AN ACTION OF BRAID GROUPS ON COMPACT

SPACES WITH CHAOTIC BEHAVIOUR

Stephen P. Humphries

Abstract. We prove that there is a certain model S_{4} of a four-times punctured 2-sphere such that a subgroup of index 4 in the mapping class group of S_{4} can be realised as a group of diffeomorphisms of S_{4}. The subgroup is an image of the braid group B_{3}. We study a similar subgroup in the case of a six-times punctured sphere. We find a dense set of periodic points on S_{4} for the action of B_{3} and show that “hyperbolic” elements act chaotically. We find an antisymmetric function which is invariant for the action of B_{3}. We show that B_{3} acts on a quotient of a ring generated by polynomials related to Chebyshev polynomials. We also show that the Weierstrass elliptic function can be naturally defined on S_{4}. We find an action of B_{n} on certain topological spaces V_{n} which generalises much of this to the cases n > 3.

§1. Introduction

Let S_{g,n} be an oriented surface of genus g with n points removed. Let D_{g,n} = D_{g,n}(S_{g,n}) denote the group of orientation-preserving diffeomorphisms of S_{g,n} and let D_{g,n}^{0} = D_{g,n}^{0}(S_{g,n}) < D_{g,n} be the normal subgroup of those elements which are isotopic to the identity. Then M_{g,n} = D_{g,n}/D_{g,n}^{0} is the mapping class group of S_{g,n}. Let π_{g,n} : D_{g,n} → M_{g,n} be the quotient homomorphism. In this paper we use methods of representation theory (from [Hu4]) to construct a certain semi-algebraic variety S_{4} which is topologically a 2-sphere with 4 distinguished (singular) points; we investigate a certain natural action of the braid group B_{3} on S_{4} that fixes the set P of singular points. We study the global periodic points for this action, showing that these periodic points are dense in S_{4} and that this action on the periodic points is related to studying simple closed curves on S_{4}. We start by realising a certain subgroup of the mapping class group of S_{4} \ P.

The Nielsen realisation problem asks whether a finite subgroup of M_{g,n} can be realised as a subgroup of D_{g,n}. This has been solved by S. Kerckhoff [K1, K2]. As a first result for this action we show that the image of B_{3} referred to above gives infinite subgroups which can be so realised:

Theorem 1. The mapping class group M_{0,4} has presentation

< w_{1}, w_{2}, w_{3}|w_{1}w_{2}w_{1} =w_{2}w_{1}w_{2}, w_{2}w_{3}w_{2} = w_{3}w_{2}w_{3}, w_{1}w_{3} = w_{3}w_{1},
w_{1}w_{2}w_{3}^{2}w_{2}w_{1}, (w_{1}w_{2}w_{3})^{4} > .

Let H = < w_{2}w_{1}w_{3}w_{2}w_{1}, w_{1}w_{3}w_{2}w_{3}w_{2} > . Then H has index 4 in M_{0,4} and there is a subgroup Ĥ < D_{0,4} such that π_{0,4}|Ĥ : Ĥ → H is an isomorphism. We also have H ≅ B_{3}/Z(B_{3}).

Typeset by AMS-TEX

Moreover there is a model S_{4} \ P ⊂ ℝ^{3} of S_{0,4} as a real semi-algebraic variety such that the action of H on points (x, y, z) ∈ S_{4} is given by polynomial functions in x, y, z.

Lastly, if C is a simple closed curve on S_{4} not homotopic to a puncture, and if T (C) is the Dehn twist about C, then T (C)^{k} ∈ H if and only if k is even.

The action of the subgroup H on S_{4} in Theorem 1 was found in [Hu4]; we will now explain how this is really an action of the braid group B_{3}. Recall that the braid group B_{n} is the group with (standard) generators σ_{1}, …, σ_{n−1} and relations [Bi, p. 18]

σ_{i}σ_{i+1}σ_{i} = σ_{i+1}σ_{i}σ_{i+1}, for i = 1, …, n − 2,
σ_{i}σ_{j} = σ_{j}σ_{i},  for |i − j| > 1.

Recall [Bi] that there is an epimorphism π : B_{n} → S_{n}, where S_{n} is the symmetric group on n objects, which is induced by sending σ_{i} to the transposition (i, i + 1) ∈ S_{n} for each i = 1, …, n − 1. The kernel of π is P_{n} the group of pure braids on n strings.

In [Hu1] we studied an action of B_{n} on a certain polynomial ring R_{n}. Here

R_{n} = R[a_{12}, a_{13}, …, a_{1n}, a_{21}, a_{23}, …, a_{2n}, …, a_{n1}, …, a_{n n−1}]

is a polynomial ring in commuting indeterminates a_{ij}, 1 ≤ i ≠ j ≤ n, over a commutative ring R; we will mostly be interested in the situation where R = ℝ.

According to [Hu1] the kernel of the action of B_{n} on R_{n} is the cyclic centre of B_{n}. Thus in the case n = 3 the action of B_{3} on R_{3} is really an action of B_{3}/Z(B_{3}) ≅ P SL(2, ℤ).

We note as in [Hu2] that there is a natural ring involution ∗ on R_{n} which commutes with the action of B_{n}, so that for α ∈ B_{n} we have α(w)^{∗} = α(w^{∗}) for all w ∈ R_{n}. This involution is determined by its action on the generators a_{ij} which is as follows:

a_{ij}^{∗} = −a_{ji}.

If we factor R_{n} by the ideal generated by all a_{ij} − a_{ij}^{∗}, then we get an action of B_{n} on a ring R′_{n}. It is possible to identify R′_{n} with the polynomial subring so that we will put R′_{n} = R[a_{21}, …, a_{n1}, a_{32}, …, a_{n2}, …, a_{nn−1}], identifying a_{ij} with a_{ij} −a_{ij}^{∗} so that a_{ij} = a_{ji}.

Define the following matrix over R′_{n}:

U_{n} = ( \begin{matrix}2 & a_{21} & … & a_{i1} & … & a_{n−1 1} & a_{n1} \\ a_{21} & 2 & … & a_{i2} & … & a_{n−1 2} & a_{n2} \\ ⋮ & ⋮ & ... & ⋮ & & ⋮ & ⋮ \\ a_{i1} & a_{i2} & … & 2 & … & a_{n−1 i} & a_{ni} \\ ⋮ & ⋮ & & ⋮ & ... & ⋮ & ⋮ \\ a_{n−1,1} & a_{n−1,2} & … & a_{n−1 i} & … & 2 & a_{n n−1} \\ a_{n1} & a_{n2} & … & a_{n i} & … & a_{n,n−1} & 2 \end{matrix} ) .

Then in [Hu4] we showed that all the determinantal ideals corresponding to U_{n} are fixed under the action of B_{n}. In particular det(U_{n}) is fixed by this action and we will consider the situation where det(U_{n}) = 0.

In order to make the first few sections as self-contained as possible we will restrict ourselves to the case n = 3, noting that in this case there are only two situations to be considered, namely when U_{3} has rank 1 or 2. We will see that U_{3} has rank 1 only at the four points P (if we take R = ℝ).

In §6 we will indicate how many aspects of this generalise to the cases n > 3 obtaining an action of B_{n} on certain topological spaces, which are manifolds except at a finite number of points.

If n = 3, then U_{3} has rank 2 implies that det(U_{3}) = 0; here

−det(U_{3})/2 = a_{21}^{2} + a_{31}^{2} + a_{32}^{3} − a_{21}a_{31}a_{32} − 4.

It follows from the above that the solution set in ℝ^{3} of this equation is fixed by B_{3}. If we restrict our attention to the cube −2 ≤ a_{21}, a_{31}, a_{32} ≤ 2, then this solution set intersects thus cube in a topological sphere which we denote S_{4}. In fact [Hu4] S_{4} is B_{3}-invariant and meets the sides of this cube along six arc segments, the same way as a standard tetrahedron sitting inside the cube would meet the faces of the cube. See Figure 1a where we have shown S_{4} and Figure 1b where we have shown the six edges where S_{4} meets the cube. We have also numbered the four corner points and labeled the six edges, so that b_{ij} is the arc segment going from point i to j.

In Figures 1a, 1b we put x = a_{21}, y = a_{31}, z = a_{32}. Similar pictures to Figure 1a may be found in [Go, RB]. The function a_{21}^{2} + a_{31}^{2} + a_{32}^{3} − a_{21}a_{31}a_{32} − 4 and its level sets have been considered in [Go, Ke, KA1, KA2, Ma, Mo, Ro, RB]. For example, in [Go] they correspond to abelian representations of a group into SL(2, ℝ). In [KA, RB] they occur as invariants of certain ‘trace maps’.

The action of the braid generators σ_{1}, σ_{2} ∈ B_{3} on R′_{3} is as follows [Hu4]:

σ_{1}(a_{21}, a_{31}, a_{32}) = (−a_{21}, a_{32} − a_{21}a_{31}, a_{31});
σ_{2}(a_{21}, a_{31}, a_{32}) = (a_{31} − a_{21}a_{32}, a_{21}, −a_{32}).  (1.1)

There is an associated action of B_{3} on R^{3}, which we write as an action on the right:

(x, y, z)σ_{1} = (−x, z − xy, y), (x, y, z)σ_{2} = (y − xz, x, −z).  (1.1’)

Then for α ∈ B_{3} one has the following: α(a_{21}, a_{31}, a_{32}) = (a_{21}, a_{31}, a_{32})α, so that the left and right actions satisfy: αβ(a_{21}, a_{31}, a_{32}) = (a_{21}, a_{31}, a_{32})αβ.

Now each of the solution sets V′_{t}, 0 < t ≤ 4 of the equation x^{2} +y^{2} +z^{2} −xyz = t (over the real numbers) is invariant under the action of B_{3}. If we restrict to the cube [−2, 2]^{3}, then we get the subset V_{t} which is also a B_{3}-invariant 2-sphere. We have V_{4} = S_{4} and note that V_{t} contains the three circles of radius \sqrt{t} centered at the origin in each of the three coordinate planes. Then the four (corner) points P of V_{4} = S_{4} are permuted by the B_{3} action because they are the only singular points on the V_{t} (to see this just consider the Jacobian). However the 2-spheres V_{t}, 0 < t < 4, also have a set of points which are permuted by the B_{3} action. They are the six points τ_{t} = \{(±\sqrt{t}, 0, 0,), (0, ±\sqrt{t}, 0), (0, 0, ±\sqrt{t}). Thus we obtain an action of B_{3} on S_{0,6} if we consider these points as deleted.

We now contrast Theorem 1 with the following:

Theorem 2. Let 0 < t < 4 and let S_{6} = S_{6}(t) = V_{t}. Then M_{0,6}(S_{6}) has presentation

< w_{1}, w_{2}, w_{3}|w_{i}w_{i+1}w_{i} =w_{i+1}w_{i}w_{i+1}, w_{i}w_{j} = w_{j}w_{i}, |i − j| > 1,

w_{1}w_{2}w_{3}w_{4}w_{5}^{2}w_{4}w_{2}w_{1}, (w_{1}w_{2}w_{3}w_{4}w_{5})^{6} > .

Then the actions of σ_{1}, σ_{2} ∈ B_{3} on the sphere S_{6} are given by

c = w_{3}w_{2}w_{1}w_{2}w_{3}w_{2}w_{5}^{−1}, d = w_{4}^{−1}w_{3}^{−1}w_{1}w_{5}^{−1}w_{4}^{−1}w_{3}^{−1}w_{5}^{−1}.

The subgroup K =< c, d > is finite.

Lastly, the action of B_{3} on S_{6} is by polynomial functions.

We now investigate the points of S_{4} whose B_{3}-orbit is finite. In particular the image of such a point under < σ_{1} > is finite. This prompts us (as we shal see later) to define the following matrix. Let

M = M(a_{21}) = ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 − 3 a_{21}^{2} + a_{21}^{4} & a_{21}^{3} − 2 a_{21} \\ 0 & 2 a_{21} − a_{21}^{3} & 1 − a_{21}^{2} \end{matrix} )  (1.2)

and define the polynomials:

τ_{k}(a_{21}) = trace(M^{k}) − 3,

so that for example we have:

τ_{0} = 0; τ_{1} = a_{21}^{2} (a_{21}^{2} − 4); τ_{2} = a_{21}^{2} (a_{21}^{2} − 4) (a_{21}^{2} − 2)^{2} ;
τ_{3} = a_{21}^{2} (a_{21}^{2} − 4) (a_{21} − 1)^{2} (a_{21} + 1)^{2} (a_{21}^{2} − 3)^{2} ;
τ_{4} = a_{21}^{2} (a_{21}^{2} − 4) (a_{21}^{2} − 2)^{2} (a_{21}^{4} − 4 a_{21}^{2} + 2)^{2} ;
τ_{5} = a_{21}^{2} (a_{21}^{2} − 4) (a_{21}^{2} − a_{21} − 1)^{2} (a_{21}^{2} + a_{21} − 1)^{2} (a_{21}^{4} − 5 a_{21}^{2} + 5)^{2} .

We will later note that these polynomials are related to the Chebyshev polynomials. Fix n ≥ 1. Let W_{n} ⊂ ℝ^{3} denote the set of solutions (x, y, z)) ∈ ℝ^{3} of

τ_{n}(x) = τ_{n}(y) = τ_{n}(z) = x^{2} + y^{2} + z^{2} − xyz − 4 = 0.

Note that since τ_{n}(x) = 0 has only a finite number of solutions, W_{n} is a finite set. We will show later that if τ_{n}(x) = 0, then σ_{1}^{2n}(x, y, z) = (x, y, z) for all y, z ∈ ℝ.

Let v_{0} = (x_{0}, y_{0}, z_{0}) ∈ W_{n}. Then we obtain a matrix group

G(v_{0}) =< T_{1}(v_{0}) = ( \begin{matrix}1 & −x_{0} & −y_{0} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) , T_{2}(v_{0}) = ( \begin{matrix}1 & 0 & 0 \\ x_{0} & 1 & −z_{0} \\ 0 & 0 & 1 \end{matrix} ) ,  (1.3)

T_{3}(v_{0}) = ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ y_{0} & z_{0} & 1 \end{matrix} ) > .

Here the T_{i}(v_{0}) are transvections [A].

Now W_{n} ⊂ V_{4} and so there is an action of B_{3} on W_{n} (see (1.1’)) i.e. we have a homomorphism ρ_{n} : B_{3} → Sym(W_{n}) whose image we will denote by Ω_{n}.

Let F_{n} =< x_{1}, x_{2}, …, x_{n} > denote a free group of rank n and let

η = η(v_{0}) : F_{3} → G(v_{0})

be defined by η(x_{i}) = T_{i}(v_{0}).

Now there is a well-known action of B_{n} on F_{n}, which comes from the fact that B_{n} acts by mapping classes on the n-times punctured disc D_{n} [Bi]. Here α ∈ Aut(F_{n}) is in the image of the braid group if and only if we have Artin’s conditions: (A1) for all 1 ≤ i ≤ n we have: α(x_{i}) is a conjugate of some x_{j}; and (A2) α(x_{1}x_{2} … x_{n}) = x_{1}x_{2} … x_{n}.

