AN EXPOSITION OF RESULTS OF FRICKE AND VOGT

WILLIAM M. GOLDMAN

Contents

1. Traces in SL(2, ℂ) 2 2. Surjectivity of characters of pairs 6 3. Injectivity of characters of pairs 8 4. Trace relations for triples 10 5. Surjectivity in rank 3 13 References 17

The purpose of this paper is to give an elementary and relatively self-contained proof of the following basic theorem, apparently due to Fricke [6, 7] and, earlier by Vogt [17] Let G = SL(2, ℂ) and let G act on H = SL(2, ℂ) × SL(2, ℂ) by

g · (ξ, η) ⟼ (gξg^{−1}, gηg^{−1}).

Let κ(x, y, z) = x^{2} + y^{2} + z^{2} − xyz − 2.

Theorem (Vogt [17],Fricke [6]). Let f : H ⟶ ℂ be a regular function which is invariant under the above G-action. Then there exists a polynomial function F (x, y, z) ∈ ℂ[x, y, z] such that

f(ξ, η) = F (tr(ξ), tr(η), tr(ξη)).

For every (x, y, z) ∈ ℂ^{3}, there exists (ξ, η) ∈ H such that

[\begin{matrix}x \\ y \\ z \end{matrix}] = [\begin{matrix}tr(ξ) \\ tr(η) \\ tr(ξη) \end{matrix}] .

Conversely, if κ(x, y, z) ≠ 2 and (ξ, η), (ξ′, η′) ∈ H satisfy

(1)  [\begin{matrix}tr(ξ) \\ tr(η) \\ tr(ξη) \end{matrix}] = [\begin{matrix}tr(ξ′) \\ tr(η′) \\ tr(ξ′η′) \end{matrix}] = [\begin{matrix}x \\ y \\ z \end{matrix}] ,

Date: September 25, 2005.

then (ξ′, η′) = g.(ξ, η) for some g ∈ G.

We regard H as a variety of representations of a free group. Let π be the group freely generated by elements X, Y, which we call letters. We call X, Y, X^{−1}, Y^{−1} symbols. An element of π is a reduced word w(X, Y), that is a sequence of symbols such that no symbol follows its inverse. The length of a reduced word w(X, Y) is denoted ℓ(w). The empty reduced word has length 0 and is the identity element 𝕀 ∈ π. Reduced words are multiplied by concatenation subject to cancellation relations of the form

XX^{−1} = X^{−1}X = Y Y^{−1} = Y^{−1}Y = 𝕀.

The Cartesian product H identifies with the set Hom(π, G) of all representations π ⟶ G under the correspondence

Hom(π, G) ⟶ H = G × G
ρ ⟼ (ρ(X), ρ(Y)).

Since π is generated by X, Y subject to no relations, this map is an isomorphism. Then the Vogt-Fricke theorem is the statement that the categorical quotient of Hom(π, G) by G is affine 3-space ℂ^{3}.

This result applies to non-Euclidean geometry. Let M be a compact surface homeomorphic to a three-holed sphere. The space of hyperbolic- or elliptic- geometric structures on M is then parametrized by the lengths of the boundary curves, which relate to the traces of the corresponding representation of the fundamental group. See Buser [3], Fenchel [5], Keen [10, 11, 12], and Harvey [8]. In [12], Keen uses Fricke moduli to parametrize the Teichmüller space of a one-holed torus and other surfaces.

1. Traces in SL(2, ℂ)

1.1. Cyclic groups. The analogous result for one generator is the following.

Theorem 1. Let f : G ⟶ ℂ be a polynomial function invariant under inner automorphisms of G. Then there exists a polynomial F (t) ∈ ℂ[t] such that f(g) = F (tr(g)). Conversely, if g, g′ ∈ G satisfy tr(g) = tr(g′) ≠ ±2, then there exists h ∈ G such that g′ = hgh^{−1}.

Proof. Suppose f is an invariant function. For t ∈ ℂ, define

ξ_{t} := [\begin{matrix}t & −1 \\ 1 & 0 \end{matrix}]

and define F (t) by

F (t) = f(ξ_{t}).

