On -torsion in degree superelliptic Jacobians over
Abstract.
We study the -torsion subgroup in Jacobians of curves of the form for irreducible over a finite field of characteristic . This is a function field analogue of the study of -torsion subgroups of ideal class groups of number fields . We establish an upper bound, lower bound, and parity constraint on the rank of the -torsion which depend only on the parameters , , and . Using tools from class field theory, we show that additional criteria depending on congruence conditions involving the polynomial can be used to refine the upper and lower bounds. For certain values of the parameters , we determine the -torsion of the Jacobian for all curves with the given parameters.
Key words and phrases:
ideal class group, superelliptic curve, Jacobian, Weil pairing, Galois cohomology2020 Mathematics Subject Classification:
Primary 11R29, 11R58; Secondary 11R34, 11R37, 11G20, 11G451. Introduction
The ideal class group of a number field is one of the central topics of interest in algebraic number theory. If we consider the collection of degree extensions with some fixed Galois group, then for all but finitely many primes , the -torsion of the class group of is conjectured to be distributed according to the Cohen-Lenstra heuristics [3]. If , however, the -torsion is expected to have qualitatively different behavior. For instance, in the case and is an imaginary quadratic field, Gauss’ genus theory completely describes the -torsion of the class group of in terms of the number of ramified primes. In general, if divides , then the -torsion structure can be considerably more mysterious.
In [12], the authors used Galois cohomology to study the -torsion of the ideal class groups of the degree number fields for prime ; see Section 1.4 for more on the history of this problem. In this paper, we study an analogous problem over global function fields, namely the divisor class groups of fields of the form for irreducible. In this setting, we are able to utilize both Galois cohomology inspired by [12] and tools from arithmetic geometry to obtain more refined constraints on the -torsion, and we encounter interesting behavior which does not occur in the number field setting.
Computing the -torsion structure of the divisor class group of a function field is typically a computationally intensive problem that requires first finding the full class group. We produce constraints on the -torsion using data that are much easier to compute, and in some cases, these constraints uniquely determine the -torsion. The full results are discussed in Section 1.1, but we give one example application here.
Theorem 1.1.
Let be prime, a prime power with , and irreducible with coprime to . The -torsion of the divisor class group of is isomorphic to if and is even, and is trivial otherwise.
If , and and are coprime to , Theorem 1.1 shows that the -torsion can be determined using no information about other than its degree. If , and and are coprime to , then we can completely determine the -torsion structure using easily computable conditions depending on (Corollary 1.7).
-torsion: | ||||||
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count: |
For , the -torsion structure is typically not fully determined by the easily computable conditions mentioned above, but we prove a parity constraint which gives us a better understanding of the -torsion. To illustrate this phenomenon, up to isomorphism there are function fields of the form with irreducible of degree . For each of these fields, the authors used Magma to compute the divisor class group and recorded the -torsion structure of each; see Table 1. While the relative distribution of curves across the possible torsion structures is a subject of future exploration, our results explain the zeroes in the table. More generally, we will see that the largest power of occurring as a subgroup of the divisor class group of must be odd whenever , is a primitive root mod , and is even and coprime to (Theorem 1.3). It seems as though this phenomenon is unique to the function field setting and does not arise for number fields.
1.1. Main results
For all the results that follow, we assume is prime, is a prime power coprime to , and is an irreducible polynomial with coprime to . Let be the smooth projective curve with affine equation given by ; such a curve is an example of a “superelliptic curve.” Let be the Jacobian of , so the degree subgroup of the divisor class group of is isomorphic to . The -torsion of can be equipped with the structure of a vector space over , and we define the -rank of to be the dimension of this vector space,
The function field of is isomorphic to , and up to isomorphism is the only smooth projective curve with this function field. We define the divisor class group of to be the divisor class group of . Then an equivalent definition for is that it is the largest power of that occurs as a subgroup of the divisor class group of the function field .
Remark 1.2.
The above definitions are valid also for , but in this case we always have , because a hyperelliptic curve has no -rational -torsion in its Jacobian when is irreducible.
Let be the multiplicative order of in , that is, the smallest positive integer such that . This is an important invariant for this problem because is the smallest extension of containing -th roots of unity, and hence the Galois closure of is a degree extension field, namely .
Theorem 1.3.
Set
Then the -rank satisfies and .
The parity constraint is proved using the Weil pairing on . This phenomenon does not appear to occur in the analogous situation in number fields, namely ideal class groups of cyclic extensions for prime discussed in [12]; see Section 1.4 for a discussion of the number field case.
Example 1.4.
Consider the case . If or if is odd, then , so Theorem 1.3 implies that . Otherwise, if is even and , we have , and we recover Theorem 1.1 for . Compare [12, Theorem 6.1.1] which addresses extensions for prime .
In one special case, we can prove a lower bound that equals the upper bound in Theorem 1.3, allowing us to construct families of curves with large -torsion subgroups in their divisor class groups.
Proposition 1.5.
If , then .
Theorem 1.1 follows immediately from Theorem 1.3 and Proposition 1.5. Both Theorem 1.3 and Proposition 1.5 can be proven with linear algebra, using linear maps on defined using endomorphisms of . The parity constraint is proved using the Weil pairing on . These topics are summarized in Section 1.2 and discussed in depth in Section 2 and Section 3. The proofs of Theorem 1.3 and Proposition 1.5 are then completed in Section 7.
If then is completely determined by Theorem 1.1, so for the remainder of this section we assume . We can compute more refined bounds on if we additionally assume . This constraint ensures that totally splits in the extension ; this is analogous to the constraint in [12] which guarantees that totally splits in . Over , splits into irreducible factors, which we label in such a way that the Frobenius automorphism on sends to for all (and to ). Set
(1) |
The polynomials are associated via Kummer theory to certain cyclic degree extensions of ; see Section 6.3 for more on how these polynomials arise.
Theorem 1.6.
Let be irreducible of degree , with coprime to and . Set
Then the -rank satisfies and .
If in addition is even and , then .
Since we have , and from we can conclude . Thus Theorem 1.6 gives both upper and lower bounds that are at least as strong as those in Theorem 1.3.
In addition to the linear algebra on discussed in Section 2 and Section 3, the proof of Theorem 1.6 requires techniques from Kummer theory and Galois cohomology. These techniques are introduced in Section 1.3 and discussed in depth in Sections 4, 5, and 6. The proof of Theorem 1.6 is then completed in Section 7.
In some cases, Theorem 1.3 and Theorem 1.6 are sufficient to determine precisely.
Corollary 1.7.
Suppose . If then , and otherwise .
Proof.
If then Theorem 1.1 implies , so the only remaining option to consider is . If then , and if then ; in both cases we must have by Theorem 1.3. So we may now assume . If , then is either or . In either case , so by Theorem 1.6. If , then , but we also have and so , so again by Theorem 1.6. ∎
For larger values of , Theorem 1.3 and Theorem 1.6 are not sufficient to determine . For example, we have the following options when :
For the first two rows, we can exhibit curves attaining both possible values of , demonstrating that the parameters are not sufficient to fully determine the value of in general. For instance, consider the case , (so ), and from the introduction, summarized in Table 1. We may categorize these function fields further by the sets associated to each. See Table 2, and note in particular the last two columns, consisting of curves with the same but different values of . 111The astute reader may notice in Table 2 that is closed under . This symmetry does hold in general, following from Theorem 1.11 and Theorem 1.13 below, and can be used to cut down the number of computations needed in order to find the set .
: | |||||
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count: |
1.2. Results from the linear algebra of Frobenius eigenvectors
The most important feature of working with function fields is that we can represent elements of the ideal class group of using geometric objects, as described in Section 1.1. This allows us to use morphisms from to itself to study .
The -power Frobenius map on induces a linear operator on the -vector space . The eigenspace of eigenvalue for this action is , so can be recovered as the dimension of this eigenspace. The primary difficulty we will encounter is that the action of on is not semi-simple in general. The action of Frobenius on the -adic Tate module is semi-simple, but this property does not descend to the mod reduction. So even though we can determine the full characteristic polynomial of acting on with relatively little work (Remark 2.10), this is not enough to determine the dimension of any particular eigenspace.
To study , we use a filtration coming from the automorphism on the space denoted as
that is preserved by , and we consider the intersection of generalized eigenspaces for with this filtration.
Definition 1.8.