For B_{3} the action on F_{3} is:

σ_{1}(x_{1}) = x_{1}x_{2}x_{1}^{−1}, σ_{1}(x_{2}) = x_{1}, σ_{1}(x_{3}) = x_{3},
σ_{2}(x_{1}) = x_{1}, σ_{2}(x_{2}) = x_{2}x_{3}x_{2}^{−1}, σ_{2}(x_{3}) = x_{2}.

The generators x_{i} represent homotopy classes of simple closed curves surrounding one of the punctures as in Figure 2, where we show the curves together with the punctures π_{1}, …, π_{n} and arcs a_{i} joining the punctures so that σ_{i} is a half-twist supported in a neighbourhood of the arc a_{i}. Here p is the base point for each of the x_{i}, which together generate the fundamental group of the punctured disc π_{1}(D_{n}, p) = F_{n} =< x_{1}, …, x_{n} >.

Figure 2

It is a fact [Bi] that B_{3} acts transitively on the set of unoriented simple closed curves on D_{n} which enclose two of the punctures. In particular this implies that if u ∈ F_{3} represents a simple closed curve enclosing two punctures, then there is α ∈ B_{3} such that α(x_{1}x_{2}) = u^{±1}.

Let e = a_{21}^{2} +a_{31}^{2} +a_{32}^{2} −a_{21}a_{31}a_{32} −4 and for n ∈ ℕ let L_{n} be the subring of R′_{3} generated by the ideal < e > of R′_{3} together with the elements τ_{n}(a_{21}), τ_{n}(a_{31}) and τ_{n}(a_{32}). Define the ideal

J_{n} =< τ_{n}(a_{21}), τ_{n}(a_{31}), τ_{n}(a_{32}), a_{21}^{2} + a_{31}^{2} + a_{32}^{2} − a_{21}a_{31}a_{32} − 4 >

of R′_{3}. We will prove:

Theorem 3. Let n ≥ 1. The ideal J_{n} is B_{3}-invariant.

The quotient R′_{3}/J_{n} has a B_{3}/Z(B_{3}) action.

There is an action of B_{3} on the quotient L_{n}/ < e > such that

σ_{1}(τ_{n}(a_{21})) = τ_{n}(a_{21});
σ_{1}(τ_{n}(a_{31})) = τ_{n}(a_{21})τ_{n}(a_{31}) + 2(τ_{n}(a_{21}) + τ_{n}(a_{31})) − τ_{n}(a_{32});
σ_{1}(τ_{n}(a_{32})) = τ_{n}(a_{31});
σ_{2}(τ_{n}(a_{21})) = τ_{n}(a_{21})τ_{n}(a_{32}) − 2(τ_{n}(a_{21}) + τ_{n}(a_{32})) − τ_{n}(a_{31});
σ_{2}(τ_{n}(a_{31})) = τ_{n}(a_{21});
σ_{2}(τ_{n}(a_{32})) = τ_{n}(a_{32}).

The B_{3} action on L_{n}/ < e > fixes the element

τ_{n}^{2}(a_{21}) + τ_{n}^{2}(a_{31}) + τ_{n}^{2}(a_{32})−2(τ_{n}(a_{21})τ_{n}(a_{31}) + τ_{n}(a_{21})τ_{n}(a_{32}) + τ_{n}(a_{31})τ_{n}(a_{32}))
− τ_{n}(a_{21})τ_{n}(a_{31})τ_{n}(a_{32}).

We will say that u ∈ F_{n} is simple if the conjugacy class of u in π_{1}(D_{n}, p) has a representative which is a simple closed curve on D_{n}.

Theorem 4. Fix n ≥ 1. Let v_{0} = (x_{21}, x_{31}, x_{32}) ∈ W_{n} and let K = K_{n}(v_{0}) = ker(ρ_{n}). Then K is a congruence subgroup of level 4n.

Also, for all α ∈ K and 1 ≤ i < j ≤ 3 the order of η(α(x_{i}x_{j})) is finite (a divisor of 2n) if and only if x_{ji}^{2} ≠ 4.

For all simple closed curves C ∈ F_{3} there is w ∈ W_{n} such that η(w)(C) has order a divisor of 2n.

The above result gives a way to check for non-simplicity of an element of F_{3}.

Theorem 5. There is a subgroup H of finite index in F_{3} such that H contains no simple elements. We may also find such an H which is normal and invariant under the action of B_{3}.

Theorem 6. For every n ∈ ℕ the Galois group of the polynomial τ_{n}(x) is in the centraliser of Ω_{n} < Sym(W_{n}).

For n ∈ ℕ the Galois group of the polynomial τ_{n}(x) is abelian.

The roots of the Galois group of the polynomial τ_{2^{n}}(x) are constructible using straightedge and compass.

We now further investigate the nature of the action of an element of B_{3} on S_{4}. Recall that a periodic point for the action of f on X is an x ∈ X such that there is n ∈ ℕ with f^{n}(x) = x. The existence of a dense set of periodic points is one half the definition of chaotic behaviour [BBCDS], the other condition being topological transitivity: for all non-empty open subsets U, V of X there is k ∈ ℕ such that f^{k}(U) ∩ V ≠ ∅.

Now B_{3}/Z(B_{3}) is isomorphic to P SL(2, ℤ) and there is the standard representation

Σ : B_{3}/Z(B_{3}) → P SL(2, ℤ) where σ_{1} ↦ ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) , σ_{2} ↦ ( \begin{matrix}1 & 0 \\ −1 & 1 \end{matrix} ) . For α ∈ B_{3}/Z(B_{3}) we

will define trace(α) to be |trace(Σ(α))|.

Theorem 7. The set W_{∞} = ⋃_{n=1}^{∞} W_{n} ⊂ S_{4}, is a dense set. Also for any α ∈ B_{3} each x ∈ W_{∞} is a periodic point for α.

For α ∈ B_{3}/Z(B_{3}) with trace(α) > 2, the action of α on S_{4} is chaotic.

We now consider invariants for the action. For (x, y, z) ∈ S_{4} we let

θ_{1}(x, y, z) = arccos(x/2) = \frac{π}{2} − \frac{x}{2} − \frac{1}{2} \frac{x^{3}}{2^{3} · 3} − \frac{1 · 3}{2 · 4} \frac{x^{5}}{2^{5} · 5} − \frac{1 · 3 · 5}{2 · 4 · 6} \frac{x^{7}}{2^{7} · 7} − …;

θ_{2}(x, y, z) = arccos(y/2) = \frac{π}{2} − \frac{y}{2} − \frac{1}{2} \frac{y^{3}}{2^{3} · 3} − \frac{1 · 3}{2 · 4} \frac{y^{5}}{2^{5} · 5} − \frac{1 · 3 · 5}{2 · 4 · 6} \frac{y^{7}}{2^{7} · 7} − ….

We also define

θ_{1}^{(2)}(x, y, z) = 2arccos(x/2), θ_{2}^{(2)}(x, y, z) = 2arccos(y/2).

Then we define the functions ω, ω^{(2)} : S_{4} × S_{4} → ℝ by ω((x, y, z), (u, v, w)) = θ_{1}(x, y, z)θ_{2}(u, v, w) − θ_{1}(u, v, w)θ_{2}(x, y, z);

ω^{(2)}((x, y, z), (u, v, w)) = θ_{1}^{(2)}(x, y, z)θ_{2}^{(2)}(u, v, w) − θ_{1}^{(2)}(u, v, w)θ_{2}^{(2)}(x, y, z).

Let B =< σ_{2}^{2}σ_{1}, σ_{1}σ_{2}, σ_{1}σ_{2}^{−1}σ_{1} >⊂ B_{3}. The 4 points P on S_{4} have the form (±2, ±2, ±2). These are the ‘corners’ of the tetrahedron in Figure 1 a.

Theorem 8. The subgroup B has index 4 in B_{3}.

The function ω : S_{4} × S_{4} → ℝ satisfies:

(i) ω is B-invariant: for all α ∈ B and p, q ∈ S_{4} we have ω(α(p), α(q)) = ω(p, q); (ii) ω is antisymmetric: for p, q ∈ S_{4} we have ω(p, q) = −ω(q, p). (iii) ω is non-degenerate except at the point (2, 2, 2) of S_{4}: ω(p, q) = 0 for all q ∈ S_{4} implies that p = (2, 2, 2) ∈ S_{4}.

The function ω^{(2)} : S_{4} × S_{4} → ℝ satisfies:

(i) ω^{(2)} is B_{3}-invariant; (ii) ω^{(2)} is antisymmetric; (iii) ω^{(2)} is non-degenerate except at the four points (±2, ±2, ±2) of S_{4}.

We now investigate certain elliptic functions that can be defined on S_{4}. Let L ⊂ ℂ denote the lattice spanned by ω_{1} = 2π, ω_{2} = 2πi and let

℘(z) = \frac{1}{z^{2}} + ∑_{ω∈L′} (\frac{1}{(z − ω)^{2}} − \frac{1}{ω^{2}} ) ,

where the sum is over all non-zero lattice points, denote the Weierstrass function of L. It is well-known [Ko] that any even elliptic function relative to L is a rational function of ℘(z).

Theorem 9. There is a composition of maps

ℂ → ℂ/L ≅ S^{1} × S^{1} → S_{4}

which we denote by ℵ. Let E be any even elliptic function for L. Then for v ∈ S_{4}, the function E(v) = E(ℵ^{−1}(v)) is well-defined as a function on S_{4}.

One can also define theta functions [La1] on S_{4} as long as they also are even. For example if σ(z, L) is the Weierstrass sigma function and a ∉ L, then σ(z+a)σ(z−a) is such a function.

In §§2 − 5 we will prove Theorems 1-9. Generalisations of these results to an action of B_{n}, n > 3 may be found in §6, which may be considered as a continuation of this introductory section.

§2 Realising mapping class subgroups

The presentations for M_{0,4} = M_{0,4}(S_{4}) and M_{0,6}(S_{0,6}(t)) given in Theorems 1 and 2 can be found in [Bi].

First we will find the action of the generators σ_{1}, σ_{2} ∈ B_{3} on S_{4}. Figure 3 shows the images of the line segments b_{ij} under σ_{1}. Some of these curves can also be seen as they lie on S_{4} in Figure 1 a (Figures 1, 3, 4, 5 were drawn using Maple).

Figure 3.

Now the generators w_{ij} for M_{0,4} are represented by half-twists [Bi] which are supported in a small neighbourhood of the arcs b_{ij}. In the notation of Theorem 1 we have w_{i} = w_{ii+1}. We will assume that one twists positively relative to the usual outward orientation of the 2-sphere S_{4}. Then, by untwisting the curves in Figure 3 using the w_{i} (so as to get back to Figure 1 b), one finds that the action of σ_{1} is represented by the element

a = w_{1}w_{3}w_{2}w_{3}w_{2},

where the action is on the left.

One similarly finds that the action of σ_{2} is represented by the element

b = w_{2}w_{1}w_{3}^{−1}w_{2}^{−1}w_{1}^{−1}.

Let H =< a, b >. A computer calculation using an implementation of the Reidemeister-Schreier process as in MAGMA [MA] for example, shows that the index of H in M_{0,4} is 4 and that it has a presentation

H =< r = w_{1}w_{2}, s = w_{2}w_{3}w_{1}^{−1}|r^{3}, (rs)^{2} > .

This is exactly the same as a presentation of B_{3}/Z(B_{3}) ≅ P SL(2, ℤ) ≅ ℤ_{2} ∗ ℤ_{3}. We see that the epimorphism B_{3}/Z(B_{3}) → H, σ_{1} ↦ a, σ_{1} ↦ b is an isomorphism, since ℤ_{2} ∗ ℤ_{3} is hopfian [MKS]. Thus the diffeomorphisms σ_{1}, σ_{2} ∈ D_{0,4} generate a group Ĥ < D_{0,4}, each non-trivial element of which represents a non-trivial mapping class in M_{0,4} and which is isomorphic to a subgroup of M_{0,4} of index 4. This proves Theorem 1. □

Now the element a ∈ M_{0,4}(S_{4}) permutes the points 1, 2, 3, 4 as the 4-cycle (1, 2, 3, 4). Thus a^{4} ∈ P_{0,4}(S_{4}). We have:

Lemma 2.1. (i) The elements a^{4}, b^{4} ∈ M_{0,4}(S_{4}) are both inverse squares of Dehn twists. (ii) The pure mapping class group is P_{0,4}(S_{4}) =< w_{1}^{2}, w_{2}^{2}, w_{3}^{2}, (w_{1}w_{2})^{3}, (w_{2}w_{3})^{3} >. We have Index(M_{0,4}, P_{0,4}) = 4!. Further P_{0,4}∩ < a, b > has index 96 and is a rank 5 free group. (iii) Let C denote the Dehn twist about γ, where S_{4} \ γ has two components each a sphere with three points removed. Then for k ∈ ℤ the element C^{k} is in < a, b > if and only if k is even.

Proof. (i) Since σ_{1}, σ_{2} ∈ B_{3} are conjugate we see that a^{4}, b^{4} are conjugate in M_{0,4}. Thus we need only show that a^{4} is the inverse square of a Dehn twist. We draw the effect of a^{4} on the six curves in Figure 1 b in Figure 4:

Figure 4.

One can see directly from this, together with the fact that a^{4} fixes the points 1, 2, 3, 4, that a^{4} is a Dehn twist. Alternatively, one has the following algebraic proof. First note that w_{1}w_{2}w_{3}^{2}w_{2}w_{1} = id shows that w_{3}^{−2} = w_{2}w_{1}^{2}w_{2} and w_{3}w_{2}w_{1} = w_{3}^{−1}w_{2}^{−1}w_{1}^{−1}. We next show that (w_{1}w_{2})^{3} = id:

id = (w_{1}w_{2}w_{3})^{4} = (w_{1}w_{2}w_{1}w_{3}w_{2}w_{3})^{2}
= w_{1}w_{2}w_{1}w_{2}(w_{3}w_{2}w_{1})w_{2}w_{1}w_{2}w_{3}w_{2}
= w_{1}w_{2}w_{1}w_{2}w_{3}^{−1}w_{2}^{−1}w_{1}^{−1}(w_{2}w_{1}w_{2})w_{3}w_{2} = (w_{1}w_{2})^{3}.

