# 5 dimensional strictly locally homogeneous Riemannian by Patrangenaru V. By Patrangenaru V.

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2c for the whiteface graph. Thus, we obtain a Tait graph by taking a checkerboard colouring and smoothing each vertex as in Fig. 2. 46 3 Twisted Duality, Cycle Family Graphs, and Embedded Graph Equivalence Fig. 2 A local move that constructs Tait graphs a b c v v v∈F v v v Following a black face Following a white face Fig. 3 The six arrow marked vertex states used to construct cycle family graphs (i) (ii) (iii) (iv) (v) (vi) We can generalise Tait graphs by removing the restriction that the smoothings of Fig.

Let F be a 4-regular abstract graph and let F˜ be any embedding of F. , ˜ Gm ∼ = F ⇐⇒ G ∈ C (F), for some embedding F˜ of F. ˜ then Gm ∼ Proof. We first show that if G ∈ C (F), = F. The underlying abstract graph ˜ can be constructed as follows. of Gm , where G is a cycle family graph of F, ˜ choosing an arrow marked state from • G is obtained by, for each vertex v of F, Fig. 3. 48 3 Twisted Duality, Cycle Family Graphs, and Embedded Graph Equivalence F= = = = = = = Fig. 5 Forming a full set of cycle family graphs, C (F) • At each pair of v-labelled arrows in the resulting arrow presentation, replace the arrows with a 4-valent vertex as follows: add a vertex; connect this vertex to the arrow presentation by adding an arc between the vertex and the tip and tail of each v-labelled arrow.

11c. While not all 4-regular embedded graphs admit checkerboard colourings, we have seen that all medial graphs do. 9. Let F be a plane graph. 1. e. all vertices have even degree). 2. If F is 4-regular, then it is checkerboard colourable. Proof. If F is checkerboard colourable, then clearly F is even, since the face colours alternate about each vertex. Now assume F is even and is drawn on the plane. Note that F can be decomposed into a set of edge disjoint cycles by removing any cycle and applying induction.