Suppose that t ≠ ±2 and tr(g) = t. Then both g and ξ_{t} have distinct eigenvalues

λ_{±} = \frac{1}{2} (t ± (t^{2} − 4)^{1/2})

and hgh^{−1} = ξ_{t} for some h ∈ G. Thus

f(g) = f(h^{−1}ξ_{t}h) = f(ξ_{t}) = F (t)

as desired. If t = ±2, then by taking Jordan normal form, either g = ±𝕀 or g is conjugate to ξ_{t}. In the latter case, f(g) = F (t) follows from invariance. Otherwise g lies in the closure of the G-orbit of ξ_{t} and f(g) = f(ξ_{t}) = F (t) follows by continuity of f.

The converse direction follows from Jordan normal form as already used above. □

1.2. The Cayley-Hamilton theorem. A principal tool in this investigation is the Cayley-Hamilton theorem. If ξ is a 2 × 2-matrix,

(2)  ξ^{2} − tr(ξ)ξ + det(ξ)𝕀 = 0.

Suppose ξ, η ∈ SL(2, ℂ). Multiplying (2) by ξ^{−1} and rearranging,

(3)  ξ + ξ^{−1} = tr(ξ)𝕀

from which follows

(4)  tr(ξ) = tr(ξ^{−1})

Multiplying (3) by η and taking traces, we obtain:

Theorem 2 (The Basic Identity). Let ξ, η ∈ SL(2). Then

(5)  tr(ξη) + tr(ξη^{−1}) = tr(ξ)tr(η).

1.3. Traces of reduced words. Here is an important special case of the Vogt-Fricke theorem. Namely, let w(X, Y) ∈ π be a reduced word. Then

H ⟶ ℂ
(ξ, η) ⟼ trw(ξ, η)

is a G-invariant function on H and by the Vogt-Fricke theorem there exists a polynomial f_{w}(x, y, z) ∈ ℂ[x, y, z] such that

(6)  trw(ξ, η) = f_{w}(trξ, trη, tr(ξη))

for all ξ, η ∈ G. We describe an algorithm for computing f_{w}(x, y, z). For notational convenience we shall write t(w(X, Y)) for

f_{w(X,Y)}(x, y, z).

For example, t(𝕀) = 2 and

t(X^{−1}) = t(X) = x,
t(Y^{−1}) = t(Y) = y,
t(Z^{−1}) = t(Z) = z,

verifying assertion (6) for words w of length ℓ(w) ≤ 1. The reduced words of length two are

X^{2}, Y^{2}, XY, XY^{−1}, Y X, Y X^{−1}

and applications of the trace identities imply:

t(X^{2}) = x^{2} − 2
t(Y^{2}) = y^{2} − 2
t(Z^{2}) = z^{2} − 2
t(XY^{−1}) = xy − z
t(Y Z^{−1}) = yz − x
t(ZX^{−1}) = zx − y

For example, for X^{2},

tr(ξ^{2}) = tr(ξξ) = tr(ξ)tr(ξ) − tr(ξξ^{−1}) = tr(ξ)^{2} − 2.

and, for XY^{−1},

tr(ξη^{−1}) = tr(ξ)tr(η) − tr(ξη),

etc. For example

(7)  t(XY X^{−1}Y) = t(XY)t(X^{−1}Y) − t(X^{2})
= z(xy − z) − (x^{2} − 2)
= 2 − x^{2} − z^{2} + xyz.

Assume inductively that for all reduced words w(X, Y) ∈ π with ℓ(w) < l, there exists a polynomial f_{w}(x, y, z) = t(w(X, Y)) satisfying (6). Suppose that u(X, Y) ∈ π is a reduced word of length ℓ(u) = l.

The explicit calculations above begin the induction for l ≤ 2. Thus we assume l > 2.

Furthermore, we can assume that u is cyclically reduced, that is the initial symbol of u is not inverse to the terminal symbol of u. For otherwise

u(X, Y) = Su′(X, Y)S^{−1},

where S is one of the four symbols

X, Y, X^{−1}, Y^{−1}

and ℓ(u′) = l − 2. Then u(X, Y) and u′(X, Y) are conjugate and

t(u(X, Y)) = t(u′(X, Y)).

