Binomial Ideals from Graphs

Binomial Ideals from Graphs Danielle Farrar University of Washington Yolanda Manzano St. Mary s University Juan Manuel Torres-Acevedo University of Puerto Rico Humacao August 10, 2000 Abstract This paper contains an examination of polynomial ideals associated to graphs and conjectures about them. We will describe how to obtain an ideal I G from a graph G. This ideal encodes the relations among all spanning trees of G. There are five conjectures we are concerned with. The first claims that I G is generated by quadratic binomials. We will discuss the relationships uncovered between different classes of graphs and their ideals. For instance, for any tree or one-circuit graph, I G = 0. Also, we can contract edges that are not part of any circuit. The number of circuits in a graph has a direct correlation to its ideal. 1 Introduction Our research concerns a special case of a conjecture made by Bernd Sturmfels of University of California, Berkeley 3]. This conjecture originates in a problem from linear algebra. From a graph, G, we can construct the matrix M G such that its rows represent the edges of G and its columns represent its spanning trees. For each column, entries of 1 represent edges that are in the spanning tree, 0 represents edges not in the spanning tree. By mapping of these elements, we can create a system of equations. Macaulay uses the Elimination Theorem to solve for the ideal of this matrix. We investigate this and other properties of the ideals from a graph, starting with specific classes, and grouping them by the number of circuits present in the graph. 35

Our research is concerned with investigating the following conjectures. 1. I G is generated by quadratic binomials, that is, differences of two monomials, each of which has degree two. 2. There is a reduced Gröbner basis of I G that is generated by quadratic binomials. 3. Any reduced Gröbner basis of I G is generated by quadratic binomials. 4. I G has a reduced Gröbner basis consisting of quadratic binomials with squarefree initial terms. 5. Any reduced Gröbner basis of I G consists of quadratic binomials with squarefree initial terms. In our paper, we will first give some algebra background, and then provide an explanation on how to create I G from G. We will prove that this I G is generated by a certain set of binomials. Then, we will give some background on graph theory, and show how we can reduce the classes of graphs we need to investigate by showing that extraneous edges of a graph do not change the ideal of that graph. We then prove that the ideal of a graph that is a ring or a tree is 0, and explain why Conjectures 1 and 2 should hold for two-circuit graphs. We will show our progress on proving/disproving the above conjectures. 2 Algebra Background In this section we summarize some definitions from abstract algebra. Definition 2.1 A subset I kx 1,, x n ] is an ideal if it satisfies: (i) 0 I. (ii) If f, g I, then f + g I. (iii) If f Iand h kx 1,..., x n ], then hf I. Definition 2.2 A set g 1,..., g s I is a Gröbner basis of I if and only if the leading term of any element of I is divisible by one of the LT (g i ). Definition 2.3 A mapping φ : R S is a ring homomorphism if φ satisfies the properties: (i) φ preserves sums : φ(r + r ) = φ(r) + φ(r ) for all r, r R. (ii) φ preserves products : φ(r r ) = φ(r) φ(r ) for all r, r R. (iii) φ maps the multiplicative identity 1 R to 1 S. 36

3 Graphs to Ideals 1 2 4 3 Figure 1: a Let G be a connected, undirected graph and let K be a field. We will construct a binomial ideal I G in a polynomial ring. We will then discuss several conjectures about the ideal I G. First we associate to G a matrix M G with integer entries. Let e 1,..., e n be the edges of G, and let s 1,..., s r be all possible spanning trees of G. Let M G = (m kl ) be the (n r) matrix whose rows are labeled by e 1,..., e n and whose columns are labeled by s 1,..., s r. Let m kl = { 1 if edge ek is part of spanning tree s l 0 otherwise. Using M G, we can define a ring homomorphism by defining it on the variables as follows: φ : Kx 1,..., x r ] Ky 1,..., y n ] x i y m 1i 1 y m 2i 2 y m ri n. Definition 3.1 Let I G = ker(φ) = {f Kx 1,..., x r ] φ(f) = 0}. Example 3.2 Let G be the graph in Figure 1.a. In G, e 1 = 1, 2 e 2 = 1, 3 e 3 = 1, 4 e 4 = 2, 3 e 5 = 3, 4 where (b,c) represent vertices of G. We can then represent the spanning trees of G in the matrix 1 1 0 0 1 0 1 1 1 1 1 1 0 0 0 0 M G = 1 0 0 1 0 1 1 1 0 0 1 1 1 1 0 1 0 1 1 0 1 1 1 0 37

