INTERSECTION-NUMBER OPERATORS FOR CURVES

ON DISCS AND CHEBYSHEV POLYNOMIALS

Stephen P. Humphries

Dedicated to Joan Birman with gratitude and respect.

Abstract. We prove the following result and show a connection with geometric intersection-number functions of curves on punctured discs and algebraic intersection functions for curves on surfaces:

Let R be an associative ring with identity and fix r ∈ Z(R), where Z(R) is the centre of R. Define polynomials p_{n} = p_{n}(r) recursively by p_{0} = −2, p_{1} = r, p_{n} = −(rp_{n−1} + p_{n−2}). Let σ : R → R be a ring homomorphism and assume that σ(r) = r. Define operators 𝔄_{n} = 𝔄_{n}(σ, r), n ≥ 0, by 𝔄_{0} = σ − 1 and for n > 0, we let 𝔄_{n} = σ^{2} + p_{n}σ + 1. Define operators 𝔅_{n} = 𝔅_{n}(σ, r), n ≥ 0 by 𝔅_{n} = 𝔄_{0}𝔄_{1} … 𝔄_{n}. Suppose that u, v ∈ R and that 𝔅_{n}(u) = 𝔅_{m}(v) = 0 for n, m ≥ 0. Then

𝔅_{n+m}(uv) = 0.

§1. Introduction.

In this paper we prove the following result and indicate connections with Chebyshev polynomials, symmetric polynomials, Dickson polynomials, the theory of geometric intersection-number functions on punctured discs and algebraic intersection functions for curves on surfaces. This result thus gives a common method for defining both geometric and algebraic intersection numbers.

Theorem 1.1. Let R be an associative ring with identity and fix r ∈ Z(R), where Z(R) is the centre of R. Define polynomials p_{n} = p_{n}(r) recursively by

p_{0} = −2, p_{1} = r, p_{n} = −(rp_{n−1} + p_{n−2}).

Let σ : R → R be a ring homomorphism and assume that σ(r) = r. Define operators 𝔄_{n} = 𝔄_{n}(σ, r), n ≥ 0, by 𝔄_{0} = σ − 1 and for n > 0, we let 𝔄_{n} = σ^{2} + p_{n}σ + 1. Now define operators 𝔅_{n} = 𝔅_{n}(σ, r), n ≥ 0, by 𝔅_{n} = 𝔄_{0}𝔄_{1} … 𝔄_{n}.

Suppose that u, v ∈ R and that 𝔅_{n}(u) = 𝔅_{m}(v) = 0 for n, m ≥ 0. Then 𝔅_{n+m}(uv) = 0.

The polynomials p_{n} are polynomials in the variable p_{1} = r that are related to Chebyshev polynomials T (n) and Dickson polynomials. Recall that the Chebyshev polynomials are defined by T_{0}(x) = 1, T_{1}(x) = x and T_{n}(x) = 2xT_{n−1}(x) − T_{n−2}(x) for n ≥ 2 [Ri, p. 35]. Then comparing the recurrence for p_{n}(x) with this recurrence we see that

p_{n}(x) = (−1)^{n+1}2T_{n}(x/2).  (1.1)

These polynomials are also related to the Dickson polynomials D_{n}(x, a) [LMT] by p_{n}(x) = (−1)^{n+1}D_{n}(x, 1).

One of the results that we use to prove Theorem 1.1 is

Typeset by AMS-TEX

Theorem 1.2. Let n ≥ m ≥ 0. Suppose that α_{n}, α_{m} are roots of the polynomials determined by 𝔄_{n} and 𝔄_{m}, respectively, so that α_{n}^{2}+p_{n}α_{n}+1 = 0 and α_{m}^{2}+p_{m}α_{m}+ 1 = 0. Then α_{n}α_{m} is a root of either 𝔄_{n+m} or 𝔄_{n−m}.

Let D_{n} denote the disc with n punctures π_{1}, …, π_{n}. Let a_{1}, …, a_{n−1} be a set of arcs in D_{n} having disjoint interiors and such that a_{i} joins the punctures π_{i} and π_{i+1}. See Figure 1 in §5. Let C^{∞} = C_{n}^{∞} denote the set of isotopy classes of positively-oriented simple closed curves on D_{n}.

Then the braid group B_{n} acts as (isotopy classes of) diffeomorphisms of D_{n} [Bi, Ch.1]. The group B_{n} has standard generators σ_{1}, …, σ_{n−1} and presentation

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.

Here the generator σ_{i} acts as a half-twist on D_{n} and has a representative diffeomorphism which is supported in a tubular neighbourhood of the arc a_{i}. For 1 ≤ i < j ≤ n let γ_{ij} be a simple closed curve isotopic to the boundary of a tubular neighbourhood of a_{i} ∪a_{i+1} ∪⋯∪a_{j−1}. Given any c, d ∈ C^{∞} we let ι(c, d) denote the geometric intersection-number of c and d. This is the minimum number of points of c′ ∩ d′, where c′ and d′ are any simple closed curves isotopic to c and d. Note that ι(c, d) is always even since D_{n} is planar.

Note that in the context of Theorem 1.1, if m > n ≥ 0 and 𝔅_{n}(u) = 0, then 𝔅_{m}(u) = 0. The first part of the following result comes from [H1]:

Theorem 1.3. For n ≥ 2 and a commutative ring R let

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

be the polynomial ring in commuting indeterminates a_{ij}, 1 ≤ i ≠ j ≤ n. Then there is an action of B_{n} on R_{n} whose kernel is the centre of B_{n}.

Let r = a_{12}a_{21} + 2 ∈ R_{n} and let σ = σ_{1} ∈ Aut(R_{n}). Then σ_{1}(r) = r. Further, there is a function ϕ : C^{∞} → R_{n} such that for all c ∈ C^{∞} there is N = N(c) ≥ 0 such that 𝔅_{n}(σ, r)(ϕ(c)) = 0. Let Ω(c) be the minimal such N(c). Then we have

Ω(c) = ι(c, γ_{12})/2.

This result thus gives a method for calculating geometric intersection-number functions that can be compared to existing such algorithms of Reinhart [R], Zieschang [Z1, Z2], Chillingworth [C1, C2], Birman and Series [BS], Cohen and Lustig [CL] and Tan [T].

The representation of B_{n} in Aut(R_{n}) will be described later, but should be thought of in the following way. Let F_{n} =< x_{1}, …, x_{n} > denote the free group of rank n, which we identify with the fundamental group of D_{n}. The Magnus expansion M of F_{n} [Ma,MKS] is defined as follows: Let P_{n} be the algebra of formal power series in non-commutative variables X_{1}, …, X_{n} over ℂ. Then M is the homomorphism M : F_{n} → P given by

M(x_{i}) = 1 + X_{i}, M(x_{i}^{−1}) = 1 − X_{i} + X_{i}^{2} − X_{i}^{3} + ….

Then M is injective, has connections with Fox’s free differential calculus and is used to define interesting representations of the braid groups [Bi]. We obtain our representation of B_{n} in Aut(F_{n}) by looking at the situation where in P_{n} we have the relations X_{i}^{2} = 0 for all i = 1, …, n. This is accomplished very concretely by representing the free group F_{n} using transvections. This is all explained in more detail in §5. A quotient of this algebra was used by Milnor [Mi1,Mi2] to study links.

We prove Theorem 1.1 in §4 after some preliminaries in §2 and §3 (where we prove Theorem 1.2). In §5 we explain the connection between our polynomials p_{n} and interscetion-number functions for curves on planar surfaces and in §6 we give a normal form for words representing simple closed curves on such surfaces. In §7 we study the combinatorial properties of a certain ‘trace’ of such words. In §8 we prove Theorem 1.3 and in §9 we show that in the context of §2 and §3 there is an associated infinite monotone family of commuting projections [AG, §33] which gives rise to an infinite family of orthogonal idempotents (Theorem 9.3). In §10 we indicate how Theorem 1.1 applies to algebraic intersection functions for curves on surfaces.

§2. Basic Results.

We note from the above definition of 𝔅_{n} = 𝔅_{n}(σ, r) that 𝔅_{n} has degree 2n + 1 when thought of as a polynomial in σ (as we often will). Define

𝔅_{n}(σ) = ∑_{i=0}^{2n+1} q_{n i}σ^{i},

where the q_{n i} are polynomials in p_{1}. Then since 𝔅_{n+1} = 𝔅_{n}(σ^{2} + p_{n+1}σ + 1) we see that the q_{n i} satisfy the following recursion:

q_{n+1 i} = q_{n i−2} + p_{n+1}q_{n i−1} + q_{n i}.  (2.1)

We also note that q_{n,0} = −1 and that q_{n,2n+1} = 1 for n ≥ 0.

Now assume that for u ∈ R we have 𝔅_{n}(u) = 0. Then since q_{n,2n+1} = 1 we have

σ^{2n+1}(u) = − ∑_{i=0}^{2n} q_{n i}σ^{i}(u)  (2.2)

and so for j ≥ 2n + 1 we can write σ^{j}(u) as a sum of multiples of σ^{i}(u) for i = 1, …, 2n. Specifically we will define

σ^{j}(u) = ∑_{i=0}^{2n} r_{n,j,i}σ^{i}(u)

where each r_{n,j,i} is a polynomial in p_{1}. Then the r_{n,j,i} satisfy the recursion given in the following result.

Lemma 2.1. We have (i) r_{n,j,i} = δ_{ij} if 0 ≤ j ≤ 2n; (ii) r_{n,2n+1,i} = −q_{n i}; (iii) r_{n,j+1,i} = −q_{n i}r_{n,j,2n} + r_{n,j,i−1}; and (iv) if 0 < i ≤ 2n + 1 then the degree of q_{ni} is (2n − i + 1)i/2.

Proof. (i) is clear, while (ii) follows from (2.2). For (iii) we note that if σ^{j}(u) = ∑_{i=0}^{2n} r_{n,j,i}σ^{i}(u), then (2.2) also gives:

σ^{j+1}(u) = ∑_{i=0}^{2n} r_{n,j,i}σ^{i+1}(u) = ( ∑_{i=0}^{2n−1} r_{n,j,i}σ^{i+1}(u)) + σ^{2n+1}(u)

= ∑_{i=0}^{2n−1} r_{n,j,i}σ^{i+1}(u) − ∑_{i=0}^{2n} q_{n i}σ^{i}(u).

The recursion (iii) follows.

The formula for the degree of q_{ni} follows by induction from the recursion (2.1). The initial cases are easily checked and then one notes that deg(p_{n+1}) = n + 1 and that since 0 ≤ i ≤ 2n + 1 one finds that (using (2.1))

(2n − i + 3)i/2 = deg(p_{n+1}q_{ni−1}) ≥ max(deg(q_{n i−2}), deg(q_{ni}))

with equality only occuring when i = 0. □

The following result indicates how we will prove Theorem 1.1.

Proposition 2.2. Theorem 1.1 will follow if we have

∑_{i=0}^{2(n+m)+1} q_{n+m i}r_{n,i,j}r_{m,i,k} = 0

for all n, m, j, k ≥ 0.

