ARTICLE SKELETON

RAVI VAKIL AND MANJUL BHARGAVA

Abstract. This is the abstract.

Contents

1. Miscellanea 1 2. Representation theory 1 3. Number theory 4 4. Class number equality 5 5. The quadratic-cubic variation 7 6. Things to think about in this write-up 10 7. Loose ends and things to think about later 10 References 11

Everything in the first four sections should be in good shape.

1. Miscellanea

1.1. Terminology. By lattice L in a real vector space V, we mean a ℤ-submodule of V with an inner product such that L ×_{ℤ} ℝ = V. The fundamental volume of L is denoted vol. If two nonzero numbers a and b have quotient that is an integral power of 2, we say a ∼ b. If two nonzero numbers a and b have quotient that is an integral power of 3, we say a ≈ b.

1.2. Acknowledgements. Thanks Jordan and Kiran.

2. Representation theory

Throughout this section, let V be a real vector space with an inner product ⟨·, ·⟩ and an S_{4}-action preserving the inner-product. Let τ be the number of standard

Date: Thursday March 16, 2:30 pm.

Partially supported by NSF Grant DMS–9970101.

1991 MATHEMATICS SUBJECT CLASSIFICATION: PRIMARY (???), SECONDARY (???)

S_{4}  S_{3}  D_{4}  K_{4}
triv  triv  triv  triv
sgn  sgn  sgn  sgn
X_{2}  std  triv ⊕ sgn triv ⊕ sgn
std  triv ⊕ std W_{1} ⊕ W_{2} triv ⊕ X_{1} ⊕ X′_{1}
std ⊗ sgn sgn ⊕ std W′_{1} ⊕ W_{2} sgn ⊕ X_{1} ⊕ X′_{1}

Table 1. Decomposing irreducible representations of S_{4} (chad)

irreducible representations in the representation V. Let L be a lattice in V such that the S_{4}-action is rational with respect to L.

We refer to the irreducible representations of S_{4} as Irr_{S_{4}} = {triv, sgn, X_{2}, std, std⊗ sgn} (with corresponding representations R_{triv}, …, R_{std⊗sgn}). (Manjul, I changed V_{2} to X_{2}.) Let V = V_{triv} ⊕ ⋯ ⊕ V_{std} be the decomposition of V into irreducible components, and let L_{λ} = V_{λ} ∩ L for λ ∈ Irr_{S_{4}}. (Thus dim V_{std} = 3τ.) Define V_{triv,X_{2}} = V_{triv} ⊕ V_{X_{2}} ⊂ V, and L_{triv,X_{2}} = V_{triv,X_{2}} ∩ L; define V_{triv,X_{2},std} and L_{triv,X_{2},std} similarly.

As all irreducible representations of S_{4} are rational, the intersection of L with any V_{λ} is a lattice in V_{λ}. The sublattice of L generated by elements in the V_{λ} is of finite index in L.

2.1. Orthogonality. By Schur’s lemma, ⟨v, w⟩ = 0 if v ∈ V_{λ}, w ∈ V_{λ′}, λ ≠ λ′. Reason: V_{λ} is (noncanonically) self-dual (as a representation), so the inner-product restricted to V_{λ} ×V_{λ′} lies in ((V_{λ} ⊗V_{λ′})^{∗})^{G} = Hom(V_{λ}, V_{λ′})^{G} = 0. Hence V = ⊕_{λ}V_{λ} is a direct-sum decomposition as inner-product spaces (i.e. the inner product on V is the direct sums of the restricted inner products on V_{λ}).

Up to scalar, there is only one S_{4}-invariant inner product on an irreducible representation of S_{4}, also by Schur.

Table 1 shows how the irreducible representations of S_{4} decompose into irreducible representations of the subgroups S_{3}, D_{4}, and K_{4} = {e, (12), (34), (12)(34)} (not the Klein 4-group).

2.2. The one-dimensional subspaces of R_{X_{2}} fixed by D_{4} and K_{4} are the same, so V_{X_{2}}^{D_{4}} = V_{X_{2}}^{K_{4}}.