For and , let denote the set of for which
Note that for , multiplication by is a well-defined scalar operator on because in . We will show that has a basis formed by taking one true eigenvector of from each , whenever such an eigenvector exists. Thus is directed related to in the following result.
Theorem 1.9.
The -rank of the divisor class group equals the number of values such that contains an eigenvector of .
For a visual interpretation, see LABEL:fig:lifting_charts. Each of the three figures represents a different possible curve. If the cell with coordinates is shaded dark gray, this means that there exists a eigenvector with eigenvalue in . The circles correspond to cells with in , so any dark gray circle corresponds to a eigenvector of eigenvalue , that is, an element of . The number of dark gray circles equals to . See Remark 2.7 for a more thorough guide to reading these diagrams.
We prove several constraints that determine when contains an eigenvector of . A number of relations follow fairly directly from linear algebra on . We say that is a “rooftop” if has a eigenvector but there is no such that has an eigenvector. This notion is justified by statement (b) of the following Proposition. We say a rooftop is “non-maximal” if .
Proposition 1.10.
Let and . We have the following:
-
(a)
has a eigenvector if and only if and .
-
(b)
If has a eigenvector, then has a eigenvector for all .
-
(c)
If is a rooftop, then has a eigenvector (that is, and ).
-
(d)
If is a non-maximal rooftop, there is no rooftop of the form for and .
These relations are proven in Section 2.5. Part (a) is obtained by constructing an explicit basis of eigenvectors for , and part (b) follows from a relation between and the map used to define the filtration. If is a non-maximal rooftop, we will show that applying to a vector in yields an eigenvector in (a “Jordan chain” of length ), giving part (c). Part (d) comes from that if two such Jordan chains exist, a linear combination of the two generalized eigenvectors will produce an eigenvector.
The remaining three constraints Theorem 1.11, Theorem 1.12, and Theorem 1.13 are less straightforward, and form the main technical contributions of this paper. The first two can be proven using the Weil pairing on , by proving a numerical relation between pairs of Jordan chains (Lemma 3.3).
Theorem 1.11.
Suppose and . Then is a rooftop if and only if is a rooftop.
When , the above result is vacuous; however in this case we will see that the numerical relation gives us the following result.
Theorem 1.12.
Suppose and . If and is even then is not a rooftop.
Both Theorem 1.11 and Theorem 1.12 are proved in Section 3. Theorem 1.11 says that defines an involution on the set of non-maximal rooftops, and Theorem 1.12 imposes a parity constraint on the fixed points of this involution; these two observations will be used together in Section 7.3 to prove the parity constraint .
1.3. Results from Galois cohomology
Another method we use to study the -rank of comes from Galois cohomology. This approach is more closely related to the approach used to study the number field version of this problem, where there is no direct analogue of the geometric -torsion subgroup (see Section 1.4). In Section 4 and Section 5 we use Kummer theory to relate the two perspectives. The culmination of these sections is Proposition 5.7, which relates the existence of an eigenvector of in to the existence of a certain cohomology class in a Selmer group associated to a -dimensional representation of .
In Section 6 we show that is a non-maximal rooftop if and only if a certain cup product of cohomology classes does not vanish. The vanishing of this cup product can be determined by a residue field calculation, leading us to the last constraint.
Theorem 1.13.
Suppose and . Then has a eigenvector if and only if (as in Eq. 1).
We note that Proposition 1.10 and Theorem 1.11 can also be proven entirely using this Galois cohomology framework: for Theorem 1.11, instead of the Weil pairing we one can use Poitou-Tate duality. On the other hand, we have not yet found a way to prove Theorem 1.12 using this framework. The key difficulty comes from determining whether a Selmer class associated to a self-dual representation lifts to a Selmer class associated to a higher-dimensional representation. These self-dual representations are quite difficult to work with compared to their non-self-dual counterparts, so our geometric proof of Theorem 1.12 using the Weil pairing illustrates a method that can be used to work with them in the function field setting. On the other hand, we were only able to prove Theorem 1.13 using cohomological techniques. Thus, using both the geometric approach (Sections 2–3) and the cohomological approach (Sections 4–6) allows us to prove stronger results than any one approach individually.
Together with Proposition 1.10 and Theorem 1.9, constraints on can be obtained by counting arguments, analyzing the various restrictions on pairs . Section 7 contains proofs of some such constraints, including all the results stated in Section 1.1.
1.4. Prior work
The study of -torsion in divisor class groups of superelliptic extensions (for some field ) has been explored in many other contexts; for some examples [14, 7]. Most of these explorations are largely independent from the content of this paper; for instance, some take instead of , and they impose different conditions on and . Further, these works typically focus on a particular subgroup of the -torsion generated by divisors supported at the ramification locus of , which is the first stage in the filtration of discussed in Section 1.2.
In the case of hyperelliptic function fields , this first stage contains the entirety of the -torsion; Cornelissen uses this to compute the -rank of for arbitrary hyperelliptic curves over (allowing to be reducible) [4]. However, for , there is more to the filtration than this first stage, and these deeper filtration stages are one of the primary focuses of this paper. The primary difficulty we face is that unlike the action of Frobenius on , the action of on as a whole is not semi-simple. See Remark 2.10 for a discussion.
The aforementioned filtration can be defined using an endomorphism on , where is some root of unity in . This endomorphism and the filtration it defines were used by Poonen–Schaefer [11] and were further explored by Arul [1].
Other authors have studied the -torsion subgroups of divisor class groups of different kinds of degree extensions of . For instance, Wittmann considered degree Galois extensions , and studied the Galois module structure of the -torsion in the divisor class group of [15].
The question this paper is exploring has a direct analogue in number fields: namely, to study the -rank of the ideal class group of for distinct primes and . Several authors have studied this question under the assumption , which is analogous to the assumption we make in Theorem 1.6. Using deformations of Galois representations, Calegari–Emerton [2] determined conditions under which the -part is cyclic (i.e. -rank ). For instance, one of their results is that if is a -th power modulo , then the -rank of is at least [2, Theorem 1.3(ii)]. These results were generalized by Wake–Wang-Erickson [13, Proposition 11.1.1]; in particular, they interpreted the congruence condition as a cup product on Galois cohomology. The techniques of Wake–Wang-Erickson were used by Karl Schaefer and the third author to prove a full converse of Calegari–Emerton’s result by imposing additional congruence conditions.
The cohomological methods used in Sections 4–6 of this paper closely follow the work of Schaefer and the third author, using the Galois cohomology framework developed by Wake–Wang-Erickson. In particular, the upper bound in Theorem 1.6 is directly analogous to [12, Theorem 1.1.1]. On the other hand, [12, Table 3] shows that there is no parity constraint on the -rank in the number field setting; the parity constraint on appears to be a phenomenon unique to function fields.
1.5. Acknowledgements
The bulk of this research was conducted while all three authors held a CRM-ISM postdoctoral fellowship. The first author was partially supported by NSF grant DMS-2302511, and the second author was partially supported by ERC Starting Grant 101076941 (‘Gagarin’).
The authors thank Patrick Allen, Jordan Ellenberg, Jaclyn Lang, Bjorn Poonen, Karl Schaefer, Jacob Stix, Yunqing Tang, Carl Wang-Erickson for conversations that pointed them in helpful directions. The first two authors want to thank the third author for suggesting this project and for introducing them to the technical details of Galois cohomology needed for this paper, and the third author wishes to thank his collaborators for seeing this project through after he left academia.
2. Structure of the -torsion subgroup
In this section, we introduce our setup and prove Theorem 1.9 and Proposition 1.10.
2.1. Notation and Setup
We use the following notation throughout the paper.
-
•
is a prime.
-
•
is a prime power coprime to .
-
•
is the multiplicative order of in .
-
•
is a fixed nontrivial root of unity.
-
•
is an irreducible polynomial with coprime to .
-
•
is the smooth projective curve with affine equation , and its base change to .
-
•
is the Jacobian of , and its base change to .
-
•
For a field extension , denotes the -points of . Elements of can be interpreted as divisors on modulo linear equivalence, and if is a finite extension, elements of correspond to divisor classes in that are invariant under .
-
•
denotes the -torsion subgroup of , and the geometric -torsion group.