If l > 2, and u is cyclically reduced, then u(X, Y) has a repeated letter, which we may assume to equal X. That is, we may write

u(X, Y) = u_{1}(X, Y)u_{2}(X, Y)

where u_{1} and u_{2} are reduced words each ending in X^{±1}. Furthermore we may assume that

ℓ(u_{1}) + ℓ(u_{2}) = ℓ(u) = l,

so that ℓ(u_{1}) < l and ℓ(u_{l}) < l. Suppose first that u_{1} and u_{2} both end in X. Then

u(X, Y) = (u_{1}(X, Y)X^{−1}) X (u_{2}(X, Y)X^{−1}) X

and each

u_{1}(X, Y)X^{−1}, u_{2}(X, Y)X^{−1}

is reduced. Then

(u_{1}(X, Y)X^{−1})(u_{2}(X, Y)X^{−1})^{−1} = u_{1}(X, Y)u_{2}(X, Y)^{−1}

is represented by a reduced word u_{3}(X, Y) satisfying ℓ(u_{3}) < l. By the induction hypothesis, there exist polynomials

t(u_{1}(X, Y)), t(u_{2}(X, Y)), t(u_{3}(X, Y)) ∈ ℂ[x, y, z]

such that, for all ξ, η ∈ G, i = 1, 2, 3,

tr(u_{i}(ξ, η)) = t(u_{i}(X, Y))(tr(ξ), tr(η), tr(η)).

By (5),

t(u(X, Y)) = t(u_{1}(X, Y))t(u_{2}(X, Y)) − t(u_{3}(X, Y))

is a polynomial in ℂ[x, y, z]. The cases when u_{1} and u_{2} both end in the symbols X^{−1}, Y, Y^{−1} are completely analogous. Since there are only four symbols, the only cyclically reduced words without repeated symbols are commutators of the symbols, for example XY X^{−1}Y^{−1}. Repeated applications of the trace identities give:

t(XY X^{−1}Y^{−1}) = t(XY X^{−1})t(Y^{−1}) − t(XY X^{−1}Y)

= y^{2} − (2 − x^{2} − z^{2} + xyz) = κ(x, y, z). (8)

The other commutators of distinct symbols also have trace κ(x, y, z) by identical arguments.

Thus every w(X, Y) determines a polynomial f_{w}(x, y, z) such that

trw(ξ, η) = f_{w}(tr(ξ), tr(η), tr(ξη))

for ξ, η ∈ SL(2, ℂ).

2. Surjectivity of characters of pairs

We first show that

τ : H ⟶ ℂ^{3}
(ξ, η) ⟼ [\begin{matrix}trξ \\ trη \\ trξη \end{matrix}] .

is surjective. Choose ζ ∈ ℂ so that

ζ + ζ^{−1} = z,

that is, ζ = \frac{1}{2}(z ± \sqrt{z^{2} − 4}). Let

(9)  ξ_{x} = [\begin{matrix}x & −1 \\ 1 & 0 \end{matrix}] , η_{(y,ζ)} = [\begin{matrix}0 & ζ^{−1} \\ ζ & y \end{matrix}] .

Then τ(ξ_{x}, η_{(y,ζ)}) = (x, y, z).

Next we show that every G-invariant regular function f : H ⟶ ℂ factors through τ. To this end we need the following elementary lemma on symmetric functions:

Lemma 3. Let R be an integral domain where 2 is invertible, and let R′ = R[ζ, ζ^{−1}] be the ring of Laurent polynomials over R. Let R′ ⟶^{σ} R′ be the involution which fixes R′ and interchanges ζ and ζ^{−1}. Then the subring of σ-invariants is the polynomial ring R[ζ + ζ^{−1}].

Proof. Let F (ζ, ζ^{−1}) ∈ R[ζ, ζ^{−1}] be a σ-invariant Laurent polynomial. Begin by rewriting R′ as the quotient of the polynomial ring R[x, y] by the ideal generated by xy − 1. Then σ is induced by the involution σ̃ of R[x, y] interchanging x and y. Let f(x, y) ∈ R[x, y] be a polynomial whose image in R′ is F. Then there exists a polynomial g(x, y) such that

f(x, y) − f(y, x) = g(x, y)(xy − 1).

Clearly g(x, y) = −g(y, x). Let

f̃(x, y) = f(x, y) − \frac{1}{2} g(x, y)(xy − 1)

so that f̃(x, y) = f̃(y, x). By the theorem on elementary symmetric functions,

f̃(x, y) = h(x + y, xy)

for some polynomial h(u, v). Therefore F (ζ, ζ^{−1}) = h(ζ + ζ^{−1}, 1) as desired. □

By definition f(ξ, η) is a polynomial in the matrix entries of ξ and η; two polynomials which differ by elements in the ideal generated by det(ξ)−1 and det(η)−1 are regarded as equal. Thus f(ξ_{x}, η_{(y,ζ)}) equals a function g(x, y, ζ) which is a polynomial in x, y ∈ ℂ and a Laurent polynomial in ζ ∈ ℂ^{∗}.