By placing the mapping of M G in Macaulay, we obtain the following ideal: I G =< x 2 x 6 x 1 x 7, x 5 x 6 x 7 x 8, x 4 x 6 x 3 x 8, x 1 x 5 x 2 x 8, x 3 x 5 x 4 x 7, x 1 x 4 x 7 x 2 x 3 x 8 > 4 Binomial Generators of I G We will use the following notation. Let Kx 1,..., x r ] = Kx], Ky 1,..., y n ] = Ky]. Theorem 4.1 The ideal I G is generated by all binomials of the form x u+ x u, where u + u = u N n such that M G u = 0. Proof: A binomial x u x v lies in I G if and only if φ(x u ) = φ(x v ). It therefore suffices to show that each polynomial in I G is a k-linear combination of binomials. Fix a term order on kx]. Suppose f I G cannot be written as a k-linear combination of binomials. We choose a polynomial f with this property such that the initial term in (f) = x u is minimal with respect to the term order. When expanding f(t a 1,, t an ) we get zero. In particular, the term t φ(u) = φ(x u ) must cancel during this expansion. That is, there must be a term φ(x v ) such that φ(x v ) φ(x u ) = 0. Hence there is some other monomial x v x u appearing in f such that φ(v) = φ(u). Also, in the polynomial f := f x u + x v, the initial term of f is less than x u. This means that f cannot be written as a k-linear combination of binomials in I G. But since in (f ) in (f), this is a contradiction. This completes the proof. Lemma 4.2 Let v, w be integer vectors of length n, such that M G v = 0 = M G w. Then x (v+w)+ x (v+w) is a linear combination of the binomials x v+ x v and x w+ x w. More generally, if M G v i = 0 for i = 1,, a, and v = v 1 + + v a, then the binomial x v+ x v is a linear combination of x (v 1) + x (v 1),, x (va)+ x (va). Then Proof: First observe that (v + w) ± = v ± + w ±. x (v+w)+ x (v+w) = x (v+ )+(w +) x (v )+(w ) This completes the proof. The general case follows by induction. = x v+ x w+ x v x w = x v+ (x w+ x w ) + x v+ x w x v x w = x v+ (x w+ x w ) + x w (x v+ x v ). Corollary 4.3 Let v 1,, v a be a generating set for the set of all v such that M G v = 0. Then the corresponding set of binomials is a generating set for the ideal I G. x (v 1) + x (v 1),, x (va)+ x (va) 38

Proof: By Theorem 4.1, I G is generated by all binomials of the form x v+ x v for all v = v + v such that M G v = 0. Let v = α 1 v 1 + + α a v a for some integers α 1,, α a. Then by Lemma 4.2, the binomial x v+ x v combination of the binomials x v i + x v i. This completes the proof. is a linear 5 Some Graph Theory Definitions Definition 5.1 A circuit is a cycle that becomes a tree upon removal of any edge. Definition 5.2 An edge is extraneous if it is not part of any circuit. Definition 5.3 If e is an edge of G, and u and v are the vertices joined by e, then contraction of e is the operation of replacing u and v by a single vertex whose incident edges are the edges other than e that were incident to u or v.2] Lemma 5.4 Let e be an edge of G, and suppose that e is extraneous. contained in every spanning tree of G. Then e is Proof: Let e be an edge not contained in any circuit of G, and connecting vertices u and v. Suppose that a spanning tree s does not contain e. Then s contains a set of edges e p,..., e q that connect u and v. The addition of e to s creates a circuit in G. This is a contradiction, since e is extraneous. Therefore, e must be included in every spanning tree of G. Lemma 5.5 Let G be the graph obtained by contracting an extraneous edge e from a graph G. The spanning trees of G are in 1 1 correspondence with the spanning trees of G, and the corresponding vector m i is obtained from m i by deleting the entry corresponding to e. Proof: A column m i represents a spanning tree in G. Since graph G is obtained by collapsing edge e, the spanning trees in G will not contain the edge e, but all other spanning tree entries of G will correspond to those of G. Likewise, any spanning tree of G can be obtained by adding e to a spanning tree of G, since this edge will be necessary to connect the final vertex of G. Therefore, the vectors m i respresenting the spanning trees in G can be obtained by deleting the entries corresponding to e in the vectors m i in G. Lemma 5.6 For a graph G with r vertices, all spanning trees of G contain r 1 edges. 39