2.3. The vector spaces R_{std}^{S_{3}} and R_{std}^{K_{4}} are both one-dimensional, but are not the same. We analyze the restriction of some lattice L_{std} (preserved by the representation) to these subspaces. Let v_{1} − v_{2}, v_{2} − v_{3}, v_{3} − v_{4} be the usual basis of R_{std}. Then R_{std}^{S_{3}} is generated by q = 3v_{1} − v_{2} − v_{3} − v_{4} (for the S_{3} corresponding to {2, 3, 4}), and R_{std}^{K_{4}} is generated by r = v_{1} + v_{2} − v_{3} − v_{4} (for the K_{4} corresponding to \{(12), (34)\}. Note that q = r + (23)r + (24)r and 2r = q + (12)q, and |r| = \sqrt{3}|q| with respect to the norm induced by any S_{4}-invariant inner product.

Hence the linear transformation e + (12) gives an isomorphism of V_{std}^{S_{3}} onto V_{std}^{K_{4}}, which scales the inner product by (2/\sqrt{3})^{2}, and (ghana)

(1) (e + (12))L_{std}^{S_{3}} ⊂ L_{std}^{K_{4}} ⊂ \frac{1}{2}(e + (12))L_{std}^{S_{3}}.

Therefore vol(L_{std}^{S_{3}})/vol(L_{std}^{K_{4}}) = \sqrt{3}/2^{s} where s = 0 or 1.

If V = V_{std} is a representation containing only standard representations, and L_{std} is a lattice preserved by the action, then (1) still holds and e + (12) still scales the inner product by (2/\sqrt{3})^{2}, so (haiti)

(2) vol(L_{std}^{S_{3}})/vol(L_{std}^{K_{4}}) = \sqrt{3}^{τ} /2^{s},

where τ is the number of standards appearing, and s ≤ τ. (Say more here.)

2.4. Theorem. — (cuba) (a) dim(V^{S_{4}}) + dim(V^{K_{4}}) = dim(V^{S_{3}}) + dim(V^{D_{4}}),

(b) \sqrt{3}^{τ} vol(L^{S_{4}}) vol(L^{K_{4}}) ∼ vol(L^{S_{3}}) vol(L^{D_{4}}).

Proof. Part (a) is immediate from Table 1our earlier discussion. For (b), from Table 1, we may ignore the irreducibles sgn and std ⊗ std, and restrict attention to L_{triv,X_{2},std}.

The projection ρ_{std} : V → V_{std} is given by \frac{1}{8} ∑_{g∈S_{4}} χ_{std}(g)g, so

L_{triv,X_{2}} ⊕ L_{std} ⊂ L_{triv,X_{2},std} ⊂ \frac{1}{8}(L_{triv,X_{2}} ⊕ L_{std}).

Thus the index of L_{triv,X_{2}} ⊕ L_{std} in L_{triv,X_{2},std} is a power of 2, so for any subgroup G of S_{4}, vol(L^{G}) ∼ vol(L_{triv,X_{2}}^{G}) vol(L_{std}^{G}).

From Table 1, L_{triv,X_{2}}^{S_{4}} = L_{triv,X_{2}}^{S_{3}} and L_{std}^{S_{4}} = L_{std}^{D_{4}} = 0. From Section 2.2, L_{triv,X_{2}}^{K_{4}} = L_{triv,X_{2}}^{D_{4}}. From (2), \sqrt{3}^{τ} vol(L_{std}^{K_{4}}) ∼ vol(L_{std}^{S_{3}}). The result follows.

2.5. Notes to ourselves. Properties of S_{4} we use: G’s irreducible representations are rational and self-dual.

More precise version of the Theorem: 2^{t}\sqrt{3}^{τ} vol(L^{S_{4}}) vol(L^{K_{4}}) ∼ 2^{s} vol(L^{S_{3}}) vol(L^{D_{4}}), where t ≤ 2 dim V_{triv,X_{2},std} and s ≤ τ. Here is an example to show that t need not be 0: consider the representation (R_{triv} ⊕ R_{std}) ⊕ (R_{triv} ⊕ R_{X_{2}}), with generators v_{1}, v_{2}, v_{3}, v_{4}, v_{12+34}, v_{13+24}, v_{14+23}, (v_{1} + v_{2} + v_{12+34})/2.