Thus id = w_{1}w_{2}w_{1}w_{2}w_{1}w_{2} = w_{1}w_{2}w_{1}w_{1}w_{2}w_{1} and so w_{1}^{−2} = w_{2}w_{1}^{2}w_{2} = w_{3}^{−2}. Now we have:

a^{4} = (w_{1}w_{3}w_{2}w_{3}w_{2}w_{1}w_{3}w_{2}w_{3}w_{2})^{2}
= (w_{1}w_{2}w_{3}w_{2}w_{2}w_{1}w_{2}w_{3}w_{2}^{2})^{2} = (w_{1}w_{2}w_{3}w_{2}w_{1}w_{2}w_{1}w_{3}w_{2}^{2})^{2}
= (w_{1}w_{2}w_{3}w_{1}w_{2}w_{3}w_{1}^{2}w_{2}^{2})^{2} = (w_{3}^{−1}w_{2}^{−1}w_{1}^{−1}w_{3}^{−1}w_{2}^{−1}w_{1}^{−1}w_{1}^{2}w_{2}^{2})^{2}
= (w_{3}^{−1}w_{2}^{−1}w_{1}^{−1}w_{3}^{−1}w_{2}^{−1}w_{1}w_{2}^{2})^{2} = (w_{3}^{−1}w_{2}^{−1}w_{3}^{−1}w_{1}^{−1}w_{2}^{−1}w_{1}w_{2}^{2})^{2}
= (w_{2}^{−1}w_{3}^{−1}w_{2}^{−1}w_{1}^{−1}w_{2}^{−1}w_{1}w_{2}^{2})^{2} = (w_{2}^{−1}w_{3}^{−1}w_{1}^{−1}w_{2})^{2}
= w_{2}^{−1}w_{3}^{−2}w_{1}^{−2}w_{2} = w_{2}^{−1}w_{1}^{−4}w_{2}.

Now w_{1}^{2} is a Dehn twist about the curve γ_{12} which is the boundary of a tubular neighbourhood of b_{12} and so we are done.

(ii) The generators for P_{0,4} can be found in [Bi]. The rest follows by a calculation using MAGMA [MA].

(iii) It will suffice to prove:

(a) the subgroup < a, b > does not contain a Dehn twist; and

(b) for any Dehn twist C we have C^{2} ∈< a, b >.

(a) Suppose that γ is a simple closed curve and that C ∈ M_{0,4} is the corresponding Dehn twist. Then for C not to be trivial we see that S_{4} \ γ must have two components each a three-times-punctured sphere. Thus there is g ∈ M_{0,4} such that g(γ) = γ_{12}, from which it follows that gCg^{−1} = w_{1}^{2}. Since < a, b > has index 4 in M_{0,4} we see there are < a, b >-coset representatives r_{1}, …, r_{4} ∈ M_{0,4} such that g ∈ r_{i} < a, b > for some i ≤ 4. We see that we must then have w_{1}^{2} ∈ r_{i} < a, b > r_{i}^{−1}. However a calculation using MAGMA shows that the coset representatives for < a, b > as a subgroup of M_{0,4} can be taken to be {id, w_{1}, w_{2}, w_{3}} and that in this case we have r_{i}^{−1}w_{1}^{2}r_{i} ∉< a, b > for i = 1, …, 4. This contradiction proves (a).

(b) Again we use the fact that there is g ∈ M_{0,4} such that gCg^{−1} = w_{1}^{2}. The result will now follow if we can show that r_{i}w_{1}^{4}r_{i}^{−1} ∈< a, b > for all coset representatives r_{i}. This is checked using MAGMA. □

We now prove Theorem 2. Let

a = (\sqrt{t}, 0, 0), b = (0, \sqrt{t}, 0), c = (0, 0, \sqrt{t}), d = −a, e = −b, f = −c.

Then a, b, c, d, e, f ∈ V_{t} are permuted by B_{3}. These points are joined by quarters of circles in V_{t} as shown in Figure 5 a.

Figure 5 a.

Here we have indicated the quarter circles Γ_{ac}, Γ_{cb}, Γ_{bd}, Γ_{df}, Γ_{fe} joining a, c, b, d, f, e (in that order). Since the complement of these arcs is a disc we see that any diffeomorphism is determined up to isotopy by the image of these curves. In Figure 5 b we show the images of these curves under σ_{1}.

Figure 5 b.

Here σ_{1} permutes the points a, c, b, d, f, e as (ad)(bc)(ef). Again we interpret each w_{i} as a half-twist fixing an arc between the points a, c, b, d, f, e, so that w_{1} fixes an arc Γ_{ab} (not shown in Figure 5 a) which is a quarter circle between a and b in the xy-plane, w_{2} fixes the arc Γ_{cb} etc. Then a calculation shows that σ_{1} gets represented as c = w_{3}w_{2}w_{1}w_{2}w_{3}w_{2}w_{5}^{−1}. Similarly we find that σ_{2} gets represented as d = w_{4}^{−1}w_{3}^{−1}w_{1}w_{5}^{−1}w_{4}^{−1}w_{3}^{−1}w_{5}^{−1}. Here σ_{2} acts on the points as (ab)(de)(cf).

We now show that K =< c, d > is isomorphic to B_{3}/Z(B_{3}). We have (indicating the use of relations by parentheses):

c^{2} = (w_{3}w_{2}w_{1}w_{2}w_{3}w_{2}w_{5}^{−1})^{2} = (w_{3}w_{2}w_{1}w_{2}w_{3}w_{2})^{2}w_{5}^{−2}
= w_{3}w_{2}w_{1}(w_{2}w_{3}w_{2})w_{3}(w_{2}w_{1}w_{2})w_{3}w_{2}w_{5}^{−2}
= w_{3}w_{2}w_{1}w_{3}w_{2}(w_{3}w_{3}w_{1})w_{2}(w_{1}w_{3})w_{2}w_{5}^{−2}
= w_{3}w_{2}w_{1}w_{3}w_{2}w_{1}w_{3}(w_{3}w_{2}w_{3})w_{1}w_{2}w_{5}^{−2}
= w_{3}w_{2}w_{1}w_{3}w_{2}w_{1}w_{3}w_{2}w_{3}(w_{2}w_{1}w_{2})w_{5}^{−2}
= w_{3}w_{2}w_{1}w_{3}w_{2}w_{1}w_{3}w_{2}(w_{3}w_{1})w_{2}w_{1}w_{5}^{−2}
= w_{3}w_{2}w_{1}w_{3}w_{2}w_{1}w_{3}w_{2}w_{1}w_{3}w_{2}w_{1}w_{5}^{−2} = (w_{3}w_{2}w_{1})^{4}w_{5}^{−2}.

Now one can check that (w_{3}w_{2}w_{1})^{4} is equal to w_{5}^{2}, which is a Dehn twist. Thus c^{2} is the identity. Since σ_{1} and σ_{2} are conjugate in B_{3}, c and d are conjugate, and so d^{2} is also the identity. Thus K is a quotient of the group with presentation < c, d|cdc = dcd, c^{2}, d^{2} > and so is finite. This concludes the proof of Theorem 2. □

§3 Invariant ideals

In this section we prove Theorem 3. Let v = (a_{21}, a_{31}, a_{32})^{T}. Note that σ_{1}^{4}(v) = Mv, where M is the matrix given in (1.2). Thus, since σ_{1}^{2}(a_{21}) = a_{21}, we also have σ_{1}^{4k}(v) = M^{k}v for all k ∈ ℤ.

Proposition 3.1. The polynomials τ_{k} satisfy:

(i) τ_{n+2}(x) = (x^{4} − 4x^{2} + 2)τ_{n+1}(x) − τ_{n}(x) + 2x^{2}(x^{2} − 4); (ii) τ_{n}(x) is a monic polynomial in x^{2} having x^{2} −4 as a factor; further degree(τ_{n}(x)) = 4n; (iii) for k, n ∈ ℕ, n ≥ k we have τ_{n+k} = τ_{n}τ_{k} + 2τ_{n} + 2τ_{k} − τ_{n−k}. (iv) if m, n ∈ ℕ and m|n, then τ_{m}|τ_{n}.

(v) τ_{2k}(x) = τ_{k}(x^{2} − 2). (vi) τ_{n}(x^{2} − 2) = τ_{n}(x)(τ_{n}(x) + 4).

Proof. We note that M = ( \begin{matrix}1 & 0 \\ 0 & N \end{matrix} ), for a 2 × 2 matrix N = N(a_{21}) with det(N) = 1.

Further for A, B ∈ SL(2, R) we have the well-known trace identity [Ma]:

trace(AB) = trace(A)trace(B) − trace(AB^{−1}).  (3.1)

Applying this to A = N^{n+1}, B = N, together with the fact that trace(N) = a_{21}^{4} − 4a_{21}^{2} + 2 gives (i) and (ii) follows easily from this.

(iii) Applying (3.1) to A = N^{n}, B = N^{k} we get

2 + τ_{n+k} = (2 + τ_{n})(2 + τ_{k}) − (2 + τ_{n−k}),

from which (iii) follows. (iv) follows from (iii) by induction on n/m.

(v) For any Q ∈ SL(2, ℂ) we let η_{k}(Q) = trace(Q^{k}). Then (3.1) shows that there are polynomials ϕ_{k}(x) such that η_{k}(P) = ϕ_{k}(trace(P)) for any P ∈ SL(2, ℂ). Thus

ϕ_{k}(trace(P)^{2} − 2) = ϕ_{k}(trace(P^{2})) = η_{k}(P^{2}) = trace(P^{2k}) = η_{2k}(P) = ϕ_{2k}(trace(P)). But τ_{k}(a_{21}) = trace(N^{k}) − 2 = ϕ_{k}(T race(N)) − 2 and so we see that τ_{2k}(x) = τ_{k}(x^{2} − 2).

(vi) From (iii) we get: τ_{2k}(x) = τ_{k}(x)(τ_{k}(x) + 4). Using this and (v) we now have

τ_{k}(x^{2} − 2) = τ_{2k}(x) = τ_{k}(x)(τ_{k}(x) + 4),

as required. □

Proposition 3.2. Let n ∈ ℕ. Assume that there are m solutions to τ_{n}(x) = 0. Then there are at least m solutions to

tr_{n}(x) = 0, τ_{n}(y) = 0, τ_{n}(z) = 0, x^{2} + y^{2} + z^{2} − xyz − 4 = 0.  (3.2)

Proof. Let w be a solution of τ_{n}(x) = 0. Let u = w and v = w^{2} − 2. The τ_{n}(w) = 0, τ_{n}(u) = 0 and by Proposition 3.1 (vi) we see that τ_{n}(v) = 0. One now checks that u^{2} + v^{2} + w^{2} − uvw − 4 = 0. One can also check that (w, w, 2) is a solution. □

Now define the following sequence of polynomials:

κ_{n}(x) = ∑_{i=1}^{n}(−1)^{i+1}\binom{n + i − 1}{2i − 1}x^{2i−1}.

Lemma 3.3. (i) The polynomials κ_{n}(x) satisfy the recursion

κ_{n+1}(x) = (2 − x^{2})κ_{n}(x) − κ_{n−1}(x).

(ii) We have the relations

(a) (x^{2} − 2)κ_{n}κ_{n−1} + κ_{n}^{2} + κ_{n−1}^{2} − x^{2} = 0;

(b) κ_{n}(x) = −\frac{1}{2}x^{2}κ_{n−1} + κ_{n−1} + \frac{1}{2}x\sqrt{(x^{2} − 4)κ_{n−1}^{2} + 4};

(c) κ_{n+2} = (a_{21}^{4}/2 − 2a_{21}^{2} + 1)κ_{n} + a_{21}(2 − a_{21}^{2})/2\sqrt{(a_{21}^{2} − 4)κ_{n}^{2} + 4}.

(iii) For all n ∈ ℕ we have

τ_{n}(x) = (x^{2} − 4)κ_{n}^{2}(x).

(iv) We have κ_{n+4}(x) = (x^{4} − 4x^{2} + 2)κ_{n+2}(x) − κ_{n}(x).

(v) κ_{2n}(x) is divisible by x(x^{2} − 2) for all n ≥ 1. (vi) κ_{n}(x) has degree 2n − 1. (vii) κ_{n}(2) = (−1)^{n+1}2n, κ_{n}(−2) = (−1)^{n}2n and κ_{n}(0) = 0. (viii) We have κ_{2n}(x) = κ_{n}(x)\sqrt{(x^{2} − 4)κ_{n}^{2}(x) + 4}.

Proof. (i) The coefficient of x^{2i−1} on the right hand side of this recurrence is

2(−1)^{i+1}\binom{n + i − 1}{2i − 1} + (−1)^{i}\binom{n + i − 2}{2i − 3} − (−1)^{i+1}\binom{n + i − 2}{2i − 1}

= (−1)^{i+1} (2\frac{(n + i − 1)!}{(2i − 1)!(n − i)!} + \frac{(n + i − 2)!}{(2i − 3)!(n + 1 − i)!} − \frac{(n + i − 2)!}{(2i − 1)!(n − i − 1)!})

= (−1)^{i+1} \frac{(n + i − 2)!}{(2i − 3)!(n − i − 1)!}×

(\frac{2(n + i − 1)}{(2i − 1)(2i − 2)(n − i)} + \frac{1}{(n + 1 − i)(n − i)} − \frac{1}{(2i − 1)(2i − 2)})

= (−1)^{i+1} \frac{(n + i − 2)!}{(2i − 3)!(n − i − 1)!} × \frac{(n + i)(n + i − 1)}{(2i − 1)(2i − 2)(n + 1 − i)(n − i)}

= (−1)^{i+1} \frac{(n + i)!}{(2i − 1)!(n − i + 1)!} = (−1)^{i+1}\binom{n + i}{2i − 1},

which is the same as the coefficient of x^{2i−1} on the left hand side of this recurrence. This proves (i).

(ii) The proof of the first formula is by induction: take (x^{2} − 2)κ_{n+1}κ_{n} + κ_{n+1}^{2} + κ_{n}^{2} − x^{2} and substitute for κ_{n+1} using the recursion from (i). Cancel terms and one immediately obtains (x^{2} − 2)κ_{n}κ_{n−1} + κ_{n}^{2} + κ_{n−1}^{2} − x^{2}.

Then (b) follows from (a) by solving the quadratic equation and (c) follows by applying (b) twice to get κ_{n+2} as function of κ_{n}.

(iii) Let ζ_{n}(x) = (x^{2} − 4)κ_{n}(x)^{2}. The proof proceeds by showing that ζ_{n}(x) has the same initial conditions as does τ_{n}(x) and that it satisfies the recursion in Proposition 3.1 (i). This latter demonstration is straight-forward, however we need to use part (ii) above.

(iv) This follows directly from (i):

(v) follows by induction from (iv) and (vi) follows directly from (iii) since τ_{n}(x) has degree 4n by Proposition 3.1 (ii).

(vii) is by induction on n using only (i) in each case.

