Geometric general solution to the U(1) anomaly equations

Costa et al. [Phys. Rev. Lett.123 (2019) 151601] recently gave a general solution to the anomaly equations for n charges in a U(1) gauge theory. ‘Primitive’ solutions of chiral fermion charges were parameterised and it was shown how operations performed upon them (concatenation with other primitive solutions and with vector-like solutions) yield the general solution. We show that the ingenious methods used there have a simple geometric interpretation, corresponding to elementary constructions in number theory. Viewing them in this context allows the fully general solution to be written down directly, without the need for further operations. Our geometric method also allows us to show that the only operation Costa et al. require is permutation. It also gives a variety of other, qualitatively similar, parameterisations of the general solution, as well as a qualitatively different (and arguably simpler) form of the general solution for n even.


Introduction
The local anomaly cancellation equations for a U(1) gauge theory with n left-handed chiral fermions of charge z i , which may be taken to be integers, are (1. 2) The first of these, (1.1), comes from a one-loop triangle diagram with two external gravitons and one external U(1) gauge boson [1], whilst (1.2) comes from the similar diagram with three external U(1) gauge bosons [2][3][4][5][6]. Although written for left-handed chiral fermions, these equations are general for a theory with both left-handed and right-handed chiral fermions since we can charge conjugate any right-handed representation, reversing the sign of its charge and giving a left-handed representation. Eq. (1.2) is a cubic diophantine equation in n variables; since it is not yet known how to solve a generic such equation even in 2 variables (corresponding to an elliptic curve [7]), one might expect that finding the general solution to (1.1)-(1.2) is a difficult problem. However, a recent paper by Costa, Dobrescu and Fox (CDF) [8] managed to do so, in the following way.

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CDF observed that given two integer solutions x := (x 1 , . . . , x n ) and y := (y 1 , . . . , y n ), of (1.1), and (1.2), a third could be constructed from a 'merger' operation, which they denoted '⊕' Some solutions to (1.1) and (1.2) are easy to find, having for each charge z i another charge z j = −z i . Using solutions of this form, which we call vector-like solutions, and the merger CDF showed that one can construct chiral sets of charges, namely those where z i + z j = 0 for all i and j. They then showed (via rather lengthy algebra) that any solution can be constructed from these chiral sets of charges by permutation of charges or concatenation with each other or with vector-like solutions. For n even the specific mergers they considered were (l 1 , k 1 , . . . , k m , −l 1 , −k 1 , . . . , −k m ) ⊕ (0, 0, l 1 , . . . , l m , −l 1 , . . . , −l m ), (1.4) where m = n/2 − 1 ≥ 2 and k i , l i ∈ Z, i ∈ {1, . . . , m}. Whilst for n odd they were where m = (n−3)/2 ≥ 1. CDF showed that if one wants to avoid zero charges or vector-like copies of charges then conditions have to be applied to k i 's and l i 's.
Here, we show that the ingenious methods of CDF have a simple geometric interpretation, corresponding to elementary constructions long known to number theorists [9]. Viewing them in this context allows a fully general solution to be written down in one fell swoop. The geometric interpretation allows us to give a variety of other, qualitatively similar, parameterisations of the general solution, as well as a qualitatively different form of the general solution for even n. It also allows us to show that to generate all solutions from CDF's parameterisation only requires permutations and not the other operations.
The paper proceeds as follows: in § 2, we review the geometric method that we employ to solve (1.1) and (1.2), generalising a number-theoretic result of Mordell to dimensions higher than 3 in the process. We detail two solutions that our method yields directly, but which require permutations of CDF's solutions, and show that for CDF's parameterisation permutations is the only operation required. We conclude in § 3. There is one potential inconvenience in our parameterisation, in that there are special solutions generated differently from others, which we circumvent in appendix A. We present the different form of the general solution for even n in appendix B.
Our geometric solution will, on the other hand, be able to generate such solutions without JHEP05(2020)065 resorting to permutations. 1 The reasoning behind this, as we shall see later, lies in our use of a geometrical approach, namely that of projective geometry over the field Q of rational numbers. Before seeing how geometry makes an appearance in the problem at hand, let us recall the basic definitions.
For a field k, the projective space Pk n−1 is the space of all lines through the origin in the affine space k n . In other words, it is (k n − {0})/ ∼ where ∼ is the equivalence relation m 1 ∼ m 2 with m 1 , m 2 ∈ k n if and only if there exists a λ ∈ k such that m 1 = λm 2 . We denote a point in Pk n−1 by the equivalence classes [a 1 , · · · , a n ] for a i ∈ k.
Within the projective space Pk n−1 we can define d-planes. By a d-plane (for d < n−1) we mean a d-dimensional projective subspace of Pk n−1 , which can be written as where [α 1 : · · · : α d+1 ] ∈ PQ d parameterise the d-plane and p i ∈ PQ n−1 are fixed. A 1-plane, for example, is just a (projective) line, homeomorphic to a circle.
To motivate the use of projective space on physical grounds, we note that the Lie algebra of the U(1) gauge group is isomorphic to R. Given that U(1) is compact, this implies that our charges z i are not only real-valued, but also commensurate, meaning that if z j = 0, then z i /z j is rational for all i. We can scale every z i by a single real parameter without changing the physics, as long as the coupling constant is also appropriately scaled. This, along with the fact that the z i 's are commensurate, allows us to undertake a scaling such that all charges are rational, viz. z i ∈ Q. 2 It also tells us that we should think of the set of all charges as living in projective space, specifically PQ n−1 and indeed, (1.1), and (1.2), being homogeneous, define loci therein.
It is convenient for us to eliminate z n in our equations from the cubic equation in (1.2) to get This equation is homogenous, meaning it is well defined on our equivalence classes in PQ n−2 , and as such it defines a cubic hypersurface (given it is co-dimension 1) of PQ n−2 .
In order to make progress in solving this equation, we review some geometric methods used in diophantine analysis.