3. Number theory

For the purpose of this section, fix the following diagram of number fields:

\begin{aligned}& & M^{24} \\ & ↙ & & ↘ \\ & & & & L^{6} \\ F^{4} & & & & ↓ \\ & & & & K^{3} \\ & ↘ & & ↙ \\ & & Z^{1} \end{aligned} (3)

(iceland) The superscripts refer to the degree of the extension of Z, Z ⊂ M is a Galois S_{4} extension, and F (resp. L, K) is the field fixed by some S_{3} (resp. K_{4}, D_{4}). Note that this diagram can be constructed given only the data of Z ⊂ F (under the assumption that the Galois group of this field extension is S_{4}), or Z ⊂ K ⊂ L (under the same assumption that the Galois group of Z ⊂ L is S_{4}). Let d be the degree of Z over ℚ.

Let r_{1} (resp. s_{1}) be the number of real (resp. complex) embeddings of Z, h_{1} the class number, R_{1} the regulator, D_{1} the discriminant, and w_{1} the number of roots of unity. Similarly, let the corresponding variables with subscript 4 (resp. 3, 6, 24) be the corresponding numbers for F (resp. K, L, M).

3.1. Roots of unity. Note that w_{1} = w_{3} = w_{4} = w_{6}, as any root of unity of M that isn’t in Z must generate the discriminant field (as the abelianlization of S_{4} is ℤ/2), which isn’t contained in Z, F, K or L.

3.2. Suppose A ⊂ B is a galois field extension, with Galois group G. If V is the vector space in which the unit lattice of B naturally sits, then V^{G} can be identified with the vector space in which the unit lattice of A usually sits.

Applying this to (3) and using Proposition 2.4 (a),

(r_{1} + s_{1}) + (r_{6} + s_{6}) = (r_{3} + s_{3}) + (r_{4} + s_{4}).

Clearly (r_{1} + r_{6}) + 2(s_{1} + s_{6}) = (r_{3} + r_{4}) + 2(s_{3} + s_{4}) (as both sides are 7d), so r_{1} + r_{6} = r_{3} + r_{4} and s_{1} + s_{6} = s_{3} + s_{4}.

By standard properties of the Artin L-function and Table 1,

ζ_{Z}(s)ζ_{L}(s) = ζ_{K}(s)ζ_{F} (s). (4)

(spain) Using the functional equations for the four ζ-functions, D_{1}D_{6} = D_{3}D_{4}. (The arguments are identical to those given in [H] in the case Z = ℚ, so we omit them.)

Comparing residues at s = 1 in (4),

\frac{2^{r_{1}}(2π)^{s_{1}}R_{1}h_{1}}{w\sqrt{D_{1}}} · \frac{2^{r_{6}}(2π)^{s_{6}}R_{6}h_{6}}{w\sqrt{D_{6}}} = \frac{2^{r_{3}}(2π)^{s_{3}}R_{3}h_{3}}{w\sqrt{D_{3}}} · \frac{2^{r_{4}}(2π)^{s_{4}}R_{4}h_{4}}{w\sqrt{D_{4}}} .

Hence, as r_{1} + r_{6} = r_{3} + r_{4} etc., (egypt)

(R_{1}h_{1})(R_{6}h_{6}) = (R_{3}h_{3})(R_{4}h_{4}). (5)

4. Class number equality

4.1. Theorem. — (iran) h_{1}h_{6} ∼ h_{3}h_{4}, or equivalently, h_{4}/h_{1} = h_{6}/h_{3}.

4.2. Remark. Up to powers of 2 and 3, it is straightforward to show that h_{1}h_{6} = h_{3}h_{4} (by the representation theory of S_{4} to finite abelian groups, for example), and even that the relative class groups of Z ⊂ F and K ⊂ L are isomorphic away from 2 and 3. Thus the main content of Theorem 4.1 is the 3-part of the statement. (Our argument does not show that the relative class groups are isomorphic at 3, only that they have the same 3-order.)

4.3. The rest of this section will be devoted to proving Theorem 4.1.

By (5), it suffices to prove that R_{1}R_{6} ∼ R_{3}R_{4}.

