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(ii) d = k + 1, if k is odd and n is even, and (iii) d = k + 2, if both k and n are odd. Here, as usual, R is the field of real numbers and the nullity of a linear transformation is the dimension of its kernel. Dixon and Pressman showed that d > k, k + 1 and k +2 in cases (i), (ii), and (iii), respectively, and verified computationally that equality holds for small values of n and k. Note that one may view Conjecture 1.1 and the Amitsur-Levitsky theorem as pointing in opposite directions: Conjecture 1.1 gives an upper bound on the generic nullity of L(A1,...,Ak) for 2 6 k 6 2n − 2, whereas the Amitsur-Levitsky theorem gives a lower bound for k > 2n − 1. The purpose of this paper is to prove Conjecture 1.1 in case (i). Our main result is the following. Theorem 1.2. Let k = 2r be a positive even integer and F be an infinite field whose characteristic does not divide 2(2r + 1)r!. Assume that n > r. Then for A1, A2,...,Ak ∈ Matn(F ) in general position, the nullity of L(A1, A2,...,Ak) is k. Our proof will rely on graph-theoretic techniques. To motivate it, let us briefly recall Swan’s proof of the Amitsur-Levitsky theorem. Since the standard polynomial [A1,...,Am] is multi-linear in A1,...,Am, it suffices to show that [Ea1b1 ,...,Eambm ] = 0 for any choice of a1,b1,...,am,bm ∈ {1,...,n}, as long as m > 2n. Here Eab denotes the elementary matrix with 1 in the (a,b)-position and 0s elsewhere. As we expand [Ea1b1 ,...,Eambm ], the term sgn(σ)Eaσ(1) bσ(1) ...Eaσ(m)bσ(m) contributes sgn(σ)Eaσ(1) bσ(m) to the sum if
(1.1) bσ(1) = aσ(2), bσ(2) = aσ(3), ... ,bσ(m−1) = aσ(m), and 0 otherwise. Conditions (1.1) can be conveniently rephrased in graph-theoretic terms. Let G be the directed graph with n vertices, 1,...,n and m edges, e1 = (a1,b1),...,em = (am,bm). Then conditions (1.1) hold if and only if eσ(1),...,eσ(m) form an Eulerian path on G. We will say that this Eulerian path is even if σ is an even permutation and odd otherwise. This way the Amitsur-Levitsky theorem reduces to the following graph-theoretic assertion.
Theorem 1.3. (Swan [7, 8]) Let G be a directed graph with n vertices and m edges. Let a and b be two of the vertices (not necessarily distinct). If m > 2n, then the number of even Eulerian paths from a to b equals the number of odd Eulerian paths from a to b.
If one were to use a similar approach to prove Conjecture 1.1, one would set n (ℓ) (1.2) Aℓ = xab Eab, a,bX=1 (ℓ) 2 2 2 for ℓ =1,...,k. Here xab are kn independent variables. Each entry of the n × n matrix of L = L(A1,...,Ak) is then a multilinear polynomial of degree k in the groups of variables, (1) (k) {xab },..., {xab }. (Here we identify the linear transformation L(A1,...,Ak): Matn → Matn with its matrix in the standard basis Eab of Matn.) The coefficient of the monomial x(1) · ... · x(k) in a given position in L can again be computed as the signed sum of a1b1 akbk Eulerian paths on a certain graph. However, for k 6 2n − 2, these signed sums will no longer be identically 0. To prove Conjecture 1.1(i) in this way, one would need to assemble these coefficients into the n2 × n2 matrix L with polynomial entries, then show that the (ℓ) nullity of L over the field F (xab ) is k (or equivalently, is 6 k). We are not able to carry A GRAPH-THEORETIC APPROACH TO A CONJECTURE OF DIXON AND PRESSMAN 3 out the computations directly in this setting; the matrix L is too complicated. To prove Theorem 1.2 we will modify this approach in the following ways. (ℓ) (1) We will specialize the matrices Aℓ by setting some of the variables xab equal to 0. For the purpose of showing that null(L) 6 k, this is sufficient. In fact, we will set n2 − n entries of each Al equal to 0; the other n entries will remain independent variables.
(2) We will choose A1,...,Ak so that L(A1,...,Ak) decomposes as a direct sum
L(A1,...,Ak)= L0 ⊕ L1 ⊕ ... ⊕ Ln−1, where each Li is represented by an n × n matrix. This simplifies our analysis of L and reduces the problem to showing that null(L0)+null(L1)+...+null(Ln−1) 6 k. The specific matrices we will use are described in Sections 3 and 6.
(3) To get a better handle on the nullities of L0,...,Ln−1, we will replace each Lj by its “matrix of initial coefficients” Ic(Lj ) with respect to a certain lexicographic monomial order on the variables xℓ,α; see Section 5. This will further simplify the computations in two ways. First, the entries of Ic(Lj ) will be integers, rather than polynomials. These integers will be obtained by counting Eulerian paths on certain graphs, as in Swan’s argument. Secondly, passing from Lj to Ic(Lj ) will allow us to focus only on the (rather special) graphs corresponding to leading monomials. We will classify these “maximal graphs” in Sections 7 and 8 and complete the proof of Theorem 1.2 in Sections 9 and 10. The last part of the proof will rely on the computations of signed counts of Eulerian paths in Section 2. The overall structure of the paper is shown in the flowchart below.
Theorem 1.2
Proposition 3.2
Proposition 6.3 Proposition 7.2
Lemma 7.1 Lemma 7.3 Lemmas 2.3, 2.4 Lemma 8.1 Proposition 9.1 Lemma 8.3 Lemma 8.2 Section 10 Lemmas 2.5, 2.6
2. Preliminaries on graphs and Eulerian paths Throughout this paper our graphs will all be directed with labeled edges and vertices. An Eulerian path in a graph Γ is a path which visits every edge exactly once. We will denote by Eula(Γ) the set of Eulerian paths on Γ which begin at a vertex a. It is easy to see that any two paths in Eula(Γ) terminate at the same vertex. For an edge • → • appearing in a graph Γ we define srcΓ(e)= a and tarΓ(e)= b to be a e b the source and target vertices of the edge e respectively. We define the outdegree outdegΓ(v) to be the number of edges in Γ whose source vertex is v, and the indegree indegΓ(v) to be 4 MATTHEWBRASSILANDZINOVYREICHSTEIN the number of edges in Γ whose target vertex is v. When the graph Γ is clear from the context we will abbreviate these terms as src(e), tar(e), outdeg(v), and indeg(v). An Eulerian path beginning and ending at the same vertex is known as an Eulerian circuit. The following fundamental theorem, due to Euler, is usually stated in terms of Eulerian circuits. In the sequel we will need a variant in terms of Eulerian paths. Theorem 2.1. Let a, b be vertices of Γ, not necessarily distinct. There exists an Euler- ian path from a to b on Γ, if and only if Γ is connected and
outdegΓ(v) = indegΓ(v) , for all v 6= a, b ,
outdegΓ(a) = indegΓ(a)+1 , if a 6= b , (2.1) outdegΓ(b) = indegΓ(b) − 1 , if a 6= b ,
outdegΓ(a) = indegΓ(a) , if a = b . Proof. If a = b, then this is the usual form of Euler’s Theorem; see, e.g., [2, Theorems 12, 13]. If a 6= b, let Γ′ be the graph obtained from Γ by adding an edge from b to a. Eulerian paths w from a to b on Γ are in bijective correspondence with Eulerian circuits w′ on Γ′. Indeed, given w, we obtain w′ by appending e at the end. Conversely, given w′, after cyclically permuting the edges, we may assume that e is the last edge in w′. Now w is obtained from w′ by removing e. Finally, observe that conditions (2.1) are equivalent to outdegΓ′ (v) = indegΓ′ (v) for every vertex v. Thus Theorem 2.1 reduces to Euler’s theorem for Γ′.
Given a labeling of the edges e1,e2,...,em in Γ, we define the signature sgn(w) of an Eulerian path w = (eσ(1),...,eσ(m)) to be the signature of the permutation σ ∈ Sm. Note that changing the initial labeling e1,e2,...,em either leaves every sgn(w) unchanged or multiplies sgn(w) by −1 for every Eulerian path w. We will be particularly interested in the sgn(w); this sum is well-defined (i.e., is independent of the labeling signed sum w∈Eula(Γ) of the edges) up to a factor of −1. P We say that a graph has no repeated edges if there are no distinct edges which share both source and target vertices. The following lemma is remarked upon by Swan; [7, page 369]. Lemma . 2.2 Let Γ be a graph with a repeated edge. Then w∈Eula(Γ) sgn(w)=0 for any vertex a of Γ. P Proof. Let e1 and e2 be a pair of repeated edges. Let us partition the Eulerian paths in Eula(Γ) into two groups, Eul1 and Eul2, as follows: w ∈ Eul1 if e1 occurs before e2 in w and w ∈ Eul2 if e2 occurs before e1. Given an Eulerian path w on Γ, we can form a new ′ Eulerian path w by interchanging e1 and e2. This way we obtain a bijective correspondence ′ between Eul1 and Eul2. Since we have performed a transposition to get from w to w , sgn(w′)= −sgn(w). This shows that
sgn(w)= sgn(w)+ sgn(w′)=0, ′ w∈EulXa(Γ) w∈XEul1 w X∈Eul2 as desired.
The remainder of this section will be devoted to computing w∈Eula(Γ) sgn(w) for sev- eral families of graphs which will arise in the sequel. P A GRAPH-THEORETIC APPROACH TO A CONJECTURE OF DIXON AND PRESSMAN 5
Lemma 2.3. If P1
P T2 2 S1 T1
S2 T3 Γ = P0 P3 S3 Tα Sα
Pα ±α! , if a =0, or then sgn(w)= (±(α − 1)! otherwise. w∈EulXPa (Γ) Proof. First assume a = 0. There are α! Eulerian paths from P0 on Γ, determined by the order in which each of the vertices P1,...,Pα are visited. Each is of the form
wτ = (Sτ(1),Tτ(1),Sτ(2),Tτ(2),...,Sτ(α),Tτ(α)) for τ ∈ Sα. It thus suffices to show that these α! Eulerian paths all have the same signature. Indeed, the edges of wτ come in groups of 2, being (S1,T1),..., (Sα,Tα). Interchanging any two of these groups results in an even permutation of the edges. Thus
sgn(w)= sgn(wτ )= ±α! .