6 A Reduction Theorem Theorem 6.1 Let G be a connected graph, and let G be the graph obtained from G by contracting all extraneous edges, that is, all edges that are not part of any circuit. Then I G = I G Proof: By induction it is sufficent to first consider the case where G has one extraneous edge e. Let e 1,..., e n be the edges of G, and suppose that e n is extraneous. By Lemma 5.4, e n is contained in every spanning tree of G, so the last row of M G has all its entries equal to 1. Let M G = (m 1,..., m r ) Then, if φ(x i ) = y m i, each y m i contains y n. By Lemma 5.5, we can say M G is obtained from M G by deleting the last row. Now we must show that I G = I G. First observe that φ = p φ where p : ky 1,..., y n ] ky 1,..., y n 1 ] is the projection sending y n to 1, so that φ is the mapping of G such that Therefore, if f kerφ = I G, then φ : kx 1,..., x r ] ky 1,..., y n 1 ]. φ (f) = p φ(f) = p(0) = 0. So I G I G. To show the reverse inclusion, let x u+ x u I G be a generator. If u = u + u, then M G u = 0. Since we need to show that i=1 u M G = M G 1 1... 1 1 u i = 0. This implies that M G u = 0, so that x u+ x u I G. From the system of equations M G u = 0, we obtain a consequence by adding up all rows. Since the columns of M G are the incidence vectors of the spanning trees of G, they all contain the same number of 1 s, since all spanning trees contain the same number of edges, say t, by Lemma 5.6. Then the equation we obtain is tu 1 + tu 2 + + tu r = 0. Dividing by t gives us the desired equation u 1 + u 2 + + u r = 0. This shows that I G I G, and the proof is complete. 40 ]

7 Proof that I G = 0 if G 1 Circuit We will that for trees and graphs with less than or equal to 1 circuit, I G = 0. From Theorem 5.4 we are able to ignore the case where a one-circuit graph contains extraneous edges. For trees, the only spanning tree of G is G. This means that M G will always be a column consisting entirely of ones where the number of entries in G is equal to the number of edges of G. The mapping for such a matrix will then be kx] ky 1, y 2,..., y n ]. It then follows that When f is of the form x y 1, y 2,..., y n. then From this we can conclude that if and only if Therefore, f = x r + a r 1 x r 1 +... + a 1 x + a 0 φ(f) = (y 1...y n ) r + a r 1 (y 1...y n ) r 1 +... φ(f) = 0 f = 0. kerφ = I G = 0. This proves that I G = 0 for trees. For graphs with exactly one circuit, the number of edges of G is equal to the number of spanning trees of G. Each spanning tree is distinguished by the edge it excludes. This creates a matrix M G such that M G is a square matrix with 0 entries in the diagonal and 1 entries elsewhere. That is M G = 0 1 1 1... 1 0 1 1... 1 1 0 1... 1 1 1 0........................ The vectors,α i of M G of this form are linearly independent. It follows that there are no integers v i such that M G v = v 1 α 1 +... + v i α i = 0. Therefore, kerφ = 0 41