(viii) By (iii) above and Lemma 3.1 (v), (vi) we have

(x^{2} − 4)κ_{2n}^{2}(x) = τ_{2n}(x) = τ_{n}(x)(τ_{n}(x) + 4) = (x^{2} − 4)κ_{n}^{2}(x) ((x^{2} − 4)κ_{n}^{2}(x) + 4) . □

Suppose that n = 2^{m}, m ∈ ℕ. Then, by Lemma 3.3 (iii), τ_{n}(x) has at most 2n + 1 roots. But by Lemma 3.1 (v) if x^{2} − 2 is a root of τ_{n/2}(x), then x is a root of τ_{n}(x). It follows that the roots of τ_{n}(x) include 0, ±2 and all

±\sqrt{2 ± \sqrt{2 ± \sqrt{2 ± ⋯ ± \sqrt{2}}}},

where there are at most n square roots. (The roots with less than n square roots are roots of τ_{n/2}(x).) This gives a total of 2n + 1 distinct roots, and so these must be all of them. Since only square roots are used the roots are constructible by straight-edge and compass and τ_{n}(x) has a solvable Galois group [DF, §14.7]. This proves the last part of Theorem 6.

Theorem 3.4. Let R = ℝ. Let J_{n} be the ideal of R_{3} generated by

τ_{n}(a_{21}), τ_{n}(a_{31}), τ_{n}(a_{32}), e = a_{21}^{2} + a_{31}^{2} + a_{32}^{2} − a_{21}a_{31}a_{32} − 4.

Then J_{n} is B_{3}-invariant.

Proof. The main step will be the second of the following identities:

Proposition 3.5. (i) For all n ≥ 1 we have

τ_{n}^{2}(a_{21}) + τ_{n}^{2}(a_{31}) + τ_{n}^{2}(a_{32})−2(τ_{n}(a_{21})τ_{n}(a_{31}) + τ_{n}(a_{21})τ_{n}(a_{32}) + τ_{n}(a_{31})τ_{n}(a_{32}))
− τ_{n}(a_{21})τ_{n}(a_{31})τ_{n}(a_{32}) ≡ 0 mod e  (3.3.)

(ii) For all n ≥ 1 we have

τ_{n}(a_{31} − a_{21}a_{32}) = τ_{n}(a_{21})τ_{n}(a_{32}) + 2(τ_{n}(a_{21}) + τ_{n}(a_{32})) − τ_{n}(a_{31}) ≡ 0 mod e.

Proof. (i) We first need to relate the polynomials τ_{n}(x) to the Chebyshev poynomials T_{x}(x) (of the first kind). These are defined [Ri] by

T_{0}(x) = 1, T_{1}(x) = x and T_{n+1}(x) = 2xT_{n}(x) − T_{n−1}(x).

We shall also use the alternative definition: T_{n}(x) = cos(nθ) where x = cos(θ). One of the properties of T_{n}(x) that we will use is [Ri, p.160]:

T_{n}(T_{m}(x)) = T_{mn}(x).  (3.4)

The relationship between T_{n}(x) and τ_{n}(x) is given by:

T_{n}(\frac{x^{4} − 4x^{2} + 2}{2}) = \frac{1}{2}τ_{n}(x) + 1.  (3.5)

This is proved by induction. If we let A_{ij} = \frac{a_{ij}^{4}−4a_{ij}^{2}+2}{2}, then the identity (3.3) reduces to:

T_{n}^{2}(A_{21}) + T_{n}^{2}(A_{31}) + T_{n}^{2}(A_{32}) − 2T_{n}(A_{21})T_{n}(A_{31})T_{n}(A_{32}) − 1 ≡ 0 mod e.

But we now note that T_{4}(x/2) = \frac{x^{4}−4x^{2}+2}{2} and so by (3.4) what we need to prove is:

T_{4n}^{2}(a_{21}/2) + T_{4n}^{2}(a_{31}/2) + T_{4n}^{2}(a_{32}/2) − 2T_{4n}(a_{21}/2)T_{4n}(a_{31}/2)T_{4n}(a_{32}/2) − 1 ≡ 0 mod e.

We will actually prove:

Lemma 3.6. For all n ∈ ℕ we have:

T_{n}^{2}(a_{21}/2) + T_{n}^{2}(a_{31}/2) + T_{n}^{2}(a_{32}/2) − 2T_{n}(a_{21}/2)T_{n}(a_{31}/2)T_{n}(a_{32}/2) − 1 ≡ 0 mod e.

Proof. For n ≥ 1 define the following matrices:

P_{n} = ( \begin{matrix}1 & T_{n}(a_{21}/2) & T_{n}(a_{31}/2) \\ T_{n}(a_{21}/2) & 1 & T_{n}(a_{32}/2) \\ T_{n}(a_{31}/2) & T_{n}(a_{32}/2) & 1 \end{matrix} ) = ( \begin{matrix}1 & cos(nθ_{21}) & cos(nθ_{31}) \\ cos(nθ_{21}) & 1 & cos(nθ_{32}) \\ cos(nθ_{31}) & cos(nθ_{32}) & 1 \end{matrix} ) ,

where cos(θ_{ij}) = a_{ij}/2. Note that what we want to prove for Lemma 3.6 is that det(P_{n}) ≡ 0 mod e, where e = det(P_{1}).

Now each P_{n} is a Gram matrix [HJ p. 407]; thus there are row vectors v_{1n}, v_{2n}, v_{3n} ∈ ℝ^{3}

such that if L_{n} = ( \begin{matrix}v_{1n} \\ v_{2n} \\ v_{3n} \end{matrix} ), then P_{n} = L_{n}L_{n}^{T}. We let L = L_{1} and v_{i} = v_{i1}, so that the v_{i}

are unit vectors with the angle between v_{i} and v_{j} being θ_{ij}.

Now suppose that e = det(P_{1}) = det(L)^{2} = 0. This occurs exactly when v_{1}, v_{2}, v_{3} are linearly dependent. We also note that in this case one of the angles θ_{ij} is the sum of the other two. Another way to see this is by solving the quadratic equation det(P_{1}) = 0; one obtains the solution θ_{21} = ±(θ_{31} ± θ_{32}). This shows that Lemma 3.7. With notation as in the above the following are equivalent:

(i) e = 0; (ii) θ_{21} = ±(θ_{31} ± θ_{32}); (iii) v_{1}, v_{2}, v_{3} lie in a subspace in ℝ^{3} of dimension at most 2. □

Now cos(θ) = cos(−θ) and we may assume θ_{31} = θ_{21} + θ_{32} (other cases being similar). Let Q = Span(v_{1}, v_{2}, v_{3}). If dim(Q) = 1, then θ_{ij} = 0, π for all ij and one checks that the result follows for these cases. Now assume that dim(Q) = 2.

Now for n ≥ 1 we can rotate v_{1} in the direction away from v_{2} to obtain a new vector v′_{2} ∈ Q whose angle with v_{2} is nθ_{21}; similarly we can rotate v_{3} away from v_{2} to obtain a vector v′_{3} ∈ Q whose angle with v_{2} is nθ_{32}. Thus the angle between v′_{1} and v′_{3} is nθ_{21} +nθ_{32} = nθ_{31}. This shows that we may take v_{1n} = v′_{1}, v_{2n} = v_{2}, v_{3n} = v′_{3} in L_{n}. It follows from P_{n} = L_{n}L_{n}^{T} that det(P_{n}) = 0 since Span(v′_{1}, v_{2}, v′_{3}) ⊂ Q. This concludes the proof of Lemma 3.6 and so also the proof of Proposition 3.5 (i).

(ii) We again convert this problem into the setting of Chebyshev polynomials. This time we need to prove:

T_{4n}((a_{31} − a_{21}a_{32})/2) ≡ 2T_{4n}(a_{21}/2)T_{4n}(a_{32}/2) − T_{4n}(a_{31}/2) mod e.  (3.6)

We will actually prove (3.6) with 2n replacing 4n. As in the above, putting a_{ij}/2 = cos(θ_{ij}) and solving e = 0 we get θ_{31} = ±(θ_{21} ± θ_{32}).

Now we also put (a_{31} + a_{21}a_{32})/2 = cos(θ), so that cos(θ) = cos(θ_{31}) − 2cos(θ_{21})cos(θ_{32}). Putting θ_{31} = (θ_{21} ± θ_{32}) we get:

cos(θ) = −cos(θ_{21})cos(θ_{32}) ∓ sin(θ_{21})sin(θ_{32}) = −cos(θ_{21} ∓ θ_{32}).

This then gives θ = π − (θ_{21} ∓ θ_{32}). Thus what we want to prove is:

T_{2n}(cos(θ)) ≡ 2T_{2n}(cos(θ_{21}))T_{2n}(cos(θ_{32})) − T_{2n}(cos(θ_{31})) mod e,

or in other words

cos(2n(π − (θ_{21} ∓ θ_{32}))) ≡ 2(cos(2nθ_{21})(cos(2nθ_{32}) − cos(2n(θ_{21} ± θ_{32})) mod e. (3.7)

But cos(2n(π − (θ_{21} ∓ θ_{32}))) = cos(2n(θ_{21} ∓ θ_{32})) and (3.7) follows from standard trigonometric identities. This proves Proposition 3.5 (ii). □

Using the action of the generators σ_{i} given in (1.1) Proposition 3.5 (ii) shows that τ_{n}(σ_{2}(a_{21})) ∈ J_{n}. Since σ_{2}(a_{31}) = a_{32} and σ_{2}(a_{32}) = −a_{32} it is clear that τ_{n}(σ_{2}(a_{31})) and τ_{n}(σ_{2}(a_{32})) are both in J_{n}. The proofs for the action of σ_{2}^{−1} and σ_{1}^{±1} are similar. This proves the first statement of Theorem 3, the second being clear.

The rest of Theorem 3 follows from Proposition 3.5 (i) and a calculation of the action of σ_{1} and σ_{2}. □

We noted in [Hu4, Theorem 7.4] (which is also true for n = 3) that there is a double cover η : S^{1} × S^{1} → S_{4} which, by the above, we can choose to be given by

η(θ_{1}, θ_{2}) = (2cos(θ_{1}), 2cos(θ_{2}), 2cos(θ_{1} + θ_{2})).

Here we let S^{1} = ℝ/(2π). Then we have η(θ_{1}, θ_{2}) = η(−θ_{1}, −θ_{2}). We further note that if (x, y, z) = (2cos(θ_{1}), 2cos(θ_{2}), 2cos(θ_{1} + θ_{2})) is one point of S_{4}, then the other point with the same x, y value is (2cos(θ_{1}), 2cos(−θ_{2}), 2cos(θ_{1} − θ_{2})).

Now we lift the action of B_{3} on S_{4} to an action on S^{1} × S^{1}. First represent (x, y, z) ∈ S_{4} as 2(cos(θ_{1}), cos(θ_{2}), cos(θ_{1} + θ_{2})). From the actions of σ_{1}, σ_{2} given in (1.1)’ we obtain the following action of B_{3} on S^{1} × S^{1}:

σ_{1}(θ_{1}, θ_{2}) = (π − θ_{1}, θ_{1} − θ_{2} − π); σ_{2}(θ_{1}, θ_{2}) = (π − 2θ_{1} − θ_{2}, θ_{1}).  (3.8)

This action thus projects, via η, to the action of B_{3} on S_{4}, however the action on S^{1} × S^{1} of the central generator (σ_{1}σ_{2})^{3} is no longer trivial, but satisfies (σ_{1}σ_{2})^{3}(θ_{1}, θ_{2}) = (−θ_{1}, −θ_{2}). Thus the representation of B_{3} on S^{1} × S^{1} is that of SL(2, ℤ) via the representation

σ_{1} ↦ s_{1} = ( \begin{matrix}−1 & 0 & π \\ 1 & −1 & −π \\ 0 & 0 & 1 \end{matrix} ); σ_{2} ↦ s_{2} = ( \begin{matrix}−2 & −1 & π \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} )

on the coordinates (θ_{1}, θ_{2}, 1)^{T}. The matrices s_{1}, s_{2} belong to SL(2, ℝ/(2π)).

Let n ∈ ℕ; then it follows that the set of points

V_{n} = {(2πk/n, 2πm/n)|k, m ∈ ℕ} ⊂ S^{1} × S^{1}

is invariant under the B_{3} action and consequently that the image η(V_{n}) is invariant under the B_{3}/Z(B_{3}) action. In fact we have the following connection:

Lemma 3.8. (i) For all n ∈ ℕ we have η(V_{4n}) = W_{n}.

(ii) The cardinality of W_{n} is 8n^{2} + 2.

Proof. (i) We need only show that τ_{n}(2cos(θ)) = 0 if θ = 2πk/(4n). From (3.5) we see that T_{4n}(x) = \frac{1}{2}τ_{n}(2x) + 1 and so

\frac{1}{2}τ_{n}(2cos(θ)) = T_{4n}(cos(θ)) − 1 = cos(4n × 2πk/(4n)) − 1 = 0,

as required for (i).

(ii) There are (4n)^{2} points in V_{4n}. The double cover restricted to V_{4n} identifies points (\frac{2πk_{1}}{4n} , \frac{2πk_{2}}{4n}) and (−\frac{2πk_{1}}{4n} , −\frac{2πk_{2}}{4n}) i.e. points with their negatives. But there are 4 such points which are their own negatives and so the number of points in W_{n} is ((4n)^{2}−4)/2+4 = 8n^{2}+2. □

Let z = (σ_{1}σ_{2})^{3}. Recall that the principal congruence subgroup of level n is the kernel of the epimorphism SL(2, ℤ) → SL(2, ℤ_{n}); it is denoted Γ(n). A subgroup containing Γ(n) is called a congruence subgroup of level n [Ko]. We will let Γ(n)′ denote the kernel of the map P SL(2, ℤ) → P SL(2, ℤ_{n}).

Lemma 3.9. The kernel of the action of B_{3}/ < z^{2} >≅ SL(2, ℤ) on V_{4n} is exactly Γ(4n). The kernel of the action of B_{3}/ < z >≅ P SL(2, ℤ) on W_{n} is Γ′(4n). Further,

Ω_{n} ≅ P SL(2, ℤ_{4n}).

Proof. We only need to prove the first part since the rest follows upon acting by η. Let I_{n} denote the n×n identity matrix. Now if α ∈ Γ(4n) < SL(2, ℤ), then α = I_{2} +4nA

for some 2 × 2 integral matrix A. Then α ( \begin{matrix}θ_{1} \\ θ_{2} \end{matrix} ) = ( \begin{matrix}θ_{1} \\ θ_{2} \end{matrix} ) if ( \begin{matrix}θ_{1} \\ θ_{2} \end{matrix} ) ∈ V_{4n}.

Conversely, if α ∉ Γ(4n), then either α ( 0^{\frac{2π}{4n}} ) ≠ ( 0^{\frac{2π}{4n}} ) or α ( 0_{\frac{2π}{4n}} ) ≠ ( 0_{\frac{2π}{4n}} ). In either

case we see that α does not belong to the kernel of the action of SL(2, ℤ) on V_{4n}. □

The matrices s_{1}, s_{2} ∈ SL(2, ℝ/(2π)) fix the antisymmetric form represented relative to the same basis by the matrix

( \begin{matrix}0 & 1 & 0 \\ −1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} ) .

This is where the function ω on S_{4} introduced in §1 comes from. We now prove Theorem 8.