w∈Eul (Γ) τ∈Sα XP0 X Now assume that a 6= 0. In this case the Eulerian paths on Γ from Pa are precisely those ′ ′ of the form w = (Ta, w ,Sa), where w is an Eulerian path from P0 on Γ \{Sa,Ta}. As we showed above, there (α − 1)! possibilities for w′, and they all have the same signature; hence, sgn(w)= ±(α − 1)!. w∈EulXPa (Γ) Lemma 2.4. Let α > 2 and e
P1 P2
T1 S2 S1 P0 T2 Γ= T3 . P3 S3 Sα T4 S4 Tα
Pα P4 Then sgn(w)= ±(α − 1)!. w∈Eul (Γ) XP1 6 MATTHEWBRASSILANDZINOVYREICHSTEIN
Proof. Let us subdivide the Eulerian paths from P1 on Γ into three groups, Λ1,Λ2 and Λ3, depending on whether the edge e occurs at the beginning, the end or the middle of the path. It follows from Lemma 2.3, that there are (α − 1)! paths in Λ1, all having the same signature, and there are (α − 1)! paths in Λ2, all having the same signature. Moreover, the signature of a path from Λ1 is the same as the signature of a path from Λ2. This can be seen directly by comparing the signatures of, say (e,T2,S1,T1,τ,S2) ∈ Λ1 and (T1,τ,S2,T2,S1,e) ∈ Λ2, where τ = (S3,T3,...,Sα,Tα) is a path from P0 to P0. A simple calculation shows that
sgn(e,T2,S1,T1,τ,S2) = sgn(T1,τ,S2,T2,S1,e),
as claimed. We now turn our attention to Λ3. Any path in Λ3 begins with edge T1 and ends with edge S2. These paths are determined by the order in which the α − 1 circuits (S1,e,T2), (S3,T3), ..., (Sα,Tα) from P0 are traversed in the subgraph
e
P1 P2
S1 P0 T2 T3 . P3 S3 Sα T4 S4 Tα
Pα P4
Since interchanging any two of these circuits results in an even permutation of the edges, all paths in Λ3 have the same signature. Moreover, a path in Λ3 has the opposite signature to the paths in Λ1 and Λ2, as illustrated by
sgn(T1,S2,T2,τ,S1,e)= −sgn(T1,S1,e,T2,τ,S2) ,
where τ = (T3,S3,...,Tα,Sα). Thus
sgn(w)= ± sgn(w1)+ sgn(w2)+ sgn(w3) w∈Eul (Γ) w ∈Λ w ∈Λ w ∈Λ XP1 X1 1 X2 2 X3 3 = ± 2(α − 1)! − (α − 1)! = ±(α − 1)!. A GRAPH-THEORETIC APPROACH TO A CONJECTURE OF DIXON AND PRESSMAN 7
Lemma 2.5. Let a,b ∈{0, 1, 2,...,α} and
P1 P2 T2
S2 S1 T1 T3 e P3 Γ= . S3 Pb P0 a
Tα Sα
Pα Then ±(α + 1)! , if a = b =0, ±2(α − 1)! , if a = b 6=0, (2.2) sgn(w)= ±α! , if a =0 and b 6=0, or a 6=0 and b =0, w∈EulPa (Γ) X ±(α − 1)! , in all other cases, where for fixed b, the sum sgn(w) is either positive for all a, or negative for all w∈EulPa (Γ) a. P Proof. Any Eulerian path on Γ is a cyclic permutation of either
(e,Sτ(1),Tτ(1),Sτ(2),Tτ(2),...,Sτ(α),Tτ(α)) if b =0, or (Sτ(1),e,Tτ(1),Sτ(2),Tτ(2),...,Sτ(α),Tτ(α)) if b 6= 0, for some permutation τ ∈ Sα. Interchanging any two blocks of the form (Si1 ,Ti1 ) and (Si2 ,Ti2 ) induces an even permutation of the edges. Cyclically permuting a path of length 2α + 1 also induces an even permutation of the edges. Hence every Eulerian path has the same fixed signature when b = 0, and every Eulerian path has the same (opposite) fixed signature when b 6= 0. If a = b = 0, the Eulerian paths from P0 are determined by the order in which the α +1 circuits e, (S1,T1), ..., (Sα,Tα) are traversed. Thus there are (α + 1)! Eulerian paths in this case. If a = b 6= 0, then e occurs either at the beginning or the end of each Eulerian path. Lemma 2.3 tells us that there are (α − 1)! Eulerian paths starting with e and (α − 1)! Eulerian paths ending with e. Thus the total number of Eulerian paths from Pa in this case is 2(α − 1)!. If a = 0 and b 6= 0, then Eulerian paths from Pa are in bijective correspondence with permutations of the α circuits, (Sb,e,Tb) and (Si,Ti), where i 6= b. If a 6= 0 and b = 0, then every path starts with Ta and ends with Sa, and the Eulerian paths from Pa are in bijective correspondence with permutations of the remaining α circuits e and (Si,Ti), where i 6= a. Finally, if a, b are distinct and non-zero, then again every path starts with Ta and ends with Sa, so the count is the same as above, except that instead we only have α − 1 circuits, (Sb,e,Tb) and (Si,Ti), i 6= a,b. 8 MATTHEWBRASSILANDZINOVYREICHSTEIN
Lemma 2.6. If
P1
P T2 2 S1 T1 e S2 Sα+1 Sα+β Γ= T3
P3 P0 Tα+1 Tα+β Pα+β S3 Tα Sα
Pα then sgn(w)= ±2α! . w∈Eul (Γ) XPα+β
Proof. Any Eulerian path from Pα+β on Γ is either of the form
(e,Tα+β,...,Tα+1,w,Sα+1,...,Sα+β) , or (Tα+β,...,Tα+1,w,Sα+1,...,Sα+β,e) , where w is an Eulerian path from P0 on the subgraph of Γ consisting of edges S1,T1,S2,T2, ...,Sα,Tα. These two Eulerian paths on Γ for fixed w are related by the even permutation cyclically permuting the 2α+2β+1 edges, and all α! Eulerian paths from P0 on the subgraph have the same signature, by Lemma 2.5. Thus the signed count of Eulerian paths on Γ from Pα+β is ±2α!.
3. Proof of Theorem 1.2: First reductions Fix integers k,n > 1. Recall from the Introduction that given a k-tuple of n × n matrices A1,...,Ak, we defined the linear transformation L(A1,...,Ak): Matn → Matn by L(A1,...,Ak)(Ak+1) = [A1,...,Ak, Ak+1]. We will identify L(A1,...,Ak) with its matrix relative to the standard basis {Eab | a,b = 1,...,n} of elementary matrices in Matn. Here Eab is the elementary matrix with 1 in the (a,b)-position and 0s elsewhere; we will sometimes write Ea,b in place of Eab. k Let Wnull > i ⊂ (Matn) be the locus of k-tuples (A1,...,Ak) ∈ Matn such that 2 null(L(A1,...,Ak)) >i or equivalently, rank(L(A1,...,Ak))
null(L(A1,...,Ak)) > k for any A1,...,Ak ∈ Matn. (c) Assume that k is even and 2 6 n 6 2n − 2. In order to prove Theorem 1.2 it suffices to show that there exists a field K containing F and k matrices A1,...,Ak ∈ Matn(K) such that
null(L(A1,...,Ak)) 6 k. A GRAPH-THEORETIC APPROACH TO A CONJECTURE OF DIXON AND PRESSMAN 9
2 2 Proof. (a) The entries of the n × n matrix L(A1,...,Ak): Matn → Matn are poly- k nomials in the entries of A1,...,Ak. By definition Wnull > i ⊂ (Matn) is the common zero locus of the determinants of the (n2 −i)×(n2 −i)-minors of this matrix. These determinants are again polynomials in the entries of A1,...,Ak, and part (a) follows. (b) Since the standard polynomial [X1,...,Xk+1] is alternating in X1,...,Xk+1, we have
L(A1,...,Ak)(Aj ) = 0 for every j =1,...,k.
In other words, the kernel of L(A1, A2,...,Ak) contains the span of A1,...,Ak. If A1,...,Ak are linearly independent, this shows that null(L(A1,...,Ak)) > k. In other words, Wnull > k−1 k contains the dense open subvariety of Matn consisting of linearly independent k-tuples of n×n matrices. Since Wnull > k−1 is Zariski closed by part (a), we conclude that Wnull > k−1 = k Matn. (c) Note that under the assumptions of Theorem 1.2, k = 2r < 2n. Hence, k 6 n2, and part (b) applies. In view of part (b), Theorem 1.2 is equivalent to the assertion k k that Wnull > k 6= (Matn) . By part (a), Wnull > k is Zariski closed in Matn. To prove that k k Wnull > k 6= (Matn) , it suffices to show that the complement (Matn) \ Wnull > k has a K- point for some field K containing F . In other words, it suffices to show that there exist matrices A1,...,Ak ∈ Matn(K) such that null(L(A1,...,Ak)) 6 k.
Our proof of Theorem 1.2 will be based on Lemma 3.1(c). Note that it is not a priori clear how to choose the matrices A1,...,Ak. Informally speaking, if they are chosen to be very general (e.g., if their entries are independent variables over F ), it becomes difficult to compute L(A1,...,Ak) explicitly enough to determine its nullity. On the other hand, in multiple examples where we chose the k-tuple A1,...,Ak in various special positions, the 1 nullity of L(A1,...,Ak) turned out to be higher than k (and usually −→ ∞ with n) . The remainder of this section will be devoted to defining a field K containing F and a k-tuple A1,...,Ak ∈ Matn(K) that will, in retrospect, turn out to be “just right”: “special enough” to make null(L(A1,...,Ak)) computable, yet “general enough”, to ensure that
(3.1) null(L(A1,...,Ak)) 6 k.
The special property of A1,...,Ak that will facilitate subsequent computations is that the 2 2 n × n matrix L(A1,...,Ak) naturally decomposes as a direct sum of n × n matrices. On the other hand, the inequality (3.1) will not be obvious at this stage; its proof will take up much of the remainder of this paper. Note also that the k-tuple A1,...,Ak we will define in this section is really a family of k-tuples that depends on the integer parameters s1,...,sk. These integer parameters will remain unspecified until Section 6. From now on we will set K = F (xℓ,α), where xℓ,α are independent variables, as ℓ ranges from 1 to k and α ranges over Z/nZ. For notational convenience, we will label rows and columns of n × n matrices by 0, 1,...,n − 1 and view these labels as integers modulo n. Let
(3.2) D1 = diag(x1,0, x1,1,...,x1,n−1),...,Dk = diag(xk,0, xk,1,...,xk,n−1) be a k-tuple of diagonal matrices in Matn(K). We will study L(A1,...,Ak) for
s1 s2 sk (3.3) A1 = D1 · C , A2 = D2 · C ,..., Ak = Dk · C
1 2 As an extreme example of this phenomenon, L(A1,...,Ak) = 0 and hence has nullity n , if k > 2 and A1,...,Ak are required to commute pairwise. 10 MATTHEWBRASSILANDZINOVYREICHSTEIN in Matn(K). Here C denotes the cyclic permutation matrix 0 1 0 ··· 0 0 0 1 ··· 0 . . . . . (3.4) C = Ei,i+1 = ...... i∈Z/nZ 0 0 0 ··· 1 X 1 0 0 ··· 0 n and the exponents s1,...,sk are integers, to be specified later. Note that C = I, where I denotes the n × n identity matrix. Moreover, j CEii = Ei−1,i−1C and Ei,i+j = EiiC for every i, j ∈ Z/nZ.
Let Vj be the n-dimensional vector space spanned by the matrices j j j (3.5) E0,j = E0,0C , E1,j+1 = E1,1C ,...,En−1,n−1+j = En−1,n−1C .