Lemma 4. Let ξ, η ∈ G such that κ(τ(ξ, η)) ≠ 2. Then there exists g ∈ G such that

g · (ξ, η) = (ξ^{−1}, η^{−1}).

Proof. Let (x, y, z) = τ(ξ, η). By a simple calculation

tr[ξ, η] = κ(x, y, z)

where [ξ, η] = ξηξ^{−1}η^{−1}.

Let L = ξη − ηξ. (Compare §4 of Jørgensen [9] or Fenchel [5].) Then tr(L) = tr(ξη) − tr(ηξ) = 0. Furthermore for any 2 × 2 matrix M, the characteristic polynomial

λ_{M}(t) := det(t𝕀 − M) = t^{2} − tr(M)t + det(M).

Thus

det(L) = det([ξ, η] − I)det(ηξ)
= det([ξ, η] − I)
= −λ_{[ξ,η]}(1)
= −2 + tr[ξ, η]
= −2 + κ(x, y, z) ≠ 0.

Choose μ ∈ ℂ^{∗} such that μ^{2}det(L) = 1 and let g = μL ∈ G.

Since tr(g) = 0 and det(g) = 1, the Cayley-Hamilton Theorem λ_{M}(M) = 0 implies that g^{2} = −𝕀. Similarly

det(gξ) = det(g) = 1
det(gη) = det(g) = 1,

and

tr(gξ) = μ(tr((ξη)ξ) − tr((ηξ)ξ))
= μ(tr(ξ(ηξ)) − tr((ηξ)ξ)) = 0

and

tr(gη) = μ(tr((ξη)η) − tr((ηξ)η))
= μ(tr((ξη)η) − tr(η(ξη)) = 0

so (gξ)^{2} = (gη)^{2} = −𝕀. Thus gξg^{−1}ξ = −gξgξ = 𝕀 whence gξg^{−1} = ξ^{−1}. Similarly gηg^{−1} = η^{−1}, concluding the proof of the lemma. □

Apply Lemma 4 to ξ = ξ_{x} and η = η_{(y,ζ)} as above to obtain g such that conjugation by g maps

ξ ⟼ ξ^{−1} = [\begin{matrix}0 & 1 \\ −1 & x \end{matrix}]

and

η ⟼ η^{−1} = [\begin{matrix}y & −1/ζ \\ ζ & 0 \end{matrix}] .

Let

h = [\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}]

Then

hgξ(hg)^{−1} = hξ^{−1}h^{−1} = [\begin{matrix}x & −1 \\ 1 & 0 \end{matrix}] = ξ

and

hgη(hg)^{−1} = hη^{−1}h^{−1} = [\begin{matrix}0 & ζ \\ −1/ζ & y \end{matrix}] .

Thus

g(x, y, ζ) = f(ξ, η)
= f(hgξ(hg)^{−1}, hgη(hg)^{−1})
= g(x, y, ζ^{−1}).

Lemma 3 implies that there exists a polynomial F ∈ ℂ[x, y, z] such that

(10)  g(x, y, ζ) = F (x, y, ζ + 1/ζ)

whenever κ(x, y, ζ + 1/ζ) ≠ 2. Since this condition defines a nonempty Zariski-dense open set, (10) holds on all of ℂ^{2} × ℂ^{∗} and f(ξ, η) = F (tr(ξ), tr(η), tr(ξη)) as claimed.

3. Injectivity of characters of pairs

Finally we show that if (ξ, η), (ξ′, η′) ∈ H satisfy (1) and κ(x, y, z) ≠ 2, then (ξ, η) and (ξ′, η′) are G-equivalent. By §1.3, the triple

[\begin{matrix}x \\ y \\ z \end{matrix}] = [\begin{matrix}trξ \\ trη \\ tr(ξη) \end{matrix}]

determines the character function

χ : π ⟶ ℂ
w(X, Y) ⟼ trw(ξ, η) = f_{w}(x, y, z).

