c_2 invariants of hourglass chains via quadratic denominator reduction
cc INVARIANTS OF HOURGLASS CHAINS VIA QUADRATICDENOMINATOR REDUCTION
OLIVER SCHNETZ AND KAREN YEATS
Abstract.
We introduce families of four-regular graphs consisting of chains of hourglasseswhich are attached to a finite kernel. We prove a formula for the c invariant of these hour-glass chains which only depends on the kernel. For different kernels these hourglass chainstypically give rise to different c invariants. An exhaustive search for these c invariants forall kernels with a maximum of ten vertices provides Calabi-Yau manifolds with point-countswhich match the Fourier coefficients of modular forms whose levels and weights are [3,36],[4,8], [4,16], [6,4], [9,4]. We also confirm the conjecture that curves (weight two modularforms) are absent in c invariants up to level 69. Introduction
Given a graph G , to each edge e associate a variable α e and define the Kirchhoff polynomial or first Symanzik polynomial to be Ψ G = (cid:88) T (cid:89) e (cid:54)∈ T α e where the sum is over all spanning trees T of G . When it converges define the Feynmanperiod of G to be the projective integral(1) P G = (cid:90) α e ≥ ΩΨ G where Ω = (cid:80) | E ( G ) | i =1 ( − i dα · · · (cid:99) dα i · · · dα | E ( G ) | . When G is a scalar Feynman diagram then P G is the residue of its Feynman integral which contributes to the β -function, so the Feynmanperiod is of physical significance. Furthermore, the Feynman period exhibits the arithmeticcontent of the Feynman integral, so it is very interesting for anyone studying the geometryand number theory underlying quantum field theory [3, 1, 17].In [5] Francis Brown introduced denominator reduction as a tool for studying Feynmanperiods algebraically. Denominator reduction describes how the denominators of the Feyn-man periods change through successive edge integrations. When these denominators fail tofactor into linear pieces denominator reduction ends. Keeping track of the numerators in thisprocess using multiple polylogarithms gives an algorithm for parametric Feynman integra-tion which fails when denominator reduction stops [4, 5]. Improvements and variants of thisintegration approach have been implemented [15], but the constraint of the non-factoringdenominators remain.Denominator reduction itself is purely algebraic, but still carries substantial informationabout these integrals since the numerators typically are of simpler geometry. In particular,denominator reduction tells us about weight drop [11] and can be used to compute the c a r X i v : . [ m a t h - ph ] F e b Figure 1.
An illustration of the type of graph we consider. The shadedsection is the kernel which is attached to the hourglass chain.invariant, an arithmetic graph invariant that predicts properties of the Feynman periods[18, 8] (for a definition see Section 2.3).In [20] one of us defined a generalized denominator reduction, called quadratic denomina-tor reduction , that can always progress at least one more step than standard denominatorreduction, and sometimes much further. At the cost of not working for the prime 2, thisquadratic denominator reduction can be used to compute c invariants and so still tells usabout the geometries underlying the Feynman periods.The geometric idea behind quadratic denominator reduction is that in certain cases adenominator with a square root still has rational geometry as it defines a projective line.As a demonstration of the power of quadratic denominator reduction, in this paper we willstudy infinite families of graphs built by attaching a chain of hourglasses to a finite kernel,see Figure 1. The c invariant for any such graph will only depend on the kernel, givingarbitrarily many infinite families of graphs for which we know the c invariant (we prove thisonly for p = 2 or modulo p if q = p k is an odd prime power).With the notions of hourglass, kernel, and hourglass chain as illustrated in Figures 1 and2 we can state our theorem. Theorem 1.
Let K be a kernel and let L ∈ G K be a graph of the type we consider with anhourglass chain of length at least 6, see Section 2.8 for definitions. Let v be a vertex of thehourglass chain that is shared by the second and third hourglasses from one end. Let K (cid:48) be K with new edges joining the vertices in the same part of the bipartition. Index these newedges by and . If q = p k is 2 or an odd prime power then (2) c ( q )2 ( L − v ) ≡ α (Ψ , K (cid:48) ) Ψ , K (cid:48) Ψ K (cid:48) , mod p. The ‘Dodgson’ polynomials on the right hand side of (2) are defined in Section 2. SeeFigure 2 for K (cid:48) .Hourglass chains (for suitable kernels) have no trivial reductions like double triangles orthree vertex splits [10]. They are the first families of such graphs for which the c invariant (cid:48) Figure 2.
The graph K (cid:48) is the kernel K with the two extra edges 1 and 2.can be calculated. In a certain sense these hourglass chains can be considered as ‘telescopes’that enables us to look into geometries of Feynman graphs at very high loop order. This hasnever been achieved before: all previous techniques were either restricted to the analysis ofsmall graphs, or they worked in a way which was fundamentally prime-by-prime [22, 12, 24]and hence did not provide non-trivial graph families with the same underlying geometries.The paper is organized as follows: in Section 2 we provide the necessary backgroundinformation on denominator reduction and the c invariant. Section 3 contains the mainpart to the proof of Theorem 1: the hourglass reductions. The prime 2 plays a special rolewhich we handle in Section 4. We prove that for q = 2 the c is always zero in hourglasschains.Finally we use Theorem 1 in Section 5 for an exhaustive search for c invariants in hour-glass chains with kernels of at most six internal vertices (vertices which are not attachedto the hourglasses, see Figure 18 and Table 1 for a maximum of five internal vertices). Wefind Legendre symbols (see Section 2) (4 /q ) and ( − /q ) along with several modular forms.Explicitly the weight and the level of the identified modular forms (see [9]) are [3,36], [4,8],[4,16], [6,4], [9,4]. The modular forms [3,36], [9,4] were not found in φ graphs of loop ordersless or equal twelve in [20]. They are new additions to the table of modular forms in φ theory.An important outcome of this article is confirming the conjecture that in the physical φ theory (corresponding to four-regular graphs) the c is free of curves (which correspond toweight two modular forms), Conjecture 26 in [9], see also [20]. This puzzling conjecture isfor the first time tested to any loop order for some non-trivial geometries. It seems to beconnected to some deep algebraic structure in quantum field theories. Note that as soon asone lifts the restriction to four regular graphs, curves in c invariants are ubiquitous.For extra support of this ‘no-curves-puzzle’ it might be worthwhile to study kernels inthe future which lead out of φ graphs. Which (non- φ ) kernels provide c invariants thatcorrespond to weight two modular forms? It is also possible to extend the c -search to theincreasing number of kernels with more than six internal vertices, see Table 1 and Question13. Acknowledgements
Both authors are deeply indebted to Dirk Kreimer for many years of encouragement andsupport. Oliver Schnetz is supported by DFG grant SCHN 1240. Karen Yeats is supportedby an NSERC Discovery grant and by the Canada Research Chairs program; during someof this work she was visiting Germany as a Humboldt fellow. . Background
Dodgsons.