Let B′ =< s_{2}^{2}s_{1}, s_{1}s_{2}, s_{1}s_{2}^{−1}s_{1} > be the image of B in SL(2, ℤ). Then one checks using MAGMA that [B_{3} : B] = 4 and that B′ has index 4 in < s_{1}, s_{2} >= SL(2, ℤ). In fact B is the stabiliser of the singular point (2, 2, 2). Furthermore one checks that the last column of each element of B′ is (0, 0, 1)^{T}.

We now show that B fixes ω. It will suffice to prove this for each of the three generators of B. Let α be such a generator. Fix ϵ > 0. Now we can find an open neighbourhood U ⊂ S_{4} such that for all (x, y, z) ∈ U we have |x|, |y| < ϵ. Note that since each θ_{i} is an inverse cosine function with θ_{i}(0) = π/2 we may also assume that ω(pα, qα) ∈ (π^{2}/4 − ϵ, π^{2}/4 + ϵ) for p, q ∈ U. In particular with this restricted domain we don’t need to worry about the fact that arccos is only defined up to multiples of 2π. Thus for p = (x, y, z), q = (u, v, w) ∈ U we may think of ω(p, q) as a power series in the variables x, y, u, v and the above shows that this power series is fixed by the action of α.

This proves Theorem 8 (i) for ω and (ii) and (iii) are easy.

The definition of ω^{(2)} is designed to get rid of the appearance of the πs in (3.8) and so gives the invariance of ω^{(2)} under the action of B_{3}. The rest follows. □

Now let ζ_{n} = ζ_{n}(a_{21}, a_{31}, a_{32}) denote the polynomial which is the left hand side of (3.3). Thus for example:

ζ_{1} =(a_{32}^{4} − 2a_{21}^{2}a_{32}^{2} − 2a_{31}^{2}a_{32}^{2} + a_{21}^{2}a_{31}^{2}a_{32}^{2} + a_{31}^{4} − 2a_{21}^{2}a_{31}^{2} + a_{21}^{4})
× (a_{32}^{2} − 4 + a_{31}^{2} − a_{21}a_{31}a_{32} + a_{21}^{2})(a_{32}^{2} − 4 + a_{31}^{2} + a_{21}a_{31}a_{32} + a_{21}^{2});
ζ_{2} =ζ_{1} × (16 + a_{21}^{4}a_{31}^{4}a_{32}^{4} − 4a_{21}^{4}a_{31}^{4}a_{32}^{2} − 4a_{21}^{4}a_{31}^{2}a_{32}^{4} + 16a_{21}^{4}a_{31}^{2}a_{32}^{2} − 4a_{21}^{2}a_{31}^{4}a_{32}^{4}
+ 16a_{21}^{2}a_{31}^{4}a_{32}^{2} + 16a_{21}^{2}a_{31}^{2}a_{32}^{4} − 64a_{21}^{2}a_{31}^{2}a_{32}^{2} − 32a_{31}^{2} + 24a_{31}^{4} − 32a_{32}^{2} + 24a_{32}^{4}
− 8a_{31}^{6} + a_{31}^{8} − 8a_{32}^{6} + a_{32}^{8} + 32a_{31}^{2}a_{32}^{2} − 8a_{31}^{2}a_{32}^{4} − 8a_{31}^{4}a_{32}^{2} + 2a_{31}^{4}a_{32}^{4} + 32a_{21}^{2}a_{31}^{2}
− 8a_{21}^{2}a_{31}^{4} − 8a_{21}^{4}a_{31}^{2} + 2a_{21}^{4}a_{31}^{4} + 32a_{21}^{2}a_{32}^{2} − 8a_{21}^{2}a_{32}^{4} − 8a_{21}^{4}a_{32}^{2}
+ 2a_{21}^{4}a_{32}^{4} − 32a_{21}^{2} + 24a_{21}^{4} − 8a_{21}^{6} + a_{21}^{8}).

Then from the above we have

Corollary 3.10. If n, m ∈ ℕ with m|n, then ζ_{m} divides ζ_{n}. □

§4 Some groups generated by transvections with some simple elements of finite order

In this section we will prove the rest of Theorem 4. Let v ∈ W_{n}. We first investigate properties of the matrices T_{i}(v) generating G(v) =< T_{1}(v), T_{2}(v), T_{3}(v) > as in (1.3).

Lemma 4.1. Fix n ∈ ℕ. Suppose that v = (x_{21}, x_{31}, x_{3}) ∈ W_{n}. Then for 1 ≤ i < j ≤ 3 the matrix T_{i}(v)T_{j}(v) has order a divisor of 4n if x_{ji}^{2} ≠ 4. If x_{ji}^{2} = 4, then T_{i}(v)T_{j}(v) has infinite order.

Proof. It will suffice to consider the case i = 1, j = 2. Let P_{1}(a_{21}) = ( \begin{matrix}1 & −a_{21} \\ 0 & 1 \end{matrix} ) , P_{2}(a_{21}) =

( \begin{matrix}1 & 0 \\ a_{21} & 1 \end{matrix} ). Then one checks that

(P_{1}P_{2})^{2} = ( \begin{matrix}1 − 3 a_{21}^{2} + a_{21}^{4} & a_{21}^{3} − 2 a_{21} \\ 2 a_{21} − a_{21}^{3} & 1 − a_{21}^{2} \end{matrix} ) = N(a_{21}),

where N = N(a_{21}) is the submatrix of M defined in the proof of Proposition 3.1. Now

T_{1}(v) = ( \begin{matrix}P_{1}(x) & \begin{matrix}−y \\ 0 \end{matrix} \\ 0 0 & 1 \end{matrix} ) , T_{2}(v) = ( \begin{matrix}P_{2}(x) & \begin{matrix}0 \\ −z \end{matrix} \\ 0 0 & 1 \end{matrix} ) .

Further, [LS, p. 168] shows that < P_{1}(±2), P_{2}(±2) > is a free group of rank 2 and so in this case T_{i}(v)T_{j}(v) has infinite order.

From now on we will assume that x_{ji}^{2} ≠ 4.

Lemma 4.2. For n ∈ ℕ we have

N^{n}(a_{21}) = ( \begin{matrix}\frac{κ_{2n+2}+(a_{21}^{2}−1)κ_{2n}}{a_{21}(a_{21}^{2}−2)} & κ_{2n} \\ −κ_{2n} & \frac{κ_{2n+2}−(a_{21}^{4}−3a_{21}^{2}+1)κ_{2n}}{a_{21}(a_{21}^{2}−2)} \end{matrix} ) .

Further, if x is a root of κ_{2n}(a_{21}) = 0, then N^{n}(x) = ( \begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix} ).

Proof. First note that by Lemma 3.3 (v), the entries of the above matrix are really polynomials in a_{21}. The result is now proved by induction on n ≥ 1: One multiplies N^{n+1} as N × N^{n}, noting that the off-diagonal entries are exactly what they should be (namely κ_{2n+2}). Then one uses Proposition 3.3 (iv) (with n replaced by 2n) to show that the (1, 1) and (2, 2) entries are what they should be.

Now if κ_{2n}(x) = 0, then Proposition 3.3 (ii) (c) shows that the diagonal entries of N^{n}(x) are both −1. □

Now we see that

(T_{1}(v)T_{2}(v))^{2n} = ( \begin{matrix}N^{n}(x) & \begin{matrix}∗ \\ ∗ \end{matrix} \\ 0 0 & 1 \end{matrix} ) = ( \begin{matrix}−1 & 0 & ∗ \\ 0 & −1 & ∗ \\ 0 & 0 & 1 \end{matrix} )

and so (T_{1}(v)T_{2}(v))^{4n} = I_{3}. This proves Lemma 4.1. □

Let T′_{1}, T′_{2} be two n × n transvections which do not commute. Since each transvection fixes a hyperplane, there is a codimension 2 subspace W fixed by < T′_{1}, T′_{2} >. There is also a complementary subspace V that is < T′_{1}, T′_{2} >-invariant. The action of T′_{1}, T′_{2} on this

subspace can be represented as t_{1} = ( \begin{matrix}1 & λ \\ 0 & 1 \end{matrix} ) , t_{2} = ( \begin{matrix}1 & 0 \\ −λ & 1 \end{matrix} ) where λ ≠ 0 (since they

don’t commute). Now trace(t_{1}t_{2}) = trace(T′_{1}T′_{2}) − (n − 2) = 2 − λ^{2}. Thus if τ_{n}(λ) = 0, then by the above (t_{1}t_{2})^{4n} = I_{2} and so (T′_{1}T′_{2})^{4n} = I_{n}.

Now if α ∈ K(v), then trace(α(T′_{1})α(T′_{2})) = trace(T′_{1}T′_{2}) and we have proved the second paragraph of Theorem 4.

If we consider all of the solutions W_{n} to (3.2), then we get a homomorphism B_{3}/Z(B_{3}) → Sym(W_{n}) with kernel K_{n} such that trace(η(w)α(T_{i}T_{j})) = trace(η(w)(T_{i}T_{j})) for all α ∈ K, 1 ≤ i ≠ j ≤ 3, and all w ∈ W_{n}.

Proposition 4.3. The kernel K = K_{n} is a finitely generated free group.

Proof. Note that K, being of finite index in B_{3}/Z(B_{3}), is finitely generated. By the Kurosh subgroup theorem [MKS] K < B_{3}/Z(B_{3}) ≅ P SL_{2}ℤ ≅ ℤ_{2} ∗ ℤ_{3} is a free product of a finite number of cyclic groups of order 2, 3, ∞, each such cyclic group of finite order being conjugate to one of the two generating subgroups of ℤ_{2} ∗ ℤ_{3}. But K is normal and ℤ_{2} ∗ ℤ_{3}/K is not cyclic and so there can be no occurences of the finite cyclic summands in the representation of K as such a free product. □

Example 4.4. For n = 1, 2, …, 8 the ranks of the free groups K_{n} are

5, 33, 97, 257, 481, 769, 1345, 2049.

These calculations were made using MAGMA.

We now prove the rest of Theorem 6:

Lemma 4.5. The Galois group of τ_{n}(x) is contained in the centraliser of Ω_{n} < Sym(W_{n}) and is abelian for all n ∈ ℕ.

Proof. Denote the centraliser of Ω_{n} in Sym(W_{n}) by C(Ω_{n}). We first show that C(Ω_{n}) is abelian. For α ∈ Ω_{n}, w ∈ W_{n}, ϕ ∈ Gal(τ_{n}) we have w = (w_{1}, w_{2}, w_{3}) and (w)α = (z_{1}, z_{2}, z_{3}) where z_{i} = z_{i}(w_{1}, w_{2}, w_{3}) are polynomials in w_{1}, w_{2}, w_{3} depending only on α. Thus we have

wαϕ = (z_{1}(w_{1}, w_{2}, w_{3}), z_{2}(w_{1}, w_{2}, w_{3}), z_{3}(w_{1}, w_{2}, w_{3}))ϕ
= (z_{1}(w_{1}ϕ, w_{2}ϕ, w_{3}ϕ), z_{2}(w_{1}ϕ, w_{2}ϕ, w_{3}ϕ), z_{3}(w_{1}ϕ, w_{2}ϕ, w_{3}ϕ))
= wϕα,

as required.

The rest of Lemma 4.5 will follow if we can show that C(Ω_{n}) is abelian. Recall that Ω_{n} = P SL(2, ℤ_{4n}) and that Ω_{n} acts on W_{n}. Now suppose that f, g ∈ C(Ω_{n}) and that fg ≠ gf. Let h = fgf^{−1}g^{−1}. Then there is w ∈ W_{n} such that (w)h ≠ w. Let O(w) denote the orbit of w under the B_{3} action. Then for all v ∈ O(w) there is α ∈ B_{3} such that v = (w)α. Thus we have

(v)h = (w)αh = (w)hα ≠ wα = v.

So that (v)h ≠ v for all v ∈ O(w).

Now write h = c_{1}c_{2} … c_{r} ∈ Sym(W_{n}) in its disjoint cycle decomposition. Then the fact that αhα^{−1} = h for all α ∈ Ω_{n} shows that each such α permutes the cycles c_{i}. We will next show that r = 1.

We may without loss reduce all coefficients mod p for some prime p which divides 4n i.e. we consider the representation Ω → P SL(2, ℤ_{4n}) → P SL(2, ℤ_{p}). Assume that c_{1} = (x_{1}, x_{2}, …, x_{s}), where x_{i} represents a non-zero element of ℤ_{p}^{2}. Note that s > 1 since h does not fix anything in O(w). Then there is a parabolic element β ∈ Ω_{n} which fixes x_{1} and which has order p. Consider the action of β on x_{i}, i > 1; since β fixes x_{1} and β and h = c_{1} … c_{r} commute we see that β(x_{i}) is in the cycle c_{1} for i = 2, …, s. Since β has order p we see that β^{k}(x_{2}), k = 1, …, p are also in the cycle c_{1}. Now we repeat this argument for each of β^{k}(x_{2}), k = 1, …, p and we see that the (arbitrary) cycle c_{1} has at least p(p − 1) elements in it. However there are only p^{2} − 1 non-zero elements of ℤ_{p} and it follows that r = 1.

So h = c_{1} = (x_{1}, x_{2}, …); however there is β ∈ Ω_{n}, β ≠ id, with β(x_{1}) = x_{1}. Then the fact that hβ = βh shows that β(x_{2}) = x_{2}. Repeating this we see that β = id, a contradiction. Thus there is no such h and so no such f, g. □

§5 Periodic points and chaos on the 2-sphere

Recall that in §3 we defined a double cover η : S^{1} × S^{1} → S_{3} and that in Lemma 3.8 we showed that η(V_{4n}) = W_{n}. From the definition V_{n} = {(2πk/n, 2πm/n)|k, m ∈ ℕ} for V_{n} we clearly see that ⋃_{n∈ℕ} V_{4n} is a dense subset of S^{1} × S^{1}. It follows that W_{∞} is a dense subset of S_{4}. This proves the first part of Theorem 7.

Fix n ≥ 1 and w ∈ W_{n}. Let C ∈ F_{3} be a non-trivial simple closed curve on S_{4} and denote by T (C) the Dehn twist about C. Then there is α ∈ B_{3} such that α(x_{1}x_{2}) = C. Now by Artin’s condition (A1) we have α(x_{1}) = Ax_{i}A^{−1}, α(x_{2}) = Bx_{j}B^{−1} for A, B ∈ F_{3}. Thus α(x_{1}), α(x_{2}) are also transvections and so η(w)(T (C)) fixes a line W and a complementary 2-dimensional subspace V ⊂ ℝ^{3}. It follows that the order of η(w)(T (C)) is completely determined by its trace. Certainly we have: if κ_{n}(trace(η(w)(T (C))) − 1) = 0, then η(w)(T (C)) has finite order (a divisor of 4n). Now as we have seen in Lemma 4.1 trace(η(w)(T (C))) − 1 = 2 − x^{2}, where τ_{n}(x) = 0 and so the only situation where we don’t have η(w)(T (C)) of finite order is when x^{2} = 4.

