The 2-Factor Polynomial Detects Even Perfect Matchings

In this paper, we prove that the 2-factor polynomial, an invariant of a planar trivalent graph with a perfect matching, counts the number of 2- factors that contain the the perfect matching as a subgraph. Consequently, we show that the polynomial detects even perfect matchings.


introduction
In [2], the first author developed a cohomology theory of a planar trivalent graph equipped with a perfect matching, and observed that the graded Euler characteristic of that cohomology is a polynomial invariant called the 2-factor polynomial. Given a planar trivalent graph G with a perfect matching M , a perfect matching drawing Γ for the pair (G, M ) is a topological embedding of the graph in the 2-sphere that marks the perfect matching edges. We define the 2-factor bracket of the perfect matching drawing, written Γ 2 as the Laurent polynomial in the variable z calculated recursively on immersed perfect matching drawings by the relations in The bold edge in Figure 1 represents a perfect matching edge (the only edges to which we apply the bracket), and the circle represents any immersed circle with no vertices.
In Section 3 of [2], it was shown that the 2-factor polynomial is independent of the choice of perfect matching drawing, and thus is an invariant of planar trivalent The second author was supported in part by NSF grant DMS-1811344. 1 graphs equipped with a perfect matching, i.e., the pair (G, M ). Thus we can define the 2-factor polynomial of the pair (G, M ) to be the 2-factor bracket of the perfect matching drawing Γ.
G : M 2 (z) = Γ 2 (z). The 2-factor polynomial is easy to calculate. For example, the 2-factor polynomial of the θ graph with perfect matching is: The polynomial depends on both the graph and the perfect matching. To see this dependence, we compute the 2-factor polynomial of the 3-prism, or circular ladder, denoted P 3 , for each of the perfect matchings in Figure 2. The calculation of these examples also serve as a way to introduce the cube of resolutions, which we will use throughout this paper (cf. [2]). Briefly, to construct a cube of resolutions for (P 3 , L), first resolve each perfect matching of the graph into one of the two resolutions shown in the first equation of Figure 1 to get a state. Next, order the states into columns from left-to-right by the number of "crossresolutions" × in each state. Finally, draw an arrow from a state in one column to a state in the adjacent column to the right if the two states are related by replacing one "open-resolution" )( with one cross-resolution × . The cube of resolutions for (P 3 , L) is given in Figure 3. To calculate the polynomial, note that each state contributes a term of the form (−z) k (z + z −1 ) ℓ where k is the number of cross-resolutions × and ℓ is the number of immersed circles in the state. From Figure 3, we can calculate the polynomial of (P 3 , L) directly to see that On the other hand, Figure 4 shows the cube of resolutions for (P 3 , C), and hence, we can calculate the polynomial directly to see that The polynomials for the two perfect matchings are different, and therefore the perfect matchings are different. The polynomials depend upon the perfect matching. It is the goal of this paper to partially characterize that dependence.
It turns out that the 2-factor polynomial incorporates some of the most interesting results to come out of knot theory in the past thirty years together with counting techniques important to graph theory. By evaluating the polynomials at 1, we see that P 3 : L 2 (1) = 0 and P 3 : C 2 (1) = 2. The main theorem of this paper says that these numbers are the count of 2-factors that contain L and C respectively. To describe the main theorem, we need to introduce some notions related to factorability.
Recall that a k-factor of G is a spanning k-regular subgraph of G. For example, the vertex set V (G) is a 0-factor, and a perfect matching (with vertices) is a 1factor. A 2-factor is a disjoint set of a cycles that span the vertex set. When G is also regular, it is natural to count the number of k-factors "between" an ℓ-factor M of G and G, i.e., the index of k-factors that contain M and are contained in G. Our main theorem in this paper says that the 2-factor polynomial counts the number of 2-factors that factor through the perfect matching: Looking back at the circular ladder examples, it is easy to check by hand that |P 3 : L| 2 = 0, while |P 3 : C| 2 = 2, just as the theorem says. Moreover, notice that in this example, P 3 \ L has two 3-cycles, while P 3 \ C contains a single 4-cycle.
(Here and throughout we think of a 1-factor as either a subgraph or a set of perfect matching edges, depending on context.) It turns out that the cycles of G \ M , and their parity in particular, provides important information about |G : M | 2 . It is easy to see that G \ M is always a set of disjoint cycles {C 1 , C 2 , . . . , C k }. Consider one of these cycles, C i , and note that every cycle of a 2-factor that factors through M must enter C i through a perfect matching edge of M and then leave C i through a different perfect matching edge of M . Thus, any 2-factor that factors through M must span an even number of vertices of C i . If any cycle C i in G \ M has an odd number of edges, i.e., if C i is an odd cycle, then |G : M | 2 = 0.