In order to define the denominator reductions we first need to give a de-terminantal expression for Ψ G and define some related polynomials.Assume for the rest of the paper that G is a connected graph.Choose an order on the edges and vertices of G and choose a direction for each edge. Thenthe signed incidence matrix of G is a | V ( G ) | × | E ( G ) | matrix with entries − , ,
1, where the i, j th entry is − j starts at vertex i , is 1 if edge j ends at vertex i , and is 0 otherwise.Let E be the signed incidence matrix with one row removed. Since G is connected the rankof the signed incidence matrix is | V ( G ) | − E is full rank.Define the expanded Laplacian to be the matrix L G = (cid:20) Λ E t E (cid:21) where Λ is the diagonal matrix with diagonal entries α e , e ∈ E ( G ), in the edge order chosenabove. This matrix is called the expanded Laplacian because it behaves very much likethe Laplacian (with a matching row and column removed), but the pieces of it have beenexpanded out into a larger block matrix. Proposition 2. Ψ G = ( − | V ( G ) |− det L G . This proposition is at its core the matrix-tree theorem ( § L G . Polynomials from minors of L G are called ‘Dodgson’ polynomials in [5]. In general, these polynomials have signs whichdepend on the sequence of edges [20]. While in classical denominator reduction signs areoften insignificant, they play an important role in quadratic denominator reduction. Bythe special structure of (2) it is sufficient for our purpose to define the sign of Dodgsonpolynomials in a trivial case. Definition 3.
For any subsets
I, J, K of the edges of a connected graph G with | I | = | J | wedefine the Dodgson polynomials Ψ I,JK as Ψ I,JG,K = ± det L I,JG (cid:12)(cid:12)(cid:12) α k =0 ,k ∈ K where L I,JG is L G with rows in I and columns in J deleted. In the case I = J the sign is ( − | V ( G ) |− . The contraction-deletion formula [5], Lemma 9 in [20],(3) Ψ
I,JG,K = α e Ψ Ie,JeG,K + Ψ
I,JG,Ke = α e Ψ I,JG \ e,K + Ψ I,JG/e,K relates Dodgson polynomials to minors. Note that in the context of Dodgson polynomialsgraphs may have multiple edges and self-loops (which contract to zero, i.e. every Dodsonpolynomial of a graph with a contracted self loop vanishes).
Example 4.
A tree has the graph polynomial 1. The Dodgson polynomial of a circle is thesum of its edge-variables. In this case Ψ e,f = ± e, f ∈ E ( G ) with a plussign if e = f .
234 5
Figure 3.
An example of a graph with a partition of some of the verticesmarked by the shape of the large vertices.We get the following vanishing cases ([5], § I,JG,K = 0 if I or J cut G ,Ψ I,JG,K = 0 if I ∪ K \ J or J ∪ K \ I contain a cycle.(4)An important Dodgson identity is [5](5) Ψ , G, Ψ , G, − Ψ , G Ψ G, = (Ψ , G ) for any two edges 1, 2 in G . Many more identities for Dodgson polynomials can be found in[5] and [20].2.2. Spanning forest polynomials.
Dodgson polynomials can also be written in terms ofspanning forests. To that end, given a partition P of a subset of the vertices of G , definethe spanning forest polynomial Φ PG = (cid:88) F (cid:89) e (cid:54)∈ F a e where the sum is over spanning forests with the property that there is a bijection betweenthe parts of P and the trees of the forest such that every vertex in a part of the partition isin the corresponding tree.For example, given the graph illustrated in Figure 3, with the partition P indicated by theshape of the large vertices, the corresponding spanning forest polynomial is α α ( α + α + α ).This technique of marking the partition by the shape of the large vertices will be used withoutfurther comment in the main hourglass chain reduction argument.We have the following proposition Proposition 5. Ψ I,JG,K = (cid:88) f k Φ P k G \ ( I ∪ J ∪ K ) where the P k run over partitions of the ends of I , J , and K , and f k ∈ {− , , } . Fur-thermore, those f k which are nonzero are exactly those where each forest in the polynomialbecomes a tree in G \ I/ ( J ∪ K ) and in G \ J/ ( I ∪ K ) ; if any forest in the polynomial has thisproperty then all of them do. This proposition is a particular interpretation of the all minors matrix tree theorem, for aproof see Proposition 8 and 12 of [11].The signs in the spanning forest polynomial expansion of a Dodgson polynomial can betricky, however, the only case we will need for the argument below is given in the followinglemma.
Lemma 6.
With notation as in Proposition 5, if P and P both have nonzero coefficients,and P and P differ by swapping two vertices which are in the same component of J viewedas a subgraph of G , then f = − f . his is a special case of Corollary 17 of [11].Another useful observation, see Proposition 21 of [11], is that if G is formed as the 2-sumof G and G , that is G and G each have a distinguished edge, e and e respectively, and G is the result of identifying e and e and then removing this new identified edge whileleaving the induced identifications on the incident vertices, then(6) Ψ G = Ψ G \ e Ψ G /e + Ψ G /e Ψ G \ e = Ψ G \ e Φ { v } , { v } G \ e + Φ { v } , { v } G \ e Ψ G \ e where v and v are the ends of e and e .Note that all Dodgson polynomials and all spanning forest polynomials are explicitly linearin all their variables.2.3. The c invariant. A general aim in the mathematical theory of Feynman periods isto understand what kinds of numbers can appear [3, 17, 19]. The interest in this topic wasrecently intensified by the (conjectural) discovery of a Galois coaction structure on thesenumbers [16, 6, 7]. In general, the Feynman period (1) is hard to analyze. Even the zerolocus of the graph hypersurface Ψ G = 0 has a complicated geometrical structure.The number contents of the Feynman period is intimately related to the motivic structureof its integral. This motivic structure lifts to all fields and for finite fields F q informationon the geometry of the Feynman period can be extracted from the number of elements onthe singular locus of the integrand. While the knowledge of this point-count for a single q isstill not very informative, its value for all (or many) q carries important information for the(‘framing’ of the) Feynman period.To this end we define(7) [ F ] q = |{ F = 0 in F q }| as the point-count of the zero locus of the polynomial F . In the context of Feynman periodsthe important information of the point-count is hidden in the first non-trivial reductionmodulo q . For any connected graph G with at least three edges we define [18](8) c ( q )2 ( G ) ≡ [Ψ G ] q q mod q as the c invariant of the Feynman graph G . The above definition implies that the point-count of the graph hypersurface is always divisible by q (the index 2 refers to this square).For a given graph G one should think of the c as the infinite sequence ( c ( q )2 ) q =2 , , , , , , , ,... of remainders modulo q .The benefit of the reduction modulo q is that the point-count is combinatorially quiteaccessible. In practice, non-trivial prime powers are still harder to come by, so that the c invariant is often studied for pure primes only. It can be conjectured that the knowledge ofthe c for all primes determines the c for all prime powers.An important ingredient which exemplifies the power of the reduction is a variant of theclassical Chevalley-Warning theorem Lemma 7.