Proof. Suppose that u, v ∈ R and that 𝔅_{n}(u) = 𝔅_{m}(v) = 0. Then, since p_{1} is central in R, the r_{n,j,i} are also central and we have

𝔅_{n+m}(uv) = ∑_{i=0}^{2(n+m)+1} q_{n+m i}σ^{i}(uv)

= ∑_{i=0}^{2(n+m)+1} q_{n+m i}σ^{i}(u)σ^{i}(v)

= ∑_{i=0}^{2(n+m)+1} q_{n+m i} ∑_{a=0}^{2n} r_{n,i,a}σ^{a}(u) ∑_{b=0}^{2m} r_{m,i,b}σ^{b}(v)

= ∑_{a=0}^{2n} ∑_{b=0}^{2m} ( ∑_{i=0}^{2(n+m)+1} q_{n+m i}r_{n,i,a}r_{m,i,b}) σ^{a}(u)σ^{b}(v).

The result follows. □

Proposition 2.3. For n, y, t ≥ 0 we have

r_{n,2n+1+t,y} = ∑_{j=1}^{t+1} ∑_{s=0}^{t} ∑_{\begin{matrix}1≤i_{1},i_{2},...,i_{s}≤t \\ ∑_{k} i_{k}=t+1−j \end{matrix}}(−1)^{s+1}q_{n y+1−j} ∏_{k=1}^{s} q_{n 2n+1−i_{k}}.

Here we interpret the case s = 0 in this formula as r_{n,2n+1+0,y} = (−1)^{0+1}q_{n y+1−1}.

Proof. This will be by induction on t ≥ 0, the case t = 0 following from Lemma 2.1 (ii) and the rest following from the recursion given in Lemma 2.1 (iii), as follows. First note that

r_{n,2n+1+t,y−1} = ∑_{j=1}^{t+1} ∑_{s=0}^{t} ∑_{\begin{matrix}1≤i_{1},i_{2},...,i_{s}≤t \\ ∑_{k} i_{k}=t+1−j \end{matrix}}(−1)^{s+1}q_{n y−j} ∏_{k=1}^{s} q_{n 2n+1−i_{k}}

= ∑_{j=1}^{t+1} ∑_{s=0}^{t+1} ∑_{\begin{matrix}1≤i_{1},i_{2},...,i_{s}≤t \\ ∑_{k} i_{k}=t+1−j \end{matrix}}(−1)^{s+1}q_{n y−j} ∏_{k=1}^{s} q_{n 2n+1−i_{k}}.

We now consider the sum in Proposition 2.3 for t + 1. We split this sum up into the cases j = 1 and j > 1 as follows:

∑_{j=1}^{t+2} ∑_{s=0}^{t+1} ∑_{\begin{matrix}1≤i_{1},i_{2},...,i_{s}≤t+1 \\ ∑_{k} i_{k}=t+2−j \end{matrix}}(−1)^{s+1}q_{n y+1−j} ∏_{k=1}^{s} q_{n 2n+1−i_{k}}

= ∑_{s=0}^{t+1} ∑_{\begin{matrix}1≤i_{1},i_{2},...,i_{s}≤t+1 \\ ∑_{k} i_{k}=t+1 \end{matrix}}(−1)^{s+1}q_{n y} ∏_{k=1}^{s} q_{n 2n+1−i_{k}}

+ ∑_{j=2}^{t+2} ∑_{s=0}^{t+1} ∑_{\begin{matrix}1≤i_{1},i_{2},...,i_{s}≤t+1 \\ ∑_{k} i_{k}=t+2−j \end{matrix}}(−1)^{s+1}q_{n y+1−j} ∏_{k=1}^{s} q_{n 2n+1−i_{k}}

= ∑_{s=0}^{t+1} ∑_{\begin{matrix}1≤i_{1},i_{2},...,i_{s}≤t+1 \\ ∑_{k} i_{k}=t+1 \end{matrix}}(−1)^{s+1}q_{n y} ∏_{k=1}^{s} q_{n 2n+1−i_{k}}

+ ∑_{j=1}^{t+1} ∑_{s=0}^{t+1} ∑_{\begin{matrix}1≤i_{1},i_{2},...,i_{s}≤t+1 \\ ∑_{k} i_{k}=t+1−j \end{matrix}}(−1)^{s+1}q_{n y−j} ∏_{k=1}^{s} q_{n 2n+1−i_{k}}

= r_{n,2n+1+t,y−1} + ∑_{s=0}^{t+1} ∑_{\begin{matrix}1≤i_{1},i_{2},...,i_{s}≤t+1 \\ ∑_{k} i_{k}=t+1 \end{matrix}}(−1)^{s+1}q_{n y} ∏_{k=1}^{s} q_{n 2n+1−i_{k}}.

Now we also have

−q_{ny}r_{n,2n+1+t,2n} = −q_{ny} ∑_{j=1}^{t+1} ∑_{s=0}^{t} ∑_{\begin{matrix}1≤i_{1},i_{2},...,i_{s}≤t \\ ∑_{k} i_{k}=t+1−j \end{matrix}}(−1)^{s+1}q_{n 2n+1−j} ∏_{k=1}^{s} q_{n 2n+1−i_{k}}

= ∑_{j=1}^{t+1} ∑_{s=0}^{t} ∑_{\begin{matrix}1≤i_{1},i_{2},...,i_{s}≤t \\ ∑_{k} i_{k}=t+1−j \end{matrix}}(−1)^{s+2}q_{n y} (∏_{k=1}^{s} q_{n 2n+1−i_{k}}) q_{n 2n+1−j}

= ∑_{s=0}^{t+1} ∑_{\begin{matrix}1≤i_{1},i_{2},...,i_{s}≤t \\ ∑_{k} i_{k}=t+1 \end{matrix}}(−1)^{s+2}q_{n y} (∏_{k=1}^{s} q_{n 2n+1−i_{k}})

Putting this together with the previous calculation and the recursion from Lemma 2.1 (iii) gives the result. □

The following result indicates a connection with symmetric polynomials.

Proposition 2.4. For n, m ≥ 0 we have

q_{n m} = ∑_{k=0}^{m}(−1)^{m+k+1}\binom{n − k}{⌊\frac{m−k}{2} ⌋} ∑_{1≤i_{1}≤i_{2}≤···≤i_{k}≤n} p_{i_{1}}p_{i_{2}} … p_{i_{k}}.

Proof. The proof is by induction on the pair (n, m), ordered lexicographically, where n, m ≥ 0, the cases m = 0 following since q_{n 0} = −1. We will use the following notation Σ_{nk} = ∑_{1≤i_{1}≤i_{2}≤···≤i_{k}≤n} p_{i_{1}}p_{i_{2}} … p_{i_{k}}. Using the recursion (2.1) for q_{n m} we have

q_{n+1,m} = q_{n,m−2} + p_{n+1}q_{n,m−1} + q_{n,m}

= ∑_{k=0}^{m−2}(−1)^{m+k+1}\binom{n − k}{⌊\frac{m−2−k}{2} ⌋}Σ_{nk} + p_{n+1} ∑_{k=0}^{m−1}(−1)^{m+k}\binom{n − k}{⌊\frac{m−1−k}{2} ⌋}Σ_{nk}

+ ∑_{k=0}^{m}(−1)^{m+k+1}\binom{n − k}{⌊\frac{m−k}{2} ⌋}Σ_{nk}.

Now Σ_{n+1 k} = Σ_{nk} + p_{n+1}Σ_{n k−1} and so the coefficient of Σ_{n s} in the above sum is

(−1)^{m+s+1}\binom{n − s}{⌊\frac{m−2−s}{2}⌋} − (−1)^{m+s+1}\binom{n − s + 1}{⌊\frac{m−s}{2}⌋} + (−1)^{m+s+1}\binom{n − s}{⌊\frac{m−s}{2}⌋}

which is equal to zero by a standard binomial coefficient identity. One similarly finds that the coefficient of Σ_{n+1 k} is (−1)^{m+k+1}\binom{n+1−k}{⌊ \frac{m−k}{2} ⌋} as required. □

Proposition 2.5. For n, t, y ≥ 0 we have

∑_{i=0}^{2n+1} q_{n i}r_{n,i+t,y} = 0.

Proof. By Proposition 2.3 and Lemma 2.1 (i) we have

∑_{i=0}^{2n+1} q_{n i}r_{n,i+t,y} = q_{n y−t} + ∑_{k=2n+1}^{2n+1+t} q_{n k−t}r_{n,k,y}

= q_{n y−t} + ∑_{u=0}^{t} q_{n 2n+1+u−t}r_{n,2n+1+u,y}

= q_{n y−t} + ∑_{u=0}^{t} q_{n 2n+1+u−t}( ∑_{j=1}^{u+1} q_{n y+1−j}

∑_{s=0}^{u} ∑_{\begin{matrix}1≤i_{1},...,i_{s}≤u+1−j \\ ∑_{k} i_{k}=u+1−j \end{matrix}}(−1)^{s+1} ∏_{k=1}^{s} q_{n 2n+1−i_{k}}).

We will show how the terms in this sum cancel in pairs. Each summand in the above sum is determined by a 4-tuple (u, j, s, (i_{1}, …, i_{s})) where the values in this 4-tuple satisfy

0 ≤ u ≤ t, 1 ≤ j ≤ u + 1, 0 ≤ s ≤ u and ∑_{k=1}^{s} i_{k} = u + 1 − j.

We will think of this latter condition as determining the value of j.

The summand corresponding to (u, j, s, (i_{1}, …, i_{s})) is

q_{n 2n+1+u−t}q_{n y+1−j}(−1)^{s+1} ∏_{k=1}^{s} q_{n 2n+1−i_{k}}

where j = u+1−∑_{k=1}^{s} i_{k}. Also note that since j ≥ 1, u ≤ t and ∑_{k=1}^{s} i_{k} = u+1−j, we must have i_{k} ≤ t for all k = 1, …, s.

Assume that s > 0. Thus, since q_{n 2n+1} = 1, to the 4-tuple (t, j, s, (i_{1}, …, i_{s})) we associate the summand

q_{n 2n+1}q_{n y+1−j}(−1)^{s+1} ∏_{k=1}^{s} q_{n 2n+1−i_{k}} = q_{n y+1−j}(−1)^{s+1} ∏_{k=1}^{s} q_{n 2n+1−i_{k}}

where j = t + 1 − ∑_{k=1}^{s} i_{k}.

Further, to the 4-tuple (t − i_{s}, j′, s − 1, (i_{1}, …, i_{s−1})) we associate the summand

q_{n 2n+1−i_{s}}q_{n y+1−j′}(−1)^{s} ∏_{k=1}^{s−1} q_{n 2n+1−i_{k}}.

But we have

j′ = (t − i_{s}) + 1 − ∑_{k=1}^{s−1} i_{k} = t + 1 − ∑_{k=1}^{s} i_{k} = j.

Then the two summands corresponding to the 4-tuples (t, j, s, (i_{1}, …, i_{s})) and (t − i_{s}, j, s − 1, (i_{1}, …, i_{s−1})) cancel if s > 0.

Now the only summands of the sum not accounted for in the above argument are the initial summand q_{n y−t} and the summand corresponding to the 4-tuple (t, t + 1, 0, ()). This latter 4-tuple gives the summand q_{n 2n+1}q_{n y−t}(−1) = −q_{n y−t} which thus cancels with q_{n y−t}. Thus all terms cancel in pairs and Proposition 2.5 follows. □

Now by repeated application of (2.1) we may write

q_{n+m i} = ∑_{k=0}^{2m} w_{n,m,k}q_{n i−k}  (2.3)

for all n, m, i ≥ 0, where w_{n,m,k} are polynomials in p_{1}.