-
•
The -torsion rank of is defined to be
We also define two morphisms of by giving their actions on geometric points . Using the root of unity chosen above, by abuse of notation we also let denote the morphism defined by
We let denote the relative Frobenius map,
Note that is an automorphism of , while is a degree endomorphism; both act invertibly on . On , we also have the relation
By further abuse of notation, we also let and denote the respective endomorphisms of induced by their namesakes, as well as the induced linear maps on considered as a vector space over .
Convention 2.1.
When not otherwise specified, an “eigenvector” will refer to an eigenvector of acting as a linear map on the –vector space , i.e. a nonzero satisfying for some . Likewise, a “generalized eigenvector” will refer to a generalized eigenvector of , i.e. a nonzero satisfying for some and ).
2.2. The Filtration of
The automorphism and endomorphism on were discussed in detail in [1, Section 2.3]. Here we use them to construct a filtration on .
Noting that the endomorphism is annihilated by the cyclotomic polynomial, we can derive the relation
in the endomorphism ring . For each , is equal to times a unit, and so is times a unit. We can conclude that the kernel of on is exactly .
Define to be the -points of ; these then give a filtration
Note that each subgroup has the structure of a -vector space.
Lemma 2.2.
For each , the endomorphism induces an isomorphism of -dimensional vector spaces over .
Proof.
We have , , and the curve has genus by Riemann-Hurwitz. So for all ,
by [8, Proposition 12.12]. These endomorphisms are all separable and so has points in . This implies . For , the kernel of the map induced by is , so is an isomorphism by dimension considerations. ∎
2.3. A modification of
Recall from Section 2.1 that the relative Frobenius map is induced by the action on . The maps and on satisfy the relation
(2) |
Since and are associates in , this identity shows that the action of on preserves the filtration stages . However, the automorphism of does not preserve the generalized eigenspaces of . To account for this, we introduce a modification of that interacts in a more predictable way with the Frobenius map.
Definition 2.3.
Let be defined by
where denotes an inverse of modulo .
While the endomorphism depends on the choice of inverses mod , the action of on is well-defined, independent of the choice of for each . We have the following two important facts about , which both capture the idea that behaves like a “logarithm” of . The first statement in the following lemma says that acts like up to higher-order terms.
Lemma 2.4.
We have , and for each , and are equal as isomorphisms .
Proof.
This follows from Lemma 2.2 and the fact that is in the ideal generated by and . ∎
The second statement about in the following lemma can be thought of as a linearization of the relation . As a note of caution, the following relation does not hold when and are considered as endomorphisms of ; we obtain the desired equality only when we restrict to the actions on .
Lemma 2.5.
As linear maps on ,
Proof.
This can be proven by formal manipulation of polynomials; see Appendix A. ∎
A consequence of this result is that if is a (generalized) eigenvector of , then so is .
2.4. Generalized eigenvectors of
Any lies in some filtration stage . Then , and by Lemma 2.5 we have
so is an eigenvector of eigenvalue . So our first goal is to identify which powers of arise as eigenvalues of acting on .
To start, we find a basis of which is most suitable for our study of the action. The dimension of is for . The following Lemma, which is well-known in the literature (see e.g. the proof of [5, Theorem 1.7]), shows that the action of on is diagonalizable over .
Lemma 2.6.
The action of on has a basis
where is an eigenvector of with eigenvalue .
(Caution: recall that is induced by the -power Frobenius map on , not an -power Frobenius map on .)
Proof.
Suppose that factors over as , where . The curve has a unique point above by the assumption , which we also call . Define the points for . We have the relation ; by [1, Proposition 2.3.1], the points form a basis for (see also [5, Proof of Theorem 1.7]).
Given satisfying and , set
(3) |
Since , the coefficients of and are distinct and so . We have
Thus we have eigenvectors with distinct eigenvalues, so these form a basis for the -dimensional vector space . ∎
For , an eigenvector in with eigenvalue lifts under to an eigenvector of in with eigenvalue (by Lemma 2.4 and Lemma 2.5). This lift is a priori only an eigenvector in the quotient space, but we show in Lemma 2.8 that we can always take the lift to be a generalized eigenvector of acting on using properties of the operator .
Recall the definition of the sets .
See 1.8
Namely is the set of generalized eigenvectors of in the -th filtration stage with eigenvalue .
Remark 2.7.
By this point we have developed a lot of notation, so it may be helpful to have a picture in mind as we proceed. Examples are provided in LABEL:fig:lifting_charts. To each curve , we associate a grid of cells with and . Roughly speaking, the shaded cells can be matched bijectively with an independent set of vectors in ; rows (indexed by ) correspond to the filtration stages ; columns (indexed by ) are -invariant subspaces; and each diagonal with constant corresponds to a distinct eigenvalue.
More precisely, we shade the cell with coordinates light gray if the set is nonempty: that is, if there exists a generalized eigenvector with eigenvalue in . Lemma 2.8(a) says that acts on the grid by shifting everything down one cell, taking to and annihilating the bottom layer . Lemma 2.8(b) tells us that we can also go backwards: any shaded cell has a shaded cell above it. Thus each column is either entirely shaded or entirely empty. Corollary 2.9 tells us precisely which columns are shaded or empty. Cells and correspond to the same generalized eigenvector if , or equivalently, if .
In Section 2.5 we will see that true eigenvectors of form “towers” in these grids, and in Section 3 we will see that the Weil pairing imposes a kind of rotational symmetry on these grids.
Lemma 2.8.
Let and . Suppose .
-
(a)
If , then .
-
(b)
If , there exists with .
Proof.
To set up the proof of (b), we first note that the action of on is semisimple (for instance by recalling from Lemma 2.6 that splits into a direct sum of eigenspaces). Let be the eigenspace of acting on (note that may be - or -dimensional), and let be the -invariant complementary subspace of in .
Now suppose and . Since maps surjectively onto , there exists with . We have
(4) |
so . Write for and . Since is preserved by and does not contain any eigenvectors of eigenvalue , there exists such that . Therefore
(5) |
so . This proves and , so (b) holds. ∎
Corollary 2.9.
Let and . Then is nonempty if and only if and .
Proof.
Remark 2.10.
One can generalize the above discussion to determine a basis for consisting of generalized eigenvectors. More precisely, for all and all with , there exists satisfying the following conditions:
-
•
.
-
•
For , .
-
•
is a generalized eigenvector with eigenvalue .
Further, any set satisfying the above conditions is a basis for . We can use this to explicitly determine all the diagonal entries of the Jordan canonical form of acting on , and hence compute the characteristic polynomial of acting on . Thus the only real obstacle remaining is the failure of to be diagonalizable over : the rest of this paper can be thought of as a study of the Jordan blocks in the Jordan canonical form of .
Since our interest lies with , we will typically restrict our attention to the subspace of generated by generalized eigenvectors with eigenvalues equal to a power of , that is, the -invariant and -invariant subspace spanned by all the sets .
2.5. Basic counts and lifting results
The primary goal of this section is to determine, given some containing a eigenvector, whether also contains a eigenvector; that is, whether the property of containing a eigenvector “lifts” from to . Having an understanding of when this lifting occurs will help us to compute because of Theorem 1.9, which we recall and prove below.
See 1.9
Proof.
For each such that contains a eigenvector, let denote such an eigenvector. We claim that the set of all such is a basis for . First observe that by definition of , has eigenvalue , so . Further, the set of all is linearly independent because each lies in a distinct filtration stage. So it just remains to prove that these vectors span .
We will prove by induction on that if then is in the span of the eigenvectors with . If , then we must have , as there is no eigenvector of eigenvalue in by Lemma 2.6. Now let . If then the result follows by the induction hypothesis, so we can assume . This means that is an eigenvector, and so must be defined. Now and define nonzero elements of the quotient , and the -eigenspace of in this quotient is at most one-dimensional by Lemma 2.4 and Lemma 2.5. So for some we have . By the induction hypothesis, is a linear combination of for , so is a linear combination of for . ∎
In the remainder of this section we will prove Proposition 1.10. We first note the following important fact about the generalized eigenvector lifts defined in Lemma 2.8(b).
Lemma 2.11.
Let , , and suppose is a eigenvector. Exactly one of the following holds:
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•
For all with , is an eigenvector.
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•
For all with , is an eigenvector.
Proof.