Equivalently, Vj is the space of matrices all of whose non-zero entries are concentrated on the main diagonal, shifted up by j units, i.e., in positions (i,i + j), where j is fixed and i ranges over Z/nZ. Now observe that Matn = V0 ⊕ V1 ⊕···⊕ Vn−1. Moreover, every term in
j s1 sk j L(A1,...,Ak)(EiiC ) = [D1C ,...,DkC , EiiC ] j+s is of the form DC˜ for some diagonal matrix D˜. Here s = s1 + ... + sk. This matrix lies in Vj+s, where j + s is viewed modulo n. In other words, the linear transformation L = L(A1,...,Ak): Matn(K) → Matn(K) naturally decomposes as a direct sum of n linear maps Lj : Vj → Vj+s, where j ranges over 0, 1, 2,...,n − 1 and dim(Vj ) = n for each j. Consequently, Ker(L) decomposes as a direct sum Ker(L0) ⊕ ... ⊕ Ker(Ln−1). In summary, we have reduced Theorem 1.2 to the following. Proposition 3.2. Assume that k = 2r is even, n > r, the base field F is infinite of characteristic not dividing 2(2r + 1)r!, and the matrices A1,...,Ak are as in (3.3). Let
Lj : Vj → Vj+s1 +...+sk be the restriction of L(A1,...,Ak) to Vj . Then for some choice of the exponents s1,...,sk ∈ Z,
null(L0)+ ··· + null(Ln−1) 6 k. The remainder of this paper will be devoted to proving Proposition 3.2.
4. A graph-theoretic description of Lj
Throughout this section we fix positive integers k and n as well as s1,...,sk and j in Z/nZ. We will continue using the following notations: D1,...,Dk will be generic diagonal matrices, as in (3.2), and n−1 sℓ Aℓ = DℓC = xℓ,αEα,α+sℓ i=0 X will be as in (3.3) for ℓ = 1,...,k. As we saw in the previous section, for this choice of A1,...,Ak, the linear transformation L(A1,...,Ak) decomposes as L0 ⊕ L1 ⊕ ... ⊕ Ln−1. k We will identify the linear transformation Lj : Vj → Vj+s, where s = ℓ=1 sℓ, with the n × n matrix representing it in the bases (3.5) of Vj and Vj+s. P In order to find the matrix of Lj , we will want to calculate the generalized commutator
L(A1,...,Ak)(Ak+1) = [A1,...,Ak, Ak+1] A GRAPH-THEORETIC APPROACH TO A CONJECTURE OF DIXON AND PRESSMAN 11
j as Ak+1 ranges over the basis Eα,α+j = EααC of Vj . Expanding the generalized com- mutator [A1,...,Ak+1], we see that each entry is a multilinear polynomial in the groups of variables {x1,α}, {x2,α},..., {xk,α}, i.e., a linear combination of monomials of the form x1,α1 x2,α2 ··· xk,αk with integer coefficients. We will now give a graph-theoretic description of the coefficients of these monomials. Notational Conventions 4.1. (a) For the rest of this paper by a graph we will mean a directed graph with n vertices labeled Pv for v ∈ Z/nZ and at most k + 1 edges, labeled eℓ with ℓ from {1, 2,...,k +1}. For ℓ =1,...,k, the labeled edge eℓ will be of the form • −→ • e Pαℓ ℓ Pαℓ+sℓ and the labeled edge ek+1 will be of the form • −→ • . Each edge eℓ will appear e Pαk+1 k+1 Pαk+1+j in a given graph at most once. (b) We will say that a graph Γ is a disjoint union of Γ′ and Γ′′ and write Γ=Γ′ ∐ Γ′ if the edge set of Γ is the disjoint union of the edge sets of Γ′ and Γ′′. Here by “disjoint” ′ ′′ we mean that eℓ cannot be an edge in both Γ and Γ for any ℓ = 1,...,k + 1. Note that ′ ′′ the vertices of Γ, Γ and Γ are assumed to be Pv, v ∈ Z/nZ, as in (a).
(c) Let G be a graph with vertex set {Pv | v ∈ Z/nZ} whose edge set is a subset of {e1,...,ek}. We define the graph Gb to be
Gb = G ∐ • −→ • . Pb ek+1 Pb+j
In other words, Gb is the graph obtained from G by adding one extra edge ek+1 having source vertex Pb and target vertex Pb+j . Let m be a monomial of the form
m = x1,α1 x2,α2 ··· xk,αk where α1,...,αk are elements of Z/nZ. We define gr(m) to be the graph with k edges (4.1) • −→ • • −→ • ···• −→ • . e e e Pα1 1 Pα1+s1 Pα2 2 Pα2+s2 Pαk k Pαk +sk
Conversely, for a graph G with vertices of the form Pv, v ∈ Z/nZ whose edges have labels e1,e2,...,ek, we define the monomial mon(G) to be
(4.2) mon(G)= x1,srcG(e1)x2,srcG(e2) ··· xk,srcG(ek ) . Note that our definitions of mon(G) and gr(m) are inverse to each other: mon(gr(m)) = m and gr(mon(G)) = G for any monomial m and graph G of our required form. We will use this correspondence between graphs and monomials to study the entries of Lj in a graph- theoretic manner. Graphs of the form gr(m) correspond to monomials m which may appear in the matrix Lj. The addition of the extra edge ek+1 in gr(m)b will help us keep track of th the coefficient of m in the b column of Lj, as is explained in the lemma below.
Lemma 4.2. Fix α1,...,αk and b to be integers defined modulo n. Define m to be the monomial
m = x1,α1 x2,α2 ··· xk,αk 12 MATTHEWBRASSILANDZINOVYREICHSTEIN
th and G = gr(m). The coefficient of m appearing in the (a,b) entry of Lj is sgn(w) . w∈EulXPa (Gb)
Here, as usual, EulPa (Gb) denotes the set of Eulerian paths on Gb originating at Pa. j j j Proof. Recall that Vj is spanned by the ordered basis {E0,0C , E1,1,C ,...,En−1,n−1C }. The matrix Lj is the restriction of L to Vj , mapping Vj to Vj+s where s = s1 + s2 + ··· + sk. th j+s j Thus the (a,b) entry of Lj is the coefficient of Ea,aC appearing in Lj (Eb,bC ) = j th j L(Eb,bC ). This is the (a,a + j + s) entry of L(Eb,bC ). j Write αk+1 = b and Ak+1 = Eb,bC . To calculate the coefficients of x1,α1 ··· xk,αk appearing in Aσ(1) ··· Aσ(k+1) we set all indeterminants xℓ,α, other than x1,α1 ,...,xk,αk , to
0. Thus the coefficients of x1,α1 ··· xk,αk appearing in Aσ(1) ··· Aσ(k+1) are the same as the coefficients of x1,α1 ··· xk,αk appearing in Bσ(1) ··· Bσ(k+1), where for ℓ =1, 2,...,k, 0 ... 0 ... 0 ...... . . . . . B = 0 ... x ... 0 · Csi = x E ℓ ℓ,αℓ ℓ,αℓ αℓ,αℓ+sℓ . . . . ...... 0 ... 0 ... 0 and Bk+1 = Ak+1 = Eb,b+j . Recall that by the definition of gr(m),
(αℓ, αℓ + sℓ) = (src(eℓ), tar(eℓ)) for ℓ =1,...,k, and (b,b + j) = (src(ek+1), tar(ek+1)).
As in Swan’s paper [7], Lj (Bk+1) = [B1,...,Bk+1] can be described by counting Eulerian paths. Using the product rule ′ Ep,q′ if q = p , and Ep,qEp′,q′ = (0 otherwise , we see that
Bσ(1) · ... · Bσ(k+1) = mEsrc(eσ(1) ),tar(eσ(1) ) ··· Esrc(eσ(k+1) ),tar(eσ(k+1) ) th has an m = x1,α1 ,...,xk,αk in the (src(eσ(1)), tar(eσ(k+1))) entry, if and only if tar(eσ(ℓ))= src(eσ(ℓ+1)) for each ℓ =1, 2,...,k. This is precisely the requirement that (eσ(1),...,eσ(k+1)) forms an Eulerian path from Psrc(eσ(1) ) to Ptar(eσ(k+1)) on Gb. All other entries of this product are zero. When (eσ(1),...,eσ(k+1)) does form an Eulerian path on Gb, this path terminates at tar(eσ(k+1)) = src(eσ(1))+ s1 + s2 + ··· + sk + j = src(eσ(1))+ j + s, as s1,...,sk and j are the differences between the source and target vertices of each edge in the path. Thus the th (a,a + j + s) entry of Bσ(1) ··· Bσ(k+1) is m = x1,α1 ··· xk,αk if and only if src(eσ(1))= Pa and (eσ(1),...,eσ(k+1)) is an Eulerian path on Gb, that is, if and only if (eσ(1),...,eσ(k+1)) is an Eulerian path from Pa on Gb.
Summing over all permutations σ ∈ Sk+1 we obtain that the coefficient of x1,α1 ,...,xk,αk th j appearing in the (a,a+j +s) entry of L(Eb,bC ) = [A1,...,Ak+1] is sgn(w), w∈EulPa (Gb) th and therefore the (a,b) entry of Lj is sgn(w). w∈EulPa (Gb) P
With this lemma we can study Lj byP considering only graphs of the form gr(m)b which admit Eulerian paths. We define a set of graphs which could possibly give rise to a nonzero th coefficient in the a row of Lj. A GRAPH-THEORETIC APPROACH TO A CONJECTURE OF DIXON AND PRESSMAN 13
Definition 4.3. Fix a, j and s1,s2,...,sk to be elements of Z/nZ and let I ⊆{1, 2,...,k}. Define U(a, j, I) to be the set of directed graphs G, as in Notational Conventions 4.1(a), satisfying the following additional conditions.
(i) The edges of G are precisely eℓ for ℓ ∈ I.
(ii) Each edge eℓ for ℓ ∈ I is of the form • −→ • for some αℓ. e Pαℓ ℓ Pαℓ+sℓ (iii) G has no repeated edges.
(iv) There is some b ∈ Z/nZ such that Gb has an Eulerian path from Pa in Gb. We abbreviate U(a, j)= U a, j, {1, 2,...,k} . and
U (a, j)= G ∈ U(a, j) | There exists b ∈ Z/nZ such that sgn(w) 6=0 . nz w∈EulXPa (Gb) Remark 4.4. We will be primarily interested in the case where I = {1,...,k}. We allow I to be a proper subset of {1, 2,...,k} to facilitate induction arguments later on. In the case where I = {1, 2,...,k}, conditions (i) and (ii) are equivalent to the re- quirement that G is of the form gr(m) for some monomial m = x1,α1 x2,α2 ...xk,αk . The reason for conditions (iii) and (iv) is that if they fail, then sgn(w) = 0 for w∈EulPa (Gb) each b ∈ Z/nZ; see Lemma 2.2. Thus by Lemma 4.2 the monomial m never appears in the P th ath row of Lj, and the graph G does not contribute anything to the a row of Lj. For the same reason we are only really interested in graphs from Unz(a, j). However, it is not always transparent which graphs lie in Unz(a, j), so as a preliminary step, it will be convenient for us to work with all graphs from U(a, j). Here is a brief example illustrating Definition 4.3.