Proposition 2.6. (i) The w_{n,m,k} satisfy the recursion

w_{n,m+1,k} = w_{n,m,k−2} + p_{n+m+1}w_{n,m,k−1} + w_{n,m,k},

with initial conditions w_{n,1,0} = w_{n,1,2} = 1, w_{n,1,1} = p_{n+1}. (ii) For N, n, m ≥ 0 we have

w_{N,n,m} = ∑_{k=0}^{m} \binom{n − k}{[\frac{m−k}{2} ]} ∑_{1≤i_{1},i_{2},...,i_{k}≤n} p_{N+i_{1}}p_{N+i_{2}} … p_{N+i_{k}}.

Proof. (i) follows immediately from (2.1), while the proof of (ii) is the same as that for Proposition 2.4 with (2.1) replaced by the recursion given in (i). □

Proposition 2.7. For t, n, j, k ≥ 0 we have

∑_{i=0}^{2(n+k)+1} q_{k+n i}r_{n,i+t,y} = 0.

Proof. This follows from Proposition 2.5 together with the formula (2.3) allowing us to express q_{k+n i} as a linear combination of q_{n i−u}’s with coefficients not depending on i. □

We remark that Propositions 2.5 and 2.7 do not rely on the polynomials p_{n} satisfying the recursion given in Theorem 1.1.

Lemma 2.8. (i) For non-negative integers n ≥ m we have

p_{n}p_{m} = −(p_{n+m} + p_{n−m}).

(ii) For non-negative integers n ≥ m we have

(p_{n+m} + 2)(p_{n−m} + 2) = (p_{n} − p_{m})^{2}.

(iii) For all m, n ≥ 0 we have

\frac{1}{4π} ∫_{−2}^{2} \frac{p_{n}p_{m}}{\sqrt{1 − p_{1}^{2}/4}}dp_{1} = δ_{mn}

(iv) For all m, n ≥ 0 we have

p_{mn} = −(p_{m}p_{m(n−1)} + p_{m(n−2)}).

Proof. (i) From [R, p. 5] we have the following relation for T_{n}(x): T_{n}(x)T_{m}(x) = \frac{1}{2}(T_{n+m}(x) + T_{|m−n|}(x)). The result follows by substituting for p_{n}(x) using (1.1). (ii) From [R, p. 5] we also have (T_{n}(x) − 1)(T_{m}(x) − 1) = (T_{n}(x) − T_{m}(x))^{2}. Our relation for p_{n} follows from this. For (iii) use the well-known orthogonality condition for Chebyshev polynomials [R, p.30]. For (iv) use (1.1) and the relation T_{n}(T_{m}(x)) = T_{mn}(x) for all m, n ≥ 0 [R, p.5]. □

INTERSECTION-NUMBER OPERATORS AND CHEBYSHEV POLYNOMIALS 9

§3 Results on roots

In this section we will prove Theorem 1.2. Proposition 3.1. We have

(i) p_{2n} − 2 = −p_{n}^{2}; (ii) p_{2n+1} − 2 = (p_{1} − 2)(1 + ∑_{i=0}^{n}(−1)^{i}p_{i})^{2}; (iii) (a) p_{4n} + 2 = −(p_{1} + 2)(p_{1} − 2)(∑_{i=1}^{n} p_{2i−1})^{2}; (iii) (b) p_{4n+2} + 2 = −(p_{1} + 2)(p_{1} − 2)(1 + ∑_{i=0}^{n} p_{2i})^{2}; (iv) p_{2n+1} + 2 = (p_{1} + 2)(1 + ∑_{i=0}^{n} p_{i})^{2}. Proof. (i) follows by putting m = n in Lemma 2.8 (i). To prove the rest of this result we will need Lemma 3.2. We have

(i) (1 + ∑_{i=0}^{n} p_{i})^{2} = 2n + 1 − ∑_{i=0}^{2n−1}(i + 1)p_{2n−i}; (ii) (1 + ∑_{i=0}^{n}(−1)^{i}p_{i})^{2} = 2n + 1 − ∑_{i=0}^{2n−1}(−1)^{i}(i + 1)p_{2n−i}; (iii) (∑_{i=1}^{n} p_{2i−1})^{2} = n − ∑_{i=1}^{2n−2}(i + 1)p_{4n−2−2i}; (iv) (1 + ∑_{i=0}^{n} p_{2i})^{2} = 2n + 1 − ∑_{i=1}^{2n} ip_{4n+2−2i}. Proof. (i) This is by induction on n ≥ 0, the cases n = 0, 1 being easily checked. Assume that (1 + ∑_{i=0}^{n} p_{i})^{2} = 2n + 1 − ∑_{i=0}^{2n−1}(i + 1)p_{2n−i}. Then

(1 + ∑_{i=0}^{n+1} p_{i})^{2} = ((1 + ∑_{i=0}^{n} p_{i}) + p_{n+1})^{2}

= (1 + ∑_{i=0}^{n} p_{i})^{2} + 2p_{n+1}(1 + ∑_{i=0}^{n} p_{i}) + p_{n+1}^{2}

= 2n + 1 − ∑_{i=0}^{2n−1}(i + 1)p_{2n−i} + 2p_{n+1}

− 2 ∑_{i=0}^{n}(p_{n+1+i} + p_{n+1−i}) − (p_{2n+2} + p_{0})

= 2n + 3 − ∑_{i=0}^{2n−1}(i + 1)p_{2n−i} − 2p_{1} − 2p_{n+1}

− 2 ∑_{i=1}^{n}(p_{n+1+i} + p_{n+1−i}) − p_{2n+2}

= 2n + 3 − ∑_{i=0}^{2n−1}(i + 1)p_{2n−i} − 2 ∑_{i=1}^{n+1}(p_{n+1+i} + p_{n+1−i})

= 2n + 3 − ∑_{i=0}^{2n+1}(i + 1)p_{2n+2−i}.

The result follows.

The proofs of (ii), (iii) and (iv) are similar to the above. □

Now we use Lemma 3.2 (i) to prove Propositon 3.1 (iv):

(p_{1} + 2)(1 + ∑_{i=0}^{n}p_{i})^{2} = (p_{1} + 2)(2n + 1 − ∑_{i=0}^{2n−1}(i + 1)p_{2n−i})

= (2n + 1)(p_{1} + 2) − ∑_{i=0}^{2n−1}(i + 1)(2p_{2n−i} − p_{2n+1−i} − p_{2n−1−i}).

One now sees that for k ≠ 0, 1, 2n + 1 the coefficient of p_{k} is

2(2n − k + 1) − (2n + 2 − k) − (2n − k) = 0,

as required. The constant term in the above is seen to be 2,the coefficient of p_{1} is 0 while the coefficient of p_{2n+1} is 1. The result follows.

The proofs of the rest of Proposition 3.1 follow similarly from Lemma 3.2. □

Let α_{0 ±1} = 1 and for n ≥ 1 let α_{n ϵ}, ϵ = ±1, be the roots of the quadratic equation 𝔄_{n}(x) = 0, where α_{nϵ} = (−p_{n} + ϵ\sqrt{p_{n}^{2} − 4})/2. By Lemma 2.8 (i) we have p_{n}^{2} − 4 = −p_{2n} − 2; thus

α_{n ϵ} = (−p_{n} + ϵ\sqrt{p_{n}^{2} − 4})/2 = (−p_{n} + ϵ\sqrt{−p_{2n} − 2})/2

and by Proposition 3.1 the discriminant −p_{2n} − 2 can be written as (p_{1}^{2} − 4)u_{n}^{2} where u_{n} is given in Proposition 3.1 (iii) (a) and (b):

u_{2m} = ∑_{i=1}^{m} p_{2i−1} and u_{2m+1} = 1 + ∑_{i=0}^{m} p_{2i}.

We now wish to specify α_{n ϵ} more explicitly by taking the square root of u_{n}^{2} so as to be able to write

α_{n ϵ} = (−p_{n} + ϵu_{n}\sqrt{p_{1}^{2} − 4})/2.

Lemma 3.3. For positive integers n ≥ m we have

(i) p_{2n} ∑_{i=1}^{m} p_{2i−1} ± p_{2m} ∑_{i=1}^{n} p_{2i−1} = −2 ∑_{i=1}^{n±m} p_{2i−1}; (ii) p_{2n}p_{2m} ± (p_{1}^{2} − 4)(∑_{i=1}^{n} p_{2i−1})(∑_{i=1}^{m} p_{2i−1}) = −2p_{2(n±m)}; (iii) (a) p_{2n}(1 + ∑_{i=0}^{m} p_{2i}) + p_{2m+1} ∑_{i=1}^{n} p_{2i−1} = 2 − 2 ∑_{i=1}^{n+m} p_{2i}; (iii) (b) p_{2n}(1 + ∑_{i=0}^{m} p_{2i}) − p_{2m+1} ∑_{i=1}^{n} p_{2i−1} = 2 − 2 ∑_{i=1}^{n−m−1} p_{2i}; (iv) p_{2n}p_{2m+1} ± (p_{1}^{2} − 4)(∑_{i=1}^{n} p_{2i−1})(1 + ∑_{i=0}^{m} p_{2i}) = −2p_{2(n±m)±1};

(v) (a) p_{2n+1}(1 + ∑_{i=0}^{m} p_{2i}) + p_{2m+1}(1 + ∑_{i=0}^{n} p_{2i}) = −2 ∑_{i=1}^{n+m+1} p_{2i−1};

(v) (b) p_{2n+1}(1 + ∑_{i=0}^{m} p_{2i}) − p_{2m+1}(1 + ∑_{i=0}^{n} p_{2i}) = 2 ∑_{i=1}^{n−m} p_{2i−1}. (vi) (a) p_{2n+1}p_{2m+1} + (p_{1}^{2} − 4)(∑_{i=0}^{n} p_{2i})(1 + ∑_{i=0}^{m} p_{2i}) = −2p_{2(n+m+1)}; (vi) (b) p_{2n+1}p_{2m+1} − (p_{1}^{2} − 4)(∑_{i=0}^{n} p_{2i})(1 + ∑_{i=0}^{m} p_{2i}) = −2p_{2(n−m)}.

Proof. All of the proofs are straightforward applications of Lemma 2.8 (i). One uses the relations:

\{ \begin{matrix}p_{n}u_{m} = −u_{n+m} − u_{m−n}, if m ≥ n and \\ p_{n}u_{m} = −u_{n+m} + u_{n−m} if n ≥ m \end{matrix}

which are also proved using Lemma 2.8. □

The following result proves Theorem 1.2.

INTERSECTION-NUMBER OPERATORS AND CHEBYSHEV POLYNOMIALS 11

Proposition 3.4. For all positive integers n ≥ m and ϵ, δ ∈ {±1} there are integers N, Δ with Δ ∈ {±1} and N = n ± m such that α_{n ϵ}α_{m δ} = α_{N Δ}. In particular we have

(i) α_{n ϵ}α_{n −ϵ} = 1, for all n ≥ 0 and ϵ ∈ {±1}. (ii) α_{2n ϵ}α_{2m δ} = α_{2(n+ϵδm) ϵ}, for all n ≥ m ≥ 1 and ϵ, δ ∈ {±1}; (iii) α_{2n ϵ}α_{2m+1 δ} = α_{2n+ϵδ(2m+1) ϵ}, for all 2n ≥ 2m + 1 ≥ 1 and ϵ, δ ∈ {±1}; (iv) α_{2n+1 ϵ}α_{2m δ} = α_{2n+1+ϵδ2m ϵ}, for all 2n + 1 ≥ 2m ≥ 1 and ϵ, δ ∈ {±1}; (iv) α_{2n+1 ϵ}α_{2m+1 δ} = α_{2n+1+ϵδ(2m+1) ϵ}, for all 2n+1 ≥ 2m+1 ≥ 1 and ϵ, δ ∈ {±1}.