Let F ∈ Z [ α , . . . , α N ] be a polynomial of degree ≤ N in the N variables withinteger coefficients. Then (9) ( − N − [ F ] p ≡ coefficient of α q − · · · α q − N mod p. he reduction modulo p for all prime powers q is a practical compromise between thereduction modulo q and the restriction to pure primes p . The proof of the lemma is (e.g.)a special case of Lemma 30 in [20] using Equation (15) (also see [22] Lemma 2.6 for anexposition).The c has been studied quite deeply in particular in the context of φ quantum field theory(Section 2.7). The focus of these studies can either be the general mathematical structure ofthe c [10, 23, 13] or the zoology of the geometries identified by c s [8, 9, 22, 12, 24, 20]. Thenature of this article is more in the latter direction, particularly when we analyze the c sof small kernels in Section 5. We would like to emphasize that although identifying c s canhave an experimental flavour it might be of high importance to understanding the algebraicstructure of quantum field theories.We can get rid of the division by q in (8) by using Dodgsons instead of the graph poly-nomial (Corollary 28 and Theorem 29 in [8])(10) c ( q )2 ( G ) ≡ − [Ψ , G Ψ , G, ] q mod q for every connected graph G with a degree three vertex v . In this version the c is amendableto further simplifications by denominator reductions and quadratic denominator reductionsas will be outlined in the next sections.2.4. Denominator reduction.
Successive integration of the Feynman period (1) leads todenominators which are linear in the next integration variable. After three initial steps thenew denominator will be the resultant of the old denominator with respect to the integrationvariable. Eventually the denominator may cease to factorize into linear pieces and then onetypically enters very complicated territory. This successive taking of resultants has a point-count version which says that under certain conditions[( Aα + B )( Cα + D )] q ≡ − [ AD − BC ] q mod q. If the resultant AD − BC factorizes in some new variable then the reduction can be re-peated. The size of the polynomial that has to be counted reduces rapidly and when nomore reduction is possible then one can still resort in brute force counting at the last step.More precisely, we obtain the following result. Definition 8 (Denominator reduction, Definition 120 and Proposition 126 in [5]) . Given aconnected graph G with at least three edges and a sequence of edges , , . . . , | E ( G ) | we define Ψ G (1 , ,
3) = ± Ψ , G Ψ , G, . Suppose n Ψ G for n ≥ factorizes as n Ψ G (1 , . . . , n ) = ( Aα n +1 + B )( Cα n +1 + D ) then we define n +1 Ψ G (1 , . . . , n + 1) = ± ( AD − BC ) . Otherwise denominator reduction terminates at step n . If it exists we call n Ψ G an n -invariantof G . Note that the n -invariants are only defined up to sign. The 4-invariant always factorizes[5](11) Ψ G = ± Ψ , G Ψ , G . herefore the 5-invariant always exists. For n ≥ n -invariants become independentof the sequence of the reduced edges (they only depend on the set of reduced variables).Denominator reduction is compatible with the c invariant. Theorem 9 (Theorem 29 in [8]) . Let G be a connected graph with at least three edges and h ( G ) ≤ | E ( G ) | / independent cycles. Then (12) c ( q )2 ( G ) ≡ ( − n [ n Ψ G ] q mod q whenever n Ψ G exists for n < | E ( G ) | . The theorem was proved for ≥ n Ψ G = 0 for some sequence of edges and some n (and hence for allsubsequent n ) then G has ‘weight drop’[11]. In this case the c invariant vanishes.2.5. Quadratic denominator reduction.
Only in a few particularly simple cases denom-inator reduction goes through to the very end where all variables are reduced. Brute forcepoint-counting after the last step of denominator reduction is very time consuming and doesnot lend itself to more theoretical understanding. Therefore it is desirable to continue thereduction as far as possible. The integration of a denominator that does not factorize pro-duces a square root. The existence of further reduction steps is suggested by the fact thateven in the presence of square roots integrals may stay rational in a geometrical sense.For the algebraic implementation of this idea we pass from point-counts to Legendre sums.
Definition 10 (Definition 27 in [20]) . Let q be an odd prime power. For any a ∈ F q theLegendre symbol ( a/q ) ∈ {− , , } is defined by (13) (cid:18) aq (cid:19) = |{ x ∈ F q : x = a }| − . For any polynomial F ∈ Z [ α , . . . , α N ] we define (14) ( F ) q = (cid:88) α ∈ F Nq (cid:18) F ( α ) q (cid:19) , where the sum is in Z . The Legendre symbol is multiplicative, ( ab/q ) = ( a/q )( b/q ) for a, b ∈ F q and trivial forsquares ( a /q ) = 1 − δ a, where the delta is the characteristic function on F q . This leads to( F ) q = q N − [ F ] q for any polynomial F in N variables. If N ≥ F ] q ≡ − ( F ) q mod q. With the above equation we can translate point-counts to Legendre sums. Quadratic de-nominator reduction knows two cases,(( Aα + Bα + C ) ) p ≡ − ( B − AC ) p mod p, (( Aα + Bα + C )( Dα + E ) ) p ≡ − ( AE − BDE + CD ) p mod p if the total degree of the polynomials on the left hand sides does not exceed twice the numberof their variables. The proof of these identities uses Lemma 7 (hence the reduction modulo p ); it is in Section 7 of [20]. Note that in the case that both quadratic reductions areapplicable one is back to the case of standard denominator reduction. Then both reductionslead to the same result. e define quadratic n -invariants n Ψ G in analogy to Definition 8. Definition 11 (Quadratic denominator reduction, Definition 31 in [20]) . Given a connectedgraph G with at least three edges and a sequence of edges , , . . . , | E ( G ) | we define (16) Ψ G (1 , ,
3) = ( Ψ G (1 , , . Suppose n Ψ G for n ≥ is of the form (17) n Ψ G (1 , . . . , n ) = ( Aα n +1 + Bα n +1 + C ) then we define n +1 Ψ G (1 , . . . , n + 1) = B − AC.
Suppose n Ψ G is of the form (18) n Ψ G (1 , . . . , n ) = ( Aα n +1 + Bα n +1 + C )( Dα n +1 + E ) then we define n +1 Ψ G (1 , . . . , n + 1) = AE − BDE + CD . Otherwise quadratic denominator reduction terminates at step n . If it exists we call n Ψ G aquadratic n -invariant of G . If n Ψ G = 0 for some sequence of edges and some n then we saythat G has weight drop. Note that quadratic n -invariants have no sign ambiguity. The connection to the c invari-ant is similar to the standard case, with the exception that it works only modulo p for oddprime powers q = p n . Theorem 12.