The method of chords
Consider a homogenous cubic in n-variables, with rational coefficients, defining a locus in Q n . Let a and b be two points in Q n on the locus. A result from antiquity 3 tells us that a 1 Though, as we indicate, utilising permutations can be useful. 2 In the end, we can scale them all so they are integer, as we previously claimed. But working with the field Q, rather than the ring Z, allows us to do geometry. 3 The result certainly goes back at least to Fermat and Newton in the 17th century and may go back even further to Diophantus in the 3rd century. A historical account is given in [10].

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chord between a and b will intersect the surface at a third point in Q n . One can understand this result as follows, let L(t) = a + t(b − a) be the chord joining a and b. Points both lying on this chord and in the cubic surface must satisfy the equation This result comes from considering the cubic along the chord and noting that a cubic has one or three (possibly degenerate) real roots. Hence within Q n , there is a third point of intersection, corresponding to t = t 0 and given by L(t 0 ). We note that this result is equally valid in projective space, PQ n . We will call this construction the 'method of chords'. Further, a rather more recent (though equally elementary) result of Mordell [9] states that all rational points in a cubic surface in PQ 2 can be constructed from chords in this way, starting from a projective line, L, and a point, p / ∈ L that both lie in the surface. It follows from the realisation that in fact any point in PQ 2 (ergo any point on the cubic) is on a chord from p 1 to a point in L. As we will see, this result generalises in a straightforward way to PQ n , but there is no analogous result in affine space. In Q 3 for example, the analogous result would have to involve two skew lines, L 1 and L 2 . However, points forming a plane with L 2 which is parallel to L 1 will be missed. In PQ 2 , there is no concept of parallel lines -pairs of lines are either disjoint or intersecting -and indeed the aforementioned points all lie on a chord connecting a point on L to p.
This simple observation, when generalised to higher n, underlies the fact that the point (0, −9, 7, −1, 8, −5) is missing from CDFs n = 6 parameterisation, but is included when we work in projective space, as we will discuss in detail in § 2.4.
Before actually using any of these results, we note that our general method will not work in the cases for n = 1, and n = 2. This is because for n = 1 and n = 2 it would require a notion of a (−1)-plane! Part of the discussion, namely that in appendix A, is also valid only for n ≥ 4. Happily, the solutions to the n = 1, 2, 3 cases can be found directly, allowing us to restrict our general discussion to n ≥ 4. Namely for n = 1 the solution is z 1 = 0. For n = 2, (2.2) results in no effective constraint (one obtains that the left-hand side is identically zero for any z 1 ) and so the solution of (1. We now consider higher n where the results above are more useful. For illustrative purposes, we will start with a rather explicit discussion of the case n = 4.