We first make the following observation: if satisfies , and is in the generalized eigenspace with eigenvalue , then either or . If with , we apply this observation to to conclude that the value of does not depend on the choice of . Since is an eigenvector we have
so we apply the same observation to to reach the desired conclusion. ∎
Recall that is a rooftop if has a eigenvector but there is no for which has a eigenvector, and that is a non-maximal rooftop if . The non-maximal rooftops are exactly the sets where the property of having a eigenvector fails to lift to .
See 1.10
Remark 2.12.
Following from Remark 2.7, we give a brief visual explanation of each of these conditions. We shade a cell dark gray if contains a true eigenvector. (a) says that in the bottom layer , if a cell is shaded at all then it is shaded dark gray. (b) says that the dark gray cells form “towers:” any cell below a dark gray cell must also be dark gray. (c) and (d) both place limitations on which cells can contain the top cells of towers (i.e. the rooftops). (c) says that if a diagonal intersects the layer in an empty cell, then the diagonal below it cannot contain any non-maximal rooftops. (d) says that no diagonal can contain two non-maximal rooftops. These constraints place some limitations on the possible “skylines” that can occur as in LABEL:fig:lifting_charts.
Proof of Proposition 1.10.
-
(a)
It follows from Corollary 2.9 and that all elements of are eigenvectors.
- (b)
-
(c)
If is a maximal rooftop (), then . Now has an eigenvector by (b) and so and by (a); this implies and . So again by (a), has an eigenvector.
Now suppose is a non-maximal rooftop, so . By Lemma 2.11, is nonempty, so by (a), and .
-
(d)
Suppose and are both non-maximal rooftops, where (the case follows by symmetry). By Lemma 2.11, there exist and for which both and map under to , the set of nonzero vectors in a one-dimensional eigenspace. In particular, there exists such that
Then is an eigenvector of eigenvalue , contradicting the assumption that is a rooftop. ∎
The remaining proofs require more setup. Lemma 2.11 tells us that the obstruction to lifting an eigenvector in to an eigenvector in is given by an element of . Our next goal is to establish relations between these obstructions for different values of (with the same ).
3. The Weil pairing
In this section, we will use the Weil pairing to prove Theorem 1.11 and Theorem 1.12. These will then be used in Section 7 to prove Theorem 1.3.
The Weil pairing is a non-degenerate alternating -equivariant bilinear form
with the property that
for any endomorphism and .
For the purposes of this paper we will only need the existence of a pairing satisfying the properties listed above; in particular we will never need to compute the pairing explicitly. For the definition of the Weil pairing and proofs of the stated properties, see for example [8, Section 16]. Note that what we call is obtained by taking what Milne calls (with the canonical principal polarization of ) and restricting to . The alternating property follows from Milne’s Lemma 16.2(e), and the endomorphism property follows from Lemma 16.2(c).
In the remainder of this section we will prove that the Weil pairing interacts with the filtration and with Frobenius in a compatible way.
Lemma 3.1.
Let , and let and .
-
(a)
If , then .
-
(b)
If , then
-
(c)
If , and and for some , then if and only if .
In light of Lemma 3.1(c), we make the following definition.
Definition 3.2.
We say the sets and are dual. If and , then we say that form a dual pair. See Fig. 1.
In terms of the visual interpretation as described in Remark 2.7, each dual pair and corresponds to cells and that are related by rotating the grid . Point (b) relates the Weil pairing of vectors at cells and to the Weil pairing of vectors at cells and , moving one cell up and the other one down. This mirroring effect of the Weil pairing will play an important role in what follows.
Proof of Lemma 3.1.
All three statements depend on the following calculation. Let . Since is an automorphism of , we have . Since and the Weil pairing is bilinear, we have
(6) |
Further, since is alternating, the same relation holds if we swap the entries on both sides.
We begin by proving (a). If , then for some . Then
because annihilates .
We now prove (b). By definition of (Definition 2.3), we can write for some . So
since by part (a). By a symmetric argument we have , so it suffices to show that
Now note that can be written as an integer polynomial in with constant term . So applying Eq. 6,
for some ; again by part (a), . Therefore
In order to deduce (c) we will prove a more general result. As in the proposition statement, let , and , meaning and as an element of the quotient . But now let (and similarly for ), and let be any vector which reduces to an eigenvector of in the quotient . In particular, the eigenvalue of does not a priori need to be a power of . The Weil pairing extends in a natural way to , and under these weaker assumptions we will show that if and only if .
Since is an endomorphism of degree , we have the following for any :
for some and . By part (a), we can eliminate and . Thus
so solving for we find
Writing the minimal polynomial of as
we have
If , we must have in . Since is irreducible but has a root in , it must be linear; hence , showing that in fact .
Conversely, assume and . First consider the case . By (a), we know for all . By the proof of the reverse direction above, we know that for any , if reduces to a eigenvector in with any eigenvalue other than , then . We also know that is spanned by its eigenspaces, which are all one-dimensional. So if , then we would have for all in a basis of , contradicting non-degeneracy of . Hence .
For , note that , and we can write for some . So applying the case and part (b),
3.1. Lifting relations with dual pairs
For all with and , fix once and for all a eigenvector (which exists by Proposition 1.10(a)). Now suppose that has an eigenvector for some and . By Lemma 2.11, there exists mapping to this eigenvector under , such that either is an eigenvector itself, or is an eigenvector. In other words, there exists a constant such that
(7) |
and is an eigenvector if and only if . In the same way, if has an eigenvector then there exists and such that
(8) |
The key observation behind the following lemma is that if we lift Eq. 8 along powers of until the preimage of reaches the very top of the filtration, then the lifts of and form dual pairs with the vectors and appearing in Eq. 7. See Fig. 1 for a summary of this setup following the visual interpretation laid out in Remark 2.7.
Lemma 3.3.
Assume and both contain eigenvectors, and let be as above. Then there exist and such that , , and
Since and are dual pairs, their Weil pairings are both nontrivial by Lemma 3.1(c). This will allow us to make conclusions about the constants and .
Proof.
By Lemma 2.8, there exist and that are preimages of and , respectively, under . Then
for some , which can be checked by showing that both sides have the same image under . Therefore,
We now apply bilinearity. Since , we can use Lemma 3.1(a) to eliminate all pairings involving , as well as the pairing of with . We obtain
which implies the desired result. ∎
With this we can now give proofs of Theorem 1.11 and Theorem 1.12.
See 1.11
Proof.
We first prove by induction on that if is a non-maximal rooftop, then so is . We must first show that contains a eigenvector. Since is a rooftop, we have that is nonempty by Proposition 1.10(c). For the base case this already establishes that has a eigenvector. Otherwise, for the sake of contradiction, suppose is a rooftop for some . By the induction hypothesis, is also a rooftop. But since is a non-maximal rooftop, this contradicts Proposition 1.10(d). Hence has an eigenvector.
Now since is a non-maximal rooftop, we can take and as in Eq. 7 with . If we assume for the sake of contradiction that is not a rooftop, then we can take and in Eq. 8. By Lemma 3.3 we have
where and are dual pairs; in particular, we have by Lemma 3.1(c). Since and we obtain a contradiction. Hence must be a rooftop. This concludes the proof for . ∎
See 1.12
Proof.
For the proofs of Theorem 1.3 and Proposition 1.5, the reader may skip ahead to Section 7. The intervening sections on Galois representations and Galois cohomology are only required for the proof of Theorem 1.6, though they can also be used to provide an alternate cohomological interpretation of some of the preceding results.
4. From Frobenius eigenvectors to Galois representations
The primary goal of this section is to show that the existence of an eigenvector of in is equivalent to the existence of a certain -dimensional representation of the absolute Galois group
where denotes the separable closure of . A precise statement is given in Proposition 4.14. In the following sections we use Galois cohomology to analyze conditions under which such a Galois representation can occur, with the goal of proving Theorem 1.13.
4.1. Automorphisms of function fields
Our first step is to translate our setup to the function field setting. The function field associated to the curve is , where satisfies the equation .Since is coprime to , the extension is separable. The Galois closure of is the field , since is generated by roots of unity. We define two automorphisms of by their actions on , on , and on :
Both of these automorphisms preserve and fix and so define elements of . More precisely, is a generator of the subgroup , and maps to a generator of the quotient group . As discussed in the following Lemma 4.1, we can pick a section and thus we will use to denote both the element in and its image in . In summary, we have the following structure.
Lemma 4.1.
The Galois group is isomorphic to , satisfying the exact sequence
and with the semi-direct product structure given by .
The following diagram summarizes the fields we are considering so far. All pictured extensions are Galois except for .
