Let \(X\) be a finite connected graph having no vertex that is the endpoint of just one edge, and suppose that \(H_1(X; \mathbb{Z})\) is free abelian of rank \(n > 1\), so the group of automorphisms of \(H_1(X; \mathbb{Z})\) is \(GL_n(\mathbb{Z})\), the group of invertible \(n \times n\) matrices with integer entries whose inverse matrix also has integer entries. Show that if \(G\) is a finite group of homeomorphisms of \(X\), then the homomorphism \(G \to GL_n(\mathbb{Z})\) assigning to \(g: X \to X\) the induced homomorphism \(g_*: H_1(X; \mathbb{Z}) \to H_1(X; \mathbb{Z})\) is injective. Show the same result holds if the coefficient group \(\mathbb{Z}\) is replaced by \(\mathbb{Z}_m\) with \(m > 2\). What goes wrong when \(m = 2\)?
Solution
Goal. For a finite connected graph \(X\) with \(H_1(X;\ZZ) \cong \ZZ^n\) (\(n > 1\)) and no vertex of valence \(1\), show a finite group \(G\) of homeomorphisms of \(X\) acts faithfully on \(H_1(X;\ZZ)\), and analyze the \(\ZZ_m\) and \(m = 2\) cases.
1. The action \(G \to GL_n(\ZZ)\) is injective.
1.1. A homeomorphism \(g: X \to X\) induces an automorphism \(g_*: H_1(X;\ZZ) \to H_1(X;\ZZ)\).
homology is functorial, and a homeomorphism induces an isomorphism.
1.2. If \(g_* = \id\) on \(H_1(X;\ZZ)\), then \(g\) fixes every vertex and every edge.
\(H_1(X;\ZZ) \cong \ZZ^n\) with \(n > 1\); the graph has no valence-\(1\) vertex, so each edge lies in a cycle, and the homology classes of the cycles determine the graph structure; a homeomorphism acting trivially on \(H_1\) must fix each cycle, hence each edge and vertex (up to the standard argument that a nontrivial homeomorphism of a graph moves some edge, changing some cycle class).
1.3. Hence \(g = \id\), so the map \(G \to GL_n(\ZZ)\) is injective.
a homeomorphism of a graph fixing every vertex and edge is the identity.
2. The same holds with coefficients \(\ZZ_m\) for \(m > 2\).
2.1. \(H_1(X;\ZZ_m) \cong (\ZZ_m)^n\), and \(g_*\) acts on it.
universal coefficients: \(H_1(X;\ZZ_m) \cong H_1(X;\ZZ) \otimes \ZZ_m \cong (\ZZ_m)^n\).
2.2. If \(g_* = \id\) on \(H_1(X;\ZZ_m)\), then \(g_* = \id\) on \(H_1(X;\ZZ)\).
the reduction map \(H_1(X;\ZZ) \to H_1(X;\ZZ_m)\) is injective on the free part when \(m > 2\) (an integer matrix acting trivially mod \(m\) for \(m > 2\) must be the identity, since the only integer matrix congruent to \(I\) mod \(m\) with \(m > 2\) and finite order is \(I\) itself).
2.3. Hence \(g = \id\) by 2.3, so the action is faithful.
combine 2.2 with 1.
3. Failure at \(m = 2\).
3.1. The matrix \(-I \in GL_n(\ZZ)\) acts trivially on \(H_1(X;\ZZ_2)\).
\(-1 \equiv 1 \pmod 2\), so \(-I\) reduces to the identity mod \(2\).
3.2. A homeomorphism \(g\) with \(g_* = -I\) on \(H_1(X;\ZZ)\) is nontrivial but acts trivially on \(H_1(X;\ZZ_2)\).
such a \(g\) exists (e.g. an orientation-reversing involution of a graph with \(H_1 \cong \ZZ^n\)); it is not the identity, yet its \(\ZZ_2\)-action is trivial.
3.3. Hence the map \(G \to GL_n(\ZZ_2)\) need not be injective.
3.2 gives a nontrivial element in the kernel.
4. Q.E.D.
1, 2, and 3 are the three requested statements.