Proof. (i) is obvious since α_{nϵ} is a root of s^{2} + p_{n}s + 1. We will now prove (ii) and (iii), the rest being similar. For (ii) we have, using Proposition 3.1 (iii) (a) and Lemma 3.3 (i) and (ii),

α_{2n ϵ}α_{2m δ} = (−p_{2n} + ϵu_{2n}\sqrt{p_{1}^{2} − 4}) /2 (−p_{2m} + δu_{2m}\sqrt{p_{1}^{2} − 4}) /2

= (p_{2n}p_{2m} + ϵδ(p_{1}^{2} − 4) ∑_{i=1}^{n} p_{2i−1} ∑_{i=1}^{m} p_{2i−1}) /4

− (ϵp_{2m} ∑_{i=1}^{n} p_{2i−1}\sqrt{p_{1}^{2} − 4} + δp_{2n} ∑_{i=1}^{m} p_{2i−1}\sqrt{p_{1}^{2} − 4}) /4

= (−2p_{2(n+ϵδm)} + ϵ2 ∑_{i=1}^{n+ϵδm} p_{2i−1}\sqrt{p_{1}^{2} − 4}) /4

= α_{2(n+ϵδm) ϵ}.

For (iii) we have to consider two cases, ϵ = ±δ, so that for ϵ = −δ we have

α_{2n ϵ}α_{2m+1 −ϵ} = \frac{1}{4} (−p_{2n} + ϵu_{2n}\sqrt{p_{1}^{2} − 4}) (−p_{2m+1} − ϵu_{2m+1}\sqrt{p_{1}^{2} − 4})

= \frac{1}{4} (p_{2n}p_{2m+1} − (p_{1}^{2} − 4) ∑_{i=1}^{n} p_{2i−1}(1 + ∑_{i=0}^{m} p_{2i}))

− \frac{1}{4} (ϵp_{2m+1} ∑_{i=1}^{n} p_{2i−1} − ϵp_{2n}(1 + ∑_{i=1}^{m} p_{2i})) \sqrt{p_{1}^{2} − 4}

= \frac{1}{4} (−2p_{2n−2m−1} + ϵ(2 − 2 ∑_{i=0}^{n−m−1} p_{2i}\sqrt{p_{1}^{2} − 4})) = α_{2n−2m−1,ϵ}.

The other case is proved similarly. □

§4 Proof of Theorem 1.1

Since the roots α_{n,j} are distinct the recursion given by Proposition 2.5, together with standard results about recursions (see for example [B] Ch. 7) allow us to conclude that for fixed n, y there are constants c_{n,y,j,k} such that

r_{n,i,y} = c_{n,y,0,0} + ∑_{j=1}^{n}(c_{n,y,j,−}α_{j,−}^{i} + c_{n,y,j,+}α_{j,+}^{i})  (4.1)

for each i ≥ 0.

Conversely, we note that for N ≥ n and for any constants c_{n,y,j,k} we have

∑_{i=0}^{2N+1} q_{N,i} (c_{n,y,0,0} + ∑_{j=1}^{n}(c_{n,y,j,−}α_{n,−}^{i} + c_{n,y,j,+}α_{n,+}^{i})) = 0.  (4.2)

But by (4.1) we see that for fixed n, m, j, k and varying i any product r_{n,i,j}r_{m,i,k} can be written as a linear sum of terms each of which is either constant or has the form α_{n,u}^{i} or α_{n,u}^{i}α_{m,v}^{i}. But by Proposition 3.4 any such term is of the form α_{N,w}^{i} where N ≤ n + m. Thus by (4.2) the sum ∑_{i=0}^{2(n+m)+1} q_{n+m,i}r_{n,i,j}r_{m,i,k} is zero, as required. □

§5 Braid group action on simple closed curves on surfaces

Let C^{m} = C_{n}^{m} denote the set of isotopy classes of oriented simple closed curves on D_{n} where each such simple closed curve surrounds m of the punctures.

In [H2] we used the action of B_{n} on C^{2} to obtain linear representations of the braid groups B_{n} over ℤ[t] by first finding a natural map from C^{2} to a polynomial algebra, and then considering the induced action of B_{n} on certain quotients of ideals in this polynomial algebra. We need to explain some of this more carefully now.

For R a commutative ring with identity we defined the ring R_{n}[a_{ij}] in §1. It will be convenient for us to let a_{ii} = 0 for all i = 1, …, 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}[a_{ij}]. Then the cycles generate a subalgebra of R_{n}[a_{ij}] 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 generated by the (finite number of) simple cycles.

Note that the action of B_{n} on D_{n} fixes the boundary of D_{n} and so there is an induced action of B_{n} on the fundamental group of D_{n} (where we choose a base point p on the boundary of D_{n}). This fundamental group is the free group F_{n} of rank n. We choose a standard set of generators x_{1}, …, x_{n} for F_{n}, where x_{i} is a simple closed curve enclosing the ith puncture π_{i} and such that x_{1}x_{2} … x_{n} is parallel to the boundary of D_{n}. See Figure 1.

In Figure 1 we have shown the arcs a_{i} and the generators x_{i}; we have also shown some cut arcs b_{1}, …, b_{n} for the generators x_{1}, …, x_{n}. Thus if c is an oriented simple closed curve in D_{n} which is based at p and which is in general position with all the arcs b_{i}, then the word w(c) in F_{n} corresponding to c is determined by the sequence of oriented intersections of c with the oriented arcs b_{1}, …, b_{n}.

The action of B_{n} on the generators x_{i} is as follows:

σ_{j}(x_{i}) = x_{i} if i ≠ j, j + 1, σ_{j}(x_{j}) = x_{j}x_{j+1}x_{j}^{−1}, σ_{j}(x_{j+1}) = x_{j}.

We now find a particularly convenient representation of the free group F_{n}. For i = 1, …, n define the following 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} )

where the non-zero off-diagonal entries occur in the ith row. Here a transvection [A] is a matrix M = I_{n} + A where I_{n} is the identity matrix, det(M) = 1 and A^{2} = 0. In particular, conjugates of transvections are transvections.

Then < T_{1}, …, T_{n} > is a free group of rank n (see [H1]). 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, 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^{∞}, 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} [H1]. Then a map ϕ : C → R_{n} is defined uniquely by

ϕ(c) = trace(c) − n.

Thus ϕ can be thought of as being defined on certain conjugacy classes of elements of F_{n} (namely those representing simple closed curves). This map 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.

Now for m ≥ n and s ∈ F_{n} we may also consider s as an element of F_{m} under the natural inclusion of F_{n} in F_{m}. In this case we note that ϕ(s) has the same value whether we consider s as an element of F_{n} or F_{m}.

Now for all i, j 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

where b_{ij} ∈ R_{n}[a_{ij}] (see [H1]; in fact this also follows from Proposition 7.1). It is also easy to see that there is a natural choice so that

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

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. Here the action of α on the a_{ij} is defined by

trace(α(T_{i})α(T_{j})) = α(a_{ij})α(a_{ji}) + n,

(see [H1] for more details) 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} > we have

ϕ(α(w)) = α(ϕ(w)).  (5.1)

For example the action of σ_{i} 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},
σ_{i}(a_{h i+1}) = a_{h i} − a_{h i+1}a_{i+1,i}, σ_{i}(a_{i j}) = a_{i+1 j},
σ_{i}(a_{i+1 j}) = a_{i j} + a_{i i+1}a_{i+1 j}, etc.

where 1 ≤ h < i, i + 1 < j ≤ n.

It follows from Theorem 2.5 and Theorem 6.2 of [H1] 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. We note as in [Hu2] that there is a natural ring involution ∗ on R_{(n)}[a_{ij}] which commutes with the action of B_{n}; this involution is determined by its action on the generators a_{ij} which is as follows: a_{ij}^{∗} = −a_{ji}. This involution has the following property:

trace(A^{−1}) = trace(A)^{∗},

for all A ∈ F_{n}. Thus ϕ(c^{−1}) = ϕ(c)^{∗}, where c^{−1} is the curve c with its orientation reversed.

If c is a simple closed curve in D_{n} surrounding k punctures, then we let Σ(c) denote the spine of γ, so that Σ(γ) is an embedded tree in D_{n} ∪ {π_{1}, …, π_{n}} whose k vertices are in {π_{1}, …, π_{n}} and such that c is isotopic to the boundary of a small tubular neighbourhood of Σ(c) (with some orientation).

Suppose that c is a simple closed curve which is disjoint from γ_{12}. Then σ_{1}(c) = c and so 𝔅_{0}(σ_{1}, a_{12}a_{21}+2)(ϕ(c)) = 0 which shows that Ω(c) = 0. It easily follows that for such a curve 2Ω(c) = ι(γ_{12}, c). The key observation that indicates a connection with the previous sections was made in [H2] and is

Lemma 5.1. Let α, β_{1}, β_{2}, γ_{1}, γ_{2} ∈ C^{2} be curves with spines as shown in Figure 2, where we assume that there are no punctures inside the heart-shaped diagram other than the three shown (π_{i}, π_{j} and π_{k}).

Then we have the following relation:

[ϕ(γ_{1}) − ϕ(γ_{2})] + (1 + a_{12}a_{21})[ϕ(β_{1}) − ϕ(β_{2})] = 0.  (5.2)

Now note that we may take γ_{1} = σ_{1}^{2}(γ_{23}), γ_{2} = σ_{1}^{−1}(γ_{23}), β_{1} = γ_{23} and β_{2} = σ_{1}(γ_{23}). (In fact the proof of Lemma 5.1 consists essentially of checking that the relation holds in this case, which is easily done.) Thus we can write (5.2) as (σ_{1}^{2} − σ_{1}^{−1} − (1 + a_{12}a_{21})(1 − σ_{1}))[γ_{23}] = 0. Upon acting on this equation by σ_{1} on the left and factoring this gives (since σ_{1}(a_{12}a_{21}) = a_{12}a_{21})

(σ_{1} − 1)(σ_{1}^{2} + (2 + a_{12}a_{21})σ_{1} + 1)ϕ(γ_{23}) = 0.  (5.3)

Thus if p_{1} = a_{12}a_{21} + 2, then the above operator is 𝔅_{1}(σ_{1}, a_{12}a_{21} + 2) and we have 𝔅_{1}(σ_{1}, a_{12}a_{21} + 2)(ϕ(γ_{23})) = 0. Now if c is any curve such that if ι(γ_{12}, c) = 2, then the pair (γ_{12}, c) is diffeomorphic to the pair (γ_{12}, γ_{23}) and we check that 𝔅_{0}(σ_{1}, a_{12}a_{21} + 2)(ϕ(γ_{23})) ≠ 0. The above gives 𝔅_{1}(σ_{1}, a_{12}a_{21} + 2)(ϕ(γ_{23})) = 0. Thus 𝔅_{0}(σ_{1}, a_{12}a_{21} + 2)(ϕ(c)) ≠ 0 and 𝔅_{1}(σ_{1}, a_{12}a_{21} + 2)(ϕ(c)) = 0.