Let q be an odd prime power and G be a connected graph with at least threeedges and h ( G ) ≤ | E ( G ) | / independent cycles. Then (19) c ( q )2 ( G ) ≡ ( − n − ( n Ψ G ) q mod p whenever n Ψ G exists. If the n -invariant n Ψ G exists we get n Ψ G = [ n Ψ G ] , generalizing (16). In many casesquadratic denominator reduction goes significantly beyond standard denominator reduction.2.6. Scaling.
Even when quadratic denominator reduction stops it is often possible to sim-plify further by scaling some variables (see Section 3.4). In the context of Legendre sumsthis technique is based on the elimination of square factors from the Legendre symbol. Forclassical point-counts scaling was already used in [18] and later in [8]. In this article we havea case where the result after scaling is amendable to further steps of quadratic denominatorreduction. This makes the scaling technique particularly powerful.2.7. φ theory. The most interesting graphs for us are the primitive 4-point φ graphs.Rephrased in a purely graph theoretic language, this means we are most interested in graphswhich can be obtained by taking a 4-regular graph and removing one vertex. The 4-regulargraph needs to be internally 6-edge connected, that is the only 4-edge cuts of the 4-regulargraph are those which separate one vertex from the rest of the graph. For graphs obtainedfrom an internally 6-edge connected 4-regular graph in this way the Feynman period isconvergent. For such graphs we have denominator reduction, Theorems 9, 12 as well asthe Chevalley-Warning theorem (Lemma 7) as extra tool. Chevalley-Warning lets us turna c calculation into a game of counting edge partitions, see Section 2.9 for a summary. igure 4. An hourglass and a bihourglass.This approach is particularly convenient for p = 2, and so nicely complements the quadraticdenominator reduction which is blind to p = 2.Furthermore, the Feynman period of a graph obtained by removing a vertex of an internally6-edge connected 4-regular graph does not depend on the choice of vertex removed. Thisis the completion invariance of the Feynman period. The analogous invariance for the c invariant is conjectural [8], but the edge partition counting approach has enabled a proofwhen p = 2 and the 4-regular graph has an odd number of vertices [23]. Upcoming workof one of us with Simone Hu will complete the p = 2 proof. In our hourglass chain graphs,we will be removing the most convenient vertex; if we assume the conjecture then this isequivalent to removing any other vertex.With the completion conjecture we can also ignore four-regular graphs with three vertexsplits (‘reducible’ graphs in [17]): By deleting one of the three split vertices the decompletedgraph inherits a two-vertex split which renders the c trivial [10]. The Feynman period ofa graph with a three-vertex split factorizes [17]. Another reduction is obtained by ignoringgraphs with double triangles (a pair of triangles with a common edge). It was shown in [10]that double triangles can be reduced to single triangles (one of the common vertices becomesa crossing) without changing the c invariant. With all reductions (internally 6-connected,3-vertex connected, double-triangle-free) we are lead to considering ‘prime ancestors’ [20].Note that for suitable kernels K the hourglass chain graphs we consider in this paper provideinfinite families of prime ancestors.2.8. Hourglasses.
By an hourglass we mean two triangles sharing one common vertex, seeFigure 4. Taking two of the degree two vertices of an hourglass which are not in the sametriangle and joining to two such vertices in another hourglass, we obtain a bihourglass , seeFigure 4. Continuing by joining a third hourglass in the same way to the remaining degree2 vertices of the second hourglass, and so on, we obtain longer hourglass chains , where thehourglass and bihourglass are the hourglass chains of length 1 and 2 respectively.Hourglass chains of any length have four degree 2 vertices and all remaining vertices ofdegree 4. Take another fixed graph K which has four degree 2 vertices and all remainingvertices of degree 4. Additionally, fix a bipartition of the degree 2 vertices of K into twoparts of size 2. We will call K (with this choice of bipartition) the kernel . Let G K be thefamily of graphs obtained by taking an hourglass chain of any length, joining the two degree2 vertices at one end of the chain to the two degree 2 vertices in one part of the bipartitiongiven with K , and joining the 2 vertices at the other end of the chain to the two degree 2vertices in the other part of the bipartition. See Figure 1 for an illustration. Note there aretwo ways to join on any hourglass chain compatible with the bipartition, differing by a halftwist. In the end, this half twist will not affect the c invariant, and so we include both in G K .2.9. The combinatorics of p = 2 . By Euler’s formula one can see that any step of ordinarydenominator reduction for a graph that is the result of removing one vertex from a 4-regular raph gives a polynomial with the appropriate degree and number of variables to applyLemma 7. Along with Theorem 9 this means that we can compute a c invariant at 2 ofsuch a graph by taking the coefficient of the monomial given by the product of all remainingedge variables in the n -invariant.Consider in a bit more detail what this means combinatorially. When the n -invariantis written as a quadratic expression in Dodgson polynomials then we can rewrite it as aquadratic expression in spanning forest polynomials by Proposition 5. Consider any term inthis expression: we have a product of two spanning forest polynomials on the same graphand want to find the coefficient of the monomial given by the product of all edge variables ofthis graph. That is, we want to find the number of partitions of the edge variables into twoparts one of which gives a monomial in the first spanning forest polynomial and the otherof which gives a monomial in the second spanning forest polynomial. Remembering thatspanning forest polynomials are sums of the edges not in the corresponding spanning forests,and swapping the parts, this is the same as counting bipartitions of the edges where onepart gives a spanning forest corresponding to the first spanning forest polynomial and thesecond part gives a spanning forest corresponding to the second spanning forest polynomial.We want to know this count modulo 2.Similar arguments work for p >
2, but it become combinatorially more intricate to workwith. One would need to count modulo p the number of partitions of p − p − p − p − Hourglass reductions
The goal of this section is to prove Theorem 1 for odd prime powers. Note that after usingthe theorem one can continue to reduce any variable of the kernel K . If K is the kernelof a four-regular hourglass chain then K (cid:48) has four degree three vertices at the ends of theextra edges 1 and 2. The topology of degree three vertices simplifies the structure of relatedDodgsons [5, 8], Lemma 17 in [20]. Particularly simple is the case of edge 2 whose variableis absent in (2) (note that the choice of labels 1 and 2 is arbitrary). Experiments suggestthat it might always be possible to reduce both edges of any vertex adjacent to edge 2. Question 13.
Let v be a degree three vertex attached to edge 2 in K (cid:48) . Let α and α bethe edges of v which are in K . Is it always possible to quadratically denominator reduce (2)with respect to α and α ? If yes, what expression does one get for the right hand side of(2) after the quadratic reduction of α and α ?In general it is quite helpful for analyzing larger kernels to have closed expressions for the c with as many reductions as possible.Figure 5 gives an overview of how the reductions will proceed. Following the specified orderin an explicit example using a computer for the reductions can also be a helpful way to followthrough the general argument.3.1. Initial reductions.