and, in turn, This completes the proof. I G = 0. 8 Two Circuit Graphs Lemma 8.1 For a graph containing two circuits, let the set of edges e 1, e 2,..., e n be all edges of one of the two circuits. The two circuits have no edges in common, therefore every edge not in one of the two circuits is not part of any circucit. Proof: By definition of extraneous, any edge not a member of one of the two circuits is extraneous. By Theorem 5.4, this edge can be contracted without altering the ideal. Contraction of all extraneous edges will give the two circuit graph the form of two single circuit graphs, or rings, joined by a single vertex. The two graphs cannot have any edges in common, because the existence of such an edge would form another circuit, which would be the set of all edges besides the one the two graphs share. Therefore the only way the graph can remain connected is by a single vertex. The matrix of any two-circuit graph can be represented in the form ] s1 s M G = 1...s 1 s 2 s 2...s 2... s a...s a t 1 t 2...t b t 1 t 2...t b... t 1...t b s i and t j represent vectors where s i has a entries, corresponding to a edges in the first circuit, and t j has b entries, corresponding to b edges in the second circuit. The vector s i consists of all 1 s, except the i th entry of s i which is 0, and the vector t j consists of all 1 s, except the j th entry of t j, which is 0. This can be done because each spanning tree can be obtained by systematically eliminating one edge from each circuit. The quadratic binomials are constructed by choosing blocks i and j in the matrix, where 1 i < j a. Then choose two columns in block i, say ] ] si t k, si Then choose two columns in block j as follows: ] ] sj t k, t l sj (s i, t k ) can be used to represent the spanning tree ] si t l These form the binomial x (si,t k ) x (sj,t l ) x (si,t l ) x (sj,t k ). Conjecture 8.2 The set of all these binomials is a Gröbner basis for I G. 42 t k

9 Quadratic Gröbner Bases of I G As seen previously, graphs with 0 and 1 circuits yield a reduced Gröbner basis of 0. The reduced Gröbner bases of graphs with 3,5,6,7, 13 and 14 circuits contain cubic terms, and the reduced Gröbner basis of a graph with 9 circuits contains quartic terms. All of these provide counterexamples to Conjecture 3. We performed these calculations with respect to graded reverse lex ordering, and must try all different term orders to further evaluate Conjecture 2. We have found that all ideals calculated from graphs are generated by quadratic binomials, by isolating the quadratic binomials from the reduced Gröbner basis and determining if the ideal generated by this set of quadratic binomials was equal to the ideal generated by the reduced Gröbner basis. In the case of a graph with 13 circuits, we found that the ideal obtained from this graph was not generated by quadratic binomials when computed with respect to graded reverse lex ordering. We calculated the ideal with respect to lex ordering, and found that the ideal was indeed generated by quadratic binomials. This leads us to believe that term ordering may be important, and we will further investigate this in the future. 10 Ideas for Future Research We will conduct further investigation into Conjectures 2 and 4, doing computation using different term orders in order to draw conclusions regarding these conjectures, as described previously. Graphs calculated with 2 and 4 circuits had ideals and Gröbner Bases consisting of quadratic binomials. This leads us to the conjecture that all graphs with 2 n have this property. Further research is necessary to prove/disprove this concept. Definition 10.1 A block of a graph G is a maximal connected subgraph of G that has no cut-vertex, meaning it does not contain an edge whose deletion increases the number of components. If G itself is connected and has no cut-vertex, then G is a block. The general graphs have been reduced to graphs without extraneous edges. We believe that it may be possible to further reduce our examination of graphs to blocks, that is, it would be sufficient to prove the conjectures for blocks. We suspect that we can collapse edges of blocks that do not alter the number of cycles of the graph. References 1] D. Cox, J. Little, & D. O Shea, Ideals, Varieties and Algorithms, 2nd Ed., Springer-Verlag, New York, 1997. 2] D. West, Introduction to Graph Theory, Prentice Hall, Upper Saddle River, 1997. 43

3] B. Sturmfels, Gröbner Bases and Convex Polytopes, American Mathematics Society, Providence, RI, 1996. 44