We briefly discuss how these function field maps relate to the morphisms from Section 2.1. Let and denote the respective maps of function fields induced by the maps and . Then we can immediately see that . On the other hand, the maps and are quite different. The arithmetic Frobenius map does not fix the base field , and so is not induced by a morphism of -varieties. The relative Frobenius map – which is the map from to itself that sends , , and fixes – is induced by a morphism of -varieties, but is not a field automorphism; the image is a degree subfield of . The composition is the absolute Frobenius map, which acts on any by . See [10, Page 93] for more on the decomposition of absolute Frobenius into relative Frobenius and arithmetic Frobenius. These observations give us the following relations for all geometric points and rational functions :
As a consequence, on the divisor classes we have
(9) |
4.2. Constructing Galois representations
We first define a few explicit representations that will be used to identify eigenvectors in . Recall that is generated by and , and there is a natural quotient map .
Definition 4.2.
Let be the representation given by the action on . That is, factors through where it acts by
Definition 4.3.
Let be the representation that factors through where it acts by
for a choice of basis .
The span of is the unique one-dimensional subrepresentation of , which is isomorphic to as can be seen by considering the upper left entry of the matrix form of . Moreover, for any , we have for some ; in particular, is a crossed homomorphism representing a class in . Using the cohomological setup we will introduce in Section 5.1, this is equivalent to saying that is the extension of by associated to the function .
For , the -th symmetric power of , , is defined as follows. Taking the basis for as in Definition 4.3, a basis for is given by formal monomials for (note that is invertible over ). For each , define
computed by expanding the right-hand side as a sum of monomials. Then factors through , and has the following matrix representation on the generators of with respect to the basis .
(10) |
The matrix is similar to a Jordan block, so there is no nontrivial decomposition of into a direct sum of two subspaces invariant under . We can conclude that is indecomposable for .
Lemma 4.4.
For any and , there is a short exact sequence of -representations
(11) |
Proof.
The matrix representation for is given by the matrix representation for with every entry multiplied by . The span of is a -dimensional invariant subspace. The action of on this subspace given by the upper left submatrix, giving an explicit isomorphism to . The corresponding quotient representation is given by the entry at the bottom-right, which is isomorphic to . ∎
Under the same assumptions as Lemma 4.4, we also have a short exact sequence of the form
(12) |
We will also compute the dual representations of the representations defined above. Let be a vector space over , and its dual space, so that there is a natural perfect pairing .
Definition 4.5.
The linear dual (or contragredient) of a representation is a representation characterized by the condition that induces a -equivariant homomorphism .
The cohomological dual of a Galois representation is defined as
If we identify the vector space with its dual via the pairing , we can check that for each , the matrix representing must be the transpose of the matrix representing , in order to guarantee that for all .
Lemma 4.6.
Let and . The linear dual of is isomorphic to .
Proof.
Since factors through , the same must be true of its linear dual, so we can restrict our attention to . Let . Using the change of basis determined by the matrix
(13) |
we compute
These are the inverses of and , respectively. The former assertion is clear, and the latter uses the observation that for all ,
We can conclude that the change of basis determined by takes to the linear dual of . ∎
4.3. as a Galois representation
Note that any eigenvector is automatically in since and
We therefore restrict our attention to the structure of as a Galois module.
We will define an action of on via Kummer theory. Let denote the subgroup of defined by the property that if and only if is a multiple of in .
Lemma 4.7.
There is a split exact sequence
Proof.
Define the map by . If is in the kernel of this map then there exists with , so for some constant . This implies that as elements of the quotient group . Since , we have .
It suffices to define a right inverse for the map . Pick an arbitrary place of with corresponding uniformizer and valuation , and define the set of “monic” rational functions
For every , there exists such that . Then is a subgroup of , and we have an exact sequence
Now for any , there exists satisfying . Thus maps to . ∎
The action of on induces actions on and , and the map from Lemma 4.7 is equivariant under these actions. This induces a quotient representation structure on . Explicitly, for each , pick a preimage . By Eq. 9, we have and as elements of . Thus the structure of as a -representation is determined by stipulating that the map factors through , where it acts by
Using this setup, we can show that each eigenvector in generates a -subrepresentation of that is isomorphic to one of the representations constructed in the previous section.
Lemma 4.8.
There exists an eigenvector of in if and only if there exists a -subrepresentation of isomorphic to .
Proof.
As was discussed above, the structure of as a -representation is determined by the fact that the representation factors through , where acts by and acts by . So to prove the lemma it suffices to consider the actions of and .
Suppose is a eigenvector. Since , the vectors are linearly independent and span a -dimensional vector space which we call . By Lemma 2.5, the matrix representing acting on the basis is
(14) |
Since , is also stable under the action of . By Definition 2.3, the actions of and satisfy the equation
on . The matrix representing acting on the basis is therefore
(15) |
Comparing the two matrices with Equations Eq. 10, we conclude that the action on (with acting via and acting via ) is isomorphic to .
Conversely, suppose is a subgroup of that is isomorphic as a -representation to . Let correspond to the vector . Then we have , , and , establishing that is a eigenvector in as desired. ∎
4.4. Galois representations from unramified extensions
Recall that is defined by the property that if . By Lemma 4.7 is a finite group. Given a subgroup , we define
Lemma 4.9.
The maximal unramified elementary -extension of is .
Proof.
By Kummer theory, subgroups of correspond bijectively with abelian extensions of of exponent by taking -th roots of all elements of the subgroup. Adjoining an -th root of results in an unramified extension if and only if for every place of , for some unit and , where is a choice of uniformizer. This is equivalent to requiring , that is, . ∎
As a consequence of Lemma 4.9, every subextension of is an abelian extension of . Let . See the following diagram for a summary of the fields involved, and the Galois groups corresponding to some of the extensions.
Proposition 4.10.
Let be a -invariant subgroup. Then we have an isomorphism of -representations, where is the cohomological dual of defined in Definition 4.5.
Proof.
By Kummer theory, we have an isomorphism of groups
where is associated to the homomorphism for any choice of -th root of . It suffices to check that this isomorphism is -equivariant. For , , and , we have for some integer , and so
the last equality following because is in and therefore fixes . ∎
Lemma 4.11.
Let be a -invariant subgroup. The map has a splitting; equivalently,
Proof.
We follow the proof of Lemma 3.1.3 of [12]. The extension
(16) |
determines a cohomology class , and it suffices to determine whether this class is . To do this, we consider the subgroup
which determines a restriction map . The image of under this restriction map corresponds to the sequence
We show that this map has a splitting. Let be a place of lying above , the place determined by the irreducible polynomial . Since is totally ramified in , but the extension is unramified by Lemma 4.9, the inertia group at is a copy of in that maps isomorphically to , defining a splitting as desired.
Hence maps to the zero class under the restriction map. But by the Lyndon-Hoschild-Serre spectral sequence we have an inflation-restriction exact sequence
Since has order (coprime to ), while is a vector space over , the first term of this sequence is . Hence the restriction map is injective, proving that as desired. ∎
Lemma 4.12.
Suppose is isomorphic as a -representation to for . Then maps isomorphically onto its image in under the map defined in Lemma 4.7.
Proof.
If contains , then spans a one-dimensional -subrepresentation of isomorphic to . The unique one-dimensional subrepresentation of is , so we can conclude . Since , there is a two-dimensional subrepresentation of isomorphic to . Hence there exists with and as elements of . This implies that the map is a homomorphism that factors through and sends and . But the map is identical, so by Kummer theory is equal to as elements of . This is a contradiction because and so . We can conclude that and so maps isomorphically to its image in . ∎
4.5. Relating eigenvectors to Galois representations
As stated in Lemma 4.8, the existence of a eigenvector in is equivalent to the existence of a -subrepresentation of isomorphic to . Recall the representation has kernel . We will transform this condition to the existence of a representation related to a field extension of .
Definition 4.13.
Let for some separable field extension , and a representation of . The kernel field of , denoted , is the fixed field in of .
If is a finite-dimensional representation over , then is finite index in , and so is a finite extension of the base field . The kernel field is a Galois extension of the base field , and by the first isomorphism theorem, descends to a faithful representation of . In the following statement, we consider the case .
Proposition 4.14.