The first step of the proof is to begin a conventional denominatorreduction on L − v , see Section 2.4.Consider the two hourglasses around v and label the edges as in Figure 6. Then beginningour denominator reduction at the 4-invariant with (see (11)) u u u u
12 345 67 8 u x n = u z n = u u u = t t = u b c d e f Figure 5.
The order in which the edges should be reduced. First decompleteat v , thus removing the red edges. Then reduce the dark blue edges withconventional denominator reduction. Next reduce the light blue edges withquadratic denominator reduction. Continue with quadratic denominator re-duction to reduce the six pink edges. The same form of denominator reappears,and so inductively we can reduce the analogous six edges in each subsequenthourglass until we reach the last hourglass. Finally reduce the remains of thelast hourglass, the five dark cyan edges, according to Section 3.4.Ψ , L − v Ψ , L − v , the reductions of 5 and 6 are forced to avoid contracting the triangles 125 and 346 in thefirst factor, and then the reductions of 7 and 8 are forced to avoid disconnecting the degreethree vertices 257 and 468 in the first factor, see (4). This yieldsΨ , L − v, Ψ , L − v, . u u u u Figure 6.
Edge labelling around v . u u u u Figure 7.
Edge labelling for the triangle containing u . u u u u u u u u u u u u u u u u u u u u u u u u − − + Figure 8.
The first factor in the expression after reducing edge 10.At this point it is more convenient to consider the situation in terms of spanning forestpolynomials via Proposition 5 and Lemma 6,Ψ , L − v, Ψ , L − v, = ± (Φ { u ,u } , { u ,u } L (cid:48) − Φ { u ,u } , { u ,u } L (cid:48) )Φ { u } , { u } L (cid:48) where L (cid:48) is L without v , the edges 1 through 8, and without the three vertices isolated bythose removals. Label the triangle of L (cid:48) containing u as in Figure 7. Reduce edge 9 by thegeneral deletion and contraction reduction formula (3). Notice that in both terms where 9was deleted, 10 is an isthmus and u , the vertex at the isolated end of 10, is not in a part byitself. Thus 10 cannot be cut in these terms, forcing 10 to be contracted in the other factors.This gives ± (Φ { u ,u } , { u ,u } L (cid:48) − u − Φ { u ,u } , { u ,u } L (cid:48) − u − Φ { u ,u } , { u ,u } L (cid:48) − u + Φ { u ,u } , { u ,u } L (cid:48) − u )Φ { u } , { u } L (cid:48) − u , where the vertices are as labelled in Figure 7.Consider which trees the vertices u and u can belong to in the first factor of the previousexpression. This factor is illustrated in Figure 8. Both trees of the forest appear in theportion of the graph including vertices u , u , u and u and so must exit from this portionto the rest of the graph via the only possible vertices, u and u . Since there are two treesand two vertices, one must use u and the other must use u . In view of the shape of thegraph, the only way this can happen is that the tree corresponding to the square verticesexits via u and the tree corresponding to the circle vertices exits via u . Then the onlything undetermined in how the trees go through the illustrated part of the graph, is the treeto which u belongs in the third and fourth terms. Summing over both possibilities we seethat one of the possibilities cancels with the first two terms, and so what remains of theentire expression is ± (Φ { u ,u ,u ,u }{ u ,u } L (cid:48) − u − Φ { u ,u ,u ,u }{ u ,u } L (cid:48) − u )Φ { u } , { u } L (cid:48) − u . u u u u u u u u
12 1314
Figure 9.
The vertex labellings for the hourglass containing u and u .In the first factor, edge 11 must always be deleted since its two ends are in different partsand so we obtain ± (Φ { u ,u ,u ,u }{ u ,u } ( L (cid:48) − u ) \ − Φ { u ,u ,u ,u }{ u ,u } ( L (cid:48) − u ) \ )Φ { u } , { u } ( L (cid:48) − u ) / . The only way that the hourglass with v and v affected this computation was to guaranteethat both trees had to leave the part of the graph illustrated on the left. In other words, allwe needed to know is that both trees appeared in the part of the graph illustrated on theright. Swapping left and right this remains true and so we can use the same argument asabove on the triangle involving u . Labelling the vertices of the hourglass including u and u as in Figure 9, this calculation gives ± (Φ { u ,u ,u ,u ,u ,u } , { u ,u } L − Φ { u ,u ,u ,u } , { u ,u ,u ,u } L )Φ { u } , { u } L where L = ( L (cid:48) − { u , u } ) \{ , } and L = ( L (cid:48) − { u , u } ) / { , } . Now comes thekey observation that the first factor can be factored. First, the triangles u , u , u and u , u , u factor off since they are only joined at a vertex and the tree to which that vertexbelongs is known. Additionally, similarly to the observations used above, both trees need topropagate through each hourglass remaining in the chain since both trees appear on bothsides. However, the trees cannot cross within an hourglass, so one tree must run down oneside of the hourglasses and the other tree down the other side; only the middle vertex couldbe in either tree.To write this down nicely we need some more systematic notation. The edges and verticesof each hourglass will be labelled as in Figure 10, where the hourglasses X i are indexed by i . When it is useful to talk about an hourglass generically we will leave out the subscripts.Additionally, let t , t , t and t be the degree 2 vertices of K with the bipartition being { t , t } , { t , t } . With this notation, the factorization observation allows us to rewrite thedenominator expression so far as(20) ± Ψ L [ u ,u ,u ] Ψ L [ u ,u ,u ] (cid:32)(cid:89) i Φ { w i ,x i } , { y i ,z i } X i (cid:33) (cid:16) Φ { t ,t } , { t ,t } K − Φ { t ,t } , { t ,t } K (cid:17) Φ { u } , { u } L , where L [ S ] for a set of vertices S indicates the induced subgraph of L given by the verticesof S , that is the subgraph with the vertices of S and all edges in L that have both endsin S . In this case the two induced subgraphs are both triangles. Diagrammatically, thisexpression can be represented as in Figure 11. The reader is encouraged to draw all thesteps diagrammatically, as this gives the most insight into the calculation.Note that due to the choice of v , one of the triangles u , u , u and u , u , u shares twovertices with K , while the other does not. Without loss of generality say that u , u , u isthe one that is adjacent to other hourglasses, if there are any. This is the one drawn pointingupwards in Figure 11, and u is the top vertex. i b i c i d i e i f i w i x i y i z i X i = Figure 10.
Labels for hourglasses. (cid:89) i −± Figure 11.
Diagrammatic representation of (20).3.2.
The two triangles.