On the other hand, if C i has an even number of edges for all i, then every 2-factor that factors through M consists of M and a perfect matching for each C i . This is because every other edge of C i must be included in the 2-factor. Since each C i has two perfect matchings, there are 2 k choices. Thus we obtain the following: The existence of an even perfect matching for a planar trivalent graph clearly shows that there is a Tait coloring for the graph, i.e., it is 3-edge colorable (cf. [6]). In fact, Theorem 2 implies more. Pick three colors {i, j, k} to label edges. By (Recall that [G : V (G)] 1 is the set of all 1-factors of G.) The polynomial T G is an invariant of G (cf. [2]). In particular, it is easy to prove the following about T G : Theorem 6. If G is a connected planar trivalent graph, then T G (1) counts the number of Tait colorings of G, i.e., For example, the theta graph θ has three perfect matchings, one for each edge of the graph. The 2-factor polynomial is the same for each perfect matching, thus and T θ (1) = 6. It is easy to verify that there are six Tait colorings of the θ graph.
Another example that we can easily calculate is P 3 . There is one perfect matching corresponding to the ladder L and three perfect matchings corresponding to C and two 120 • rotations of C. Therefore, and T P3 (1) = 6, which is the number of Tait colorings of P 3 . Note that the perfect matching L does not contribute any colorings.
Of course, saying that a planar trivalent graph G has a Tait coloring is equivalent to saying that the faces are 4-colorable (cf. [6], [4], [7]). Hence, the four color theorem is equivalent to the following: While all known proofs of the four color theorem rely on checking many cases by computer, it may be possible to prove Theorem 7 directly, thereby obtaining a non-computer aided proof of the four color theorem (cf. [1], [3], [5], [12]).

Preliminaries
The proof of Theorem 2 rests upon the use of IH-moves to transform a given bridgeless, planar trivalent graph with a perfect matching (G, M ) into one where we know Theorem 2 holds. Definition 8. An IH-move is a local replacement of the configuration shown on the left in Figure 5 with the one shown on the right. Note that we only perform an IH-move on a perfect matching edge.
Definition 9. Given two graphs, G and G ′ , and perfect matchings for each M and M ′ respectively, we say that (G, M ) is IH-equivalent to (G ′ , M ′ ) if there is a sequence of IH-moves taking one to the other.
Before we proving the main theorem, we need to establish a series of lemmas about IH-moves.  Figure 6. The non-perfect matching edges of the second picture of Equation 2.2 will then be parts of two different cycles C 2 and C 2 in G 2 (V 2 , E 2 \ M 2 ), and so on as displayed in Figure 6. Let C 4 be the cycle corresponding to C 1 in the fourth picture. Clearly the number of cycles in G 1 \ M 1 is equal to the number of cycles in G 4 \ M 4 , and since C 4 has two fewer edges than C 1 , both cycles have the same parity. Therefore, by Proposition 3, |G 1 : Suppose that C 2 and C 2 have k and k edges respectively. The corresponding cycles C 3 and C 3 have k − 1 and k − 1 edges respectively. If both k and k are even, then k − 1 and k − 1 are odd, implying that |G 3 : M 3 | 2 = 0. Also, in this case, Proof. The result for |G : M | 2 follows from Proposition 3. For the 2-factor polynomial, first note that if G has a bridge, then it must be a perfect matching edge (cf. [2]). Resolving this perfect matching edge shows that the 2-factor polynomial has a (z − 1) factor, and therefore G : M 2 (1) = 0 (again, see [2]). Proof. Suppose that M has k edges, and that G has n vertices (n = 2k). Since G(V, E \ M ) is a cycle, it has an inside and an outside. Either the inside or outside of C contains at least k/2 edges of the perfect matching M . Without loss of generality, assume the inside of C contains at least k/2 edges of M . Call this inside region D. Since at least k/2 edges of M lie in D, those edges split D into at least 1 + k/2 disk faces. The boundary of each of these disks will contain some number of edges of M , and some number of vertices that do not belong to any edge of M . If such a disk has m edges from M and ℓ vertices not incident to any edge of M on the inside of the cycle C, then we label that disk D ℓ m . The configurations in Figure 7 are D 0 1 , D 1 1 , D 2 1 , D 0 2 , D 1 2 and D 0 3 . By way of contradiction, assume no such configuration appears in D.
