Simplicial map of the circle $S^1$

Question

I want to construct a simplicial map $S^1 \to S^1$ of degree $n>0$ by giving a simplicial structure on $S^1$ with minimal number of vertices. (Here, the simplicial structure for the domain $S^1$ and the target $S^1$ need not be the same.) Since a face of a simplcial complex is uniquely determined by its vertices, a simplicial structure of $S^1$ must have at least $3$ vertices. Thus for $n=1$, the identity map will be an answer.

More generally, if we give the simplicial structure on the domain $S^1$ having $3n$ vertices (more precisely, the $3n$-th roots of unity as vertices) and give the simplicial structure on the target $S^1$ having $3$ vertices, then the map $S^1\to S^1$, $z\mapsto z^n$ is a simplicial map of degree $n$. My guess is that this map should be a simplicial map of degree $n$ with minimal vertices.

However, I can't find a way to prove that the map above is a simplicial map of degree $n$ with minimal number of vertices. How can we show that for a simplicial map $S^1\to S^1$ of degree $n$, there must be at least in total $3n+3$ vertices for the domain $S^1$ and the target $S^1$?

Maybe there should be a better lower bound if my guess is wrong, but still I can't think of an idea to prove that such a lower bound would be the best that we can have.

Answer 1

Let $p : \mathbb R \to S^1, p(t) = e^{2\pi i t}$, be the standard covering map. Any triangulation $\tau$ of $S^1$ (which of course has a minimum of $3$ vertices) induces a unique (infinite) triangulation $\bar \tau$ of $\mathbb R$ such that $p$ becomes simplicial. If $\tau$ has $k$ vertices $v_1, \ldots, v_k$ which we may assume to be arranged counterclockwise on $S^1$ starting with $v_1$ and $w_1$ is a vertex of $\bar \tau$ such that $p(w_1) = v_1$, then we have $k+1$ vertices $w_1,\ldots,w_{k+1}$ of $\bar \tau$ such that $w_1 < w_2 < \ldots < w_k < w_{k+1} = w_1 + 1$ such that $p(w_i) = v_i$ for $i \le k$.

Let $\tau, \tau'$ be triangulations of $S^1$ and $f : (S^1,\tau) \to (S^1,\tau')$ be a simplicial map, where $\tau'$ has $l$ vertices $v'_1, \ldots,v'_l$ and we may assume that $f(v_1) = v'_1$. It lifts to a simplicial map $\bar f : (\mathbb R, \bar \tau) \to (\mathbb R, \bar \tau')$. We may assume that $\bar f(w_1) = w'_1$. Then $\bar f(w_{k+1}) = w'_1 + n$ for some $n \in \mathbb Z$ because $p\bar f(w_{k+1}) = f p(w_{k+1}) = f(v_1) = f p(w_1) = p \bar f (w_1)$.

It is well-known that the degree of $f$ is the number $\bar f(t+1) - f(t)$, where $t \in \mathbb R$ is arbitrary. Thus $\deg(f) = \bar f(w_{k+1}) - \bar f(w_1) = w'_1 + n - w'_1 = n$.

Now assume that $n > 0$. The line segment from $w'_1$ to $w'_1 + n$ contains exactly $nl + 1$ vertices of $\bar \tau'$. All these vertices must be images of $w_1,\ldots,w_{k+1}$, otherwise $\bar f ([w_1,w_{k+1}])$ could not contain the full linterval $[w'_1,w'_1+n]$ as assured by the IVT. Thus we have $k+1 \ge nl+1$, i.e $k \ge nl$. This shows the following:

1. For $\deg(f) = n > 1$ we cannot use the the same triangulation $\tau = \tau'$ on domain and range.

2. $\tau$ must have at least $n$-times as many vertrices a $\tau'$. Since the minimal number of vertices of $\tau'$ is $3$, $\tau$ must have a minimum of $3n$ vertices. Obviously there are solutions with $3n$ vertices.

Answer 2

(parts of this argument still feel a bit sketchy to me, but I think the idea works.)

A simplicial structure on $S^1$ is a graph consisting of one non-degenerate cycle, and as you said there need to be at least $3$ vertices for an edge to be determined by its endpoints. Denote by $\tau$ a minimal simplicial structure, where the vertices are $\{1, \zeta_3, \zeta_3^2\}$.

Suppose we have another simplicial structure $\sigma$ on $S^1$, and let's represent it by a sequence of points $(z_1, z_2, \dots, z_k)$ distributed counter-clockwise around the circle (where it's implied that $z_i$ and $z_{i+1}$ span an edge). Any simplicial map $\varphi\colon(S^1, \sigma) \to (S^1, \tau)$ needs to send endpoints of an edge in $\sigma$ to endpoints of an edge in $\tau$, and since all vertices in $\tau$ are adjacent a simplicial map is actually equivalent to choosing an arbitrary sequence of vertices $(v_1,\dots,v_k)$ in $\tau$ and letting $\varphi(z_i) = v_i$. Conversely, given a sequence of $k \geq 3$ vertices in $\tau$ then the simplicial structure on $S^1$ given by the $k$-th roots of unity admits a simplicial map defined by $\varphi(\zeta_k^i) = v_i$. Since we're looking for minimal simplicial structures on the domain WLOG we can assume that $v_i \neq v_{i+1}$, and then given such a sequence $(v_1,\dots,v_k)$ in $\tau$ we cannot realize it with a simplicial map from a structure having less than $k$ vertices.

Now, for $n>0$ a degree $n$ map $S^1 \to S^1$ wraps the circle counter-clockwise around itself $n$ times. The shortest possible choice of vertices that wraps $n$ times around is $(1, \zeta_3, \zeta_3^2,\dots, 1, \zeta_3, \zeta_3^2)$ where the triple of vertices is repeated $n$ times (prove by induction, you might have to formalize what I mean by "wraps around" in this context). This puts a lower bound on the size of a simplicial structure admitting a degree $n$ map $S^1 \to S^1$, namely it must have at least $3n$ vertices. As you mention in your question, the simplicial structure $\sigma = (1, \zeta_n, \dots, \zeta_n^{n-1})$ and the map $z^n \colon (S^1, \sigma) \to (S^1, \tau)$ realize this lower bound.

Answer 3

Say $K$ and $L$ are simplicial structures of $S^1$, and $L$ has three edges $e_1$, $e_2$, and $e_3$ ordered and oriented counterclockwise about $S^1$. If $K$ has $k$ ordered and oriented edges $\tilde e_1, \ldots, \tilde e_k$, then the simplicial homology groups $H_1^K(S^1) \cong \mathbb Z$ and $H_1^L(S^1) \cong \mathbb Z$ with respect to each simplicial structure are respectively generated by the cycles $\tilde e_1 + \cdots + \tilde e_k$ and $e_1 + e_2 + e_3$.

Suppose $f : K \to L$ is a simplicial map of degree $n$. Then the induced map $f_* : H_1^K(S^1) \to H_1^L(S^1)$ is given by: $$ f_*\left( \tilde e_1 + \cdots + \tilde e_k\right) = n(e_1 + e_2 + e_3) = ne_1 + ne_2 + ne_3 $$ Since the map is simplicial, each edge $\tilde e_i$ gets sent homeomorphically to one of $e_1, e_2, e_3$, so each edge $e_j$ in $L$ has at least $n$ disjoint preimages in $K$ (there may be more). So $K$ has at least $3n$ edges. The map you've constructed has degree $n$, so the bound is sharp.