Now we note that up to diffeomorphism there are only two ways in which we can have ι(γ_{12}, c) = 4. These are (i) w(c) = x_{2}(x_{3} … x_{u})x_{2}^{−1}(x_{u+1}x_{u+2} … x_{v}); and (ii) w(c) = x_{2}(x_{3} … x_{v})x_{1}(x_{3} … x_{u})^{−1} where 2 < u < v ≤ n. One easily checks that 𝔅_{2}(σ_{1}, a_{12}a_{21} + 2)(ϕ(c)) = 0 and 𝔅_{1}(σ_{1}, a_{12}a_{21} + 2)(ϕ(c)) ≠ 0 in each case. Thus we have proved

Lemma 5.3. If c is a simple closed curve on D_{n} with ι(γ_{12}, c) ≤ 2, then 2Ω(c) = ι(γ_{12}, c). □

This indicates the initial connection of the operators 𝔅_{n}(σ, a_{12}a_{21} + 2) with intersection-number functions. We now prove:

Proposition 5.4. Let U_{n} be the subring of R_{n} which is generated by all elements of the form a_{ij} for 2 < i, j ≤ n and all

a_{i1}a_{1j}, a_{i2}a_{2j}, a_{12}a_{21}, a_{i1}a_{12}a_{2j}, a_{i2}a_{21}a_{1j}

for 2 < i, j ≤ n. Then U_{n} contains Y_{n} and for u ∈ U_{n} there is a positive integer N = N(u) such that 𝔅_{N}(σ_{1}, a_{12}a_{21} + 2)(u) = 0.

Proof. As we noted before Y_{n} is generated by the simple cycles and it is easily seen that each such simple cycle is in U_{n}. Next we check that for 2 < i, j ≤ n we have 𝔅_{0}(a_{ij}) = 0 and

𝔅_{1}(a_{i1}a_{1j}) = 0, 𝔅_{1}(a_{i2}a_{2j}) = 0, 𝔅_{0}(a_{12}a_{21}) = 0,
𝔅_{1}(a_{i1}a_{12}a_{2j}) = 0, 𝔅_{1}(a_{i2}a_{21}a_{1j}) = 0.  (5.4)

The result now follows from Theorem 1.1. □

For u ∈ U_{n} we will let Ω(u) denote the minimal N such that 𝔅_{N}(σ_{1}, a_{12}a_{21} + 2)(u) = 0.

We note that the operators 𝔅_{m}(σ_{1}, a_{12}a_{21} + 2) act on some R_{n} and so depend on the number n. However it will be convenient for us to ignore this dependency, so that 𝔅_{m}(σ_{1}, a_{12}a_{21} +2) will be considered as acting on all of the R_{n}, n ≥ 2. This is consistent with considering D_{n} as a subspace of D_{n+1} for all n.

§6 A normal form for curves on D_{n}

Let c be a simple closed curve on D_{n} and let w(c) ∈ F_{n} represent c. Note that we may (and will) choose w(c) to be cyclically reduced. We will say that a cyclically reduced word w ∈ F_{n} is 1,2-reduced if w is cyclically conjugate to a word of the form w′ = A_{1}B_{1} … A_{r}B_{r}, where

A_{i} ∈< T_{3}, T_{4}, …, T_{n} > and B_{i} ∈ {T_{1}^{±1}, T_{2}^{±1}, (T_{1}T_{2})^{±1}}

with A_{i} ≠ id for i = 1, … r.

In this section we prove the following result which gives a normal form for an element of F_{n} representing a simple closed curve on D_{n} and derive some relevant consequences.

Theorem 6.1. Let c be a simple closed curve on D_{n}. Then there is k ∈ ℤ such that w(σ_{1}^{k}(c)) is 1, 2-reduced.

Proof. Considering Figure 1 in §5 we see that the arcs a_{1}, b_{1}, b_{2} cut off a simplyconnected subset E_{1} of D_{n}. See Figure 3. Let O_{1} denote a tubular neighbourhood of the union of E_{1} and the interior of the curve x_{1} in D_{n} such that O_{1} contains only the punctures π_{1}, π_{2} together with the base point p.

Let c be a simple closed curve on D_{n}. Then w(c) certainly has a cyclic conjugate w′ which can be written w′ = A_{1}B_{1} … A_{r}B_{r}, where A_{i} ∈< T_{3}, T_{4}, …, T_{n} >, B_{i} ∈< T_{1}, T_{2} > and A_{i}, B_{i} ≠ id for i = 1, … r. Fix 1 ≤ i ≤ r and let T_{u}^{±1}, T_{v}^{±1} be the terminal and initial letters of A_{i} and A_{i+1} respectively. Since u, v ≥ 3 we see that there is a subarc c′ of c, with end points on the cut arcs b_{u} and b_{v}, representing the subword T_{u}^{±1}B_{i}T_{v}^{±1}. Now note that we can homotope c′ so that the end points go to p, while all the points of c′ ∩ E_{1} are held fixed during the homotopy. Call the resulting curve c′′ = c′′_{i}. We may also require that c′′ is in O_{1}. Now O_{1} is a twice-puctured disc and c′′ is a simple closed curve in O_{1} based at p. Further, the fundamental group of O_{1} based at p is generated by x_{1}, x_{2}. It follows that either (i) c′′ = σ_{1}^{k}(x_{1}^{±1}) for some k = k(i) ∈ ℤ, or (ii) c′′ = γ_{12}^{±1}. These two cases occur when the interior of c′′ has one or two punctures inside it (respectively).

If for all i = 1, …, r we have case (ii), then w′ is 1, 2-reduced and the Theorem follows. If, on the other hand, there is some 1 ≤ i ≤ r such that we have case (i), then we replace c by σ_{1}^{−k}(c). This means that we can assume that A_{1} = T_{1}^{±1} (cyclically permute w′ as necessary). Now if for any 1 < j ≤ r we now have k(j) ≠ 0, then the curve c′′_{j} can be homotoped so as to meet x_{1} only at the base point p. It follows that c′′_{j} is either (a) x_{1}^{±1}; (b) x_{2}^{±1}; (c) γ_{12}^{±1}; or (d) the curve σ_{1}(x_{1}), this being the curve x_{1}x_{2}x_{1}^{−1}. The first three cases are covered by Theorem 6.1, while if we have the fourth case, then we act on c by σ_{1}^{−1} so that now we have A_{1} = T_{2}^{±1} and A_{j} = T_{1}^{±1}.

Now suppose that we are in this latter case where A_{1} = T_{2}^{±1} and A_{j} = T_{1}^{±1}. Let 1 < i ≤ r, i ≠ j. Then c′′_{i} can be homotoped so as to meet x_{1} and x_{2} only at the point p. One easily sees that either (i) c′′_{i} = x_{1}^{±1}; or (ii) c′′_{i} = x_{2}^{±1}; or (iii) c′′_{i} = (x_{1}x_{2})^{±1}, and so w′ is 1, 2-reduced. □

For c and k as in Theorem 6.1 we will call w(σ_{1}^{k}(c)) the normal form for c. We note that the normal form is not unique.

Lemma 6.2. Let k ∈ ℤ and let c be a simple closed curve on D_{n} and let c′ = σ_{1}^{k}(c); for example c′ could be the normal form for c. If 𝔅_{m}(σ_{1}, a_{12}a_{21} + 2)(ϕ(c)) = 0, then 𝔅_{m}(σ_{1}, a_{12}a_{21} + 2)(ϕ(σ_{1}^{k}(c))) = 0. In particular, we have Ω(c′) = Ω(c).

Proof. Since σ_{1}(a_{12}a_{21}) = a_{12}a_{21} it follows that σ_{1} commutes with the operator 𝔅_{m}(σ_{1}, a_{12}a_{21} + 2). Also, we have ϕ(σ_{1}^{k}(c)) = σ_{1}^{k}(ϕ(c)) by (5.1). Thus

𝔅_{m}(σ_{1}, a_{12}a_{21} + 2)(ϕ(σ_{1}^{k}(c))) = σ_{1}^{k}(𝔅_{m}(σ_{1}, a_{12}a_{21} + 2)(ϕ(c))) = 0.

The rest follows from this. □

We now show the connection of the above normal form with intersection-numbers.

Theorem 6.3. Let c be a simple closed curve on D_{n} and let c′ = σ_{1}^{k}(c) be its normal form. Suppose that w(c′) = A_{1}B_{1} … A_{r}B_{r}, where A_{i} ∈< T_{3}, T_{4}, …, T_{n} >, B_{i} ∈ {T_{1}^{±1}, T_{2}^{±1}, (T_{1}T_{2})^{±1}} and A_{i} ≠ id for i = 1, … r. Let h be the number of i such that B_{i} ∈ {T_{1}^{±1}, T_{2}^{±1}}. Then h = ι(γ_{12}, c′) = ι(γ_{12}, c).

Proof. It is clear that ι(γ_{12}, c′) = ι(γ_{12}, c) from Lemma 6.2. Now note that in a neighbourhood of E_{1} (oriented) arcs coming from any B_{i} are as shown in Figure 3.

Now any B_{i} = T_{1}^{±1} corresponds to an arc of c ∩ E_{1} which crosses the arc b_{1} once and does not cross b_{2}. It thus crosses a_{1} and contributes 1 to the intersection number ι(γ_{12}, c). Similarly for any B_{i} = T_{2}^{±1}. Lastly, any B_{i} = (T_{1}T_{2})^{±1} corresponds to an arc of c ∩ E_{1} which does not cross a_{1}. The result follows. □

§7 Combinatorial results on traces

In this section we will introduce a way of grouping together terms of ϕ(c) and use this to prove that Ω(c) ≤ ι(c, γ_{12})/2.

Let w = T_{i_{1}}^{e_{1}}T_{i_{2}}^{e_{2}} … T_{i_{r}}^{e_{r}} ∈ F_{n} =< T_{1}, …, T_{n} > where e_{i} = ±1 for i = 1, …, r. Then by a subword of w we mean any w′ = T_{i_{a}}^{e_{a}}T_{i_{b}}^{e_{b}} … T_{i_{z}}^{e_{z}} where 1 ≤ a < b < ⋯ < z ≤ r. In fact we will really be thinking of w′ as being the sequence 1 ≤ a < b < ⋯ < z ≤ r, but it will be convenient to write w′ = T_{i_{a}}^{e_{a}}T_{i_{b}}^{e_{b}} … T_{i_{z}}^{e_{z}}. We will need the following generalisation of [H1, Lemma 2.3].

Proposition 7.1. Let w = T_{i_{1}}^{e_{1}}T_{i_{2}}^{e_{2}} … T_{i_{r}}^{e_{r}} ∈ F_{n}. Then trace(w) − n is equal to the sum of all terms of the form

(e_{h}e_{i}e_{j}e_{k} … e_{m})a_{hi}a_{ij}a_{jk} … a_{mh},

where T_{h}^{e_{h}}T_{i}^{e_{i}}T_{j}^{e_{j}}T_{k}^{e_{k}} … T_{m}^{e_{m}} is a non-empty subword of w.

Proof. We first prove the following result by induction. The induction hypothesis is: the hk entry of the matrix w is the sum of all terms of the kind

(e_{j_{1}}e_{j_{2}}e_{j_{3}}e_{j_{4}} … e_{j_{m}})a_{j_{1}j_{2}}a_{j_{2}j_{3}}a_{j_{3}j_{4}} … a_{j_{m}k},

where T_{j_{1}}^{e_{j_{1}}} … T_{j_{m}}^{e_{jm}} is a subword of w and j_{1} = h. This is easily proved (similar to the proof of Lemma 2.3 of [H1]) and Lemma 7.1 follows directly from this. □

Thus, for example, if n = 4 and w = T_{1}T_{2}^{2}T_{3}^{3}T_{2}^{−1}, then

trace(w)−4 = a_{12}a_{21}+3a_{13}a_{31}+3a_{23}a_{32}−3a_{13}a_{32}a_{21}+6a_{12}a_{23}a_{31}−6a_{12}a_{21}a_{23}a_{32}.