Let . There exists an eigenvector of in if and only if there exists a representation satisfying the following conditions:
-
(a)
there is an exact sequence of -representations
where denotes the one-dimensional trivial representation;
-
(b)
the kernel field of is an unramified extension of with .
Proof.
Recall . Given a Frobenius eigenvector , let be the span of the -orbit of , and let be the image of this subgroup under the map from the proof of Lemma 4.7. Then as -representations we have by Lemma 4.8, and hence by Proposition 4.10 and Lemma 4.6.
We can use the splitting Lemma 4.11 to construct a representation . We assert that this representation factors through and for an arbitrary element we define
where is the -representation on . Since , the kernel field contains , which is equal to since ; hence is an extension of . By construction, satisfies the desired exact sequence in condition (a). The fact that factors through implies is contained in , which is an unramified extension of by Lemma 4.9. In fact because acts faithfully on , so .
Conversely, suppose there exists with the given properties. Let be the subrepresentation isomorphic to . The exact sequence implies that with respect to an appropriate basis, can be written in the form
for some . Condition (b) says that is an unramified elementary -extension, so by Lemma 4.9, for some .
From the matrix form for we see that for with mapping into , we have , so is isomorphic to as a -representation. So by Proposition 4.10, is isomorphic as a -representation to , which maps isomorphically to a subrepresentation of by Lemma 4.12. The structure of implies existence of a vector such that and spans ; by considering the dimension of we can conclude and therefore . ∎
In the next section we will relate the existence of this representation to the existence of a certain cohomology class in .
Remark 4.15.
A version of Proposition 4.14 holds also for , but this requires a different set of conditions on . First, the fixed field of the kernel of is not , but rather some subfield of depending on the value of ; thus the fixed field of may not be an extension of . We must replace condition (b) with the condition that is unramified, and then we may continue the proof as above but with . Second, if then there may exist a representation satisfying all the conditions, but for which the fixed field is the degree base field extension ; then does not correspond to any nontrivial subspace of . To obtain the desired equivalence we must impose the condition that is not unramified over . While it is possible to keep track of these additional constraints, we consider only for convenience, since we already have a criterion for the existence of a eigenvector in by Proposition 1.10(a).
5. Galois cohomology
Throughout this section and the next we will rely on many facts about Galois cohomology of global fields, most of which can be found in Neukirch, Schmidt, and Wingberg [9].
5.1. Cohomology classes and kernel fields
Let be a group. Given an -dimensional representation , an element can be represented by a crossed homomorphism satisfying for . Any different representative for the same cohomology class differs from by a coboundary. That is, there exists an element such that for all .
Definition 5.1.
With a representation and a crossed homomorphism as above, define a dimensional representation by
We say that is the extension of by associated to .
This definition can be summarized using matrix notation:
(21) |
If is a different crossed homomorphism which represents the same class , then we have
where is the vector satisfying . Thus, different representatives of a cocycle class give rise to representations that are equivalent up to conjugation. From now on, we will denote this representation class by since it only depends on and the cocycle class .
It is straightforward to check that is a -representation and fits into an exact sequence
of -representations, where represents the one-dimensional trivial representation. This exact sequence induces a long exact sequence in cohomology which begins
and we have . Conversely, for any exact sequence of -representations there is a class with .
Recall the definition of the kernel field of a Galois representation in Definition 4.13 and for , we will denote the kernel field of as . Since , is necessarily an extension of . One can check that the kernel field is well-defined on cohomology classes (in particular, if is a coboundary then , though the converse does not necessarily hold), and in fact that the kernel field is invariant under scaling:
Lemma 5.2.
If and then .
5.2. Selmer conditions
Let denote the set of all places of . For each , let denote the localization of at , and pick once and for all an inclusion of separable closures. This is equivalent to picking a prime above in , or equivalently a compatible system of one place above in each finite extension of . We define
and the inclusion induces an inclusion by restriction. The image of is the decomposition group of the prime above in .
Given a Galois representation , we define to be its restriction to the decomposition group . For convenience we will define the notation
By restriction to the decomposition group , we obtain a map
For any , let denote the residue field of at the place , so that is the quotient of by the inertia group above . Let , where denotes the place determined by the irreducible polynomial defining the curve , and is the place determined by .
For each we define a subgroup :
(22) |
The subgroup for is the “unramified subspace”
which is equal to the group defined in [9, Definition 7.2.14] by the inflation-restriction exact sequence. The unramified subspace is so called because kernel fields of classes in the unramified subspace introduce no new ramification at : if (so the kernel field of is unramified over at ), and if satisfies , then in fact we also have (the kernel field of is unramified over at ). With this setup, we can define the Selmer group
5.3. A basis for local cohomology groups
Recall that is the order of in and . In Section 4.2 we defined -representations and which factor through .
Let
and let denote the -dimension of cohomology group . Since is similar to a Jordan block as is described in Eq. 10, the dimension of depends only on whether the top-left entry of equals . That is,
(23) |
We now consider , the restriction of to the decomposition group , for . The polynomial splits over into factors which are cyclically permuted by . The decomposition group fixes these factors, so the only powers of lying in the image of are powers of . On the other hand has a unique prime above it in , so is in the image of . We can conclude that
(24) |
By Tate duality [9, (7.2.6)] and Lemma 4.6, this implies
(25) |
Finally, since is coprime to , the local Euler characteristic of at any is trivial [9, (7.3.2)] and so
(26) |
In particular, we observe that is at most -dimensional for .
We now give an explicit basis for and discuss their corresponding kernel fields.
Lemma 5.3.
Let or , and let or respectively. Let for , and the restriction to .
-
(a)
If then there exists a nonzero element such that the kernel field is the degree unramified extension of .
-
(b)
If then there exists a nonzero element such that the kernel field is .
Further, has a basis consisting of whichever of and it contains.
Here means the completion of at the place above determined by the decomposition group .
Remark 5.4.
If then the , so when exists in , is not a nontrivial extension of ; this serves as a caution that a nontrivial cohomology class may define a trivial extension of kernel fields. However, if then is a subfield of , so is a nontrivial extension in this case.
Proof.
We start by defining an unramified class . Since is the trivial representation, a class in is represented by a group homomorphism. Let be defined by the property that it factors through , where is represented by given by sending the Frobenius map to .
If , then following Eq. 23 is a subrepresentation of , and the image of under we also denote by . The kernel field of is the compositum ; as is the composite of an unramified extension with a ramified extension, it never contains . Thus is a nontrivial extension and so the class is nonzero.
If , then , so the localizations at of and are isomorphic. The representation is an extension of by following Lemma 4.4, so we can define to be the class corresponding to this extension by , or equivalently the image of under in the long exact sequence
The injection is an isomorphism because both groups have the same dimension, and therefore . The kernel field of is equal to the fixed field of . Note that this equals the fixed field of provided , which is why we can’t use the kernel field to deduce that defines a nonzero cohomology class.
If both and , then and define independent classes in because they define distinct kernel fields over (Lemma 5.2). So whichever of the elements and exist in , they span a subspace that matches the dimension computed above, and therefore they must form a basis. ∎
5.4. Selmer groups of characters
Using the computations from Section 5.3, we can compute the dimension of global Selmer groups of characters.
Lemma 5.5.
Recall denotes the dimension of the cohomology group . We have
Proof.
We begin by defining an alternate Selmer group. Let be a -representation and its cohomological dual (Definition 4.5). For each place we define a subgroup of by
where is the unramified subspace as defined in Eq. 22. For all places of , the subgroup is precisely the annihilator under the local Tate pairing of the subgroup as defined in Eq. 22 [9, Theorem 7.2.15]. Using these subgroups we define
In this setting, the Greenberg-Wiles formula [9, Theorem 8.7.9] reduces to
because for all , we have (see the proof of [9, (8.7.9)]). The right-hand side of this equation can be determined using the dimension computations in Section 5.3, and equals the right-hand side of the statement of the lemma. So it suffices to show that is trivial.
Now suppose there exists a nonzero class and let denote the kernel field of . For , the Selmer condition implies that the -representations and have the same kernel field, so the extension is totally split at all places over . Since is not a coboundary, the extension of determined by must have a nontrivial unipotent element in its image, so there is an element of order in . Since has order coprime to , this implies that is a extension of some subfield of that is unramified everywhere and split at and . No such extensions exist, so in fact is trivial. Hence the dimension of is exactly as predicted by the statement of the lemma. ∎
5.5. From eigenvectors to cohomology
Using the cohomology computations above, we can now complete the work we began in Section 4 of relating the existence of eigenvectors of in to the existence of certain cohomology classes.