Application for n = 4
Let us consider the cubic anomaly-free surface in PQ 2 , such that l 1 = 1. At this stage, however, we refrain from doing so, preferring a slightly redundant parameterisation in order to stay closer to our analysis of the higher n cases below. We then construct a line passing through a generic point on each of Γ 1 and Γ 2 as The homogeneous parameter [α 1 : α 2 ] ∈ PQ 1 parameterises L 1 , which must intersect the cubic surface at a third point, assuming that L 1 is not wholly in the cubic surface. On substituting the chord into (2.3) we obtain the constraint on α 1 and α 2 at intersections of the line and the cubic surface: . This, combined with (2.5), tells us that every such rational solution to the cubic equation can be found by considering lines between points on Γ 1 and Γ 2 . What we have done here is apply Mordell's result to solve the n = 4 case of our problem.

Arbitrary n ≥ 4
To consider arbitrary values of n ≥ 4 we must generalise Mordell's result to an arbitrary cubic hyper surface X in PQ n−2 . The generalisation is immediate and gives the following Proof. The result is obvious if p ∈ Γ 2 . If p / ∈ Γ 2 , then p and Γ 2 define a (d 2 + 1)-plane, which intersects Γ 1 in a point p 1 . The line through p and p 1 intersects Γ 2 in a point p 2 , yielding a chord.

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or, if the terms on the right-hand side both vanish, we have that every rational point on L is on X. Lines which lie in X may be regarded as slightly awkward to deal with. Happily, it is possible, as we show in appendix A, to find every solution on such a line by a permutation of the coordinates of a solution arising as the unique third point of intersection on a line not lying in X. A comparison of (2.7) with (1.3) shows that the 'merger' operation is really nothing but the finding of the third rational point starting from two others.

Comparison with CDF
The parameterisations of CDF, in contrast to ours, have k me+1 = −l 1 and l mo+1 = k 1 .
We have already discussed above that CDF's solution misses the point (0, −9, 7, −1, 8, −5), for n = 6 and that for them this has to be found by permuting another solution, for example that generated with k 1 = 14, k 2 = 2, l 1 = −18, l 2 = −9 after scaling. In our parameterisation (0, −9, 7, −1, 8, −5) can be obtained directly with, for example, k 3 = 0, We can be more specific and ask: given the planes in (2.8) where we force k me+1 = −l 1 and l m 0 +1 = k 1 to retrieve CDF's solution, what points don't lie on lines between them? It is easy to see that for even n this would require either k me+1 or l 1 to be zero and for odd n either l m 0 +1 or k 1 , but not both. Thus, for the point [a 1 : · · · : a n ] to not lie on such a line, we need, for even n, a 1 + · · · + a n−1 = 0 or a 1 + a n−2 = 0, (2.10)

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or, for odd n, a 1 + · · · + a n−1 = 0 or a m 0 +2 = 0. (2.11) For a non-zero solution we can always rearrange the charges so that none of these conditions are satisfied. The only other points CDF miss are those where the line between the two planes in (2.8) lies within X. For example for n = 4 setting k 2 = k 1 gives a line L which lies in X. As an explicit example, consider k 1,2 , l 1 = 1. This line is given by can never land on such solutions. Nevertheless, CDF's parameterisation can get these by permutations, for the same reason that the parameterisation given here can, as we discuss in appendix A. The above two points not only show when CDF's parameterisation fails to reach a specific point but also proves that their parameterisation produces every point up to permutations.