We introduce the following notation: Fix a word w = T_{i_{1}}^{e_{1}}T_{i_{2}}^{e_{2}} … T_{i_{r}}^{e_{r}} ∈ F_{n}, which we will think of as a cyclic word. Then by Lemma 7.1 for any subword s = T_{h}^{e_{h}}T_{i}^{e_{i}}T_{j}^{e_{j}} … T_{m}^{e_{m}} of w we obtain the contribution

η(s) = (e_{h}e_{i}e_{j}e_{k} … e_{m})a_{hi}a_{ij}a_{jk} … a_{mh}

to the trace of w. Such a contribution we call a term. Note that by Lemma 7.1 trace(w) − n is the sum of all such terms η(s) for all such non-empty subwords s of w. We now look for a way of grouping together some of these terms in a way that will be compatible with the action of the operators 𝔅_{n}(σ_{1}, a_{12}a_{21} + 2).

Let s be a subword of w. Then we can write η(s) uniquely as

η(s) = Ca_{1}b_{1}a_{2}b_{2} … a_{u}b_{u}a_{u+1},

where C is a constant, a_{i} is either 1 or is a monomial in the generators a_{rs} where r, s ≥ 3 and b_{i} is a monomial of the form a_{ri_{1}}a_{i_{1}i_{2}} … a_{i_{t−1}i_{t}}a_{i_{t}s} where r, s ≥ 3 and i_{j} ∈ {1, 2} for j = 1, …, t. Note that Ω(a_{i}) = 0 for all i and so by Theorem 1.1 we see that

Ω(a_{1}b_{1}a_{2}b_{2} … a_{u}b_{u}a_{u+1}) ≤ Ω(b_{1}) + Ω(b_{2}) + ⋯ + Ω(b_{u}).  (7.1)

Now we define a 12 -reduction of s to be any subword s′ of s which can be obtained from s by deleting any (or all) of the letters T_{1}^{±1} or T_{2}^{±1} occuring in s.

Let ψ_{ˆ1ˆ2}(s) denote the sum of all of the η(s′) where s′ is a 12 -reduction of s. From the above we see the following:

Lemma 7.2. Let s be a subword of w as above and write a = A_{1}B_{1} … A_{u}B_{u}A_{u+1} where A_{i} ∈< T_{3}, …, T_{n} > and B_{i} ∈< T_{1}, T_{2} > \{id}. Then the element ψ_{ˆ1ˆ2}(s) has a factorization

ψ_{ˆ1ˆ2}(s) = E_{1}G_{1}E_{2}G_{2} … G_{u−1}E_{u}G_{u}E_{u+1}

where E_{i} = 1 or E_{i} = ψ_{ˆ1ˆ2}(A_{i}) ∈ R[a_{rs}|3 ≤ r ≠ s ≤ n] and G_{i} = ψ_{ˆ1ˆ2}(t_{i}) where t_{i} is the subword T_{y_{i}}^{p}B_{i}T_{z_{i}}^{q} for some p, q = ±1 and y_{i}, z_{i} ≥ 3. □

Example 7.3. Let s = T_{3}T_{1}T_{2}T_{4}T_{6}T_{4}T_{2}^{−1}T_{5}T_{1}T_{5}. Then

ψ_{ˆ1ˆ2}(s) = (a_{31}a_{12}a_{24} + a_{31}a_{14} + a_{32}a_{24} + a_{34})a_{46}a_{64}(−a_{42}a_{25} + a_{45})(a_{51}a_{15})a_{53}.

We now calculate the G_{i} = ψ_{ˆ1ˆ2}(t_{i}) in the above Lemma.

Lemma 7.4. Choose integers 3 ≤ r, s ≤ n and u ≥ 0. Then

𝔅_{0}(σ_{1}, a_{12}a_{21} + 2)(ψ_{ˆ1ˆ2}(T_{r}(T_{1}T_{2})^{u}T_{s})) = 0, and
𝔅_{0}(σ_{1}, a_{12}a_{21} + 2)(ψ_{ˆ1ˆ2}(T_{r}(T_{2}^{−1}T_{1}^{−1})^{u}T_{s})) = 0,

while if y ∈< T_{1}, T_{2} >, then

𝔅_{1}(σ_{1}, a_{12}a_{21} + 2)(ψ_{ˆ1ˆ2}(T_{r}yT_{s})) = 0.

Proof. Now any monomial summand of ψ_{ˆ1ˆ2}(T_{r}yT_{s}) is of the form Ca_{ri_{1}}a_{i_{1}i_{2}} … a_{i_{u}s} where C is a constant and i_{1}, i_{2}, …, i_{u} = 1, 2. Any such word can be factored as C(a_{12}a_{21})^{k}Q where Q is one of a_{r1}a_{1s}, a_{r1}a_{12}a_{2s}, a_{r2}a_{2s}, a_{r2}a_{21}a_{1s}. Now the second part of the result follows from Theorem 1.1 and equations (5.4) which show that 𝔅_{0}(σ_{1}, a_{12}a_{21} + 2)((a_{12}a_{21})^{k}) = 0 and 𝔅_{1}(σ_{1}, a_{12}a_{21} + 2)(Q) = 0.

Now consider the element w = T_{r}(T_{1}T_{2})^{k}T_{s}. Then by Lemma 7.1 we have

trace(w) − n = ψ_{ˆ1ˆ2}(w)a_{sr} + U,

where each monomial summand μ of U is a product of a_{ij}’s where the set of such subscripts i, j for this fixed μ contains only one of r, s.

But w represents a closed curve c on D_{n} which, since r, s > 2, does not meet the arc a_{1}. Thus we have σ_{1}(c) = c which implies (by (5.1)) that σ_{1}(trace(w) − n) = trace(w) − n. Since σ_{1}(a_{sr}) = a_{sr} we see that we must have σ_{1}(ψ_{ˆ1ˆ2}(w)) = ψ_{ˆ1ˆ2}(w) from which it follows that 𝔅_{0}(σ_{1}, a_{12}a_{21} + 2)(ψ_{ˆ1ˆ2}(w)) = 0. □

Refering back to Lemma 7.2 we now have

Lemma 7.5. Let w ∈ F_{n}. Let η(s) = A_{1}B_{1}A_{2}B_{2} … B_{u}A_{u+1}, for a subword s of w.

Suppose that k of the B_{i} are of the form T_{1}^{u} or T_{2}^{u}. Then

𝔅_{k}(σ_{1}, a_{12}a_{21} + 2)(ψ_{ˆ1ˆ2}(s)) = 0.

Proof. This follows from Lemma 7.2, Lemma 7.4 and equation (7.1). □

Let c be a simple closed curve and let w = w(c). Let η(w) = A_{1}B_{1} … B_{u}A_{u+1}. Let w′ be the word w′ = w′(w) = A_{1}A_{2} … A_{u+1}. Now to every subword of w′ there corresponds naturally a subword of η(w); conversely, we can project any subword s of η(w) to a subword ζ(s) of w′ by just deleting from the subword s all T_{1}^{j}’s and T_{2}^{j}’s that are in it.

Now let S denote the set of all subwords of η(w) (including the empty subword). Let S′ denote the set of all subwords of η(w′). Then we have a natural map β : S → S′, which consists of deleting from any element of S all elements of the form T_{1}^{j} or T_{2}^{j}. We partition the elements of S into preimages of β. This in turn gives a way of collecting together terms of ϕ(c) = trace(w)−n into sums each of which has the form E_{1}G_{1}E_{2}G_{2} … G_{u−1}E_{u}G_{u}E_{u+1} indicated in Lemma 7.2. Now if k is as defined in Lemma 7.5, then at most k of the G_{i} in E_{1}G_{1}E_{2}G_{2} … G_{u−1}E_{u}G_{u}E_{u+1} can satisfy 𝔅_{0}(σ_{1}, a_{12}a_{21} + 2)(G_{i}) ≠ 0. So by Theorem 1.1 we have

𝔅_{k}(σ_{1}, a_{12}a_{21} + 2)(E_{1}G_{1}E_{2}G_{2} … G_{u−1}E_{u}G_{u}E_{u+1}) = 0,

as required to prove that Ω(c) ≤ ι(c, γ_{12})/2.

Example 7.6. For n = 4 consider the curve c represented by the conjugacy class w = w(c) = T_{1}T_{2}T_{3}T_{2}T_{3}^{−1}T_{2}^{−1}T_{4}. One easily sees that ι(γ_{12}, c) = 2. Also

S′ = {∅, T_{3}, T_{3}^{−1}, T_{4}, T_{3}T_{3}^{−1}, T_{3}T_{4}, T_{3}^{−1}T_{4}, T_{3}T_{3}^{−1}T_{4}}.

For example we have

β^{−1}(T_{3}T_{3}^{−1}T_{4}) = {T_{3}T_{3}^{−1}T_{4}, T_{1}T_{3}T_{3}^{−1}T_{4}, T_{2}T_{3}T_{3}^{−1}T_{4}, T_{1}T_{2}T_{3}T_{3}^{−1}T_{4}, T_{3}T_{2}T_{3}^{−1}T_{4},
T_{1}T_{3}T_{2}T_{3}^{−1}T_{4}, T_{2}T_{3}T_{2}T_{3}^{−1}T_{4}, T_{1}T_{2}T_{3}T_{2}T_{3}^{−1}T_{4}, T_{3}T_{3}^{−1}T_{2}^{−1}T_{4},

T_{1}T_{3}T_{3}^{−1}T_{2}^{−1}T_{4}, T_{2}T_{3}T_{3}^{−1}T_{2}^{−1}T_{4}, T_{1}T_{2}T_{3}T_{3}^{−1}T_{2}^{−1}T_{4}, T_{3}T_{2}T_{3}^{−1}T_{2}^{−1}T_{4},

T_{1}T_{3}T_{2}T_{3}^{−1}T_{2}^{−1}T_{4}, T_{2}T_{3}T_{2}T_{3}^{−1}T_{2}^{−1}T_{4}, T_{1}T_{2}T_{3}T_{2}T_{3}^{−1}T_{2}^{−1}T_{4}}.

This gives rise to the product

π = (a_{32}a_{23})(−a_{34} + a_{32}a_{24})(a_{43} + a_{42}a_{23} + a_{41}a_{13} + a_{41}a_{12}a_{24}).

which occurs as a summand of trace(w) − 4 of the kind indicated in Lemma 7.2. Now

𝔅_{1}(σ_{1}, a_{12}a_{21} + 2)(a_{32}a_{23}) = 0,
𝔅_{1}(σ_{1}, a_{12}a_{21} + 2)(−a_{34} + a_{32}a_{24}) = 0, and
𝔅_{0}(σ_{1}, a_{12}a_{21} + 2)(a_{43} + a_{42}a_{23} + a_{41}a_{13} + a_{41}a_{12}a_{23}) = 0.

This shows that 𝔅_{2}(σ_{1}, a_{12}a_{21} + 2)(π) = 0. At the other extreme one has that for the trivial subword ∅ ∈ S′ we see that β^{−1}(∅) contributes a_{12}a_{21}. One similarly calculates the contributions made by each of the other members of S′.