Example 4.5. Let k = 2, n = 3, j = 1 and s1 = 1, s2 = −1. For fixed a, U(a, j) consists of 4 graphs, being
e e1 G1 = 1 , G2 = , Pa Pa+1 e2 Pa+2 Pa Pa+1 e2 Pa+2 e1 e G3 = and G4 = 1 . Pa−1 e2 Pa Pa+1 Pa−1 e2 Pa Pa+1
The only graphs of the form Gb which admit Eulerian paths from Pa, for G ∈ U(a, j), are
e3 e1 1 e1 2 e3 Ga+1 = , Ga = , Pa Pa+1 e2 Pa+2 Pa Pa+1 e2 Pa+2
e1 e3 3 e3 4 e1 Ga = and Ga−1 . Pa−1 e2 Pa Pa+1 Pa−1 e2 Pa Pa+1
Each of the above graphs has a unique Eulerian path from Pa and so Unz(a, j)= U(a, j). th For fixed a ∈ {0, 1, 2}, by Lemma 4.2, the only monomials appearing in the a row of L1 are the following. th (1) x1,ax2,a+2 appears with coefficient −1 in the (a,a + 1) entry, corresponding to path (e1,e3,e2). 14 MATTHEWBRASSILANDZINOVYREICHSTEIN
th (2) x1,a+1x2,a+2 appears with coefficient 1 in the (a,a) entry, corresponding to path (e3,e1,e2). th (3) x1,a−1x2,a appears with coefficient −1 in the (a,a) entry, corresponding to path (e2,e1,e3). th (4) x1,ax2,a appears with coefficient 1 in the (a,a − 1) entry, corresponding to path (e2,e3,e1). We obtain
x1,1x2,2 − x1,2x2,0 −x1,0x2,2 x1,0x2,0 L = x x x x − x x −x x . 1 1,1 2,1 1,2 2,0 1,0 2,1 1,1 2,0 −x1,2x2,1 x1,2x2,2 x1,0x2,1 − x1,1x2,2 5. The matrix of initial coefficients In order to prove Proposition 3.2 (and thus Theorem 1.2), we need to bound the nul- lities of L0,L1,...,Ln−1 from above. We will not be able to work with the matrices Lj directly; their entries are too complicated (recall that these entries are polynomials in the variables xℓ,α). Our approach will be to consider the matrices Ic(Lj ) arising from the initial coefficients of Lj with respect to a suitably defined lexicographic monomial order on the variables xℓ,α. The matrix Ic(Lj ) will turn out to be more manageable than Lj and as we shall soon see, its nullity will give us an upper bound on the nullity of Lj .
Definition 5.1. Let R = F [x1,...,xt] be a polynomial ring. (a) We define a lexicographic order ≻ on R to be a total order on monomials from R induced by an order on the variables x1,...,xt. (b) Let f ∈ R and write f = i∈I cimi for nonzero coefficients ci ∈ F and distinct monomials mi ∈ R. If m is the maximal monomial from {mi | i ∈ I} with respect to ≻, then we define the initial monomial of f Pto be In(f)= m. The leading coefficient of f is defined to be the coefficient of In(f) in f. (c) We define the matrix Ic(M) ∈ Matn(F ) of initial coefficients of an n × n matrix th M ∈ Matn(R) as follows. For each a let ma be the largest monomial occurring in the a row of M, that is, ma = max(In(Ma,0),..., In(Ma,n−1)). Then the entry Ic(M)a,b of Ic(M) in position (a,b) is defined to be the coefficient of ma in Ma,b. Lemma 5.2. Let R be a polynomial ring as above, and M be an n × n matrix with coefficients in R. Then null(M) 6 null(Ic(M)). Proof. Suppose r = rank(Ic(M)). Then there exists a non-singular r × r submatrix N0 of Ic(M). Let us say that N0 is obtained from Ic(M) by removing rows a1,...,an−r and columns b1,...,bn−r. Let N be the r × r submatrix of M obtained by removing the same rows a1,...,an−r and columns b1,...,bn−r. Clearly, det(N0) is the leading coefficient of det(N). Hence, det(N) 6= 0 and consequently, rank(M) > r = rank(Ic(M)). Equivalently, null(M) 6 null(Ic(M)) = n − r. Recall that a graph G ∈ U(a, j) determines a monomial mon(G) by the source vertices th of its edges, as in (4.2). To determine the a row of Ic(Lj ) we need only consider the maximal graph from Unz(a, j), where we define G1 ≻ G2 if mon(G1) ≻ mon(G2).
Lemma 5.3. Assume Unz(a, j) is nonempty, and let G be the maximal graph from th Unz(a, j). Then the (a,b) entry of Ic(Lj) is
(Ic(Lj ))a,b = sgn(w) . w∈EulXPa (Gb) A GRAPH-THEORETIC APPROACH TO A CONJECTURE OF DIXON AND PRESSMAN 15
th Proof. Let m be the maximal monomial appearing in the a row of Lj with nonzero coefficient. Then gr(m) is the maximal graph in Unz(a, j). Thus mon(G) = m, and the th th (a,b) -entry of Ic(Lj ) is the coefficient of mon(G) appearing in the (a,b) -entry of Lj. By Lemma 4.2, this coefficient is
(Ic(Lj ))a,b = sgn(w) . w∈EulXPa (Gb)
Example 5.4. Let k = 2, n = 3, j = 1 and s1 = 1, s2 = −1, as in Example 4.5. Define an order on graphs from U(a, j) to be the lexicographic order induced by the order on edges
≻ ≻ ≻ ≻ ≻ . P e1 P e2 e1 P e2 P P e1 P P e2 P 2 P0 2 P0 P0 1 P0 1 2 1 2 1 x1,2 x2,0 x1,0 x2,1 x1,1 x2,2
The correspondence between variables xℓ,α and pairs (eℓ, Pα) of labeled edges with source vertices determines a monomial order. The graph order defined here corresponds to the lex- icographic monomial order on the polynomial ring F [x1,0, x1,1, x1,2, x2,0, x2,1, x2,2] induced by x1,2 ≻ x2,0 ≻ x1,0 ≻ x2,1 ≻ x1,1 ≻ x2,2. Our maximal graph from U(a, 1) is then
e1 G = P2 P1 e2 P0 for any choice of a = 0, 1, 2. This graph G does lie in Unz(a, 1) when a = 0, 1, however when a = 2, the only graph of the form Gb admitting Eulerian paths is G2. But G2 has a repeated edge from P2 to P0. By Lemma 2.2 we have sgn(w) = 0 and hence w∈EulP2 (G2) G/∈ U (2, 1). We consider the next largest graph from U(2, 1) and see that the maximal nz P graphs from Unz(a, 1) are
e1 e1 , and . P2 P1 P2 P1 P2 e1 e2 P1 e2 P P 0 e2 P0 0 a =0 a =1 a =2
For each of the above 3 graphs, there is precisely one placement of the edge e3 such that the graph Gb = G ∐ • → • admits an Eulerian path from Pa. These are Pb e3 Pb+1
e3 e e 1 e 3 , 1 and . P2 e3 P1 P2 P1 P2 e1 P1 e2 P P e2 0 e2 P0 0 a =0,b =0 a =1,b =1 a =2,b =0 By Lemma 5.3 we obtain −1 0 0 Ic(L1)= 0 10 −1 0 0 from which we see that null(Ic(L1)) = 1 and hence null(L1) 6 1 by Lemma 5.2. 16 MATTHEWBRASSILANDZINOVYREICHSTEIN
In our description of L1 in Example 4.5, the largest monomials in each row are x1,2x2,0, x1,2x2,0 and x1,2x2,1 respectively. Lemma 5.3 allows us to extract the coefficients of these monomials by considering the associated order on graphs from U(a, j).
6. Specialization of the lexicographic order and the exponents si From now on we will assume that k =2r is even and 1 6 r 6 n − 1. We will now choose our exponents s1,...,sk. These exponents will be fixed for the remainder of the proof sℓ of Theorem 1.2. The matrices Aℓ = DℓC , ℓ = 1,...,k and the linear transformations L(A1,...,Ak) and Lj, j =0, 1,...,n − 1 defined in Section 3, will also be fixed. We define i s = ⌈ ⌉, for i =1, 2,...,r, and i 2 i s = s = −s = −⌈ ⌉, for i =1, 2, . . . , r . r+i −i i 2 In other words,
(s1,s2,s3,...,sk) = (s1,s2,s3,...,sr,s−1,s−2,...,s−r) 1 2 3 r 1 2 r = ⌈ ⌉, ⌈ ⌉, ⌈ ⌉,..., ⌈ ⌉, −⌈ ⌉, −⌈ ⌉,..., −⌈ ⌉ 2 2 2 2 2 2 2 r r = (1, 1, 2,..., ⌈ ⌉, −1, −1,..., −⌈ ⌉) 2 2 and our matrices A1,...,Ak specialize to 2 ⌈ r ⌉ A1 = D1C, A2 = D2C, A3 = D3C , ... , Ar = DrC 2 , −1 −1 −⌈ r ⌉ Ar+1 = Dr+1C , Ar+2 = Dr+2C , ... , Ak = DkC 2 . Notational Conventions 6.1. In the sequel [a ←→ b] will denote an ‘interval’ in Z/nZ, i.e., the set of successive integers a,a+1,... in Z/nZ up to the first integer congruent to b modulo n. Definition 6.2. For v ∈ Z/nZ we let |v| denote the absolute value of the representative n−1 n−1 of v in [−⌈ 2 ⌉ ←→⌊ 2 ⌋]. For two vertices Pv1 , Pv2 , v1, v2 ∈ Z/nZ we define n − 1 P ≻ P , if |v | < |v |, or if v = −v and v ∈ [1 ←→ ⌊ ⌋]. v1 v2 1 2 1 2 2 2 Our order on vertices becomes
(6.1) P0 ≻ P−1 ≻ P1 ≻ P−2 ≻ P2 ≻ ....
This defines a lexicographic order on unordered pairs of vertices. If max(Pv1 , Pv2 ) ≻ max(P ′ , P ′ ), or max(P , P ) = max(P ′ , P ′ ) and min(P , P ) ≻ min(P ′ , P ′ ), then v1 v2 v1 v2 v1 v2 v1 v2 v1 v2 we define (P , P ) ≻ (P ′ , P ′ ). v1 v2 v1 v2 Next we define an order on pairs of labeled edges with their source vertices. For ℓ1,ℓ2 ∈ {1, 2,...,k} and v1, v2 ∈ Z/nZ we define
(eℓ1 , Pv1 ) ≻ (eℓ2 , Pv2 ), if (Pv1 , Pv1+sℓ ) ≻ (Pv2 , Pv2+sℓ ) , (6.2) 1 2 (e , P ) ≻ (e , P ), if (P , P ) = (P , P ) , and ℓ <ℓ . ℓ1 v1 ℓ2 v2 v1 v1+sℓ1 v2 v2+sℓ2 1 2 The equality in the second line of this definition is of unordered pairs. Note that for n > r, the equality (P , P ) = (P , P ) can only hold if s = ±s . v1 v1+sℓ1 v2 v2+sℓ2 ℓ1 ℓ2 This order on pairs of labeled edges and vertices determines an order on graphs from U(a, j, I). A graph G ∈ U(a, j, I) is determined by the pairs (eℓ, srcG(eℓ)) for ℓ ∈ I. Our order on U(a, j, I) is the lexicographic order induced by the order on these pairs in (6.2). A GRAPH-THEORETIC APPROACH TO A CONJECTURE OF DIXON AND PRESSMAN 17
In a graph of the form G = G(m), an edge eℓ will satisfy tarG(eℓ) = srcG(eℓ)+ sℓ. Thus when comparing edges e ∈ G and e′ ∈ G′, the order given in (6.2) first compares the ′ ′ unordered pairs (srcG(e), tarG(e)) and (srcG′ (e ), tarG′ (e )), with ties broken by comparing edge labels. Our global pictures of the n vertices used in the graphs of U(a, j), when n is even and odd respectively, are
n P n−1 P− P n−1 2 P n −1 − 2 n 2 2 P n− P1− 1 −1 2 2 P n− 1− 1 2
(6.3) and .