Let W_{24} be the inner product space with standard basis associated to the r_{24}+s_{24} infinite places of M. If U is the unit group of M, let ϕ : U → W_{24} be the group homomorphism

u ↦ (log |u^{(1)}|, …, log |u^{(r_{24}+s_{24})}|).

Let L_{24} be the lattice in W_{24} generated by the image of U, and (1, …, 1). The regulator is given ([L] p. 109) by

R_{24} = \frac{2^{r_{24}}}{r_{24} + 2s_{24}} vol(L_{24}).

The galois group S_{4} of Z ⊂ M acts on R_{24} by permuting the coordinates.

Define W_{i}, L_{i} similarly for i = 1, 3, 4, 6. There are natural linear maps ψ_{i} : W_{i} → W_{24} restricting to L_{i} → L_{24}. Note that R_{1}R_{6} ∼ R_{3}R_{4} ⇔ vol(L_{1}) vol(L_{6}) ∼ vol(L_{3}) vol(L_{4}); we will prove the latter.

4.4. Assumption R. We’ll need to assume that the index of ψ_{i}(L_{i}) in L_{24}∩W_{24}^{G} (where G is the appropriate subgroup of S_{4}) is a power of 2.

It will suffice to assume that the discriminant extension of Z isn’t Z[ω], where ω is a cube root of unity. Details, should we need it, are in an e-mail, currently posted on Manjul’s page in Ravi’s account.

4.5. By Theorem 2.4, \sqrt{3}^{τ} vol(L_{24}^{S_{4}}) vol(L_{24}^{K_{4}}) ∼ vol(L_{24}^{S_{3}}) vol(L_{24}^{D_{4}}), where τ is the number of standard irreducible representations in W_{24}. From Table 1, τ = dim(W_{4}) − dim(W_{1}) = r_{4} + s_{4} − r_{1} − s_{1}. Hence, by (Assumption R), (nepal)

(6) \sqrt{3}^{r_{4}+s_{4}−r_{1}−s_{1}} vol(ψ_{1}(L_{1})) vol(ψ_{6}(L_{6})) ∼ vol(ψ_{3}(L_{3})) vol(ψ_{4}(L_{4})).

We next compute the scaling factors vol(ψ_{i}(L_{i}))/ vol(L_{i}). Note that W_{i} is naturally a direct sum of components coming from each place of Z, so we consider four cases.

Case s_{1}. Each of the s_{1} complex places of Z splits into 3, 4, and 6 complex places of K, F, and L respectively.

The splitting of the real places of Z over K, L, and M turns out to be determined by the splitting over F, which can happen in 3 ways.

Case α. Suppose α of the real places of Z split into 4 real places of F. Then each one splits into 3, 6, and 24 real places of K, L, and M respectively.

Case β. Suppose β of the real places of Z split into 2 real and 1 complex place of F. Then each one splits into 1 real and 1 complex place of K, 2 real and 2 complex places of L, and 12 complex places of M.

Case γ. Suppose γ of the real places of Z split into 2 complex places of F. Then each one splits into 3 real places of K, 2 real and 2 complex places of L, and 12 complex places of M.

Note that (france)

r_{4} + s_{4} − r_{1} − s_{1} = (4α + 2β) + (β + 2γ + 4s_{1}) − (α + β + γ) − s_{1}
= 3α + 2β + γ + 3s_{1}.

We now compute the scaling factors contributed by each of the four cases.

Scaling factor in Case s_{1}. As each complex place of Z splits into 24 complex places of M, the corresponding basis element of W_{1} is sent to the sum of 24 basis elements of M. Hence the contribution to vol(ψ_{1}(L_{1}))/ vol(L_{1}) is \sqrt{24}^{s_{1}} ∼ \sqrt{6}^{s_{1}}. Each of the correponding three complex places of F splits into 8 complex places of M, so the contribution to vol(ψ_{3}(L_{3}))/ vol(L_{3}) is (\sqrt{8}^{3})^{s_{1}} ∼ \sqrt{2}^{s_{1}}. Similarly, the contribution to vol(ψ_{4}(L_{4}))/ vol(L_{4}) is (\sqrt{6}^{4})^{s_{1}} ∼ \sqrt{3}^{4s_{1}}, and the contribution to vol(ψ_{6}(L_{6}))/ vol(L_{6}) is (\sqrt{4}^{6})^{s_{1}} ∼ 1.