Discussion
The pioneering work of CDF finds solutions to the local U(1) anomaly cancellation constraints. This allows the construction of the general solution, provided one allows permutations. Our geometric method provides the general solution directly without having to perform additional steps. The geometric method also explains how some of the otherwise obscure features of CDF's construction (particularly the 'merging' procedure of two solutions) come about. Due to an immediate generalisation of a theorem by Mordell, the geometric method is guaranteed to find all rational solutions for a fixed number of charges n. Therefore (after clearing all denominators), it finds all integer solutions.
Two further remarks are in order. Firstly, as we have seen, our parameterisation of the general solution is somewhat distasteful, in that occasionally the chord L joining points on Γ 1,2 lies in X, and so yields not one, but infinitely many solutions. Another way to find these solutions is to permute the coordinates z i of solutions arising as the unique third intersection of a line L which is not in X, as shown in appendix A. Secondly, in the case where n is even, a completely different, and arguably even simpler, construction of a general solution is possible. Indeed, in such cases, the cubic hypersurface has double points, where both the left-hand side of (2.2) and its partial derivatives vanish (e.g. the rational point

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Acknowledgments This work was supported by STFC consolidated grants ST/P000681/1 and ST/S505316/1. We thank other members of the Cambridge Pheno Working Group for helpful discussions.

A Any solution via permutations
Here, we give a proof of the statement that any solution sitting on a line in X between the d-planes defined in (2.8) can be found by the permutation of the coordinates of a solution which is on a line not in X. The proof of this statement follows similar reasoning to the proof regarding permutations of solutions in [8]. We must distinguish between n odd and even, so we do them each in turn.

A.1 Even n ≥ 4
We redefine variables such that for i = 1, for i = 2, · · · , m e + 1, The d-planes in (2.8) are defined in our new variables by y i = 0 for Γ e 1 and x i = 0 for Γ e 2 . Consider a point p = [x i : There is a unique line through p, p 1 ∈ Γ e 1 and p 2 ∈ Γ e 2 . Under the permutation φ e : z me+1 ↔ z 2me+1 , only y me changes and L φ e (p) = αp 1 + βφ e (p 2 ). (A.3) A necessary condition for L p to be in X is that where the dots indicate terms which are independent of y me . Thus if L p is in X, for a solution p with coordinates permuted such that then L φ e (p) will not be in X. The only case where this cannot be done is where all |z i | are equal, but such solutions already occur in Γ e 1 after permutations of the z i .

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A.2 Odd n ≥ 4 Here, x i = z i−1 + z m 0 +1+i , for i = 2, · · · , m o + 1, Again, Γ o 1 is simply defined by y i = 0 and Γ o 2 is defined by x i = 0. Similar to the even n case, we take a point p = [x i : There is again a unique line A necessary condition for L p to be in X is then where now the dots indicate terms which are independent x 2 . If L p is in X for a solution p with coordinates permuted such that then L φ o (p) will not be in X. We may use this construction for all solutions and n odd.
B Alternative solution for n-even Consider a line through our double point d, L = γ 1 d + γ 2 r, for r ∈ PQ n−2 a fixed point and [γ 1 : γ 2 ] specifying the position along the line. Any point in PQ n−2 lies on such a line, and further every such line is either in the hypersurface X (defined by (2.2)) or passes through that hypersurface at exactly one other point.

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This other point of intersection can be found by substituting L into (2.2):