Lemma 5.6.
Let for some , and let be the kernel field of . Then is an unramified extension of .
Proof.
The kernel field contains the fixed field of , which is because . The Selmer condition ensures is unramified over at all , so it suffices to check the ramification at . By Lemma 5.3, for , is a linear combination of and (allowing the coefficient to be if the corresponding class is not in ). Since these classes both define unramified extensions of , the corresponding extensions of both vanish on the inertia group of , so the same is true of any linear combination. This implies that the kernel field of is an unramified extension of . Since the kernel field of is the completion at a prime above of the kernel field of , we can conclude that is unramified over at , and also at all by the Selmer condition. ∎
Proposition 5.7.
Let and . The following are equivalent:
-
•
There exists an eigenvector of in .
-
•
, and there exists a class that maps to a nonzero class under the map induced by Eq. 11.
Proof.
Given a Frobenius eigenvector , we obtain a representation as in Proposition 4.14, which is an extension of by and therefore corresponds to a class . Letting denote the kernel field, by Proposition 4.14 we have unramified and therefore . If we write in matrix form by picking a cocycle as
(27) |
then maps to a matrix of the form . Since descends to a faithful representation of , the map is an isomorphism onto by Proposition 4.14. So letting be the class given by restriction of to the bottom entry, there exists for which ; this proves that the kernel field of is strictly larger than , so must define a nonzero class in .
We also have . For if , then and by Lemma 5.5. This implies that up to scalar multiple, is the class defining as an extension of by , and therefore has kernel field , a contradiction.
Conversely, suppose we are given an element satisfying the described condition with . This class corresponds to an extension of by which we denote . Let denote the kernel field of . By Lemma 5.6, is an unramified extension of .
We can pick a cocycle such that the matrix form has the form as in Eq. 27 where the last entry of represents the nonzero class given by the assumption.
Now we claim that with this matrix form, there must exist for which . If not, the fact for all implies that the kernel field of is contained in . Thus the representation factors through . Since
and by assumption, we must have , so the representation factors through which is a cyclic group. This contradicts the assumption that is a nonzero class in .
Hence there exists for which has nonzero bottom entry. Now for arbitrary we have , so every vector in the orbit of under the -representation is obtained as for some . The orbit under of a vector with nonzero bottom entry has full dimension , so the restriction of to has image isomorphic to ; in particular this implies . We thus obtain a representation as in Proposition 4.14, from which we obtain an eigenvector in . ∎
6. Cup products
Suppose and we want to determine the existence of eigenvectors in for . In light of Proposition 5.7, we are led to consider when contains an element that maps to a nontrivial element of under the map induced by Eq. 11. The map factors through the map coming from Eq. 12, so a necessary condition is that contains a class that maps to a nontrivial element of . So we will now assume we are given a class in , and want to know when it lifts to an element in . We will see that this condition is equivalent to the vanishing of a local cup product, and explain how to determine this vanishing condition explicitly in the case .
6.1. Lifting cohomology classes
Let denote the maximal separable extension of that is unramified outside . Then contains , and so for any -representation that factors through , descends to a representation of . For the Selmer conditions defined in Section 5.2, we can write the corresponding Selmer group as a full cohomology group:
This will allow us to locate within a long exact sequence in cohomology. We also define for all .
Let . Take the short exact sequence of representations
from Eq. 12, but now considered as representations of . The corresponding long exact sequence has a portion given by
(28) |
From this we see that a class lifts to a class if and only if the image of under the connecting homomorphism to is . We will show that this condition can be detected by the vanishing of a cup product
induced by the dual pairing together with the calculation in Lemma 4.6.
Note that is an extension of by , unramified away from , and therefore corresponds to a class in . Under the map as in Eq. 11, this class maps to the class that determines as an extension of , as discussed in Section 4.2. We therefore denote this class by . As was discussed in Section 5.1, we can also identify this class as the image of under the connecting homomorphism induced by the following short exact sequence obtained from Eq. 11 by taking ,
Lemma 6.1.
Proof.
This follows from the formal properties of cup products and connecting homomorphisms in group cohomology. To be precise, if we let then we have a -equivariant map of short exact sequences
The left vertical arrow is induced by the dual pairing; the right vertical arrow is the isomorphism ; then there is a unique choice of middle arrow that makes the diagram commute. Then [9, (1.4.3.i)] says that the following diagram commutes:
Note in particular that the top cup product sends . Therefore we have
6.2. Local cup product
In the previous section we showed that a class in lifts to a class in if and only if its cup product with vanishes. Our next goal is to show that the vanishing of the cup product of and a class in can be detected locally at the place (Proposition 6.4).
Lemma 6.2.
Recall denotes the dimension of the cohomology group . We have
Proof.
Lemma 6.3.
The restriction map is an isomorphism.
Proof.
By Eq. 25 and Lemma 6.2, if then both groups are trivial, so we may suppose . Then both groups have dimension , so it suffices to show that is nonzero. We will extract this from the end of the Poitou-Tate exact sequence [9, (8.6.10)]. Recall since is finite and unramified, we have ([9, page 387]). The relevant portion of the sequence is
(29) |
Our ramification set contains only two places, and . We know the dimensions of every term appearing in this sequence: the global by Lemma 6.2, the middle terms by Eq. 25, and the term has dimension when and dimension otherwise.
When the only non-zero terms are and . Exactness of this sequence implies that the map is not the zero map, and is therefore an isomorphism.
When the groups , , and are all -dimensional. The second map
is by definition (e.g. [9, page 495]) the linear dual of the restriction map
with the isomorphism following by Tate duality [9, (7.2.6)] and Lemma 4.6. Since the image of this restriction map has non-zero projection in both local factors, the kernel of the linear dual map necessarily also has non-zero projection on both local factors. From this, together with exactness of Eq. 29, we conclude that the map necessarily has non-trivial image in , and so again must be an isomorphism. ∎
Proposition 6.4.
Let . A class lifts to a class in if and only if the local cup product
vanishes.
Proof.
Since we’ve reduced our lifting question to one about the vanishing of local cup products, it is now in our interest to do a detailed analysis of these local cup products. Recall from Lemma 5.3 that contains a nonzero element when , and contains a nonzero element when , and whichever of these classes exist form a basis of . Note that maps under to .
Lemma 6.5.
Let . Under the cup product pairing
we have
whenever each exist in their respective cohomology group. In particular we have that if and only if is in the span of .
Proof.
We divide into cases depending on which of exist in each of the two groups and .
-
•
If the groups are -dimensional and spanned by the same class (for example and , but and ), then the representations and have the same restriction to . The cup product is alternating on [9, (1.4.4)], so the cup product of a class with itself equals zero.
-
•
If the groups are -dimensional but spanned by different classes (for example and , but and ), then the restrictions of and to are cohomological duals of each other. In this case the cup product is the local Tate pairing [9, (7.2.6)], which is non-degenerate; hence the cup product of the respective generators is nonzero.
-
•
If both groups are -dimensional, then the representations and have the same self-dual restriction to , so the cup product is an alternating perfect pairing: classes pair with themselves to be zero, and independent elements pair to be nonzero.
Note that it is impossible for one group to be -dimensional and the other to be -dimensional; for instance, if , , and , then
Hence the only case remaining is when or is -dimensional, in which case the claim is vacuously true. ∎
6.3. Detecting vanishing of local cup product
Suppose , and factors in as , arranged so that for all . For define
and set
Then is a -extension of , Galois over , and unramified away from and . Note that and so . Also,
is an -th power in because . Hence , so is well-defined for . The fields can also be produced using explicit class field theory for the rational function field , as in [6] for example.
Lemma 6.6.
Suppose . For all , there exists a nonzero cocycle with kernel field .
Proof.
We have
Writing for some integer , we have . We therefore have for some integer . If we let denote the automorphism sending then
and since , we can conclude .
Analogously to Definition 4.3, let be the representation defined by
for as above; this gives a well-defined representation because . This extends to a representation of that factors through . Then is an extension of by corresponding to a cocycle with kernel field . Since is unramified away from and we in fact have . ∎
We finally reach the proof of Theorem 1.13. Recall that for we have
See 1.13
Proof.