Lemma 7.7. For any simple closed curve c with η(w) = A_{1}B_{1} … B_{u}A_{u+1} where w = w(c) as in the above we have

ϕ(c) = trace(w) − n = ∑_{s} ψ_{ˆ1ˆ2}(s)

where the sum is over all subwords s ∈ S′. Moreover each ψ_{ˆ1ˆ2}(s) has the form E_{1}G_{1} … E_{u}G_{u}E_{u+1} given in Lemma 7.2 where each G_{i} is one of

± a_{r1}a_{1r}, ±a_{r2}a_{2r}, ±a_{rs}a_{sr} ± a_{r1}a_{1s}, ±a_{r2}a_{2s} ± a_{r2}a_{2s},
±(a_{uv} + a_{u2}a_{2v} + a_{u1}a_{1v} + a_{u1}a_{12}a_{2v}), ±(a_{uv} − a_{u2}a_{2v} − a_{u1}a_{1v} + a_{u2}a_{21}a_{1v}),

where u, v, r ≠ s ≥ 3.

Proof. The exact nature of the G_{i} is the only thing that hasn’t been noticed and this follows from the fact that B_{i} has one of the forms

T_{r}^{±1}T_{1}^{±1}T_{s}^{±1}, T_{r}^{±1}T_{2}^{±1}T_{s}^{±1}, T_{u}^{±1}T_{1}T_{2}T_{v}^{±1}, T_{u}^{±1}T_{2}^{−1}T_{1}^{−1}T_{v}^{±1}

where r, s, u, v ≥ 3. A calculation of each such G_{i} shows that those given in Lemma 7.7 are the only possibilities. □

§8 Proof of Theorem 1.3

Let x = a_{r1}a_{1s} or x = a_{r2}a_{2s} for some 3 ≤ r, s ≤ n. Then using the action of σ_{1} given in §5 we see that for any k ∈ ℤ, k ≥ 0, we can write

σ_{1}^{k}(x) = c_{k11}(a_{r1}a_{1s}) + c_{k12}(a_{r1}a_{12}a_{2s}) + c_{k21}(a_{r2}a_{21}a_{1s}) + c_{k22}(a_{r2}a_{2s}), (8.1)

where the c_{kij} = c_{kij}(x) are polynomials in a_{12}a_{21} = p_{1} − 2.

Lemma 8.1. The polynomials c_{kij} satisfy the following recursion:

c_{kij} + p_{1}c_{k−1 ij} + c_{k−2 ij} = d_{ij}  (8.2)

for all k ≥ 2 where d_{ij} = d_{ij}(x) are polynomials in p_{1} also. In particular for x = a_{r1}a_{1s}, a_{r2}a_{2s} we have

d_{11}(x) = 2, d_{11}(x) = 1, d_{21}(x) = −1, d_{22}(x) = 2.

We also have

c_{k−1 11} = c_{k 22}, c_{k12} = −c_{k21}, c_{k11}+2a_{12}a_{21}c_{k12} = a_{12}a_{21}c_{k−1 11}+c_{k−211}. (8.3)

Proof. This will be by induction on k ≥ 2. We will only deal with the case x = a_{r1}a_{1s}, the other case being similar. A calculation shows that we have

c_{011} = 1, c_{012} = 0, c_{021} = 0, c_{022} = 0, c_{111} = −a_{12}a_{21}, c_{112} = 1, c_{121} = −1, c_{122} = 1,

c_{211} = (a_{12}a_{21} + 1)^{2}, c_{212} = −a_{12}a_{21} − 1, c_{221} = a_{12}a_{21} + 1, c_{222} = −a_{12}a_{21}.

This allows one to check the first case (k = 2).

Now assume that for some k ≥ 2 we have (8.1). Then, since σ_{1}(c_{kij}) = c_{kij} for all k, i, j, we have (using the action of σ_{1} given in §5)

σ_{1}^{k+1}(x) = c_{k11}(a_{r2} + a_{r1}a_{12})(a_{2s} − a_{21}a_{1s}) + c_{k12}(a_{r2} + a_{r1}a_{12})a_{21}a_{1s}
+ c_{k21}a_{r1}a_{12}(a_{2s} − a_{21}a_{1s}) + c_{k22}a_{r1}a_{1s}.

From this we obtain:

c_{k+1 11} = −c_{k11}a_{12}a_{21} + c_{k12}a_{12}a_{21} − c_{k21}a_{12}a_{21} + c_{k22};
c_{k+1 12} = c_{k11} + c_{k21}; c_{k+1 21} = −c_{k11} + c_{k12}; c_{k+1 22} = c_{k11}.

Using these latter equations we can check that the equations (8.3) for k + 1 follow from (8.3) for k. Then one uses (8.3) to likewise prove (8.2) for k + 1. □

Using the recurrence given in Lemma 8.1 we see that standard results for solving non-homogeneous recurrence equations [B] allow one to conclude that

c_{kij} = e_{ij+} (\frac{−p_{1} + \sqrt{p_{1}^{2} − 4}}{2})^{k} + e_{ij−} (\frac{−p_{1} − \sqrt{p_{1}^{2} − 4}}{2})^{k} ,

where

e_{ij−} = \frac{1}{\sqrt{p_{1}^{2} − 4}} ((c_{0ij} − \frac{d_{ij}}{2 + p_{1}} ) (\frac{−p_{1} + \sqrt{p_{1}^{2} − 4}}{2}) − c_{1ij} + \frac{d_{ij}}{2 + p_{1}} ) ,

and e_{ij+} = c_{0ij} − \frac{d_{ij}}{2+p_{1}} − e_{ij−}. One then finds that

e_{11−} = \frac{p_{1} + \sqrt{p_{1}^{2} − 4}}{2(p_{1} + 2)}, e_{11+} = \frac{p_{1} − \sqrt{p_{1}^{2} − 4}}{2(p_{1} + 2)},

and so

c_{k11} = \frac{−1}{p_{1} + 2} ((\frac{−p_{1} + \sqrt{p_{1}^{2} − 4}}{2})^{k+1} + (\frac{−p_{1} − \sqrt{p_{1}^{2} − 4}}{2})^{k+1} − 2) . (8.4)

Lemma 8.2. For an integer k ≥ 0 the degree of

(\frac{−p_{1} + \sqrt{p_{1}^{2} − 4}}{2})^{k} + (\frac{−p_{1} − \sqrt{p_{1}^{2} − 4}}{2})^{k}

is k.

Proof. Expanding the kth powers in this sum using the binomial theorem we see that all terms involving the square root cancel and the term of highest degree is ±p_{1}^{k}. □

We use the above ideas to prove:

Proposition 8.3. Let n, m ≥ 0, r_{i}, s_{i}, t_{i}, u_{i} ≥ 3 and

w = ∏_{i=1}^{n} a_{r_{i}1}a_{1s_{i}}, w′ = ∏_{i=1}^{m} a_{t_{i}2}a_{2u_{i}}

Then Ω(w) = n, Ω(w′) = m and Ω(ww′) = n + m.

Proof. We will prove the third of these results, the proofs of the first and second being similar. We will prove this result by induction on n+m ≥ 0, the cases n+m ≤ 1 following from equations (5.4). By Theorem 1.1 we see that 𝔅_{n+m}(ww′) = 0 so that Ω(ww′) ≤ n + m. We will show that 𝔅_{n+m−1}(ww′) ≠ 0.

Now

𝔅_{n+m−1}(ww′) = ∑_{i=0}^{2(n+m−1)+1} q_{n+m−1 i}σ_{1}^{i}(ww′)

= ∑_{i=0}^{2(n+m)−1} q_{n+m−1 i}σ_{1}^{i} (∏_{j=1}^{n} a_{r_{j}1}a_{1s_{j}} ) × σ_{1}^{i} (∏_{j=1}^{m} a_{t_{j}2}a_{2u_{j}} ) .

But by (8.1) and (8.3) we have

σ_{1}^{i}(a_{r_{j}1}a_{1s_{j}}) = c_{i11}a_{r_{j}1}a_{1s_{j}} + … and σ_{1}^{i}(a_{t_{j}2}a_{2u_{j}}) = c_{i−1 11}a_{t_{j}1}a_{1u_{j}} + ….

Thus the coefficient of ∏_{j=1}^{n} a_{r_{j}1}a_{1s_{j}} × ∏_{j=1}^{m} a_{t_{j}2}a_{2u_{j}} in q_{n+m−1 i}σ_{1}^{i}(ww′) is

q_{n+m−1 i} ∏_{j=1}^{n} c_{i11} ∏_{j=1}^{m} c_{i−1 11} = q_{n+m−1 i}c_{i11}^{n}c_{i−1 11}^{m}.

Now using Lemma 2.1, Lemma 8.2 and equation (8.4) (where we can ignore the −2 and the initial factor of \frac{−1}{p_{1}+2}) we see that this term has degree

d(i) = (2(n + m − 1) − i + 1)i/2 + n(i + 1) + mi.

Now as a function of i (for fixed n, m) d(i) is a quadratic function with a maximum value at i = 2(n + m) − 1/2 and so on the interval [0, 2(n + m) − 1] the function d(i) is increasing. Thus for integer values of i in this interval the maximum value is attained at the unique value i = 2(n + m) − 1. Thus the sum for 𝔅_{n+m−1}(ww′) cannot be zero. This proves that Ω(ww′) = n + m as required. □

We now conclude the proof of Theorem 1.3. Let c be a simple closed curve on D_{n} and let w = w(c). Let η(w) = A_{1}B_{1} … A_{u}B_{u}A_{u+1}. Recall the decomposition of trace(w) − n given in Lemma 7.7 relative to the set S′. Then the set S′ has a maximal element i.e. there is an element S of S′ such that every other element of S′ is a subword of S. Thus every monomial in ψ_{ˆ1ˆ2}(S) has more indices in its a_{rs} factors with r > 2 or s > 2 than does any other ψ_{ˆ1ˆ2}(s′) for s′ ∈ S′, s′ ≠ S. Suppose that k of the B_{i} are T_{1}^{±1} or T_{2}^{±1}. Theorem 1.3 will follow if we can show that 𝔅_{k−1}(σ_{1}, a_{12}a_{21} + 2)(ψ_{ˆ1ˆ2}(S)) ≠ 0.

Now ψ_{ˆ1ˆ2}(S) = E_{1}G_{1} … E_{u}G_{u}E_{u+1} where the E_{i} are monomials in a_{rs} with r, s ≥ 3 and the G_{i} are in the list given in Lemma 7.7. Now 𝔅_{0}(E_{i}) = 0, 𝔅_{0}(a_{rs} + a_{r2}a_{2s} + a_{r1}a_{1s} + a_{r1}a_{12}a_{2s}) = 0 and 𝔅_{0}(a_{rs} − a_{r2}a_{2s} − a_{r1}a_{1s} + a_{r2}a_{21}a_{1s}) = 0 and so we may ignore all terms that come from E_{i} and from terms of the latter two forms. What is left is a product

u = ∏_{i=1}^{n} a_{r_{i}1}a_{1s_{i}} × ∏_{i=1}^{m} a_{t_{i}2}a_{2u_{i}}

for some n, m ≥ 0 with k = n + m. The fact that 𝔅_{k−1}(σ_{1}, a_{12}a_{21} + 2)(u) ≠ 0, which implies that 𝔅_{k−1}(σ_{1}, a_{12}a_{21} + 2)(ψ_{ˆ1ˆ2}(S)) ≠ 0, now follows from Proposition 8.3. This proves Theorem 1.3. □.

Remark 8.4. We here remark that our method gives a way of finding intersection numbers of any two simple closed curves c, c′ on D_{n}. If one of the curves (say c) surrounds 2 punctures, then we can find a diffeomorphism α so that α(c) = γ_{12}. Thus ι(c, c′) = ι(γ_{12}, α(c′)). One then finds the intersection number of γ_{12} and α(c′) using the operators B_{n}(σ_{1}, a_{12}a_{21} + 2).