Now let T_{i}(C, w) = η(w)α(T_{i}), for i = 1, 2, 3 and α(w) = (x, y, z). Then we have

trace(T_{1}(C, w)T_{2}(C, w)) = 3 − x^{2}, trace(T_{1}(C, w)T_{3}(C, w)) = 3 − y^{2},
and trace(T_{2}(C, w)T_{3}(C, w)) = 3 − z^{2}.

If x^{2} ≠ 4, then T_{1}(C, w)T_{2}(C, w) has finite order. So suppose that x^{2} = 4. Now

σ_{1}(σ_{2}([a_{21}, a_{31}, a_{32}])) = [a_{32}, −a_{21}, −a_{31}]

and so replacing w by w′ = σ_{1}σ_{2}(w) we see that T_{1}(C, w′)T_{2}(C, w′) has finite order unless y^{2} = 4. Thus we may assume that x^{2} = y^{2} = 4. If we replace w′ by w′′ = σ_{1}σ_{2}(w′), then we see that T_{1}(C, w′′)T_{2}(C, w′′) has finite order unless z^{2} = 4. But for all n ≥ 1 we can always find a w ∈ W_{n} (Lemma 3.2) not all of whose coordinates are ±2. Thus we have proved the last sentence of Theorem 4.

We next prove Theorem 5. For each w ∈ W_{n} there is a projection η(w) : F_{3} → G(w). Now G(w) is a finitely generated linear group and so has a torsion-free subgroup of finite index [Se]; call it L(w). Let L′(w) = η(w)^{−1}(L(w)). Then L′(w) is of finite index in F_{3}. Let H = ⋂_{w∈W_{n}} L′(w). Then H is of finite index in F_{3} since W_{n} is finite. Let u ∈ F_{3} be a non-trivial simple element. Then by the last part of Theorem 4 there is w′ ∈ W_{n} such that η(w′)(u) has finite order. But L(w′) is torsion free; thus η(w′)(u) ∉ L(w′), showing that u ∉ L′(w′) and so u ∉ H.

Now there are only two conjugacy classes in F_{3} which represent simple closed curves surrounding all three punctures, namely the classes of x_{1}x_{2}x_{3} and its inverse. We now show that we may obtain such an H which also does not contain any of these simple elements. Since F_{3} is residually finite [MKS] there is a finite index subgroup U of F_{3} which does not contain either of (x_{1}x_{2}x_{3})^{±1}. Then the core of U is of finite index in F_{3} and does not contain any conjugate of these two elements (since it is normal in F_{3}). Now replace H by H ∩ core(U) and we have a subgroup whose only simple elements are the conjugates of the generators x_{1}^{±1}, x_{2}^{±1}, x_{3}^{±1}.

To get rid of these we repeat the above: since F_{3} and its subgroups are residually finite there is a subgroup E of finite index in F_{3} which does not contain x_{1}^{±1}, x_{2}^{±1}, x_{3}^{±1}. Then core(E) does not contain any conjugate of x_{i}^{±1} and is of finite index in F_{3}. Now replace H by H ∩ core(E), which is also of finite index in F_{3}.

To show that one can take H to be normal just replace H by core(H). To show that we can take H to be invariant under the action of B_{3} we just note that there are only finitely many subgroups of B_{3} of index [B_{3} : H] and that [B_{3} : α(H)] = [B_{3} : H] for all α ∈ B_{3}. Thus we can replace H by the B_{3}-invariant subgroup ⋂_{α∈B_{3}} α(H), this being a finite intersection by the above. □

We now consider Theorem 9. The composition of maps ℵ : ℂ → ℂ/L ≅ S^{1} × S^{1} → S_{4} referred to therein has indeterminacy in the first arrow corresponding to points separated by lattice points. However ℘ is periodic relative to L. There is also indeterminacy in the last map, where a point and its negative are identified; however ℘ is an even function [Ko; p. 18]. This proves Theorem 9. □

Similarly, theta functions can be defined on S_{4} as long as they are even.

Lemma 5.1. If α ∈ B_{3}/Z(B_{3}) and trace(α) > 2, then α acts chaotically on S_{4}.

Proof. We need only prove topological transitivity. Since |trace(Σ(α))| > 2 the matrix Σ(α) is diagonalisable; the eigenvector for the eigenvalue of absolute value greater than 1 represents a line on S^{1} × S^{1} of irrational slope which is fixed by the action of α. Since the slope is an algebraic number any such line is dense in S^{1} × S^{1} and projects to a dense invariant set L in S_{4}. The action of α on S_{4} fixes L. Any open set U ⊂ S_{4} meets L. Let V ⊂ S_{4} also be open and non-empty; it also meets L. In fact for any p ∈ L ∩ U the open set V will meet the positive (or negative) ray of L starting at p. Since the action of α on L is to stretch by a factor greater than 1 we see that, for N sufficiently large, α^{N}(U) will meet V. □

§6 Generalisations to arbitrary braid groups

As indicated in §1 we have suppressed the generalisations of Theorem 1-9 to arbitrary B_{n} due to the necessity of understanding the basics of [Hu1-Hu4]; these we now explain.

In [Hu1] we have shown that by representing the free group π_{1}(D_{n}) using transvections and looking at certain traces we obtain an action of B_{n} on the ring R_{n} (or on R′_{n}) i.e. we have a homomorphism

ψ_{n} : B_{n} → Aut(R_{n});

the kernel of ψ_{n} is the centre of B_{n} [Hu1]. This is understood as follows: For fixed n ≥ 2 we let T_{1}, …, T_{n} be certain n × n matrices (transvections):

T_{i} = ( \begin{matrix}1 & 0 & … & 0 & … & 0 & 0 \\ 0 & 1 & … & 0 & … & 0 & 0 \\ ⋮ & ⋮ & ... & ⋮ & … & ⋮ & ⋮ \\ a_{i1} & a_{i2} & … & 1 & … & a_{in−1} & a_{in} \\ ⋮ & ⋮ & … & ⋮ & ... & ⋮ & ⋮ \\ 0 & 0 & … & 0 & … & 1 & 0 \\ 0 & 0 & … & 0 & … & 0 & 1 \end{matrix} ) .

Here the non-zero off-diagonal entries of T_{i} occur in the ith row. One way of defining a transvection [A] is as a matrix M = I_{n} + A where I_{n} is the n × n identity matrix, det(M) = 1, rank(A) = 1 and A^{2} = 0. In particular, conjugates of transvections are transvections. That < T_{1}, …, T_{n} > is a free group of rank n is shown in [Hu2]. This allows us to identify x_{i} and T_{i} for i = 1, …, n and so to identify F_{n} and < T_{1}, …, T_{n} > .

Now, since we are identifying x_{i} ∈ F_{n} with T_{i} ∈< T_{1}, …, T_{n} >, we see, by Artin’s characterisation (A1) and (A2) above, that the matrix Π_{n} = T_{1}T_{2} … T_{n} is invariant under the action of B_{n}. It follows that the B_{n}-action fixes the various determinantal ideals I_{n,r,λ} determined by Π_{n} + λI_{n} [Hu4]. Here I_{n,r,λ} is the ideal generated by the determinants of all (r + 1) × (r + 1) submatrices of Π_{n} + λI_{n}. These are the points at which Π_{n} + λI_{n} has rank r (except in the characteristic 2 case, which won’t really concern us). For general properties of such determinantal rings see [BV, DEP1, DEP2]. We will put I_{n,r} = I_{n,r,1}.

In [Hu2] we showed that Π_{n} − I_{n} and U_{n} are row equivalent and so have the same set of determinantal ideals (for this to make sense in the present context we need to consider the matrix U_{n} of [Hu2] as a matrix over R′_{n}).

For i, j, k, …, r, s ∈ {1, 2, …, n} let c_{ijk...rs} denote the cycle a_{ij}a_{jk} … a_{rs}a_{si} ∈ R_{n}. Then the cycles generate a subalgebra of R_{n} denoted Y_{n}. A cycle c_{ijk...rs} will be called simple if i, j, k, ..., r, s are all distinct. The ring Y_{n} is the trace ring of the group < T_{1}, …, T_{n} > [Hu1, Hu4], and is generated by the (finite number of) simple cycles and is B_{n}-invariant.

Now, since Π_{n} is invariant under the action of B_{n}, it follows that the characteristic polynomial

χ_{n}(x) = ∑_{i=0}^{n} c_{ni}x^{i}

of the matrix Π_{n} has coefficients c_{ni} = c_{ni}(a_{12}, …, a_{n n−1}) ∈ R_{n} which are invariant under the action of B_{n} (this was first noted in [Hu2, Theorem 2.8]). We there also noted that the c_{ni} are non-homogeneous polynomials of degree n for 1 ≤ i < n. If n < 6, then they generate the ring of invariants for the action of B_{n} on the subring Y_{n} of R_{n} [Hu3]. Note that we have c_{nn} = 1 and c_{n0} = ±1.

Note that the action of B_{n} on D_{n} fixes the boundary of D_{n}. This fact gives Artin’s condition (A2) above. Let C_{n} denote the set of all oriented simple closed curves on D_{n}.

Now, from the above, B_{n} acts by automorphisms on F_{n} in such a way that for α ∈ B_{n} the matrix α(T_{i}) is a conjugate of some T_{j}, 1 ≤ j ≤ n i.e. α(T_{i}) is also a transvection. Further, if c ∈ C_{n}, then c represents a conjugacy class in F_{n} and so its trace is well-defined (the trace of the corresponding product of transvections in F_{n} =< T_{1}, …, T_{n} >). In fact one easily sees that trace(c) ∈ Y_{n} [Hu1]. Then a map ϕ = ϕ_{n} : C_{n} → R′_{n} is defined by

ϕ_{n}(c) = trace(c) − n.

Thus ϕ_{n} can be thought of as being defined on certain conjugacy classes of elements of F_{n} (namely those representing simple closed curves). The map ϕ can be extended to act on all of F_{n}, by the requirement that for s ∈ F_{n} we have ϕ(s) = trace(s) − n. It is easy to see that if w = T_{i_{1}}^{e_{1}} … T_{i_{r}}^{e_{r}} ∈< T_{1}, …, T_{n} > is cyclically reduced as written with e_{i} ≠ 0, i_{k} ≠ i_{k+1}, i_{r} ≠ i_{1}, and r > 1, then ϕ(w) is a polynomial in Y_{n} of degree r.

A fundamental property of the transvections T_{i} is that for all 1 ≤ i, j ≤ n we have trace(T_{i}T_{j}) = a_{ij}a_{ji} + n and in general if A, B ∈ F_{n}, then

trace(AT_{i}A^{−1}BT_{j}B^{−1}) = b_{ij}b_{ji} + n,  (6.1)

where b_{ij} ∈ R_{n} (see [Hu1]). It is also easy to see that there is a natural choice so that

b_{ij} = ±a_{ij} + terms of higher degree.

For example trace((T_{1}T_{2}T_{1}^{−1})T_{3}) = (a_{23} − a_{21}a_{13})(a_{32} + a_{31}a_{12}) + n. This is explained in more detail in §2 of [Hu4].

Now for α ∈ B_{n} the image α(T_{i}) is a conjugate AT_{j}A^{−1} for some A ∈ F_{n} and 1 ≤ j ≤ n (by Artin’s condition (i) above). Here the action of α on the a_{ij} is defined by

ϕ(α(T_{i})α(T_{j})) = α(a_{ij})α(a_{ji}),

so that it has the following naturality property (with respect to the action of B_{n} on F_{n}): for all w ∈< T_{1}, …, T_{n} >, α ∈ B_{n}, we have

ϕ(α(w)) = α(ϕ(w)).

For example the action of the generator σ_{i}, 1 ≤ i < n, on R_{n} is given by

σ_{i}(a_{i i+1}) = a_{i+1 i}, σ_{i}(a_{i+1 i}) = a_{i i+1},
σ_{i}(a_{h i}) = a_{h i+1} + a_{hi}a_{i i+1}, σ_{i}(a_{h i+1}) = a_{h i},  (6.2)
σ_{i}(a_{i h}) = a_{i+1 h} − a_{i+1 i}a_{ih}, σ_{i}(a_{i+1 h}) = a_{i h},

where 1 ≤ h ≤ n and h ≠ i, i + 1.

It follows from [Hu1, Theorem 2.5 and Theorem 6.2] that the kernel of the action of B_{n} on R_{n} is the centre of B_{n} and that if B_{n} and R_{n} are thought of as sub-objects of B_{n+1} and R_{n+1} (respectively), then the action of B_{n} on R_{n+1} is faithful. The action on R′_{n} is the induced action on the quotient.

The generalisation that we seek concerns the ideal generated by all 3 × 3 minors of U_{n}. Now given any distinct i < j < k ∈ {1, 2, …, n} the principal minor with rows and columns indexed by i, j, k gives the condition a_{ji}^{2} + a_{kj}^{2} + a_{ki}^{2} − a_{ji}a_{kj}a_{ki} − 4 = 0. Putting k = n we can use this equation to solve for a_{ji} as a function of a_{ni}, a_{nj}:

a_{ji} = a_{ji}^{ϵ_{ij}}(a_{n1}, a_{n2}, …, a_{n n−1}) = a_{nj}a_{ni}/2 + ϵ_{ij}\sqrt{(a_{nj}^{2} − 4)(a_{ni}^{2} − 4)}/2,  (6.3)

where ϵ_{ij} = ϵ_{ji} = ±1. In [Hu4 Lemma 7.4] we proved:

Lemma 6.1. For any z = (a_{n1}, a_{n2}, …, a_{n n−1}) ∈ [−2, 2]^{n−1} and

ϵ = (ϵ_{21}, ϵ_{31}, …, ϵ_{n−1 n−2}) ∈ \{±1\}^{\binom{n−1}{2}}

the point f_{ϵ}(z) = (a_{21}^{ϵ_{21}}(z), a_{31}^{ϵ_{31}}(z), …, a_{n n−1}^{ϵ_{n n−1}}(z)) ∈ ℝ^{\binom{n}{2}} with the a_{ij}^{ϵ_{ij}}, 1 ≤ i < j < n, given by (6.3), is in V_{n2} if and only if the following conditions are satisfied:

ϵ_{ij}ϵ_{jk}ϵ_{ik} = −1 for all distinct 1 ≤ i, j, k < n. □  (6.4)

Using this result we obtained the following result. Let T^{n−1} be the (n − 1)-torus S^{1} × ⋯ × S^{1} (n − 1 times). Here we again represent S^{1} as ℝ mod 2π. Let α : T^{n−1} → T^{n−1} be the antipodal map α(x) = −x for all x ∈ T^{n−1}. We will think of α as acting on all the T^{n−1} for all n.