The plan of attack now is to reduce the edges in the two trian-gles u , u , u and u , u , u . After the first such edge, we will need to pass to quadraticdenominator reduction. Note that the only factors in (20) containing edge variables fromthe triangle u , u , u are Ψ L [ u ,u ,u ] and Φ { u } , { u } L . Let the edge between u and u be 15and let 16 and 17 be the other two edges of the triangle u , u , u . Reduce 15 in the usualway to obtain ( α + α )Φ { u } , { u } L \ − Φ { u } , { u } L / times the factors not involving the triangle u , u , u . Expanding out α and α we get( α + α )Φ { u } , { u } L \ − Φ { u } , { u } L / = ( α + α ) (cid:16) α Φ { z n } , { u } L + α Φ { x n } , { u } L + α α Ψ L (cid:17) − α α Ψ L = α Φ { z n } , { u } L + α Φ { x n } , { u } L + 2 α α Φ { x n ,z n } , { u } L + ( α + α ) α α Ψ L times the factors not involving the triangle u , u , u , where • n is the index of the hourglass adjacent to the triangle u , u , u with the vertices x n and z n being the same vertices as u (which has also been contracted with theoriginal u ) and u respectively, •
16 the edge from x n = u to u and 17 the edge from z n = u to u , • L = ( L − u ) \
15 and L = ( L − u ) /
16 17 16 1716 17factors notcontaining α , α = factors notcontaining α , α α + α +2 α α +( α + α ) α α Figure 12.
Diagrammatic rendition of the result of reducing edge 15.different parts of the vertex partition defining the spanning forest polynomial, and whenthere was originally only a single tree, there is no need to consider how the parts interactwith any existing parts. We also used that if a vertex is not in a part of a partition for aspanning forest polynomial then we can sum over all possibilities for putting that vertex intoa part.Now we need to move to quadratic denominator reduction, see Section 2.5. Reducing α according to quadratic denominator reduction (17) we obtain (cid:16) − α Φ { x n ,z n } , { u } L + α Ψ L (cid:17) − (cid:16) Φ { z n } , { u } L + α Ψ L (cid:17) α Φ { x n } , { u } L times the square of the factors not involving the triangle u , u , u . Note that α factorsout of this expression and so following (18) quadratic denominator reduction of α gives4 (cid:16) (Φ { x n ,z n } , { u } L ) − Φ { z n } , { u } L Φ { x n } , { u } L (cid:17) times the square of the factors not involving the triangle u , u , u .Next consider the u , u , u triangle. Unfortunately we cannot simply use the same argu-ment as above since we are not starting in conventional denominator reduction this time.This part of the argument is rather gruesome, but fortunately, it is the last bit of messywork before we get to the systematic part.Let 18 be the edge between u and u , and let 19 and 20 be the edges from u to t and t respectively. Reducing edge 18 we obtain( α + α ) (cid:16) (Φ { x n ,z n } , { u } L \ ) − Φ { z n } , { u } L \ Φ { x n } , { u } L \ (cid:17) − ( α + α ) (cid:16) { x n ,z n } , { u } L \ Φ { x n ,z n } , { u } L / − Φ { z n } , { u } L \ Φ { x n } , { u } L / − Φ { x n } , { u } L \ Φ { z n } , { u } L / (cid:17) + (cid:16) (Φ { x n ,z n } , { u } L / ) − Φ { z n } , { u } L / Φ { x n } , { u } L / (cid:17) times the square of the factors not involving either triangle. Expanding out all the α and α explicitly and freely using the observations on spanning forest polynomials used before, here are many nice cancellations and we obtain( α + α ) · (cid:16) α ((Φ {{ x n ,z n } , { t } M n ) − Φ { z n } , { t } M n Φ { x n } , { t } M n ) + α ((Φ { x n ,z n } , { t } M n ) − Φ { z n } , { t } M n Φ { x n } , { t } M n )+ α α (2Φ { x n ,z n } , { t } M n Φ { x n ,z n } , { t } M n − Φ { z n } , { t } M n Φ { x n } , { t } M n − Φ { z n } , { t } M n Φ { x n } , { t } M n + Φ { x n } , { z n } M n Φ { t } , { t } M n ) − α α ( α + α )Ψ M n Φ { x n } , { z n } (cid:17) times the square of the factors not involving either triangle and where M n = ( L − u ) \ n refers to the index of the last remaining hourglass). This calculation uses the factthat Φ { z n } , { t } M n + Φ { x n } , { t } M n − { x n ,z n } , { t } M n = Φ { z n } , { t } M n + Φ { x n } , { t } M n − { x n ,z n } , { t } M n = Φ { x n } , { z n } M n which can be seen to be true by expanding all the spanning forest polynomials so that eachof x n , z n , t , t is in each partition.Because of the factor of ( α + α ) we can proceed to reduce edge 19 to obtain α ((Φ { x n ,z n } , { t } M n ) − Φ { z n } , { t } M n Φ { x n } , { t } M n − α Ψ M n Φ { x n } , { z n } M n ) − α ( α (2Φ { x n ,z n } , { t } M n Φ { x n z n } , { t } M n − Φ { z n } , { t } M n Φ { x n } , { t } M n − Φ { z n } , { t } M n Φ { x n } , { t } M n + Φ { x n } , { z n } M n Φ { t } , { t } M n ) − α Ψ M n Φ { x n } , { z n } )+ α ((Φ { x n ,z n } , { t } M n ) − Φ { z n } , { t } M n Φ { x n } , { t } M n )times the square of the factors not involving either triangle. There is a factor of α in thisexpression, so we can reduce α to obtain(Φ { x n ,z n } , { t } M n ) − Φ { z n } , { t } M n Φ { x n } , { t } M n − { x n ,z n } , { t } M n Φ { x n ,z n } , { t } M n + Φ { z n } , { t } M n Φ { x n } , { t } M n + Φ { z n } , { t } M n Φ { x n } , { t } M n − Φ { x n } , { z n } M n Φ { t } , { t } M n + (Φ { x n ,z n } , { t } M n ) − Φ { z n } , { t } M n Φ { x n } , { t } M n times the square of the factors not involving either triangle. Expanding over all possibilitiesfor assigning whichever of x n , z n , t , t are not in the partition in each term, cancelling andthen recollecting terms we can simplify the expression above to(21) (Φ { x n ,t } , { z n ,t } M n ) − Φ { x n } , { z n } M n Φ { t } , { t } M n times the square of the factors not involving either triangle. This expression is more sym-metric than the notation makes it seem. Figure 13 shows the symmetry better. Note thatother than to choose the labelling of the vertices, we have not used the fact that all theremaining hourglasses are on one side of K . If we followed the same calculation beginningwith v in the middle of the hourglasses, then at this point we would have hourglasses oneach side of K making the expression nicely symmetric.To proceed, we can apply a Dodgson identity to the last factor. Temporarily supposethere was an edge labelled 1 joining z n and x n and an edge labelled 2 joining t and t andlet M (cid:48) n be that graph. Then (21) would be (Ψ , M (cid:48) n ) − Ψ , ,M (cid:48) n Ψ , ,M (cid:48) n , so by the Dodgson identity(5) we find that (21) is also − Ψ M (cid:48) n , Ψ , M (cid:48) n .Putting back in the factors we have been ignoring we get(22) − (cid:32)(cid:89) i Φ { w,x } , { y,z } X i (cid:33) (cid:16) Φ { t ,t } , { t ,t } K − Φ { t ,t } , { t ,t } K (cid:17) Ψ M (cid:48) n , Ψ , M (cid:48) n . actors notcontaining α , α − Figure 13.