P−3 P−3 P2 P2
P−2 P−2 P1 P1 P−1 P−1 P0 P0 In these pictures, the lower vertices have larger weight, corresponding to our order on vertices P0 ≻ P−1 ≻ P1 ≻ P−2 ≻ ... . The largest edges will be those incident on P0 and, more generally, the lower an edge appears in the above picture, the larger weight it has. The graphs G ∈ U(a, j) are defined so that there exists an Eulerian path from Pa on Gb for some b. Conversely, a graph in U(a, j) can be defined by a path (eσ(1),eσ(2),...,eσ(k+1)) on the n vertices. In the largest graph from U(a, j), this path will reach the lowest possible vertex in (6.3). Intuitively, if Pa is to the left of P0 in (6.3), we would suspect an Eulerian path on Gb from Pa on a maximal graph G ∈ U(a, j) to traverse down the left side of the picture as low as possible. If such a path can reach P0, then the remaining edges will be filled in to be incident to P0. This intuition will be formalized in the next section. The order given in Example 5.4 is equivalent to the order defined above when k = 2 and n = 3. The unordered pairs of vertices satisfy
(P0, P−1) ≻ (P0, P1) ≻ (P−1, P1) , and if the two edges share both source and target vertices we break ties by declaring e1 ≻ e2. Our proof of Proposition 3.2 will be based on the following. Proposition 6.3. Assume that n > r and that the base field F is infinite of character- istic not dividing r!. (a) If j 6=0 then
null(Ic(Lj)) 6 δj , where 2, if both j and −j lie in [⌈r/2⌉ ←→⌊r/2⌋], δj = 1, if exactly one of j, −j lies in [⌈r/2⌉ ←→⌊r/2⌋], 0, if neither j nor −j lie in [⌈r/2⌉ ←→⌊r/2⌋].
(b) Assume further that the characteristic of F does not divide 2(2r + 1)r!. Then the n × n matrix Ic(L0) is non-singular. 18 MATTHEWBRASSILANDZINOVYREICHSTEIN
Here Ic(Lj ) is the matrix of initial coefficients of Lj with respect to the order on graphs described in Definition 6.2. To see that Proposition 6.3 implies Proposition 3.2 (and thus Theorem 1.2), assume for a moment that Proposition 6.3 is established. Then by Lemma 5.2, n−1 null(L0)+null(L1)+... null(Ln−1) 6 null(Ic(L0))+null(Ic(L1))+... null(Ic(Ln−1)) 6 δj . j=1 X n−1 Each 0 6= a ∈ [⌈r/2⌉ ←→⌊r/2⌋] contributes exactly 2 to the sum j=1 δj , one when j = a and one when −j = a. (Note that a contributes 2 to this sum even if a = −a in Z/nZ.) P Since there are exactly r non-zero elements a in the interval [⌈r/2⌉ ←→⌊r/2⌋], we conclude n−1 n−1 that j=1 δj =2r = k. Substituting k for j=1 δj into the above inequality, we obtain P null(L0) + null(L1P)+ ... null(Ln−1) 6 k, and Proposition 3.2 follows.
7. Maximal graphs As a first step towards proving Proposition 6.3, we will now describe the maximal graph of U(a, j) for fixed a and j under the ordering defined in Section 6. This is a purely graph- theoretic problem. The answer is given by Proposition 7.2, whose proof will be completed in the next section. In the definition of U(a, j) our k edges were given labels e1,...,ek. In the sequence it will be convenient for us to use the following alternative labels with negative indices:
e−1 = er+1,e−2 = er+2,...,e−r = ek .
The graphs of U(a, j) will then be defined by the source vertices vi of the 2r edges of the form • −→ • Pvi ei Pvi+si i i as i ranges over {±1, ±2,..., ±r}, where si = ⌈ 2 ⌉ and s−i = −⌈ 2 ⌉ for i =1, 2,...,r. For each t > 0 let Ht be the graph consisting of the t pairs of edges of the form
e1−2i e−2i if i< 0, and if i> 0, e2i−1 e2i Pi P0 P0 Pi as i ranges over [−⌈t/2⌉ ←→ ⌊t/2⌋] \{0}. When t = 0, H0 is the empty graph with no edges, and for each t, Ht+1 is the disjoint union of Ht with a pair of edges connecting P0 to the next largest vertex. For example, H5 is the graph
e5
e−4 e−5 P0 e3 P−3 . e4 e−3 e e1 −2 P2
P−2 e−1 e2 P1 P−1
The following lemma shows that the maximal graph from U(a, j) will contain Ht for the largest possible t. A GRAPH-THEORETIC APPROACH TO A CONJECTURE OF DIXON AND PRESSMAN 19
′ ′ Lemma 7.1. Let G, G ∈ U(a, j). If there exists t such that G ⊃ Ht and G 6⊃ Ht, then G ≻ G′. Proof. Our order on edges follows the lexicographic order on pairs of vertices, with
(P0, P−1) ≻ (P0, P1) ≻ (P0, P−2) ≻ (P0, P2) ≻ ... being the maximal pairs. There are at most 2 edges connecting any pair of vertices in a graph with no repeated edges, and if two edges connect the same pair of vertices, ties are broken using the following order on the edge labels:
e2α+1 ≻ e2α+2 ≻ e−2α−1 ≻ e−2α−2 for any α > 0. We conclude that the largest 2t edges that could possibly appear in a graph G ∈ U(a, j) are those in Ht, and the lemma follows. Our next goal is to prove the following proposition, which describes the maximal graph of U(a, j). Proposition 7.2. Let t be the largest integer such that there exists G ∈ U(a, j) con- taining Ht. Let Pa′ = max(Pa, Pa+j ). ′ n−1 ˆ (a) If a ∈ [−⌈ 2 ⌉ ←→ 0], then define G to be the graph
er ′ er−1 Pa e−r ′ Pa +sr e−r+1 et+1
′ Pa +sr +···+st+1 e−t−1
Ht
. ′ In other words, Gˆ = Ht ∐ Gˆ where
er ′ ′ er−1 Gˆ = Pa . e−r ′ Pa +sr e−r+1 et+1
′ Pa +sr +···+st+1 e−t−1
′ n−1 ˆ (b) If a ∈ [1 ←→ ⌊ 2 ⌋], then define G to be the graph
e−r e−r+1 Pa′ er e−t+1 e ′ r−1 Pa −sr
et−1 ′ Pa −sr−···−st+1 Ht
. 20 MATTHEWBRASSILANDZINOVYREICHSTEIN
′ In other words, Gˆ = Ht ∐ Gˆ where
e−r ′ e−r+1 Gˆ = Pa′ . er e−t+1 e ′ r−1 Pa −sr
et−1 ′ Pa −sr −···−st+1
Then Gˆ is the maximal graph from U(a, j).
This result is a key step in our proof of Proposition 6.3 and thus of Theorem 1.2. The remainder of this section will be devoted to proving a preparatory lemma, Lemma 7.3. It de- scribes the maximal graphs in U(a, j, I), where I is a subset of {1, 2,...,r}∪{−1, −2,..., −r}, under certain conditions on a, j and I. In the subsequent application {ei | i ∈ I} will be the set of edges that are not used in Ht. We will use Lemma 7.3 to complete the proof of Proposition 7.2 in Section 8.
Lemma 7.3. Let I = {i1,i2,...,iα, −i1, −i2,..., −iα}⊆{1, 2,...,r}∪{−1, −2,..., −r} have corresponding edges labeled ei1 ,ei2 ,...,eiα ,e−i1 ,e−i2 ,...,e−iα for some iα > ··· > ′ i1 > 0. Assume that si1 + si2 + ··· + siα 6 |a|, |a + j|, Pa = max(Pa, Pa+j ), and G is the maximal graph in U(a, j, I). ′ n−1 (a) If a ∈ [−⌈ 2 ⌉ ←→ 0], then G is the graph
eiα ei − Pa′ α 1 e−iα P ′ a +siα e e−iα−1 i1 Pa′+s +···+s e iα i1 −i1 .
′ n−1 (b) If a ∈ [1 ←→ ⌊ 2 ⌋], then G is the graph
e−iα
e−iα−1 Pa′ eiα
e−i1 eiα−1 ′ Pa −siα
ei1 Pa′−s −···−s iα i1 .
′ ′ Proof. Let Pv = max(Pa +s, Pa −s), where s = si1 + si2 + ··· + siα . We claim that the edges ei1 and e−i1 appear in G, incident on Pv, as in the picture above. That is, if ′ n−1 ′ a ∈ [−⌈ 2 ⌉ ←→ 0], then v = a + s and
ei1
e−i1 P P v−si1 v A GRAPH-THEORETIC APPROACH TO A CONJECTURE OF DIXON AND PRESSMAN 21
′ n−1 ′ appears in G, and if a ∈ [1 ←→ ⌊ 2 ⌋], then v = a − s and
e−i1
ei1 P P v v+si1 appears in G. Let b ∈ Z/nZ be such that Gb = G ∐ • → • has an Eulerian path from Pa. Such Pb ek+1 Pb+j a b exists, by the definition of U(a, j, I). Our order on edges is determined by our order on vertices. The largest edge which appears in some graph from U(a, j, I) is incident on the largest vertex reachable from Pa using edges of the form ei for i ∈ I and ek+1. We will show that Pv is the largest such vertex. The vertices reachable from Pa in Gb using edges of the form ei for i ∈ I and ek+1 are necessarily of the form Pλ for some
λ ∈ [a−si1 −···−siα ←→ a+si1 +···+siα ]∪[a+j −si1 −···−siα ←→ a+j +si1 +···+siα ].
By our assumption of s 6 |a|, |a+j|, where s = si1 +si2 +··· siα , the intervals [a−s ←→ a + s] and [a + j − s ←→ a + j + s] are of the form
a˜
a˜ + s
a˜ − s
0 fora ˜ = a or a+j. In particular, 0 does not lie in the interior of either interval. The maximal vertex of this form is Pv0 , where
|v0| = min(|a|− si1 − si2 −···− siα , |a + j|− si1 − si2 −···− siα ) .
For Pv0 to be reachable from Pa in Gb, v0 − a must be expressible as a partial sum of the integers ±si1 , ±si2 ,..., ±siα and j. The only way this can happen is if
v0 = a ± (si1 + si2 + ··· + siα ), or v0 = a + j ± (si1 + si2 + ··· + siα ) .
