the convex hull of the $$n+1$$ standard basis vectors $$e^{n+1}_0,\ldots, e^{n+1}_n$$ of $$\mathbb R^{n+1}$$. The indices $$i = 0,\ldots,n$$ give a natural vertex ordering of $$\Delta^n$$. This allows to define $$n+1$$ face-emdeddings $$\delta_i : \Delta^{n-1} \to \Delta^n$$, $$i = 0,\ldots,n$$, by mapping $$e^n_j$$ to $$e^{n+1}_j$$ for $$i < j$$ and to $$e^{n+1}_{j+1}$$ for $$j \ge i$$. Then we define $$\partial_n(\sigma) = \sum_{i=0}^n (-1)^i \sigma \delta_i .$$

You see that the face-emdeddings $$\delta_i$$ do depend on the vertex ordering. One could consider "ordered singular simplices" in form of pairs $$(\sigma : \Delta^n \to X, o_n)$$ with a vertex ordering $$o_n$$ of $$\Delta^n$$. This can be described in form of a sequence $$e^{n+1}_{i_0},\ldots, e^{n+1}_{i_n}$$. Any such ordering induces the following vertex ordering $$\partial o_n$$ of $$\Delta^{n-1}$$: Letting $$i_k = n$$, we take $$e^n_{i_0}, \ldots, e^n_{i_{k-1}},e^n_{i_{k+1}},\ldots e^n_{i_n}$$. Moreover, one could define face embeddings $$(\Delta^{n-1},\partial o_n) \to (\Delta^n, o_n)$$ which depend on $$o_n$$. This would produce a variant of the singular complex. I am not going to further explore this construction, but I am sure that the resulting homology groups agree with the usual singular homology groups.

In other words: You get a more complicated construction, but do not have a benefit.

• Thanks. I also thought that the homology groups could be the same however I construct a chain complex. But it is indeed quite troublesome to check that.
– zxcv
Dec 6 '21 at 10:28
• @Paul Frost I think the ordering still doesn't matter in the case OP mentions. The space being mentioned is just a single point. Any map going to it, is just going to be the constant map. However, I agree that we might have to care about how the vertices are being ordered, if the space is anything but the single point set. Dec 6 '21 at 15:38
• @AspiringMathematician You are right, the standard singular simplices have nothing to with a vertex ordering. I wanted to explain that the vertex ordering plays a role for the boundary operators. This is normally overlooked because the $\Delta^n$ have a natural vertex ordering. But we could also work with more general orderings. Dec 6 '21 at 23:10

By definition, the "standard $$n$$-simplex" is one specific topological space (usually defined as $$\{(x_0,\dots,x_n)\in[0,1]^{n+1}:\sum x_i=1\}$$). So, a singular $$n$$-simplex is literally just a map from this one specific topological space to $$X$$. There is no other data involved--we don't pick an ordering of its vertices. (Instead, just as the standard $$n$$-simplex is a specific topological space, it also has a specific canonical ordering of its vertices, given by the ordering of the coordinates in $$[0,1]^{n+1}$$.) So, there is just one singular $$n$$-simplex in a singleton space.