Scalaing factor in Case α. Similarly, the contributions in this case are \sqrt{6}^{α}, \sqrt{2}^{α}, \sqrt{3}^{4α}, and 1.

Scaling factor in Case β. Each such real place of Z splits into 12 complex places of M, so this case contributes \sqrt{12}^{β} ∼ \sqrt{3}^{β} to vol(ψ_{1}(L_{1}))/ vol(L_{1}). The corresponding real place of K splits into 4 complex places of M, and the complex place of F splits into 8 places, contributing (\sqrt{4}\sqrt{8})^{β} ∼ \sqrt{2}^{β} to vol(ψ_{3}(L_{3}))/ vol(L_{3}). The 2 corresponding real places of F each split into 3 places of M, and the corresponding complex place of F splits into 6 places of M, so this case contributes (\sqrt{3}^{2}\sqrt{6})^{β} to vol(ψ_{4}(L_{4}))/ vol(L_{4}). The 2 corresponding real places of L each split into 2 places of M, and the 2 corresponding complex place of L splits into 4 places of M, so this case contributes (\sqrt{2}^{2}\sqrt{4}^{2})^{β} ∼ 1 to vol(ψ_{6}(L_{6}))/ vol(L_{6}).

Scaling factor in Case γ. By a similar argument, this case contributes \sqrt{12}^{γ} ∼ \sqrt{3}^{γ} to vol(ψ_{1}(L_{1}))/ vol(L_{1}), (\sqrt{4}^{3})^{γ} ∼ 1 to vol(ψ_{3}(L_{3}))/ vol(L_{3}), (\sqrt{6}^{2})^{γ} ∼ \sqrt{3}^{2γ} to vol(ψ_{4}(L_{4}))/ vol(L_{4}), and (\sqrt{4}^{2}\sqrt{8}^{2})^{γ} ∼ 1 to vol(ψ_{6}(L_{6}))/ vol(L_{6}).

Therefore, from (6) and (7),

\sqrt{3}^{3α+2β+γ+3s_{1}} vol(ψ_{1}(L_{1})) vol(ψ_{6}(L_{6})) ∼ vol(ψ_{3}(L_{3})) vol(ψ_{4}(L_{4}))

from which

\sqrt{3}^{3α+2β+γ+3s_{1}} · \sqrt{6}^{s_{1}} \sqrt{6}^{α}\sqrt{3}^{β}\sqrt{3}^{γ} vol(L_{1}) · vol(L_{6})
∼ \sqrt{2}^{s_{1}} \sqrt{2}^{α}\sqrt{2}^{β} vol(L_{3}) · \sqrt{3}^{4s_{1}} \sqrt{3}^{4α}\sqrt{54}^{β}\sqrt{3}^{2γ} vol(L_{4})
⇔ vol(L_{1}) vol(L_{6}) ∼ vol(L_{3}) vol(L_{4}),

as desired.

5. The quadratic-cubic variation

For this section, suppose you have a diagram of fields

F^{6}
↙  ↘
F^{2}  F^{3}
↘  ↙
F^{1}

5.1. Assumptions. (1) Suppose every root of unity of F^{6} actually lies in F^{1}.

(2) Suppose F^{2} can’t be obtained from F^{1} by adjoining a trace 0 root of unity.

5.2. Define r_{i}, s_{i} similarly as before. Suppose α is the number of real places of F^{1} that split completely in F^{6}, and β is the remaining number of real places.

Let τ = 2s_{1} + 2α + β. Note that τ > 0; it is also the number of standards in the representation of S_{3} on the unit lattice of F^{6}.

5.3. Theorem. — h_{F^{6}/F^{2}} = 2^{τ} h_{F^{3}/F^{1}}^{2}.

The rest of this section will be spent proving this.