′ n−1 ′ ′ If a ∈ [−⌈ 2 ⌉ ←→ 0], then v0 = a + si1 + si2 + ··· + siα . Otherwise v0 = a − si1 − si2 −
···− siα . That is, we have v = v0, and Pv is the largest vertex which may appear in a graph from U(a, j, I) with nonzero degree. As G is maximal, v0 6= a,a + j, and there exists an Eulerian path from Pa to Pa+j on
Gb, there must be at least 2 edges incident on Pv0 in Gb. The maximal edges which can be incident on Pv0 are e±i1 . Hence, in a maximal graph these edges must be incident on Pv0 . This proves the claim. We will now complete the proof of Lemma 7.3 by induction on α. Let us assume ′ n−1 ′ n−1 a ∈ [−⌈ 2 ⌉ ←→ 0]; when a ∈ [1 ←→ ⌊ 2 ⌋] the proof is symmetric. In this case, the 22 MATTHEWBRASSILANDZINOVYREICHSTEIN maximal graph G ∈ U(a, j, I) must include the edges
ei1 .
e−i1 P P v−si1 v
By induction, the maximal graph in U(a, j, I \{i1, −i1}) is
eiα ′ ei − G = Pa′ α 1 . e−iα P ′ a +siα e e−iα+1 i2 P v−si1 e−i2
ei1 We see that G′ ∐ • ⇄ • lies in U(a, j, I), and as no graph from U(a, j, I) can be Pv−s − Pv i1 e i1 larger than this graph, it must be maximal.
8. Conclusion of the proof of Proposition 7.2 ′ Let G = G ∐ Ht be the maximal graph from U(a, j). By Lemma 7.1 no G˜ ∈ U(a, j) ′ can contain Ht′ for t >t. As G ∈ U(a, j), there exists b ∈ Z/nZ such that
Gb = G ∐ • → • Pb ek+1 Pb+j admits an Eulerian path from Pa to Pa+j . For t ∈{0, 1,...,r} define + It = {t +1,t +2,...,r} , − It = {−t − 1, −t − 2,..., −r} , + − It = It ∪ It . Furthermore, let
Rt(a)= {a + p | p is a partial sum of si,i ∈ It and j} ′ be the set of indices v such that Pv is reachable from Pa using only edges from G , i.e., a subset of {ei | i ∈ It}∪{ek+1}. Recall that
(s1,s2,s3,...,sk) = (s1,s2,s3,...,sr,s−1,s−2,...,s−r) 1 2 3 r 1 2 r = ⌈ ⌉, ⌈ ⌉, ⌈ ⌉,..., ⌈ ⌉, −⌈ ⌉, −⌈ ⌉,..., −⌈ ⌉ 2 2 2 2 2 2 2 r r = (1, 1, 2,..., ⌈ ⌉, −1, −1,..., −⌈ ⌉). 2 2
If we set sk+1 = j, then Rt(a)= {a + λ∈Q sλ | Q is a subset of It ∪{k +1}}.
Lemma 8.1. Assume n > r. ThereP exists G ∈ U(a, j) containing Ht (for t> 0) if and only if Rt(a) ∩ supp(Ht) 6= ∅. Here t t t t t {−⌈ 2 ⌉, −⌈ 2 ⌉ +1,..., ⌊ 2 ⌋} = [−⌈ 2 ⌉ ←→⌊ 2 ⌋] , if t> 0 supp(Ht)= (∅ , if t = 0. A GRAPH-THEORETIC APPROACH TO A CONJECTURE OF DIXON AND PRESSMAN 23 denotes the support of Ht, i.e., the set of subscripts i such that Pi is adjacent to at least one edge from Ht. For example, supp(H0)= ∅, supp(H1)= {−1, 0}, supp(H2)= {−1, 0, 1}, supp(H3)= {−2, −1, 0, 1}, etc. Note also that It = {±1, ±2,..., ±r}\ supp(Ht).
Proof. Suppose that there exists G ∈ U(a, j) containing Ht with t > 0. Then there is an Eulerian path (ei1 ,ei2 ,...,eik+1 ) from Pa in Gb for some b ∈ Z/nZ. As this path traverses Gb, it must traverse Ht. Let Pv be the first vertex on this path with v ∈ supp(Ht) and m be the first integer with tar(eim ) = Pv (with m = 0 if Pa = Pv). Then, setting m 6 6 sk+1 = j, we have v = a + h=1 sih . Each eih with 1 h m has src(eih ) ∈/ Ht and so ih ∈ It ∪{k +1} by the definition of Ht. Thus v ∈ Rt(a) as required. P ′ Conversely, suppose that Rt(a) ∩ supp(Ht) 6= ∅ and let I be a minimal subset of It ∪{k +1} such that v = a + i∈I′ si lies in supp(Ht). Note that at most one of i and −i can lie in I′ \{k +1} for any i by the minimality of I′ (otherwise we can remove both from I′). Now consider two cases. P ′ Case 1. k +1 6∈ I . We claim that there exists a graph G with edges e±1,...,e±r containing Ht such that G has no repeated edges and G has an Eulerian path starting at Pa. If we can prove this claim, then appending ek+1 to this Eulerian path at the end, we obtain an Eulerian path for Ga starting at Pa. Thus G ∈ U(a, j), and the proof in Case 1 will be complete. ′ ′′ ′′ To prove the claim, set G = Ht ∐ G where G is a path from Pa to Pv with edge ′ ′ labels ei, i ∈ I and back to Pa with edge labels e−i, i ∈ I . Since Ht is an Euler circuit, ′ ′ G admits an Eulerian path. Note that by our construction, edges in G come in pairs eℓ ′ and e−ℓ so that src(eℓ) = tar(e−ℓ) for each ℓ. If G has edges for all the labels e±1,...,e±r, then we can set G = G′ and our proof is complete. If not, then we can construct a G from ′ G recursively, by attaching missing edges in pairs, eℓ and e−ℓ, as follows
eℓ . e−ℓ
Pc Pc+sℓ Here c ∈ supp(G′). Note that for a given d ∈ Z/nZ, there are at most two ℓ ∈ {±1,..., ±r} ′ such that sℓ = d. If we want to add eℓ and e−ℓ to G , and eℓ′ and e−ℓ′ with sℓ = sℓ′ are not ′ ′ present in G , then we can place eℓ and e−ℓ at any Pc, where c ∈ supp(G ), as above. This way the extended graph will have no repeated edges and will retain an Eulerian circuit. If eℓ′ ′ ′ and e−ℓ′ with sℓ′ = sℓ are present in G , then the same will be true if we choose c ∈ supp(G ) so that Pc 6= src(eℓ′ ). This completes the proof of the claim and thus of Lemma 8.1 in Case 1. Case 2. k +1 ∈ I′. Applying the claim in Case 1 with a replaced by a + j, we see that there exists a graph G with edges e±1,...,e±r containing Ht such that G has no repeated edges and G has an Eulerian path starting at Pa+j . Appending ek+1 to this Eulerian path at the beginning, we obtain an Eulerian path for Ga starting at Pa. Thus G ∈ U(a, j) and Ht is contained in G. This completes the proof in Case 2. Lemma 8.2. Let 0 6 t 6 r be the largest integer such that there exists G ∈ U(a, j) containing Ht. Set s = + si. Then i∈It (a) s 6 |a|, |a + j|. P + (b) If t > 0 and v ∈ Rt(a) ∩ supp(Ht) then either every si for i ∈ It or every si for − i ∈ It must appear as a summand of v. 24 MATTHEWBRASSILANDZINOVYREICHSTEIN
Proof. Throughout the proof, a′ will denote either a or a + j. ′ 6 n−1 n−1 6 ′ 6 (a) We will assume that 0 < a ⌊ 2 ⌋; the case where −⌈ 2 ⌉ a 0 is symmet- ′ rical. We argue by contradiction. Suppose sr + ... + st+1 > a . Let t +1 6 λ 6 r be the smallest integer such that ′ (8.1) sr + sr−1 + ... + sλ+1 + sλ >a . We may assume without loss of generality that λ 6 r − 1. Indeed, suppose λ = r. Then 6 ′ r ′ 0 a
v ∈ Rt+1(a) ∩ supp(Ht) ⊂ Rt+1(a) ∩ supp(Ht+1). By Lemma 8.1, this contradicts the maximality of t. It remains to show that if ǫt+1 6= 0 and ǫi 6= ǫt+1 for some t +1 |a′| − |s′| > |s| − |s′| > ⌈ ⌉, 2 and part (b) will follows. We we will prove the inequality (8.2) in two steps. Step 1. First we will show that (8.2) holds if ǫλ = 0 for any λ = t +2,...,r. (Recall that we know that ǫt+1 6= 0.) Indeed, λ t |s′| 6 |ǫ |s 6 s − s = s −⌈ ⌉ t + 2. Step 2. We are now ready to complete the proof of (8.2). By Step 1 we may assume that ǫλ = ±1 for every λ = t +1,t +2 ...,r. In this case ′ |s | = sλ − sµ λX∈A µX6∈A for some proper subset ∅ 6= A ( {t +1,t +2,...,r}. Here A = {λ | sλ =1} or A = {λ | sλ = −1}. Note that A is a proper subset of {t +1,t +2,...,r} because we are assuming that ′ ǫi 6= ǫt+1 for some t+1
Lemma 8.3. Let 0 6 t 6 r be the largest integer such that there exists a graph in U(a, j) containing Ht. Denote the maximal graph in (U(a, j) by G. (Recall that by Lemma 7.1, G ′ contains Ht.) Let G be the graph with vertices Pv, v ∈ Z/nZ obtained from G by removing ′ the edges e±1,...,e±t, i.e., by removing the edges which occur in Ht. Then G ∈ U(a, j, It).
′ Proof. If t = 0 then G = G ∈ U(a, j) = U(a, j, I0). So we may assume t > 0. As G ∈ U(a, j) there exists b ∈ Z/nZ such that
Gb = G ∐ • → • Pb ek+1 Pb+j
′ ′ admits an Eulerian path from Pa to Pa+j . Let Gb be the graph obtained from G by ′ adding the edge ek+1 originating at Pb. Equivalently, Gb is the graph obtained by removing e±1,...,e±t from Gb. It suffices to show that there exists an Eulerian path from Pa to Pa+j ′ in Gb. As there exists an Eulerian path on Gb from Pa to Pa+j , by Theorem 2.1(a), the degree of the vertices in Gb must satisfy
outdeg(Pv) = indeg(Pv) , for all v 6= a,a + j ,
outdeg(Pa) = indeg(Pa)+1 , if j 6=0 , (8.3) outdeg(Pa+j ) = indeg(Pa+j ) − 1 , if j 6=0 ,
outdeg(Pa) = indeg(Pa) , if j =0 .