Let ρ and ρ′ denote the representations π ⟶ G taking X, Y to ξ, η and ξ′, η′ respectively and let χ, χ′ denote their respective characters. Then our hypothesis (1) implies that χ = χ′.

Lemma 5. Let ρ : π ⟶ SL(2, ℂ) be a representation and (x, y, z) be as above. Then ρ is irreducible if and only if κ(x, y, z) ≠ 2.

Proof. Suppose first that ρ is reducible. If X, Y generate a representation with an invariant subspace of ℂ^{2} of dimension one, this representation is conjugate to one in which ρ(X) and ρ(Y) are upper-triangular. Denoting their diagonal entries by a, a^{−1} and b, b^{−1} respectively, the diagonal entries of ρ(XY) are ab^{±1}, a^{−1}b^{∓1}. Thus

x = a + a^{−1},
y = b + b^{−1},
z = ab^{±1} + a^{−1}b^{∓1}.

By direct computation, κ(x, y, z) = 2.

Conversely, suppose that κ(x, y, z) = 2. Let 𝔄 ⊂ M_{2}(ℂ) denote the linear span of 𝕀, ξ, η, ξη. Identities derived from the Cayley-Halmilton theorem (2) such as (3) imply that 𝔄 is a subalgebra of M_{2}(ℂ). For example, ξ^{2} is the linear combination −1 + xξ and

ηξ = (z − xy)1 + yξ − xη − ξη.

In the basis of M_{2}(ℂ) by elementary matrices, the map

ℂ^{4} ⟶ M_{2}(ℂ)

[\begin{matrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix}] ⟼ x_{1}𝕀 + x_{2}ξ + x_{3}η + x_{4}ξη

has determinant 2 − κ(x, y, z) = 0 and is not surjective. Thus 𝔄 is a proper subalgebra of M_{2}(ℂ) and the representation is reducible. □

Thus ρ and ρ′ are irreducible representations on ℂ^{2}. Burnside’s Theorem (see Lang [13], p.445) implies the corresponding representations (also denoted ρ, ρ′ respectively) of the group algebra ℂπ into M_{2}(ℂ) are surjective. Since the trace form

M_{2}(ℂ) × M_{2}(ℂ) ⟶ ℂ
(A, B) ⟼ tr(AB)

is nondegenerate, the kernel K of ρ : ℂπ ⟶ M_{2}(ℂ) consists of all

∑_{α∈π} a_{α}α ∈ ℂπ

such that

0 = tr(( ∑_{α∈π} a_{α}ρ(α))ρ(β))

= ∑_{α∈π} a_{α}trρ(αβ)

= ∑_{α∈π} a_{α}χ_{αβ}

for all β ∈ π. Thus the kernels of both representations of ℂπ are equal, and ρ and ρ′ induce algebra isomorphisms

ρ̃, ρ̃′ : ℂπ/K ⟶ M_{2}(ℂ)

respectively. The composition ρ̃′∘ρ^{−1} is an automorphism of the algebra M_{2}(ℂ), which must be induced by conjugation by g ∈ GL(2, ℂ). (See, for example, Corollary 9.122, p.734 of Rotman [16].) In particular ρ′(γ) = gρ(γ)g^{−1} as desired.

4. Trace relations for triples

Finally, we consider the case of triples (ξ, η, ζ) ∈ G, or equivalently representations of the free group of rank three. Unlike the rank two case, the quotient is no longer an affine space. Rather, the coordinate ring of the quotient has dimension six, but generated by the eight functions

f_{X}, f_{Y} , f_{Z}, f_{XY} , f_{Y Z}, f_{ZX}, f_{XY Z}, f_{XZY}

subject to the two relations expressing the sum and product of traces of the length 3 monomials in terms of traces of monomials of length 1 and 2:

f_{XY Z} + f_{XZY} = f_{XY} f_{Z} + f_{Y Z}f_{X} + f_{ZX}f_{Y} + f_{X}f_{Y} f_{Z}
f_{XY Z} f_{XZY} = (f_{X})^{2} + (f_{Y})^{2} + (f_{Z})^{2}
+ f_{XY}^{2} + f_{Y Z}^{2} + f_{ZX}^{2}

− (f_{X}f_{Y} f_{XY} + f_{Y} f_{Z}f_{Y Z} + f_{Z}f_{X}f_{ZX})

+ f_{XY} f_{Y Z}f_{ZX} − 4

We call these two identities Fricke’s Sum and Product Relations respectively. Fricke’s Sum Relation.