On the other hand, if c does not surround two punctures, then we think of D_{n} as a subset of D_{n+2} and consider the curve d such that w(d) = w(c)x_{n+1}w(c)^{−1}x_{n+2}. Then d surrounds two punctures and we can calculate the intersection ι(d, c′) = 2ι(c, c′) using the operators B_{n}(σ_{1}, a_{12}a_{21} + 2). See Figure 4 for the relationship between c and d, where we have drawn the curve c and the spine Σ(d) in a particular example.

§9 A matrix interpretation

For n ≥ 1 let R_{n} be the infinite matrix whose ij entry is r_{nij} for i, j ≥ 0. Then by Lemma 2.1 R_{n} has the following block form

( \begin{matrix}I_{2n+1} & 0 \\ U_{n} & 0 \end{matrix} )  (9.1)

where I_{2n+1} is the identity matrix of rank 2n + 1 and each 0 is a matrix of zeros.

Proposition 9.1. For n ≥ 1 we have

(i) R_{n}^{2} = R_{n}; (ii) R_{n}R_{m} = R_{n} if n ≤ m; (iii) R_{n}R_{m} = R_{m}R_{n} for n, m ≥ 1.

Proof. (i) and (ii) are obvious from the block form (9.1) for R_{n} and R_{m}. (iii) will follow from the following result:

Lemma 9.2. For all m ≥ n ≥ 1 and for all j, k ≥ 0 we have

∑_{i=0}^{2m} r_{mji}r_{nik} = r_{njk}.

Proof. The proof will be by induction on j ≥ 0. If j = 0, then we note from Lemma 2.1 that r_{mji} = δ_{ji} and so ∑_{i=0}^{2m} r_{m0i}r_{nik} = r_{n0k} as required. Now assume that ∑_{i=0}^{2m} r_{mji}r_{nik} = r_{njk}. Then since r_{mj 2m+1} = 0 we have (using Lemma 2.1)

∑_{i=0}^{2m} r_{m j+1 i}r_{nik} = ∑_{i=0}^{2m+1} r_{m j+1 i}r_{nik}

= ∑_{i=0}^{2m+1}(−q_{mi}r_{m j 2m} + r_{mj i−1})r_{nik}

= −r_{mj i−1} ∑_{i=0}^{2m+1} −q_{mi}r_{nik} + ∑_{i=0}^{2m+1} r_{mj i−1}r_{nik}.

Note that by Proposition 2.7 the first sum is equal to 0. Thus the above is equal to

∑_{i=0}^{2m+1} r_{mj i−1}r_{nik} = ∑_{i=0}^{2m+1} r_{mj i−1}(−q_{nk}r_{n i−1 2n} + r_{n i−1 k−1})

= −q_{nk} ∑_{i=0}^{2m+1} r_{mj i−1}r_{n i−1 2n} + ∑_{i=0}^{2m+1} r_{mj i−1}r_{n i−1 k−1}

= −q_{nk} ∑_{i=0}^{2m} r_{mj i}r_{n i 2n} + ∑_{i=0}^{2m+1} r_{mj i}r_{n i k−1}

= −q_{nk}r_{n j 2n} + r_{n j k−1} = r_{n j+1 k},

as required. In this last calculation we used Lemma 2.1 and the inductive hypothesis. □

This now proves Proposition 9.1 □

Let Q_{n} be the infinite diagonal matrix diag(q_{n0}, q_{n1}, …, q_{n 2n+1}, 0, 0, …) and let (1) denote the infinite all 1 vector. Let T denote transpose.

Theorem 9.3. a) The set {R_{n}}_{n≥1} is an infinite family of independent commuting idempotents.

b) If P_{1} = R_{1} and P_{n} = R_{n} − R_{n−1} for n > 1, then {P_{n}}_{n≥1} is an infinite family of independent orthogonal idempotents.

c) For n, m ≥ 1 we have R_{m}^{T}Q_{n+m}R_{n} = 0.

d) For 1 ≤ m ≤ n we have R_{m}^{T}Q_{n}(1) = 0.

Proof. a) Follows from Proposition 9.1. b) Using Proposition 9.1 one easily checks that P_{n}^{2} = P_{n} and P_{n}P_{m} = 0 if n ≠ m. The commutativity of P_{n} and P_{m} follows from Proposition 9.1.

Now c) is a direct consequence of the fact that ∑_{i=0}^{2(n+m)+1} q_{n+m i}r_{n,i,j}r_{m,i,k} = 0 for all n, m ≥ 1 and j, k ≥ 0 which was what we proved on the way to proving Theorem 1.1 (see Proposition 2.2 and §4). Lastly, d) follows from Proposition 2.7. □

§10 An application to algebraic intersection numbers of curves on surfaces

Let S_{g} be a closed orientable surface of genus g ≥ 1 and let M_{g} denote the mapping class group of S_{g} [Bi]. Let a_{1}, …, a_{g}, b_{1}, …, b_{g} be a symplectic basis [MKS] for the first homology group H_{1} = H_{1}(S_{g}, ℤ) ≅ ℤ^{2g}, which we will write multiplicatively. Let <, > denote the symplectic form or ‘algebraic intersection number function’, this being well-defined on homology classes. Thus a_{1}, …, a_{g}, b_{1}, …, b_{g} satisfy < a_{i}, b_{i} >= 1 for i = 1, …, g all other products being zero. Let RH_{1} denote the group algebra of H_{1} over the commutative ring with identity R.

Now an element of H_{1} can be represented by a simple closed curve if and only if it is primitive i.e. if and only if its coordinates relative to some (any) basis are relatively prime. For any element c ∈ H_{1} we let

r(c) = −c − c^{−1} ∈ RH_{1}.

We also define the element T_{c} ∈ Aut(H_{1}) = Sp_{2g}(ℤ) by

T_{c}(x) = xc^{<c,x>}

for all x ∈ H_{1}. Then in the situation where c is primitive and represents the simple closed curve γ we note that T_{c} is the symplectic transvection [A] which is the image of the Dehn twist Γ ∈ M_{g} about γ under the canonical map π_{g} : M_{g} → Sp_{2g}(ℤ). Here π_{g} just gives the action of M_{g} on H_{1} [MKS]. Then T_{c} extends to an automorphism of the group ring RH_{1}.

Theorem 10.1. Let c ∈ H_{1} be primitive and let d ∈ H_{1}. Let 𝔅_{n}(T_{c}, r(c)), n > 0, be the operators on the group ring RH_{1} defined in Theorem 1.1. Then | < c, d > | is equal to the minimal number m ≥ 0 such that 𝔅_{m}(T_{c}, r(c))(d) = 0.

Proof. Up to an action of Sp_{2g}(ℤ) we can assume that c = a_{1} (since M_{g} acts transitively on non-bounding simple closed curves in S_{g}). Then we can write d = a_{1}^{e_{1}}b_{1}^{f_{1}} … a_{g}^{e_{g}}b_{g}^{f_{g}} where < c, d >= f_{1}. Now note that T_{a_{1}}(a_{i}) = a_{i} for i = 1, …, g and that T_{a_{1}}(b_{j}) = b_{j} for j = 2, …, g. It follows that we may as well take d = b_{1}^{k} where k =< c, d >. Now if the polynomials p_{n} = p_{n}(r(c)) are as defined in Theorem 1.1, then it is easy to check that

p_{n}(r(c)) = −c^{n} − c^{−n}

for all n ≥ 0. We also have T_{a_{1}^{m}}(d) = T_{a_{1}^{m}}(b_{1}^{k}) = (a_{1}^{m}b_{1})^{k} for all m, k ∈ ℤ. Now 𝔅_{0}(T_{a_{1}}, r(a_{1}))(b_{1}) = a_{1}b_{1} − b_{1} ≠ 0 and

𝔅_{1}(T_{a_{1}}, r(a_{1}))(b_{1}) = (T_{a_{1}}^{2} − (a_{1} + a_{1}^{−1})T_{a_{1}} + 1)(T_{a_{1}} − 1)(b_{1})
= (T_{a_{1}}^{2} − (a_{1} + a_{1}^{−1})T_{a_{1}} + 1)(a_{1}b_{1} − b_{1})
= a_{1}^{3}b_{1} − a_{1}^{2}b_{1} − (a_{1} + a_{1}^{−1})(a_{1}^{2}b_{1} − a_{1}b_{1}) + a_{1}b_{1} − b_{1}
= 0,  (10.1)

and similarly that 𝔅_{1}(T_{a_{1}}, r(a_{1}))(b_{1}^{−1}) = 0 as required. It follows from Theorem 1.1 that 𝔅_{|k|}(T_{a_{1}}, r(a_{1}))(b_{1}^{k}) = 0 for all k ∈ ℤ. Theorem 10.1 will follow if we can show that 𝔅_{k}(T_{a_{1}}, r(a_{1}))(b_{1}^{k+1}) ≠ 0 for all k ≥ 0 (the cases where k < 0 are similar).

Lemma 10.2. Let 1 ≤ n < j. Then

𝔅_{n}(T_{a_{1}}, r(a_{1}))(b_{1}^{j}) = (∑_{i} μ_{i}(n, j)a_{1}^{i})b_{1}^{j},  (10.2)

where μ_{i}(n, j) ∈ R and μ_{j(2n+1)}(n, j) = 1 and μ_{i}(n, j) = 0 for i > j(2n + 1).

Proof. Since the action of T_{a_{1}}^{k} only multiplies each element of H_{1} by some power of a_{1} and since 𝔅_{i}(T_{a_{1}}, r(a_{1})) is a sum of such operators with coefficients in R[a_{1}] one sees that we do have (10.2). We prove the rest by induction on n for all values of j > n. The case n = 1 follows (for all j > n) from a calculation similar to (10.1). Now assume that we for some n we have 𝔅_{n}(T_{a_{1}}, r(a_{1}))(b_{1}^{j}) = (∑_{i} μ_{i}(n, j)a_{1}^{i})b_{1}^{j} where μ_{j(2n+1)}(n, j) = 1 and μ_{i}(n, j) = 0 for i > j(2n + 1). Note that according to the theorem, in order to consider the n + 1 case, we must assume j > n + 1 and so we have

𝔅_{n+1}(T_{a_{1}}, r(a_{1}))(b_{1}^{j}) = 𝔄_{n+1}(T_{a_{1}}, r(a_{1}))𝔅_{n}(T_{a_{1}}, r(a_{1}))(b_{1}^{j})
= (T_{a_{1}}^{2} − (a_{1}^{n+1} + a_{1}^{−(n+1)})T_{a_{1}} + 1)(a_{1}^{j(2n+1)}b_{1}^{j} + …)

where … indicates terms of lower degree. Thus the terms of highest degree in the above are

T_{a_{1}}^{2}(a_{1}^{j(2n+1)}b_{1}^{j}) = a_{1}^{j(2n+1)}(a_{1}^{2}b_{1})^{j} and
a_{1}^{n+1}T_{a_{1}}(a_{1}^{j(2n+1)}b_{1}^{j}) = a_{1}^{j(2n+1)}a_{1}^{n+1}(a_{1}b_{1})^{j}.

But since j > n+1 one checks that the first of these has the largest degree. Lemma 10.2 and Theorem 10.1 now both follow. □

Remark 10.3. We remark that Theorem 10.1 can easily be extended to the situation of any orientable surface. In fact it can be generalised to any compact triangulated orientable homology (4n+2)-manifold X (n ≥ 0), where we consider the symplectic structure on the cohomology group H^{2n+1}(X; ℚ) given by the cup product.

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