As every vertex has its indegree equal to its outdegree in Ht, the equation (8.3) also holds ′ ′ in Gb. We claim that Gb is connected. If we can prove this claim, then Theorem 2.1(a) will ′ ′ tell us that Gb admits an Eulerian path from Pa to Pa+j and consequently, G ∈ U(a, j, It), as desired. ′ To prove the claim, we will argue by contradiction. Assume Gb is not connected. Then ′ ′ 1 2 1 there exists a decomposition of Gb as Gb = Γ ∐ Γ for two non-empty graphs Γ and 2 1 1 Γ , such that Pa has nonzero degree in Γ , Γ is connected, and there is no vertex having 1 2 nonzero degree in both Γ and Γ . Thus for any vertex Pv, either all edges incident on Pv ′ 1 2 in Gb are present in Γ or none of them are (and similarly for Γ ). The sum of the indegrees of the vertices in Γ1 equals the sum of the outdegrees. By 1 2 (8.3) we must then have Pa+j lies in Γ and so every vertex having nonzero degree in Γ has its indegree equal to its outdegree. There is a path from Pa to some Pv ∈ Ht in Gb. By + Lemma 8.2 (b), this path necessarily uses all edges of the form eλ for λ ∈ It , or all edges − 1 + of the form eλ for λ ∈ It . Thus Γ contains either every edge eλ for λ ∈ It , or every edge − 1 + eλ for λ ∈ It . We assume that Γ contains every edge eλ for λ ∈ It , the proof in the other case is symmetric. 2 The set of edges of Γ is a subset of {e−t−1,e−t−2,...,e−r}∪{ek+1}. Let Pv′ be a vertex having nonzero degree in Γ2. As every vertex has its indegree equal to its outdegree in Γ2, 2 there exists a closed circuit in Γ originating (and terminating) at Pv′ . This implies that a nonempty partial sum of {s−t−1,s−t−2,...,s−r}∪{j} is zero modulo n. By Lemma 8.2(a),
|s−t−1 + s−t−2 + ··· + s−r| 6 |a| ··· + sλα ) for some λ1, λ2,...,λα ∈{t +1,t +2,...,r}. As j appears in this partial sum corresponding to a closed circuit in Γ2 we know that 2 1 1 ek+1 is an edge in Γ and therefore not in Γ . Thus the path from Pa to Pv in Γ does not use the edge ek+1 and hence by Lemma 8.2(b) we have v = a + (st+1 + st+2 + ··· + sr). 26 MATTHEWBRASSILANDZINOVYREICHSTEIN Using the equality 0 = j − (sλ1 + sλ2 + ··· + sλα ) we can write v as v = a + (st+1 + st+2 + ··· + sr)+ j − (sλ1 + sλ2 + ··· + sλα ) = a + j + si . + i∈It \{Xλ1,...,λα} This contradicts Lemma 8.2 (b), as v ∈ Rt(a) ∩ supp(Ht) and we have written v as a sum + that does not include si for every i ∈ It . The proof of the claim (and thus of Lemma 8.3) is now complete. We are now ready to finish the proof of Proposition 7.2. Conclusion of the proof of Proposition 7.2. Let t be the largest integer such that there exists a graph G ∈ U(a, j) containing Ht, and G be the maximal graph in U(a, j). ′ ′ By Lemma 7.1, G must contain Ht. Write G = G ∐ Ht. By Lemma 8.3, G ∈ U(a, j, It). By Lemma 7.3 (whose assumptions are met by Lemma 8.2 (a)), the largest graph in U(a, j, It) is the conjectured subgraph Gˆ′ in the proposition. We want to show that G′ = Gˆ′. As Gˆ′ ′ is maximal in U(a, j, It), it suffices to show that Gˆ = Gˆ ∐ Ht lies in U(a, j). Clearly Gˆa has an Eulerian circuit. Thus we only need to check that Gˆ has no repeated ′ n−1 edges. As st+1 + st+2 + ··· + sr ≤ |a |≤⌈ 2 ⌉, n is sufficiently large so that the vertices of ′ the graph Gˆ are all distinct. Thus the edges of Gˆ are all distinct, and the edges of Ht are all distinct. If a repeated edge were present, we would have more than one vertex having ′ ′ nonzero degree in both subgraphs Gˆ and Ht. On the other hand, Gˆ and Ht have exactly ′ ′ one vertex in common, namely Pa +sr +...+st+1 in part (a) and Pa −sr +...−st+1 in part (b). ′ Any other vertex is reachable from a using the edges from It+1 and hence, cannot lie in Ht (or even in Ht+1 by the minimality of t; see Lemma 8.1). This shows that the edges of Gˆ are distinct and hence, Gˆ ∈ U(a, j), completing the proof. 9. Proof of Proposition 6.3(a) In this section we will prove the following. Proposition 9.1. Let 1 6 r (Ic(Lj))a,b =0 for any b 6= a and (Ic(Lj ))a,a 6=0 in F . Note that Proposition 6.3(a) readily follows from Proposition 9.1. Indeed, fix 0 6= j ∈ Z/nZ. By the definition of δj (see the statement of Proposition 6.3), there are exactly δj values of a ∈ Z/nZ such that (a, j) does not satisfy the conditions of Proposition 9.1. (Note that this can only happen if a ∈{0, −j}, so in each case δj = 0, 1 or 2.) If we remove the th th a row and the a column from Ic(Lj ) for every such a, we will be left with a diagonal matrix with non-zero diagonal entries. In other words, if we remove δj rows and δj columns from In(Lj ), the remaining (n − δj ) × (n − δj) matrix is non-singular. This shows that null(In(Lj )) 6 δj , as desired. The remainder of this section will be devoted to proving Proposition 9.1. Lemma 9.2. Let j 6=0 and G be the maximal graph in U(a, j). Suppose that there exists an Eulerian path from Pa on Gb. Then b = a. A GRAPH-THEORETIC APPROACH TO A CONJECTURE OF DIXON AND PRESSMAN 27 Proof. The maximal graph G in U(a, j) is described in Proposition 7.2. It is clear from this description that the indegree matches the outdegree for each vertex of G. Thus in the graph Gb = G • → • Pb ek+1 Pb+j a the outdegree of Pb is 1 larger than its indegree and the outdegree of Pb+j is 1 smaller than its indegree. Since j 6= 0 in Z/nZ, we have Pb 6= Pb+j . Thus every Eulerian path on Gb starts at Pb and ends at Pb+j ; see Theorem 2.1. Since we assumed there exists an Eulerian path from Pa on Gb we conclude that b = a. Lemma 9.3. Let a, j ∈ Z/nZ be as in the statement of Proposition 9.1 and G be the maximal graph in U(a, j). Then the sum of the signatures of Eulerian paths on Ga from Pa to Pa+j is ±α! for some α 6 r. Proof. Let 0 6 t 6 r be the largest integer such that a graph from U(a, j) contains Ht. Recall that H0 = ∅, so t is well defined. We will consider three cases. Case 1. Pa+j ≻ Pa and a does not lie in supp(Ht). (Note that if a lies in supp(Ht), then ′ so does a + j; hence r = t and we are in Case 3 below.) By Proposition 7.2, Ga = Ht ∐ Ga, ′ for Ga of the form e±r e±(r−1) e±(t+1) ′ Ga = . ek+1 e∓r e∓(r−1) e∓(t+1) Pa Pa+j Pa+j±sr Pa+j±s where s = st+1 + ··· + sr. Since a does not lie in supp(Ht), any Eulerian path from Pa on Ga is of the form (ek+1,e±r,...,e±(t+1),w,e∓(t+1),...,e∓r), where w is an Eulerian path on Ht. By Lemma 2.3, the sum of the signatures of these Eulerian paths is ±(t − 1)! or ±t!, depending on whether Pa+j+st+1 +···+sr = P0 or not. Case 2. Pa ≻ Pa+j and a + j does not lie in supp(Ht). (Once again, if a + j lies in ′ ′ supp(Ht), then we are in Case 3.) By Proposition 7.2, Ga = Ht ∐ Ga, where Ga is of the form e±r e±(r−1) e±(t+1) ′ Ga = . e k+1 e∓r e∓(r−1) e∓(t+1) Pa+j Pa Pa±sr Pa±s where s = st+1 + ··· + sr. Again, since a + j does not lie in supp(Ht), the sum of the signatures of the Eulerian paths will be determined by the sum of the signatures of the Eulerian paths on Ht, being either ±(t − 1)! or ±t!. Case 3. Both a and a + j lie in supp(Ht). By the definition of t, this forces t = r. By our assumptions on a and j we have a 6= 0, j 6= 0 and a+j 6= 0. Thus Ga is of the same form as the graph considered in Lemma 2.4, where we showed there that sgn(w)= w∈EulPa (Ga) ±(r − 1)!. P Proof of Proposition 9.1. Let G be the maximal graph in U(a, j). The key point is that since char(F ) does not divide r!, Lemma 9.3 tells us under our assumptions on a and j, G ∈ Unz(a, j). In other words, G is the maximal graph in Unz(a, j). Thus by Lemma 5.3, (Ic(Lj ))a,b is the sum of the signatures of the Eulerian paths on Gb from Pa to Pa+j . If 28 MATTHEWBRASSILANDZINOVYREICHSTEIN b 6= a, then Lemma 9.2 tells us that there are no such paths, so (Ic(Lj ))a,b = 0. On the other hand, (Ic(Lj ))a,a 6= 0 in F by Lemma 9.3. Remark 9.4. In those cases where the pair (a, j) does not safisfy the conditions of Proposition 9.1, the maximal graph G of U(a, j) does not lie in Unz(a, j). If it did, the nullity of In(Lj ) would be lower than the value given by Proposition 6.3 for at least one j. In view of Lemma 5.2, the nullity of L = L0 ⊕ L1 ⊕ ... ⊕ Ln−1 would be lower than k, contradicting Lemma 3.1(b). To illustrate this point more concretely, let us revisit Example 5.4. Here k = 2 (so, r = 1), n = 3, and j = 1. The maximal graph from U(a, 1) does not lie in Unz(a, 1) when a = 2, though it does when a = 0 or 1. After removing the row and column corresponding to a = 2 (i.e., the last row and the last column) from the 3 × 3 matrix Ic(L1) we are left with a nonsingular diagonal 2 × 2 matrix, showing that null(Ic(L1)) = 1. 