S_{3} S_{2}  A_{3}  (1)
triv triv  triv  triv
sgn sgn  triv  triv
std triv ⊕ sgn χ_{1} ⊕ \overline{χ}_{1} triv ⊕ triv

Table 2. Decomposing irreducible representations of S_{3}

5.4. Representation theory.

We first make a table showing how irreducibles of S_{3} restrict to subgroups.

Notice that there are two possible lattices L in the standard representation of S_{3}: the usual triangular lattice, and the triangular lattice with centroids added. The fundamental domains differ in area by 3.

Thus

\frac{\sqrt{3}}{4} vol(L_{std}^{S_{2}})^{2} ≈ vol(L_{std}.

From this we get the analog of Theorem 2.4 (b):

5.5. Theorem. —

\frac{vol(L)}{vol(L^{A_{3}})} ≈ (\frac{\sqrt{3}}{4})^{τ} (\frac{vol(L^{S_{2}})}{vol(L^{S_{3}})})^{2} .

Proof. This is true if the lattice splits into irreducibles. For the general case, note that the projection to the standard component of a representation is given by \frac{1}{3} ∑ χ_{std}(g)g, so it is true for all lattices.

5.6. Number theory.

From degrees of extension over ℚ, we get:

(r_{6} + 2s_{6}) − (r_{2} + 2s_{2}) = 2((r_{3} + 2s_{3}) − (r_{1} − 2s_{1})).

Using the above table on the unit lattice, we get:

(r_{6} + s_{6} − 1) − (r_{2} + s_{2} − 1) = 2((r_{3} + s_{3} − 1) − (r_{1} − s_{1} − 1)).

Hence r_{6} − r_{2} = 2(r_{3} − r_{1}) and s_{6} − s_{2} = 2(s_{3} − s_{1}).

The ζ-function argument is as before: use group theory to show that ζ_{6}/ζ_{2} = (ζ_{3}/ζ_{1})^{2}. Thus

(\frac{R_{6}h_{6}}{R_{2}h_{2}} ) = (\frac{R_{3}h_{3}}{R_{1}h_{1}} )^{2}

assuming there is no roots of unity issue. Hence the relative class number is always a perfect square, up to 2 or 3 (which wasn’t hard to begin with).

5.7. Now let’s check unit lattice issues to see if problems come up at 2. Suppose (as mentioned earlier) that w_{6} = w_{1}. The groups S_{3} and ℤ/2 have abelianization ℤ_{2}. So if gu = αu, then α = −1,

which means u has trace 0. (Hence the assumption that F^{3} isn’t obtained from F by adding a unit with trace 0.)

The group A_{3} has abelianization ℤ_{3}. So if gu = αu, then u^{3} is fixed by g and thus lies in F^{2}. So we have no 2-issues here.

5.8. Now we consider splittings of places of F^{1}. Case s_{1}: A complex place in F^{1} splits completely into complex places in the

other number fields..

Case α: where a real place in F^{1} splits into 3 real places of F^{3}. Then (r_{2}, s_{2}) = (4, 0) and (r_{6}, s_{6}) = (0, 3).

Case β: where a real place in F^{1} splits into one real and one complex case in F^{3}. The (r_{2}, s_{2}) = (0, 1) and (r_{6}, s_{6}) = (0, 3).

From the table, τ = r_{3} + s_{3} − r_{1} − s_{1}. Hence we get the analog of (6):

vol(L_{6}) vol(ψ_{1}(L_{1}))^{2} ≈ (\frac{\sqrt{3}}{4})^{r_{3}+s_{3}−r_{1}−s_{1}} vol(ψ_{2}(L_{2})) vol(ψ_{3}(L_{3}))^{2}.

Now r_{3} + s_{3} − r_{1} − s_{1} = (3α + β) + (3s_{1} + β) − α − β − s_{1} = 2α + β + 2s_{1}.

Case s_{1}. vol(ψ_{1}(L_{1}))/ vol(L_{1}) = \sqrt{6}^{s_{1}}, vol(ψ_{2}(L_{2}))/ vol(L_{2}) = (\sqrt{3}^{2})^{s_{1}} ≈ 1, vol(ψ_{3}(L_{3}))/ vol(L_{3}) = (\sqrt{2}^{3})^{s_{1}} = \sqrt{2}^{3s_{1}}.