Theorem 6.2, [Hu4, 7.6]. For all n ≥ 3 the orbifold quotient T^{n−1}/ < α > is homeomorphic to V_{n2}. There are 2^{n−1} singular points S′_{n} of V_{n2} that are non-manifold points (if n > 3) whose links are all homeomorphic to ℝℙ^{n−2}. They are the points at which U_{n} has rank 2. The open manifold V_{n2} \ S′_{n} is acted upon by B_{n}.

In the case n = 3 the singular points are the four point (2, 2, 2), (2, −2, −2), … and in this (and only this) case V_{3n} is a manifold. This also makes the n = 3 case different.

Other than the fact that B_{n} acts on V_{n2} by diffeomorphisms Theorems 1 and 2 have no analogue if n > 3. It would be nice to know what the mapping class group of V_{n2} is.

We now consider the generalisation of Theorem 3. Fix n > 3, m ≥ 1 and let J_{nm} denote the ideal of R′_{n} generated by all the τ_{m}(a_{ij}) and all the 3 × 3 minors of U′_{n}. This includes all the a_{ji}^{2} + a_{kj}^{2} + a_{ki}^{2} − a_{ji}a_{kj}a_{ki} − 4 = 0 and guarantees that if β ∈ B_{n}, then β(a_{ji})^{2} + β(a_{kj})^{2} + β(a_{ki})^{2} − β(a_{ji})β(a_{kj})β(a_{ki}) − 4 = 0. Then

Theorem 6.3. The ideal J_{nm} is invariant under the action of B_{n}/Z(B_{n}) for all m ∈ ℕ.

Proof. This is the same as for Theorem 3 since we need only deal with three of the indices i, j, k at a time. □

Fix n > 3, m ∈ ℕ. We will let L_{nm} denote the subring of R′_{n} generated by I_{nm} and all the τ_{m}(a_{ji}) for 1 ≤ i < j ≤ n. Then as in Theorem 3 there is an action of B_{n} on L_{nm}/I_{nm}.

We now consider Theorem 4. Let W_{nm} denote the set of all solutions to the equations in L_{nm}. Then certainly |W_{nm}| < ∞ and B_{n} acts on W_{nm}.

As in the n = 3 case we have the double cover T^{n−1} → V_{n2} corresponding to η, only here we have a number of them, one for each choice of vector ϵ as in Lemma 6.1, satisfying (6.2): η_{n} = η_{n}(ϵ) : T^{n−1} → V_{n2} given specifically as

η_{n}(θ_{1}, θ_{2}, …, θ_{n−1})(a_{ni}) = 2cos(θ_{i}), for 1 ≤ i < n;
η_{n}(θ_{1}, θ_{2}, …, θ_{n−1})(a_{ij}) = 2cos(θ_{i} + ϵ_{ij}θ_{j}) for 1 ≤ j < i < n.  (6.5)

Thus, for n = 4, for example, we have

U_{4} = ( \begin{matrix}2 & 2cos(θ_{1} + ϵ_{12}θ_{2}) & 2cos(θ_{1} + ϵ_{13}θ_{3}) & 2cos(θ_{1}) \\ 2cos(θ_{1} + ϵ_{12}θ_{2}) & 2 & 2cos(θ_{2} + ϵ_{23}θ_{3}) & 2cos(θ_{2}) \\ 2cos(θ_{1} + ϵ_{13}θ_{3}) & 2cos(θ_{2} + ϵ_{23}θ_{3}) & 2 & 2cos(θ_{3}) \\ 2cos(θ_{1}) & 2cos(θ_{2}) & 2cos(θ_{3}) & 2 \end{matrix} ) .

As in the n = 3 case it is easy to see that \frac{1}{2}U_{n} is a Gram matrix: take v_{n} = (1, 0) ∈ ℝ^{2} and for i < n let v_{i} = (cos(θ_{i}), ϵ_{in}sin(θ_{i})). Then \frac{1}{2}U_{n} is the Gram matrix of the vectors v_{1}, v_{2}, …, v_{n}. Note that U_{n} has rank r if and only if dim(Span(v_{1}, …, v_{n})) = r by [HJ, 7.2.10 (d)].

For a fixed ϵ the action of B_{n} on V_{n2} lifts to the following action on T^{n−1}, which we give as n × n-matrices:

σ_{1} ↦ m_{1} = ( \begin{matrix}2 & ϵ_{12} & 0 & 0 & … & π \\ −ϵ_{12} & 0 & 0 & 0 & … & 0 \\ 0 & 0 & 1 & 0 & … & 0 \\ ⋮ & ⋮ & ⋮ & ... & … & ⋮ \\ 0 & 0 & 0 & … & 1 & 0 \\ 0 & 0 & 0 & … & 0 & 1 \end{matrix} ) ,

σ_{2} ↦ m_{2} = ( \begin{matrix}1 & 0 & 0 & 0 & … & 0 \\ 0 & 2 & ϵ_{23} & 0 & … & π \\ 0 & −ϵ_{23} & 0 & 0 & … & 0 \\ ⋮ & ⋮ & ⋮ & ... & … & ⋮ \\ 0 & 0 & 0 & … & 1 & 0 \\ 0 & 0 & 0 & … & 0 & 1 \end{matrix} ) , …,

σ_{n−2} ↦ m_{n−2} = ( \begin{matrix}1 & 0 & 0 & … & 0 & 0 \\ 0 & 1 & 0 & … & 0 & 0 \\ ⋮ & ⋮ & ... & … & ⋮ & ⋮ \\ 0 & … & 0 & 2 & ϵ_{n−2,n−1} & π \\ 0 & … & 0 & −ϵ_{n−2,n−1} & 0 & 0 \\ 0 & 0 & 0 & … & 0 & 1 \end{matrix} ) ,

σ_{n−1} ↦ m_{n−1} = ( \begin{matrix}1 & 0 & 0 & … & ϵ_{1n−1} & 0 \\ 0 & 1 & 0 & … & ϵ_{2n−1} & 0 \\ ⋮ & ⋮ & ... & … & ⋮ & ⋮ \\ 0 & 0 & … & 1 & ϵ_{n−2n−1} & 0 \\ 0 & 0 & 0 & … & 1 & π \\ 0 & 0 & 0 & … & 0 & 1 \end{matrix} ) .

Here we are acting on the coordinates (θ_{1}, θ_{2}, …, θ_{n−1}, 1)^{T} and the matrices belong to the group GL(n, ℝ/(2π)). Let us denote the group that they generate by Ψ_{n} = Ψ_{n}(ϵ). As in the n = 3 case the matrices m_{i} are only defined up to a common sign.

As in the case n = 3 if, for some m ∈ ℕ, all the θ_{i} have the form 2πk_{i}/m, then we obtain a finite set of vectors R_{nm} which is permuted by the B_{n} action. We have η_{n}(ϵ)(R_{nm}) = W_{nm}, the proof of this fact being the same as in the n = 3 case (Lemma 3.8). Let M_{n0} = M_{n0}(ϵ) denote the subgroup of P GL(n−1, ℤ) generated by the principal (n−1)×(n−1) submatrices of the above matrices. Since π is transcendental it is easy to see that the isomorphism class of M_{n0} is independent of whether we think of the matrices as belonging to GL(n−1, ℝ/(2π)) or as in GL(n − 1, ℤ).

Let M_{nm} = M_{nm}(ϵ) denote the image of M_{n0} in P GL(n − 1, ℤ_{m}). Then the action of B_{n}/Z(B_{n}) on W_{nm} is isomorphic to M_{nm} and the kernel is the kernel of the map B_{n}/Z(B_{n}) → M_{n0} → M_{nm} ⊂ P GL(n − 1, ℤ_{m}).

For n even or odd (respectively) we define symmetric and antisymmetric (respectively) forms Q_{n} = Q_{n}(ϵ) (given as (n − 1) × (n − 1)-matrices):

Q_{n} = ( \begin{matrix}1 & ϵ_{12} & −ϵ_{13} & ϵ_{14} & … \\ ϵ_{12} & 1 & ϵ_{23} & −ϵ_{24} & … \\ −ϵ_{13} & ϵ_{23} & 1 & ϵ_{34} & … \\ ϵ_{14} & −ϵ_{24} & ϵ_{34} & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) , Q_{n} = ( \begin{matrix}0 & ϵ_{12} & −ϵ_{13} & ϵ_{14} & … \\ −ϵ_{12} & 0 & ϵ_{23} & −ϵ_{24} & … \\ ϵ_{13} & −ϵ_{23} & 0 & ϵ_{34} & … \\ −ϵ_{14} & ϵ_{24} & −ϵ_{34} & 0 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

Lemma 6.4. For all n ≥ 3 and all ϵ the forms Q_{n} are fixed by the action of M_{n0}. Further, we have rank(Q_{2r}) = 1 and rank(Q_{2r+1}) = 2r.

Proof. For n = 2r + 1 the matrix Q_{n} is antisymmetric and all the off-diagonal entries are ±1. We thus need only prove:

Lemma 6.5. Let C be a 2r × 2r antisymmetric matrix over ℤ, all of whose off-diagonal entries are odd. Then det(C) is odd; in particular C is non-degenerate over ℝ or S^{1}.

Proof. A general 2r×2r antisymmetric matrix C = (c_{ij}) is well-known to have a determinant which is a square: det(C) = P f_{2r}(c_{ij})^{2}, where P f_{2r}(c_{ij}) = P f_{2r}(C) is a homogeneous integral polynomial in the entries c_{ij} of C which is called the Pfaffian [A]. We thus need to show that P f_{2r}(C) is odd. From [DW] we found the following recursive formula for P f_{2r}(C):

P f_{2}(( \begin{matrix}0 & c_{12} \\ −c_{12} & 0 \end{matrix} )) = c_{12}; P f_{2r}(C) = ∑_{j=2}^{2r}(−1)^{j}c_{1j}P f_{2r−2}(C^{(1j)}) for 2r ≥ 4.

Here C^{(1j)} is the matrix obtained from C by deleting the rows and columns labeled 1 and j.

We now proceed by induction, the case n = 2r = 2 being given above. We note that in the above recursive formula there are an odd number of summands being added together, and by induction each such summand (−1)^{j}c_{1j}P f_{n−2}(C^{(1j)}) is odd. The result follows. □

This concludes the proof of the n = 2r + 1 case of Lemma 6.4.

Now if n = 2r, then we need only show that every 2×2 minor of Q_{n} is zero. Without loss we may assume that this minor involves the rows 1 and 2 and the columns i and j where i < j. There are various cases to be considered; for example if i, j > 2, then the minor is

(−1)^{i+j+1}(e_{1i}e_{2j} − e_{1j}e_{2i}) = (−1)^{i+j+1}e_{ij}(e_{1i}e_{ij}e_{2j} − e_{1j}e_{2i}e_{ij})
= (−1)^{i+j+1}e_{ij}(−e_{1j}e_{2j} + e_{1j}e_{2j}) = 0.

The other cases are similar.

The proof that these forms are invariant for the given matrix group is straighforward. □

The analogue of Theorem 5 is

Theorem 6.6. Fix n ∈ ℕ. There is a subgroup H of finite index in F_{n} such that H contains no simple elements. We may also find such an H which is normal and invariant under the action of B_{n}.

Proof. This is by induction on n ≥ 3, the case n = 3 having been considered in §5. Assume that the result is true for n ≥ 3 and let H = H_{n} be the subgroup containing no simple elements. Let π : F_{n+1} → F_{n} be the epimorphism determined by

π(x_{i}) = x_{i}, for i ≤ n; π(x_{n+1}) = id.

Let H′ = π^{−1}(H). Let u ∈ F_{n+1} be a simple element. If π(u) ≠ id, then π(u) is simple in F_{n} and so π(u) ∉ H. Thus u ∉ H′.

Now we have to consider the situation where π(u) = id. It follows that u is in the normal subgroup << x_{n+1} >> of F_{n+1} generated by x_{n+1}. Now any simple closed curve surrounds m of the punctures of D_{n+1} and since u ∈<< x_{n+1} >> we must have m = 1 for u. Thus u is a conjugate of x_{n+1}. Now by the fact that F_{n+1} is residually finite there is a finite index subgroup E which does not contain x_{n+1}^{±1}. Then H′ ∩ core(E) contains no simple elements and has finite index. The rest of the argument follows as in the n = 3 case. □

We remark that we really didn’t need to use Theorem 5 to prove this result; one can easily see that it is true for n = 2 and then one can use the above argument to prove Theorem 6.6 inductively.

Define an n × n matrix M to be Anosov if it has a real eigenvalue λ > with eigenvector v = (v_{1}, …, v_{n}) such that v_{i}/v_{j} is irrational for all 1 ≤ i ≠ j ≤ n. We will say that α ∈ B_{n}/Z(B_{n}) is Anosov if the image of α in Ψ_{n} is Anosov.

For Theorem 7 we have the following generalisation:

Theorem 6.7. The set W_{nm} is dense in V_{n2} and each point of W_{nm} is a periodic point for the action of B_{n}.

If α ∈ B_{n}/Z(B_{n}) is Anosov, then the action of α on V_{n2} is chaotic.

Proof. The set of points ⋃_{m≥1} R_{nm} is certainly dense in T^{n−1} and so the density part of the result follows from the fact η_{n}(ϵ)(R_{nm}) = W_{nm} that we have already observed.

The finite set R_{nm} is certainly permuted by B_{n}; it follows that W_{nm} is also. This means that each point of W_{nm} is a periodic point for the action of B_{n}.

Now assume that α ∈ B_{n}/Z(B_{n}) is Anosov with eigenvalue λ > 1 and λ-eigenvector v as in the definition of Anosov. Note that the entries of v can be chosen to be algebraic integers. We see from the fact that v_{i}/v_{j}, i ≠ j, are irrational that the line through the origin in T^{n−1} that v determines is dense in T^{n−1}. The rest of the argument is as in Lemma 5.1. □

For Theorem 8 we have already seen that each Ψ_{n} fixes a certain form Q_{n}. Thus for u = (u_{1}, …, u_{n−1}), v = (v_{1}, …, v_{n−1}) ∈ V_{n2} the quantity uQ_{n}v^{T} is fixed by the action of B_{n}. Just as we obtained the function ω^{(2)} from the antisymmetric form represented by the

matrix ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) we may obtain a function ω_{n}^{(2)} : V_{n2} × V_{n2} → ℝ using the matrix Q_{n}.

This gives:

Theorem 6.8. There is a function ω_{n}^{(2)} : V_{n2} × V_{n2} → ℝ which satisfies:

(i) ω_{n}^{(2)} is B_{n}-invariant; (ii) ω_{n}^{(2)} is symmetric if n is even and antisymmetric if n is odd; (iii) if n is odd, then the only degenerate points of ω_{n}^{(2)} are the singular points of V_{n2} (the points all of whose coordinates are ±2). □

We may also define abelian functions [La2] on the V_{n2} as long as they are even.

§7 The group Ψ_{n}(ϵ).