10. Proof of Proposition 6.3(b) As we showed in the previous section, for j 6= 0 in Z/nZ the matrix Ic(Lj) is close to being diagonal. In this section we will see that Ic(L0) has a more complicated structure. For ease of visualizing the matrix Ic(L0), we will reorder the rows and columns. In the matrix Ic(L0), our rows and columns corresponding to a and b respectively range from 0 to n − 1. ′ We define the n × n matrix Ic(L0) to be the matrix Ic(L0) but with the rows and columns n − 1 n − 1 corresponding to a and b ranging from −⌈ ⌉ to ⌊ ⌋. Since permuting rows and 2 2 columns does not effect the nullity of a matrix, we have ′ null(Ic(L0) ) = null(Ic(L0)). ′ Thus our goal is to show that Ic(L0) is a non-singular matrix. We will do this by proving ′ that Ic(L0) is of the form U ∗ 0 ′ (10.1) Ic(L0) = 0 N 0 , 0 ∗ L where • U, N and L are square matrices, • U is upper triangular with non-zero diagonal matrices, • L is lower triangular with non-zero diagonal entries, and • N is the (r + 1) × (r + 1) submatrix corresponding to rows and columns labeled by elements of supp(Hr)= {−⌈r/2⌉, −⌈r/2⌉ +1,..., ⌊r/2⌋}, • N is non-singular. ′ This will imply that Ic(L0) is a non-singular matrix and hence, so is Ic(L0), thus completing the proof of Proposition 6.3(b). Proofs of these assertions will be carried out ′ in Lemmas 10.1, 10.2, 10.3 and 10.4. The idea is to read off the entries of Ic(L0) from Proposition 7.2 using their graph-theoretic interpretation given by Lemma 5.3. A GRAPH-THEORETIC APPROACH TO A CONJECTURE OF DIXON AND PRESSMAN 29 n − 1 r Lemma 10.1. The matrix consisting of the first ⌈ ⌉−⌈ ⌉ rows of Ic(L )′ has the 2 2 0 structure ∗ ∗ ∗ ... ∗ ∗ ∗ ∗ ... ∗ 0 0 ... 0 0 ∗ ∗ ... ∗ ∗ ∗ ∗ ... ∗ 0 0 ... 0 0 0 ∗ ... ∗ ∗ ∗ ∗ ... ∗ 0 0 ... 0 U ∗ 0 = ...... . ...... 0 0 0 ... ∗ ∗ ∗ ∗ ... ∗ 0 0 ... 0 0 0 0 ... 0 ∗ ∗ ∗ ... ∗ 0 0 ... 0 n−1 r The first ⌈ 2 ⌉−⌈ 2 ⌉ columns form an upper triangular block U, where the terms on the 6 n−1 r diagonal are ±2α! for some α r. The last ⌊ 2 ⌋−⌊ 2 ⌋ columns are zero. Proof. By Lemma 5.3, it suffices to prove the following. n − 1 r Fix −⌈ ⌉ 6 a 6 −⌈ ⌉− 1. Let G be the maximal graph in U(a, 0) and b ∈ Z/nZ 2 2 be such that there exists an Eulerian path from Pa to Pa on Gb = G ∐ • → • . Then Pb ek+1 Pb r r (a) b ∈{a, a +1,..., −1, 0, 1,..., ⌊ ⌋− 1, ⌊ ⌋}. 2 2 (b) the sum of the signatures of the Eulerian paths on Ga from Pa to Pa is ±2α! for some α 6 r. ′ ′ Recall that by Proposition 7.2(a), G = Ht ∐ G for some 0 6 t 6 r, where G is of the form er er−1 et+1 G′ = e−r e−r−1 e−t−1 Pa Pa+ssr Pa+sr +···+st+1 and sr + sr−1 + ··· st+1 6 |a|. The graph Gb is obtained from G by appending the extra edge ek+1 at Pb. Since j = 0, ek+1 is a loop. If Gb has an Eulerian path, it has to be connected. In other words, the loop ek+1 has to be appended at one of the vertices that has non-zero degree in G, i.e., at Pa, Pa+sr , ..., Pa+sr +...+st+1 or at one the vertices from Ht. This shows that b ∈ [a ←→ ⌊r/2⌋], thus proving (a). To prove (b), note that that the graph Ga is of the form ek+1 P a Pa+sr +···+st+1 P0 where the (possibly empty) flower on the right corresponds to Ht. Here the vertices Pa+sr +···+st+1 and P0 may be distinct (as in the above diagram) or not (in which case ′ the two edges connecting them should be removed), depending on whether G intersects Ht 30 MATTHEWBRASSILANDZINOVYREICHSTEIN at P0. In either case, Ga will have the shape of the graph in Lemma 2.6. By Lemma 2.6, sgn(w)= ±2α! , w∈EulXPa (Ga) where α is the number of vertices to the right of P0. As Ga has 2r +1 edges, there are at most r vertices to the right of P0 in the picture and hence α 6 r. n−1 r ′ Lemma 10.2. The matrix consisting of the last ⌊ 2 ⌋−⌊ 2 ⌋ rows of Ic(L0) has the structure 0 0 ... 0 ∗ ∗ ... ∗ ∗ 0 0 ... 0 0 0 0 ... 0 ∗ ∗ ... ∗ ∗ ∗ 0 ... 0 0 0 0 ... 0 ∗ ∗ ... ∗ ∗∗ ∗ ... 0 0 0 ∗ L = ...... . ...... 0 0 ... 0 ∗ ∗ ... ∗ ∗∗ ∗ ... ∗ 0 0 0 ... 0 ∗ ∗ ... ∗ ∗∗ ∗ ... ∗ ∗ n−1 r The last ⌊ 2 ⌋−⌊ 2 ⌋ columns form a lower triangular block L where the terms on the 6 n−1 r diagonal are ±2α! for some α r. The first ⌈ 2 ⌉−⌈ 2 ⌉ columns are zero. Proof. The proof mirrors that of Lemma 10.1, except that here we appeal to Proposi- tion 7.2(b) for the the structure of the maximal graph G instead of Proposition 7.2(a). ′ Lemmas 10.1 and 10.2 show that the upper and lower portions of Ic(L0) have the structure as in (10.1), and that U and L are nonsingular under our characteristic assumption. It remains to consider the rows corresponding to a ∈ [−⌈r/2⌉ ←→ ⌊r/2⌋]. By Proposition 7.2 the maximal graph in U(a, 0) for a in this range is Hr. ′ Lemma 10.3. The middle rows of Ic(L0) corresponding to a ∈ [−⌈r/2⌉ ←→⌊r/2⌋] are 0 N 0 , up to multiplication of each column by ±1, where N is the (r + 1) × (r + 1) matrix 2 1 1 ··· 1 r 1 ··· 1 1 1 1 2 1 ··· 1 r 1 ··· 1 1 1 1 1 2 ··· 1 r 1 ··· 1 1 1 ...... ...... 1 1 1 ··· 2 r 1 ··· 1 1 1 (10.2) N = (r − 1)! · r r r ··· r r(r + 1) r ··· r r r . 1 1 1 ··· 1 r 2 ··· 1 1 1 ...... ...... 1 1 1 ··· 1 r 1 ··· 2 1 1 1 1 1 ··· 1 r 1 ··· 1 2 1 1 1 1 ··· 1 r 1 ··· 1 1 2 th ′ Proof. Let a ∈ [−⌈r/2⌉ ←→⌊r/2⌋]. The (a,b) entry of Ic(L0) is sgn(w), w∈EulPa (Gb) where G is H with an additional edge e , being a loop, placed at P . For the graph G b r k+1 bP b to be connected, we require b ∈ [−⌈r/2⌉ ←→ ⌊r/2⌋]. This shows that the (a,b)th entry of ′ Ic(L0) is zero when b∈ / [−⌈r/2⌉ ←→⌊r/2⌋]. Assume b ∈ [−⌈r/2⌉ ←→⌊r/2⌋]. Lemma 2.5 with α = r gives us A GRAPH-THEORETIC APPROACH TO A CONJECTURE OF DIXON AND PRESSMAN 31 ±(r + 1)! , if a = 0 and b =0 ±2(r − 1)! , if a = b and a,b 6=0 N(a,b) = sgn(w)= ±r! , if a = 0 and b 6=0, or a 6= 0 and b =0 w∈EulPa (Gb) X ±(r − 1)! , otherwise, where for fixed b, either all entries N(a,b) are positive, or all entries N(a,b) are negative. This shows that N has the required form. Lemma 10.4. Suppose the characteristic of F does not divide (2r+1)r!. Then det(N) 6= 0 in F . Proof. We perform elementary row operations on the matrix N. After dividing N by (r − 1)! and dividing the row corresponding to 0 by r, we obtain 2 1 1 ··· 1 r 1 ··· 1 1 1 1 2 1 ··· 1 r 1 ··· 1 1 1 1 1 2 ··· 1 r 1 ··· 1 1 1 ...... ...... 1 1 1 ··· 2 r 1 ··· 1 1 1 1 1 1 ··· 1 r +1 1 ··· 1 1 1 . 1 1 1 ··· 1 r 2 ··· 1 1 1 ...... ...... 1 1 1 ··· 1 r 1 ··· 2 1 1 1 1 1 ··· 1 r 1 ··· 1 2 1 1 1 1 ··· 1 r 1 ··· 1 1 2 Next we subtract the row corresponding to 0 from every other row. Our matrix is trans- formed to 1 0 0 ··· 0 −1 0 ··· 0 0 0 0 1 0 ··· 0 −1 0 ··· 0 0 0 0 0 1 ··· 0 −1 0 ··· 0 0 0 ...... ...... 0 0 0 ··· 1 −1 0 ··· 0 0 0 1 1 1 ··· 1 r +1 1 ··· 1 1 1 . 0 0 0 ··· 0 −1 1 ··· 0 0 0 ...... ...... 0 0 0 ··· 0 −1 0 ··· 1 0 0 0 0 0 ··· 0 −1 0 ··· 0 1 0 0 0 0 ··· 0 −1 0 ··· 0 0 1 32 MATTHEWBRASSILANDZINOVYREICHSTEIN Finally, we subtract every other row from the row corresponding to 0. Our matrix becomes 1 0 0 ··· 0 −1 0 ··· 0 0 0 0 1 0 ··· 0 −1 0 ··· 0 0 0 0 0 1 ··· 0 −1 0 ··· 0 0 0 ...... ...... 0 0 0 ··· 1 −1 0 ··· 0 0 0 0 0 0 ··· 0 2r +1 0 ··· 0 0 0 , 0 0 0 ··· 0 −1 1 ··· 0 0 0 ...... ...... 0 0 0 ··· 0 −1 0 ··· 1 0 0 0 0 0 ··· 0 −1 0 ··· 0 1 0 0 0 0 ··· 0 −1 0 ··· 0 0 1 which has determinant 2r +1 6= 0. Thus N is nonsingular under our assumption on the characteristic. Acknowledgments The authors are grateful to Omer Angel, John Dixon, Irwin Pressman and Bruce Shep- herd for helpful comments. References [1] A. S. Amitsur and J. Levitzki. Minimal identities for algebras. Proc. Amer. Math. Soc., 1:449–463, 1950. [2] B. Bollob´as. Modern graph theory, volume 184 of Graduate Texts in Mathematics. Springer-Verlag, New York, 1998. [3] J. D. Dixon and I. S. Pressman. Generalized commutators and a problem related to the Amitsur-Levitzki theorem. Linear Multilinear Algebra, 66(11):2199–2207, 2018. [4] C. Procesi. On the theorem of Amitsur-Levitzki. Israel J. Math., 207(1):151–154, 2015. [5] J. P. Razmyslov. Identities with trace in full matrix algebras over a field of characteristic zero. Izv. Akad. Nauk SSSR Ser. Mat., 38:723–756, 1974. [6] S. Rosset. A new proof of the Amitsur-Levitski identity. Israel J. Math., 23(2):187–188, 1976. [7] R. G. Swan. An application of graph theory to algebra. Proc. Amer. Math. Soc., 14:367–373, 1963. [8] R. G. Swan. Correction to “An application of graph theory to algebra”. Proc. Amer. Math. Soc., 21:379– 380, 1969. Department of Mathematics, University of British Columbia, Vancouver, CANADA Email address: [email protected], [email protected]