(11)  f_{XY Z} + f_{XZY} = f_{XY} f_{Z} + f_{Y Z}f_{X} + f_{ZX}f_{Y} + f_{X}f_{Y} f_{Z}

To prove this formula, apply the Basic Identity three times:

(12)  t(XY Z) + t(XY Z^{−1}) = t(XY)t(Z)
(13)  t(Z^{−1}XY) + t(Z^{−1}XY^{−1}) = t(Z^{−1}X)t(Y)

= (t(Z)t(X) − t(ZX))t(Y)

(14) t(Y^{−1}Z^{−1}X) + t(Y^{−1}Z^{−1}X^{−1}) = t(Y^{−1}Z^{−1})t(X)
= t(Y Z)t(X)

Now subtract (13) from the sum of (12) and (14), using the facts

t(XY Z^{−1}) = t(Z^{−1}XY)
t(Z^{−1}XY^{−1}) = t(Y^{−1}Z^{−1}X)
t(Z^{−1}X^{−1}Y^{−1}) = t(Y^{−1}Z^{−1}X^{−1}) = t(XZY)

to obtain

t(XY Z) + t(XZY) = t(XY)t(Z)
− t(Z)t(X)t(Y) + t(ZX)t(Y)
+ t(Y Z)t(X)

from which (11) follows. Fricke’s Product Relation.

(15) f_{XY Z} f_{XZY} = (f_{X})^{2} + (f_{Y})^{2} + (f_{Z})^{2}
+ f_{XY}^{2} + f_{Y Z}^{2} + f_{ZX}^{2}

− (f_{X}f_{Y} f_{XY} + f_{Y} f_{Z}f_{Y Z} + f_{Z}f_{X}f_{ZX})

+ f_{XY} f_{Y Z}f_{ZX} − 4

We derive this formula in several steps. A direct application of the Basic Identity (5) yields:

(16)  t(ZXZY) = t(ZX)t(ZY) − t(XY^{−1})
= t(ZX)t(ZY) − (t(X)t(Y) − t(XY))
= t(ZX)t(ZY) − t(X)t(Y) + t(XY)

By (7) applied to X, Z^{−1}:

(17) t(XZ^{−1}X^{−1}Z^{−1}) = t(X)t(Z)t(ZX) − t(ZX)^{2} − t(X)^{2} + 2

(18) t(XY ZXZY) = t(XY)t(ZXZY)
− t(XZ^{−1}X^{−1}Z^{−1})
= t(XY) (t(ZX)t(ZY) − t(X)t(Y) + t(XY))

− (t(X)t(Z)t(ZX) − t(ZX)^{2} − t(X)^{2} + 2) (by (16) and (17))

= t(XY)t(ZX)t(Y Z)

− t(X)t(Y)t(XY) − t(Z)t(X)t(ZX)
+ t(XY)^{2} + t(X)^{2} − 2

Finally, appplying (18) and the Commutator Identity (8) to Y, Z:

t(XY Z)t(XZY) = t(XY ZXZY) + t(Y ZY^{−1}Z^{−1})
= (t(XY)t(ZX)t(Y Z) − t(X)t(Y)t(XY)

− t(Z)t(X)t(ZX) + t(XY)^{2} + t(ZX)^{2} + t(X)^{2} − 2)

+ (t(Y)^{2} + t(Z)^{2} + t(Y Z)^{2} − t(Y)t(Z)t(Y Z) − 2)

from which (15) follows.

4.1. The coordinate ring is a quadratic extension. If π is freely generated by X, Y, Z, then the ring of G-invariant polynomials on Hom(π, G) is a quadratic extension of the polynomial ring

ℂ[f_{X}, f_{Y} , f_{Z}, f_{XY} , f_{Y Z}, f_{ZX}].

The algebraic generator λ (which corresponds to f_{XY Z} or f_{XZY}) satisfies the quadratic equation

λ^{2} − f_{Σ}λ + f_{Π} = 0

where

f_{Σ} = f_{XY} f_{Z} + f_{Y Z}f_{X} + f_{ZX}f_{Y} − f_{X}f_{Y} f_{Z}

and

f_{Π} = (f_{X}^{2} + f_{Y}^{2} + f_{Z}^{2})
+ (f_{XY}^{2} + f_{Y Z}^{2} + f_{ZX}^{2})
− (f_{X}f_{Y} f_{XY} + f_{Y} f_{Z}f_{Y Z} + f_{Z}f_{X}f_{ZX})
+ f_{XY} f_{Y Z}f_{ZX} − 4

(Compare Magnus [15].)