Let n be odd. We have seen in the last section (Lemma 6.4) that Ψ_{n}(ϵ) is a subgroup of the symplectic group Sp(n − 1, ℤ), since the antisymmetric form that Ψ_{n}(ϵ) fixes is nondegenerate. Recall [A] that the symplectic transvection associated to the element e of the symplectic space (V, ·) is the linear map T (e) on V defined by

T (e)(v) = v + (e · v)e.

Usually e will be a primitive element relative to a certain basis for V.

We will also need to know that if e, f ∈ V with e · f ∈ ℤ, then (i) e · f = 0 if and only if T (e)T (f) = T (f)T (e); (ii) e · f = ±1 if and only if T (e)T (f)T (e) = T (f)T (e)T (f); (iii) |e · f| > 1 if and only if T (e) and T (f) generate a free subgroup of Sp(V).

We will let m′_{i} denote the matrix obtained from m_{i} by deleting the last row and column. One can check that each of the m′_{i} is a symplectic transvection T (e). Let (V, ·) denote the corresponding symplectic space and let m′_{i} = T (e_{i}) for all i = 1, …, n − 2 define elements e_{i} ∈ V. Here e_{i} is defined up to a sign ±1. Since σ_{i} ↦ m′_{i} defines a representation B_{n} → Sp(n − 1, ℤ) (i), (ii) and (iii) above show that we can choose the above signs so that

e_{i} · e_{i+1} = 1, for i = 1, …, n − 2; e_{i} · e_{j} = 0 for |i − j| > 1.

It follows that e_{1}, …, e_{n−1} is a basis for V, which we will denote by B = B_{n}. We will let ℤB denote the ℤ-submodule of V generated by B. Then Sp(n − 1, ℤ) acts on ℤB.

This shows in particular that the isomorphism class of Ψ_{n}(ϵ) is independent of ϵ (in fact all such are conjugate in Sp(n − 1, ℤ)). It follows from results on the mapping class group M_{g,0} [Hu5] that Sp(n − 1, ℤ) is generated by < T (e_{1}), …, T (e_{n−1}) > together with T (e_{1} + e_{3}). The reference [Hu5] concerns a minimal set of Dehn twists generating M_{g,0}, however there is a standard epimorphism M_{g,0} → Sp(n − 1, ℤ) where the image of a Dehn twist is a symplectic transvection [MKS]. We now prove

Theorem 7.1. The group Ψ_{n} =< T (e_{1}), …, T (e_{n−1}) > contains the principal congruence subgroup ker(Sp(n − 1, ℤ) → Sp(n − 1, ℤ_{2})) and is of finite index

|Sp(n − 1, ℤ_{2})|/n!

in Sp(n − 1, ℤ).

Proof. Let T_{i} = T (e_{i}) for i = 1, …, n. The image of Ψ_{n} in Sp(n − 1, ℤ_{2}) is the symmetric group. This follows from the fact that the T_{i} satisfy the standard braid relations and that their images in Sp(n − 1, ℤ_{2}) have order 2. Thus if we can prove the first statement of Theorem 7.1, then we are done.

Now by results of Mennicke [Me] the level 2 principal congruence subgroup of Sp(n−1, ℤ) is generated by the squares of all the symplectic transvections. Thus we need to show that if e ∈ V is a primitive element (i.e. the coordintates of e relative to the basis B are coprime), then T^{2}(e) ∈ Ψ_{n}.

We do this by showing that there is α ∈ Ψ_{n} such that α(e) = f, where T^{2}(f) ∈ Ψ_{n}; for then we would have T^{2}(e) = αT^{2}(f)α^{−1} ∈ Ψ as required. The method that we use to accomplish this is similar to that used in [BH], the difference being that in those papers the coefficient ring was a field F (the cases |F | = 2 and |F | > 2 being very different).

We first note that if i < j and j − i is even, then

(T_{i}T_{i+1} … T_{j})^{j−i+2} = T^{2}(e_{i} + e_{i+2} + e_{i+4} + ⋯ + e_{j}).  (7.1)

To prove this let P = T_{i}T_{i+1} … T_{j} and note that P (e_{k}) = −e_{k+1} for k = i, i+1, …, j −1, that P (e_{j}) = e_{i} + e_{i+1} + ⋯ + e_{j} and that P (e_{j+1}) = e_{i} + e_{i+1} + ⋯ + e_{j+1} with P (e_{k}) = e_{k} for all other k. Now iterate P.

As in [BH] we associate to each element e ∈ ℤB a labeled graph G(e): write e = ∑_{i=1}^{n−1} λ_{i}e_{i}; then G(e) has vertices v_{i} for each λ_{i} ≠ 0 and each vertex is labeled by λ_{i}. We also have an edge between v_{i} and v_{i+1} if λ_{i}λ_{i+1} ≠ 0. Properties of the graph G(e) will be attributed to the elements e of ℤB, so that for example we will say that e ∈ ℤB is connected if G(e) is connected; that e ∈ ℤB is discrete if each connected component of G(e) is a vertex etc.

Lemma 7.2. Each e ∈ ℤB is in the Sp(n − 1, ℤ)-orbit of a discrete element.

Proof. The proof will be by induction on n ≥ 3. If n = 3, then < T_{1}, T_{2} >= SL(2, ℤ) and any element e = (λ_{1}, λ_{2}) ∈ ℤ^{2} is in the orbit of (λ_{3}, 0) where λ_{3} = gcd(λ_{1}, λ_{2}). This does the n = 3 case.

Now suppose that n > 3 is odd and let e = (c_{1}, …, c_{n−1}) ∈ ℤB. Suppose that c_{1}c_{2} ≠ 0.

Then the action of T_{1} on the coordinates (c_{1}, c_{2}) is given by the matrix M_{1} = ( \begin{matrix}1, & 1 \\ 0 & 1 \end{matrix} )

and the action of T^{2}(e_{2} + e_{4} + ⋯ + e_{n−1}) (which is in Ψ_{n} by (7.1)) is given by the matrix

M_{2} = ( \begin{matrix}1 & 0 \\ −2 & 1 \end{matrix} ). Now one checks [MA] that < M_{1}, M_{2} >⊂ SL(2, ℤ) is of index 3 and that

the < M_{1}, M_{2} >-orbit of each vector (c_{1}, c_{2}) ∈ ℤ^{2} contains the vector (g, 0) or (0, g) where g = gcd(c_{1}, c_{2}). Thus we may (up to an action of < T_{1}, …, T_{n−1} >) assume that c_{1}c_{2} = 0. If c_{2} = 0, then induction applied to (c_{3}, …, c_{n}) gives the result. Thus we may assume that c_{1} = 0, c_{2} ≠ 0. Similarly we may assume that c_{n−1} = 0. The presence of these two zeros allows us to use the inductive hypothesis on the coordinates c_{2}, …, c_{n−2} and so we are done. □

Lemma 7.3. Each primitive e ∈ ℤB is in the Sp(n − 1, ℤ)-orbit of a discrete element all of whose coordinates are 1.

Proof. By Lemma 7.2 we may assume that e is discrete. We first show how we can ‘move labels of e around’. Suppose that e = (…, 0, c_{i}, 0, 0, …). Then T_{i}^{−1}T_{i+1}^{−1}(e) = (…, 0, 0, c_{i}, 0, …), where c_{i} is now in the i + 1 position. We also have

T_{n−2}^{−1}T_{n−1}^{−1}(…, 0, 0, c_{n−2}, 0) = (…, 0, 0, c_{n−2}) and T_{i}T_{i+1}(e) = (…, 0, 0, −c_{i}, 0, …)

etc. If G(e) has a single connected component, then we are done since the fact that e is primitive says that the non-zero coordinate is ±1. By the above we are also done if all the components have labels ±1.

So assume that G(e) has more than one component, that all of the labels are positive and that some of the labels are not 1. Then by the above we may assume that e = (…, 0, c_{i}, 0, c_{i+2}, 0, 0, …), where i may be 1, i + 2 ≤ n − 3, and c_{i}, c_{i+2} > 0 and also that c_{i+2} > c_{i} (the other case is similar). Then we have

(…, 0, c_{i}, 0, c_{i+2}, 0, 0, …) \underrightarrow{T_{i+3}^{−1}} (…, 0, c_{i}, 0, c_{i+2}, c_{i+2}, 0, …)
\underrightarrow{T^{2}(e_{i} + e_{i+1})} (…, 0, c_{i} − 2c_{i+2}, 0, −c_{i+2}, c_{i+2}, 0, …)
\underrightarrow{T_{i+3}T_{i+2}^{2}} (…, 0, c_{i} − 2c_{i+2}, 0, c_{i+2}, 0, 0, …).

We note that |c_{i} − 2c_{i+2}| < c_{i} and so continuing we may eventually reduce all labels to 1 or 0. □

From Lemma 7.3 and the way we have of moving labels around we see that any primitive e ∈ ℤB is in the orbit of f = e_{1} + e_{3} + e_{5} + ⋯ + e_{2r−1} for some r ≥ 1. If r = 1, then T (f) ∈ Ψ_{n}, while if r > 1, then T^{2}(f) ∈ Ψ_{n} (by (7.1)) as required for the proof of Theorem 7.1. □

A proof of the first part of this result also follows from [Ja].

We now consider the situation where n is even, this being more involved than the n odd case. In this case we also see that different choices of ϵ give conjugate groups. Thus we will from now on choose ϵ_{ij} = −1 for all i, j. Define the following matrix:

W_{n} = ( \begin{matrix}−1 & 1 & −1 & 1 & −1 & … & −1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & … & 0 & 1 & −1 \\ 1 & 0 & 0 & 0 & 0 & … & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 & … & 0 & 0 & −1 \\ 0 & 0 & 1 & 0 & 0 & … & 0 & 0 & 1 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & 0 & … & 1 & 0 & 1 \end{matrix} ) .

Let W_{n} is non-singular and if we let p_{i} = w^{−1}m′_{i}w, then the p_{i} all have their last row equal to (0, 0, …, 0, −1). We also have that W_{n} is the matrix which diagonalises Q_{n}, so that W_{n}^{−1}Q_{n}W_{n} is the diagonal matrix with diagonal entries 0, 0, …, 0, n. Here

p_{1} = ( \begin{matrix}1 & 0 & 0 & … & 0 & 0 \\ 0 & 1 & 0 & … & 0 & 0 \\ 0 & 0 & 1 & … & 0 & 0 \\ ⋮ & ⋮ & ⋮ & … & 0 & 0 \\ −1 & 1 & −1 & … & 1 & 2 \\ 0 & 0 & 0 & … & 0 & 0 \end{matrix} ) , p_{2} = ( \begin{matrix}0 & 0 & 0 & … & 1 & −2 \\ 0 & 1 & 0 & … & 0 & 0 \\ 0 & 0 & 1 & … & 0 & 0 \\ ⋮ & ⋮ & ⋮ & … & ⋮ & ⋮ \\ −1 & 0 & 0 & … & 2 & −2 \\ 0 & 0 & 0 & … & 0 & 1 \end{matrix} ) ,

p_{3} = ( \begin{matrix}2 & −1 & 0 & … & 0 & 2 \\ 1 & 0 & 0 & … & 0 & 2 \\ 0 & 0 & 1 & … & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ \\ 0 & 0 & 0 & … & 1 & 0 \\ 0 & 0 & 0 & … & 0 & 1 \end{matrix} ) , p_{4} = ( \begin{matrix}1 & 0 & 0 & 0 & … & 0 \\ 0 & 2 & −1 & 0 & … & −2 \\ 0 & 1 & 0 & 0 & … & −2 \\ 0 & 0 & 0 & 1 & … & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & 0 & 0 & … & 1 \end{matrix} ) ,

…

p_{n−2} = ( \begin{matrix}1 & … & 0 & 0 & 0 & 0 \\ ⋮ & ... & ⋮ & ⋮ & ⋮ & ⋮ \\ 0 & … & 2 & −1 & 0 & −2 \\ 0 & … & 1 & 0 & 0 & −2 \\ 0 & … & 0 & 0 & 1 & 0 \\ 0 & … & 0 & 0 & 0 & 1 \end{matrix} ) , p_{n−1} = ( \begin{matrix}1 & … & 0 & −1 & 0 & −1 \\ 0 & ... & 0 & −1 & 0 & −1 \\ 0 & … & ⋮ & ⋮ & ⋮ & ⋮ \\ 0 & … & 0 & 1 & 0 & 0 \\ 0 & … & 0 & −1 & 1 & −1 \\ 0 & … & 0 & 0 & 0 & 1 \end{matrix} ) .

One can now check that the p_{i} fix the singular antisymmetric form represented by the (n − 1) × (n − 1) matrix

A_{n} = ( \begin{matrix}0 & 1 & 1 & … & 1 & −1 & 0 \\ −1 & 0 & 1 & … & 1 & −1 & 0 \\ −1 & −1 & 0 & … & 1 & −1 & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ \\ −1 & −1 & −1 & … & 0 & −1 & 0 \\ 1 & 1 & 1 & … & 1 & 0 & 0 \\ 0 & 0 & 0 & … & 0 & 0 & 0 \end{matrix} ) .

By this we mean that we have p_{i}A_{n}p_{i}^{T} = A_{n} for all i = 1, …, n − 1. Let A′_{n} denote the (n − 2) × (n − 2) matrix obtained from A_{n} be deleting the last row and column. Then by Lemma 6.6 we see that A′_{n} is non-singular and so that A_{n} has rank n − 2. Let (V, ·) denote the corresponding symplectic space. Then, as in the n odd case, we see that each p_{i} is a symplectic transvection. So we put p_{i} = T (e_{i}) where e_{i} · e_{i+1} = 1 for all i = 1, …, n − 1 and e_{i} · e_{j} = 0 for |i − j| > 1. It follows that e_{1}, …, e_{n−2} are linearly independent and one checks that e_{n−1} is independent of these. Thus V is spanned by B = {e_{1}, …, e_{n−1}} and the symplectic form relative to this basis is given by the matrix

( \begin{matrix}0 & 1 & 0 & 0 & … & 0 \\ −1 & 0 & 1 & 0 & … & 0 \\ 0 & −1 & 0 & 1 & … & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & 0 & … & −1 & 0 \end{matrix} ) .

Recall that the radical [A] of (V, ·) is rad(V) = {v ∈ V |v · w = 0 for all w ∈ V }. Since rank(A_{n}) = n − 1 we see that dim(rad(V)) = −1 and so that rad(V) is the span of e_{1} + e_{3} + e_{5} + ⋯ + e_{n−1}. Each symplectic transvection T (f), f ∈ V fixes rad(V) and so any group generated by such symplectic transvections fixes rad(V). In particular Ψ_{n} fixes e_{1} + e_{3} + e_{5} + ⋯ + e_{n−1} and so belongs to the stabiliser 𝔖_{n} = Stab(Sp(V, ·), e_{1} + e_{3} + e_{5} + ⋯ + e_{n−1}).

Now [Ja] shows that Ψ_{n} contains all the squares of all transvections and so gives:

Theorem 7.4. For n ≥ 4 even, Ψ_{n} has finite index in 𝔖_{n}. □

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