Note that does not have a eigenvector by Proposition 1.10(a) and (b). Since we assume , has a eigenvector by Proposition 1.10(a); if it were a rooftop then would contain a eigenvector by Proposition 1.10(c), but this contradicts Proposition 1.10(a). Hence has a eigenvector. Correspondingly, we have and .
Now suppose . By Lemma 5.5, is one-dimensional, and by Lemma 6.6 it is spanned by a cocycle with kernel field . By Proposition 5.7, has an eigenvector if and only if has a nontrivial element mapping to . By Proposition 6.4, such an element exists if and only if , and by Lemma 6.5 it suffices to check whether is in the span of . By Lemma 5.3, we know that is spanned by , which defines an extension with nontrivial residue field degree, and by , which has kernel field . Hence we can detect whether is in the span of by checking whether and have the same completion at a prime over . Taking the place above in , this is equivalent by Kummer theory to checking that is an -th power in the residue field . ∎
Remark 6.7.
Proposition 1.10 and Theorem 1.11 may also be proven using an argument similar to that of Theorem 1.13, that is, by combining Proposition 5.7 with the cohomological lifting conditions developed in Section 6.2. While the current proof of Theorem 1.11 invokes the Weil pairing, the cohomological proof instead uses the Poitou-Tate exact sequence [9, (8.6.10)] together with the observation that and are cohomological duals.
7. Consequences of lifting conditions
Using the relations set up in the previous sections, we are now reduced to an essentially combinatorial problem: under the constraints described in Section 1.2, what are the possibilities for the set of pairs such that has an eigenvector? In particular, following Theorem 1.9, we are interested in the -rank of the divisor class group , which equals the number of for which contains a eigenvector. In this section we prove all of the constraints on mentioned in Section 1.1.
For , let denote the smallest integer such that for all , there is no eigenvector in . In other words, if then is a rooftop, and if then there is no value of for which has a eigenvector. We will say that is the “rooftop height” over . If we take an integer representative , then the number of with is equal to
Following the visual interpretation as described in Remark 2.7, counts the number of dark grey circles in column . Since by Proposition 1.10(a), we have
by Theorem 1.9, so counts the number of contributions to from . For all and , we have ; specifically, if then , and if then .
The proofs of Theorem 1.3 and Theorem 1.6 have many similar features: we will first prove the upper bounds in both theorems, then the lower bounds, then the parity constraints.
7.1. Upper bounds
We first prove the upper bound in Theorem 1.3, namely
We have if and only if and by Proposition 1.10(a). There are such values of , and for each of these we have . For all other values of we have and therefore . Combining these bounds gives the desired result. ( counts the set of pairs for which is non-empty and ; in the visual interpretation of Remark 2.7, this corresponds to circles in cells that are either light or dark gray.)
We can similarly prove the upper bound in Theorem 1.6: if then
We have if and only if by Theorem 1.13, and for each of these we have as above. We have as before, and for all other we have and so that . Combining these bounds gives the desired result.
7.2. Lower bounds
We begin by proving a general lemma that will be useful in producing lower bounds. For fixed , the diagonal containing is the set of all for with . Equivalently, each diagonal is determined by a constant value of . We have strong constraints on how rooftops can be arranged across diagonals: Proposition 1.10(d) states that any given diagonal can contain at most one non-maximal rooftop, and Proposition 1.10(c) limits which diagonals are allowed to contain rooftops at all. So if we can find many sets that contain eigenvectors among a small collection of diagonals right below the main diagonal (the circles in LABEL:fig:lifting_charts), the pigeonhole principle will ensure that only a few of them can be rooftops; the rest must all lift past this main diagonal and hence contribute to .
Lemma 7.1.
Let , and suppose there are values of such that has a eigenvector. Then .
Proof.
If this is immediate from Theorem 1.9 (we have no eigenvectors in by Proposition 1.10(a)), so from now on assume . Define the sets
Finally, let be the set of all for which , where is the rooftop height over ; equivalently, the set consists of all with . We will prove that .
To this end, let . By Proposition 1.10(c) we cannot have , so in fact . We also have : for this follows from , and for this follows from and . Again by Proposition 1.10(c) we have . Taken together, we can conclude that . In other words, all lie on one of a set of diagonals.
Now for each , we have : this follows by Proposition 1.10(a) if , and by if . So consider the rooftop . Since and we have , so is a non-maximal rooftop. By Proposition 1.10(d), it is impossible to have for any distinct . So by the pigeonhole principle, there are at most elements in .
In conclusion, there are at least elements in satisfying , so . We can conclude that . ∎
Lower bound of Theorem 1.3: We will show that implies . If then , so there exists with and . For this value of , has an eigenvector by Proposition 1.10(a). Taking in Lemma 7.1 proves .222Alternatively, simply use the fact that any generalized eigenspace must contain at least one true eigenvector.
First lower bound of Theorem 1.6: Suppose . If , then since we have by the lower bound of Theorem 1.3. So it is sufficient to assume
and prove . Note that the assumption implies has an eigenvector for all by Proposition 1.10(a). We always have by definition of , which implies by Theorem 1.13 that has an eigenvector.
First consider the case , so that is a rooftop for all . Since , we can apply Theorem 1.12 to conclude that is not a rooftop and thus has a eigenvector. Now since but we can conclude . So for each , if were a rooftop, then it would be a non-maximal rooftop; since is also a non-maximal rooftop, this contradicts Proposition 1.10(d). Therefore has an eigenvector, but is not a rooftop by Proposition 1.10(c). Hence , which implies and hence .
Now consider the case , so has an eigenvector for some . Since and both have eigenvectors, we have by Lemma 7.1, taking . Together with the previous case, we see that whenever .
Second lower bound of Theorem 1.6: Assuming is even and , we will show that . Since , we can conclude that has a eigenvector by Theorem 1.13. Since , we are in the setting of Theorem 1.12, and can conclude that is not a rooftop: thus also has a eigenvector. Further, since is not a rooftop, neither is by Theorem 1.11. Hence , , and all contain eigenvectors, so by Lemma 7.1.
7.3. Parity
We first prove the parity constraint of Theorem 1.3. Let
By Proposition 1.10(a), it is equivalent to say that is the set of with . For each , let denote the unique value in satisfying . By Theorem 1.11 we have for all , so is an involution on and for all . We can write
since in the first equality we only remove terms with , and in the second equality we only remove terms with . If , then and contribute equal quantities to the sum. So to prove , it suffices to show that the terms indexed by fixed points of the involution are even.
Suppose , so that . By definition of we have
Since we additionally have we can conclude
so that in fact . We can therefore write
By Theorem 1.12, is odd, and therefore is odd. Thus is even as desired.
We now assume and prove the parity constraint of Theorem 1.6, namely that . Following Theorem 1.3, it suffices to prove that . If is odd, then and are both even because is even. So suppose instead that is even. By Theorem 1.11, is a rooftop if and only if is a rooftop; since is even, we always have , and so the rooftops with always come in pairs. Now counts the number of for which contains an eigenvector, which equals minus the number of rooftops with . Hence . Since we are assuming we have , which implies .
7.4. Proof of Proposition 1.5
Finally, we assume and prove that . By Proposition 1.10(a), has an eigenvector only for . We will prove that the rooftop height over is , which will imply as desired.
Suppose is such that is a rooftop. By Proposition 1.10(c), this implies and ; equivalently, is a multiple of . But then and is even, so is not a rooftop by Theorem 1.12. Hence is not a rooftop for any .
Appendix A Proof of Lemma 2.5
We prove that .
Lemma A.1.
For all integers and ,
Proof.
For , this follows from the fact that the alternating sum of for is equal to . So from now on we assume . We will first prove the identity
(30) |
as an equality in . First consider the value of
for . This equals the coefficient of in . But is a multiple of and so is a multiple of , and hence the coefficient of is zero. Hence only terms with contribute to the left-hand side of Eq. 30, so both sides of the desired identity are polynomials of degree at most .
If , we can compute the coefficient of in , obtaining
Since , and for , we can sum from to without changing the value, giving us the desired equality for these specified values of . We therefore have two polynomials of degree at most in with equal values at points, and therefore the polynomials are equal.
Now compare the coefficient of in each side of Eq. 30. The coefficient of in is if and if , so we obtain the desired result. ∎
Lemma A.1 is an identity over , but the only denominators that occur are coprime to ; it therefore descends to an identity over . It is in this form that we apply it below.
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