We form another graph T by placing a vertex in each face of D and joining two vertices when their associated faces are separated by an edge of M . It is easy to see that the graph T is a tree, and the degree of each vertex in T is the number of perfect matching edges M in the boundary of that disk face. Let δ m denote the number of faces of the form D ℓ m for any ℓ. Hence δ m is the number of degree m vertices in the tree T . A standard way to count leaves in a tree results in the following formula: Define δ ≥4 = ∞ m=4 δ m . The above equality implies that δ 1 ≥ 2 + δ 3 + 2δ ≥4 , or equivalently that δ ≥4 ≤ 1 2 (δ 1 − δ 3 ) − 1. Therefore Since the faces in Figure 7 are assumed not to appear in C, each disk of the form D ℓ 1 must have ℓ ≥ 3, each disk of the form D ℓ 2 must have ℓ ≥ 2, and each disk of the form D ℓ 3 must have ℓ ≥ 1. There are also at most k vertices of G not incident to an edge of the matching M inside C. Therefore Having arrived at a contradiction, we conclude that some face from Figure 7 appears in C. Proof. An IH-move either merges or splits two cycles of G(V, E \M ). In a connected graph where G(V, E \ M ) contains more than one cycle, one can perform an IHmove that merges two cycles of G(V, E \ M ). Thus there is a sequence of IH-moves transforming (G, M ) into the pair ( Lemma 14 implies that a face as in Figure 7 must appear either inside or outside of the cycle C in G 1 . Performing the IH-move(s) in Figure 7  Proof. Assume that M has k edges and G has n vertices (n = 2k). Observe that if 2 vertices of the cycle are also connected by a perfect matching edge, then the third vertex must be an endpoint of a perfect matching edge that is a bridge, and therefore, by Lemma 13, we are done. Thus, we may assume that the three perfect matching edges incident with the 3-cycle are distinct. For the same reason, we may assume that all three perfect matching edges are all outside of the 3-cycle or are all inside the 3-cycle. Thus, we may assume, without loss of generality, that the 3-cycle, with distinct perfect matching edges at each vertex, appears as in Figure 8. Because the polynomial is calculated recursively, the cube of resolutions for G : M 2 (1) may be decomposed as follows. First, resolve every perfect matching edge that is not incident to the 3-cycle to get 2 k−3 "prestates." Each prestate P will consist of a number of immersed circles and a graph G ′ whose vertices are spanned only by the three perfect matching edges of the 3-cycle. Note: the graph G ′ may be immersed away from the 3-cycle and the three perfect matching edges. If there are ℓ immersed circles in the prestate P , then by the relations in Figure 1, We will show that in all possible cases for G ′ , G ′ 2 (1) = 0. Since G : M 2 (1) is made up as sums of terms of the form ± P 2 (1), and each of these terms is zero, Resolving all three perfect matching edges of the 3-cycle using all open-resolutions )( produces the picture shown in Figure 9. Thus, the "all-open-resolution" of the graph G ′ of a prestate P corresponds to a choice of connecting the six endpoints of Figure 9 with three arcs. Up to symmetry of the disk, there are seven ways to connect these six endpoints up to form a set of immersed circles. These seven configurations are shown in Figure 10. (The actual all-open-resolution of a particular G ′ may have more circle intersections, of course, but these extra intersections do not effect any of the calculations.) Thus, to prove G ′ 2 (1) = 0 for all possible graphs G ′ of the prestates, we need only show that it is zero for these seven configurations. The cube of resolutions corresponding to the first all-open-resolution configuration in Figure 10 is shown in Figure 11. For a G ′ corresponding to this configuration, we see that G ′ 2 (1) = 2 − 3 · 2 + 3 · 2 − 2 = 0. Figure 11. For any G ′ corresponding to the first configuration in Figure 10, The remaining six cases are handled in the Appendix. In each case, G ′ 2 (1) = 0, which proves the lemma.

Proof of Theorem 2
Now we are ready to prove Thoerem 2. It is easy to see that we need only prove Theorem 2 for the case when G is a connected graph. Furthermore, by Lemma 13, the theorem is trivially true if G has a bridge. Therefore we need only prove the theorem for connected, bridgeless planar trivalent graphs.
The proof proceeds by induction on the number of edges in the perfect matching M . If the perfect matching has a single edge, then (G, M ) is the θ graph, and the theorem is easily verified by inspection.
Assume the theorem is true for any planar trivalent graph with a perfect matching having n − 1 edges. Let (G, M ) be a connected, bridgeless planar trivalent graph with n perfect matching edges. By Lemma 15, the graph is IH-equivalent to (G ′ , M ′ ), where G ′ (V ′ , E ′ \ M ′ ) has a cycle of length at most 3. Lemma 10, together with the inductive hypothesis, implies that we need only verify that the result holds for (G ′ , M ′ ).
If G ′ (V ′ , E ′ \ M ′ ) has a cycle of length 3, then Proposition 3 and Lemma 16 imply the result. If G ′ (V ′ , E ′ \ M ′ ) has a cycle of length 1, then G ′ has a bridge and Lemma 13 implies the result. Lastly, if G ′ (V ′ , E ′ \ M ′ ) has no cycle of length 1 or 3, but does have a cycle of length 2, then Lemma 11 (the Bubble Lemma), together with the inductive hypothesis, implies the result.