Diagrammatic representation of (21). (cid:89) i −− Figure 14.
Diagrammatic representation of (22).For an illustration of this equation see Figure 14.3.3.
Systematic hourglass reduction.
Notice that the two right hand factors of (22) arethe same except that x n and z n , the top two vertices as illustrated, are identified or not andsame for t and t , the bottom two vertices. This will be important, so it will be useful tohave a compact notation for this; write T n A for Ψ , M (cid:48) n , , that is the top vertices together ( T ) and the bottom two vertices apart ( A ), and similarlyfor T n T = Ψ M (cid:48) n , A n A = Ψ , M (cid:48) n A n T = Ψ , M (cid:48) n , . The next order of business is to start reducing the top hourglass X n . Recall the hourglassnotation as in Figure 10. = b dc e f B = b dc e f C = b dc e f D = b dc e fF = b dc e f G = b dc e f H = b dc e f I = b dc e f E = b dc e fJ = b dc ef Figure 15.
The hourglass pieces that occur.First we reduce a n . We’re going to need to keep track to all the polynomials that comeabout from the remains of X n as they appear in each term after reducing a n . Writinggenerically for every hourglass for the moment, define A = deB = de ( b + c ) + bc ( d + e + f ) C = d + e + fD = bd + ( b + d )( e + f ) E = cd + ( c + d )( d + f ) F = bd ( c + e ) + cd ( b + d ) + f ( c + e )( b + d ) G = ( b + c )( d + e + f ) H = bcd + ( bc + bd + cd )( d + f ) I = bce + ( bc + be + ce )( d + f ) J = f ( bcd + bce + bde + cde ) . These are the Kirchhoff polynomials and spanning forest polynomials corresponding to thegraphs shown in Figure 15.Using (6) along with the notation above, reducing a n gives − (cid:32)(cid:89) i Finally, we can get rid of the last hourglass. With (24) we achievedthe following situation: we have two sets of variables, the five variables b , c , d , e , f fromthe hourglass X and some variables α i in the Dodgson polynomials with the kernel K (cid:48) . Theexpression (24) does not depend on the two variables α and α which are associated to theextra edges edges 1,2 in K (cid:48) . Let d be the degree of Ψ , K (cid:48) . Then Ψ , K (cid:48) , , Ψ , K (cid:48) , and Ψ , K (cid:48) , havedegree d + 1, while Ψ K (cid:48) , has degree d + 2 (see e.g. [20]). The total degree of (24) is 4 d + 14which equals twice the total number of its variables.Quadratic denominator reduction stops at (24). To obtain further reductions we usea scaling technique which was first used in [18] and later adopted in [8] to exhibit a K3structure in φ theory at loop order eight. Considering (24) as a denominator of an integrandit is clear that the two sets of variables separate under a scaling transformation of all α i by S = Y / [ b c ( d + e + f )]. Because S has total degree zero homogeneity of (24) is preserved.If S ∈ F × q the variables in the Legendre sum can be multiplied by S yielding4 b ( d + e + f ) Y Z (Ψ , K (cid:48) ) (Ψ , K (cid:48) , + c Ψ , K (cid:48) )(Ψ K (cid:48) , + c Ψ , K (cid:48) , ) S d +4 . The last factor is a non-zero square and can be dropped from the Legendre sum. Thissuggests that(25) (expression (24)) q = (4 b ( d + e + f ) Y Z (Ψ , K (cid:48) ) (Ψ , K (cid:48) , + c Ψ , K (cid:48) )(Ψ K (cid:48) , + c Ψ , K (cid:48) , )) q . orking over finite fields, however, we cannot ignore the singular locus of the scalingtransformation. We need the following lemma. Lemma 14. Let P be a homogeneous polynomial of odd degree and let q be an odd primepower. Then ( P ) q = 0 .Proof. Because q is odd there exists an x ∈ F × q which is not a square (half the elements in F × q are non-squares). Scaling all variables by x gives( P ) q = ( P x D ) q = ( P ) q (cid:18) xq (cid:19) D = − ( P ) q , where we used that the degree D of P is odd. (cid:3) We observe that in any situation where S is singular (i.e. some of the Y , b , c , d + e + f are zero) both expressions (24) and the right hand side of (25) are either zero or the productof two factors in separate variables which are homogeneous of odd degree. The validity of(25) follows from the Lemma with an inclusion-exclusion argument.The term on the right hand side of (25) is homogeneous of degree 4 d + 14 which equalstwice the total number of variables. We may use quadratic denominator reduction in thevariables f , e , d , b (in this sequence) yielding(expression (24)) q ≡ (4 c (Ψ , K (cid:48) ) (Ψ , K (cid:48) , + c Ψ , K (cid:48) )(Ψ K (cid:48) , + c Ψ , K (cid:48) , )) q mod p. By contraction-deletion (3) the right hand side is α (Ψ , K (cid:48) ) Ψ , K (cid:48) Ψ K (cid:48) , where we renamed c to α .For odd prime powers Theorem 1 follows from Theorem 33 in [20] because the total numberof reduced variables is odd. 4. The prime p = 2 we need to use a different technique, but fortunately the edge partition countingtechnique discussed at the end of Section 2 is particularly easy to deal with when p = 2.To prove the main theorem, it remains to show that with L and v be as in Theorem 1, c (2)2 ( L − v ) ≡ { u } , { u } L is one of the spanning forest polynomials and all the otherfactors in the expression give the other spanning forest polynomials. We need to find theparity of the number of ways to partition the edges of L among all these factors.Consider first an hourglass X i . For Φ { w i ,x i } , { y i ,z i } X i there are two essentially different possi-bilities. One possibility is that one of the two trees pops in to the centre of the hourglassand then back out, while the other runs along the top or bottom. The other possibility isthat the two trees of the forest go along the top and bottom of the hourglass and any oneadditional edge goes to the middle vertex.For the first possibility, the complementary edges for this hourglass form the same shapebut reversed. This means that in Φ { u } , { u } L , two branches are going through hourglass X i connecting w i to x i and y i to z i . There are exactly two ways that the edges can be biparti-tioned giving this same connectivity across X i in both polynomials depending on which tree igure 17. A way to pair up edge bipartitions in the proof that c (2)2 ( L − v ) ≡ { u } , { u } L while the black edgescontribute to the forests for the other factors.went to the centre of the hourglass in Φ { w i ,x i } , { y i ,z i } X i . This means that there is an even numberof edge bipartitions corresponding to the second possibility, and this possibility contributesnothing modulo 2. This is illustrated on the left in Figure 17.If any hourglass has its trees according to the first possibility then we’re done, so we maynow consider only those partitions of the edges so that we are left with all hourglasses in thesecond possibility.For the second possibility, the complementary edges for this hourglass are three of thecentral edges. This means that in Φ { u } , { u } L three of the corners of the hourglass are linkedthrough this hourglass. Since L has at least two hourglasses remaining, take an X i whichhas another hourglass between it and the kernel. The other side of X i may have anotherhourglass or may have the triangle u , u , u .Three of the corners are linked through X i in Φ { u } , { u } L . Suppose two of these cornersare in the same hourglass neighbouring X i . Then, since this neighbouring hourglass is alsoin the second possibility but cannot make a cycle in Φ { u } , { u } L , its configuration must againjoin, in Φ { u } , { u } L , the vertices heading to the next hourglass (or to the kernel if it is the lasthourglass). Either edge joining this hourglass to X i can appear. This again means that thereis an even number of bipartitions of this type. This is illustrated in the middle in Figure 17It remains to consider when the corners are linked through X i in Φ { u } , { u } L are u and u .