Case α. By identical calculations, we get vol(ψ_{1}(L_{1}))/ vol(L_{1}) = \sqrt{6}^{α}, vol(ψ_{2}(L_{2}))/ vol(L_{2}) = (\sqrt{3}^{2})^{α} ≈ 1, vol(ψ_{3}(L_{3}))/ vol(L_{3}) = (\sqrt{2}^{3})^{α} = \sqrt{2}^{3α}.

Case β. vol(ψ_{1}(L_{1}))/ vol(L_{1}) = \sqrt{3}^{β}, vol(ψ_{2}(L_{2}))/ vol(L_{2}) = \sqrt{3}^{β}, vol(ψ_{3}(L_{3}))/ vol(L_{3}) = \sqrt{2}^{β}.

Hence \sqrt{2}^{8α+4β+8s_{1}} vol(L_{6})\sqrt{6}^{2s_{1}+α}\sqrt{3}^{2β} vol(L_{1})^{2} ≈ \sqrt{3}^{2α+β+2s_{1}} \sqrt{3}^{β} vol(L_{2})\sqrt{2}^{6s_{1}+6α}\sqrt{2}^{2β} vol(L_{3})^{2}

\sqrt{2}^{10α+10s_{1}+4β}\sqrt{3}^{2α+2s_{1}+2β}v_{6}v_{1}^{2} = \sqrt{2}^{6α+6s_{1}+2β}\sqrt{3}^{2α+2s_{1}+2β}v_{2}v_{3}^{2}

2^{2α+2s_{1}+β}v_{6}v_{1}^{2} = v_{2}v_{3}^{2}.

From [L], vol(L_{i}) = R_{i}(r_{i} + 2s_{i})/2^{r_{i}}, so

2^{2α+2s_{1}+β}R_{6}R_{1}^{2} × 6 ≈ R_{2}R_{3}^{2} × 2 × 3^{2}
2^{2α+2s_{1}+β}R_{6}R_{1}^{2} ≈ R_{2}R_{3}^{2}.

Thus

h_{6}/h_{2} = 2^{2α+s_{1}+β}(h_{3}/h_{1})^{2} = 2^{τ} (h_{3}/h_{1})^{2}.

Ravi has double-checked this, although he may have introduced typos.

6. Things to think about in this write-up

Perhaps mention as an extended remark that our lattice results gives D_{1}D_{6} ∼ D_{3}D_{4} immediately, but the equality at 2 indicates that something very special is happening with the lattices.

Manjul will think about the sieve.

S_{6}-statement; outer automorphism. D_{4n} statements. W D_{4}: arithmetic Donagi. Check what is happening in binary-cubic situation. Make sure everything is kosher for d > 1.

March 16: Look at A_{4} case; this will be trickier. By e-mail: divisibility by 3 issue.

6.1. Things possibly to add. The character table of S_{4}:

S_{4}  (1) (12) (123) (1234) (12)(34)
triv  1  1  1  1  1
sgn  1 -1  1  -1  1
X_{2}  2  0  -1  0  2
std  3  1  0  -1  -1
std ⊗ sgn 3 -1  0  1  -1

7. Loose ends and things to think about later

7.1. 2-census. Why is there no τ in the discriminant case?

7.2. Recillas-Donagi. Our lattice argument is partially bogus, and yet the result is still true.

7.3. Order question.

References

[B] A. M. Bailey, On the density of discriminants of quartic fields. J. Reine Angew. Math. 315 (1980), 190–210. (Cited.)

[H] H. Heilbronn. On the 2-classgroup of cubic fields. 1971 Studies in Pure Mathematics (Presented to Richard Rado) pp. 117–119 Academic Press, London. (Cited.)

[L] S. Lang, Algebraic Number Theory, GTM 110.

Dept. of Mathematics, MIT, Cambridge MA 02139

   E-mail address: vakil@math.mit.edu

Dept. of Mathematics, Princeton University, Fine Hall, Washington Road, Princeton NJ 08544

   E-mail address: bhargava@math.princeton.edu