What is the minimum number of edges in simplicial complex structures \(K\) and \(L\) on \(S^1\) such that there is a simplicial map \(K \to L\) of degree \(n\)?
Solution
Let \(e(K)\) and \(e(L)\) denote the numbers of edges of the two simplicial circles.
1. Every simplicial complex structure on \(S^1\) has at least three edges.
A simplicial complex cannot have a loop edge or two distinct edges with the same pair of endpoints. Hence a simplicial circle needs at least three vertices and therefore at least three edges. The boundary of a \(2\)-simplex realizes three.
2. If a simplicial map \(f:K\to L\) has degree \(n\ne0\), then\[ e(K)\ge |n|e(L). \]
Orient \(K\) and \(L\). Let \[ z_K=\sum_{e\subset K}\epsilon_e e, \qquad z_L=\sum_{a\subset L}\epsilon_a a \] be their fundamental simplicial \(1\)-cycles. Each edge of \(K\) is sent either to a vertex or linearly onto one edge of \(L\), with coefficient \(0\) or \(\pm1\) on chains. Since \[ f_*(z_K)=n z_L, \] each target edge must receive total signed coefficient \(n\). Thus at least \(|n|\) domain edges must map nondegenerately to each of the \(e(L)\) target edges, giving the inequality.
3. The bound is sharp: for \(n\ne0\) take \(L\) to be a triangle and \(K\) a polygon with \(3|n|\) edges, mapping its edges successively around the three edges of \(L\) exactly \(|n|\) times, preserving orientation if \(n>0\) and reversing it if \(n<0\).
The vertex map is periodic with period three and sends adjacent vertices to adjacent vertices, so it extends to a simplicial map. The induced map on the fundamental \(1\)-cycle is multiplication by \(n\).
4. For \(n=0\), both \(K\) and \(L\) can be triangles and the constant simplicial map has degree zero.
A constant map sends all \(1\)-chains to zero, hence has degree zero.
Therefore the minimum numbers are \[ \boxed{e(L)=3,\qquad e(K)=3\max\{1,|n|\}.} \]