Ψ L [ u ,u ,u ] needs exactly two of the edges of the triangle, leaving one for Φ { u } , { u } L . This onecannot be the edge between u and u as this would make a cycle in Φ { u } , { u } L , and so thereis again an even number of possibilities. This is illustrated on the right in Figure 17.In this way we can pair up all the edge bipartitions giving (20) and so c (2)2 ( L − v ) ≡ Kernels In this section we study kernels which lead to four-regular hourglass chains. This impliesthat the kernel K is internally four-regular while every external vertex has two incident edges(see Figure 18).5.1. Trivial kernels. Assume the kernel K has two or more edges between external vertices.Then every hourglass chain L ∈ G K splits if one cuts the four edges of K which have exactlyone external vertex. If K has more than one internal vertex then L has a subdivergence,its period diverges and the c vanishes [10]. The same holds true if K has an non-trivialinternal four vertex split.So, we may restrict ourselves to the case that K has at most one edge between externalvertices. If such an edge e shares its vertices with edge 1 or 2 in K (cid:48) then any L ∈ G K hasa double-triangle (a pair of triangles which share an edge [17]). After two ‘double-triangle , , , , , , , , , , , , , , , 10 5 , 11 5 , 12 5 , , 14 5 , 15 5 , 16 5 , Figure 18. The kernels with up to five internal vertices. Here, edges 1 and2 in K (cid:48) would be vertical edges on either sides of the depicted kernels K .reductions’ we are left with a graph which has a three vertex split. The ‘ancestor’ of L factors into a non-trivial power of the complete graph with five vertices and a factor whichonly contains edges of K . Assuming the completion conjecture (Section 2.7), every productancestor has trivial c [10, 13]. We are left with the case that e joins the vertices of thehourglasses from opposite ends of the chain. By Theorem 1 it is irrelevant how the verticesof edges 1 and 2 in K (cid:48) are attached to their corresponding hourglasses. The case of one edgebetween external vertices of K thus reduces to a single setup (which becomes relevant forkernels with ≥ K has no edge between external vertices then there exist three potentially distinct caseshow to glue the kernel K into the hourglass chain (corresponding to the 2,2 set partitions ofthe external vertices). M¨obius twists of the chain can be ignored as the lead to equal c s.In general, periods which admit a three-vertex split are products. Assuming the completionconjecture, their c invariants vanish. So, we skip kernels which have a three-vertex split.If a kernel has a double triangle then the c invariant is equal to the c of a smaller graphwhere the double triangle is reduced [17, 10]. We also exclude these cases because their c sare found in smaller kernels.Finally, we also ignore kernels which have an external hourglass in such a way that it addsto the chain (with the exception that K is an hourglass).5.2. Small kernels. We generate all effectively different kernels with up to ten internalvertices. We use the fact that every kernel K can be made four-regular by adding a squareto the external edges. Non-trivial kernels with at least one internal vertex can be found infour-regular graphs which are internally six-connected and do not have a three vertex split.Such four-regular graphs are called irreducible primitive in [17]. From opening these graphsalong all of their squares we obtain the number of effectively different kernels given in Table 1(graphs were generated by nauty [14]). See Figure 18 for the cases with at most five internalvertices. internal vertices 0 1 2 3 4 5 6 7 8 9 10 Table 1. The number of effectively different kernels in φ theory increasesrapidly with the number of internal vertices.kernel c invariant0 Legendre symbol ( − /q )1 Legendre symbol (4 /q )2 Legendre symbol (4 /q )3 modular form of weight 4 and level 84,1 Legendre symbol ( − /q )4,2 unidentified sequence 0 , , , , , , , , , , , . . . /q )4,4 unidentified sequence 0 , , , , , , , , , , , , . . . , , , , , , , , , . . . /q )5,3 modular form of weight 9 and level 45,4 unidentified sequence 0 , , , , , , . . . , , , , , , , . . . , , , , , , , . . . , , , , , , . . . , , , , , , , . . . , , , , , , , , . . . , , , , , , , , , . . . , , , , , , , . . . , , , , , , , . . . , , , , , , , . . . , , , , , , , . . . Table 2. The c invariants for kernels with up to five internal vertices. Thefirst column gives the number of internal vertices and its label in Figure 18.The c invariants for kernels with up to five internal vertices can be found in Table 2. Notethat the results are mostly obtained by identifying finite prefixes of the infinite c sequence(indexed by prime powers), and so are not proven. All calculations were performed usingthe Maple package HyperlogProcedures by the first author [21].The modular forms [3,36] and [9,4] in kernels 5,16 and 5,3 (respectively) were not found in c invariants of φ graphs with loop order ≤ 12 (see [20]). By the time of writing these newlyidentified modular forms are confirmed up to prime 19. We will test the correspondence for igher primes in the following weeks. No new modular forms were found in hourglass chainsof kernels with six internal vertices.A proved lower bound for the level of weight 2 modular forms (corresponding to point-counts of curves) in hourglass chains of kernels with at most six internal vertices is 69.Ongoing calculations will test the no-curves-conjecture to much higher levels.The analysis of kernels with seven (or more) internal vertices requires significantly morecomputing power. We did not pursue this here.Full reductions were possible for the Legendre symbol ( − /q ) in kernel 4,1 and for theLegendre symbol (4 /q ) in kernels 5,2 and 6,7. For these hourglass families of φ ancestorsthe c is proved. References [1] S. Bloch, H. Esnault, D. Kreimer , On Motives Associated to Graph Polynomials , Comm. Math.Phys. 267, 181-225 (2006).[2] M. Borinsky , Feynman graph generation and calculations in the Hopf algebra of Feynman graphs ,Computer Physics Comm. 185, no. 12, 3317-3330 (2014).[3] D. Broadhurst, D. 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