5. Surjectivity in rank 3

Let V be the SL(2)-character variety of the (rank three) free group π = ⟨A_{1}, A_{2}, A_{3}⟩. In trace coordinates

t_{I}([ρ]) := tr(ρ(A_{i_{1}}A_{i_{2}} … A_{i_{k}}))

where

I = (i_{1}, i_{2}, … i_{k}),

V ⊂ ℂ^{8} is the codimension two subvariety defined by the two equations:

(19) t_{123} + t_{132} = t_{12}t_{3} + t_{13}t_{2} + t_{23}t_{1} + t_{1}t_{2}t_{3}
(20)  t_{123} t_{132} = (t_{1}^{2} + t_{2}^{2} + t_{3}^{2}) + (t_{12}^{2} + t_{23}^{2} + t_{13}^{2}) +
(t_{1}t_{2}t_{12} + t_{2}t_{3}t_{23} + t_{3}t_{1}t_{13}) + t_{12}t_{23}t_{13} − 4.

By eliminating t_{132} in (19) as

t_{132} = t_{12}t_{3} + t_{13}t_{2} + t_{23}t_{1} + t_{1}t_{2}t_{3} − t_{123},

the variety V may be expressed as the hypersurface in ℂ^{7} consisting of all

(t_{1}, t_{2}, t_{3}, t_{12}, t_{23}, t_{13}) ∈ ℂ^{7}

satisfying

t_{123} (t_{12}t_{3} + t_{13}t_{2} + t_{23}t_{1} + t_{1}t_{2}t_{3} − t_{123}) =
(t_{1}^{2} + t_{2}^{2} + t_{3}^{2}) + (t_{12}^{2} + t_{23}^{2} + t_{13}^{2}) +(t_{1}t_{2}t_{12} + t_{2}t_{3}t_{23} + t_{3}t_{1}t_{13}) + t_{12}t_{23}t_{13} − 4.

Proposition 6. The projection

V ⟶^{t} ℂ^{6}

[ρ] ⟼ [\begin{matrix}t_{1}(ρ) \\ t_{2}(ρ) \\ t_{3}(ρ) \\ t_{12}(ρ) \\ t_{23}(ρ) \\ t_{13}(ρ) \end{matrix}]

is surjective.

The polynomial ring

ℂ[t] := ℂ[t_{1}, t_{2}, t_{3}, t_{12}, t_{23}, t_{13}]

is the coordinate ring of affine space ℂ^{6}. Projection V ⟶^{t} ℂ^{6} is a 2-1 map corresponding to the quadratic extension

ℂ[V] ≅ ℂ[t][z] / (z^{2} − P (t)z + Q(t))

where P, Q are the polynomials appearing above:

P (t) = t_{12}t_{3} + t_{13}t_{2} + t_{23}t_{1} + t_{1}t_{2}t_{3}
Q(t) = (t_{1}^{2} + t_{2}^{2} + t_{3}^{2}) + (t_{12}^{2} + t_{23}^{2} + t_{13}^{2})+
(t_{1}t_{2}t_{12} + t_{2}t_{3}t_{23} + t_{3}t_{1}t_{13}) + t_{12}t_{23}t_{13} − 4.

Proof. By the Vogt-Fricke theorem in rank two, there exist A_{1}, A_{2} ∈ SL(2) such that

tr(A_{1}) = t_{1},
(21)  tr(A_{2}) = t_{2},
tr(A_{1}A_{2}) = t_{12}.

We seek A_{3} ∈ SL(2) such that

tr(A_{3}) = t_{3},
tr(A_{2}A_{3}) = t_{23},
(22)  tr(A_{1}A_{3}) = t_{13}.

To this end, consider the affine subspace W of M_{2}(ℂ) consisting of matrices W satisfying

tr(W) = t_{3},
(23)  tr(A_{2}W) = t_{23},
tr(A_{1}W) = t_{13}.

Since the bilinear pairing

M_{2}(ℂ) × M_{2}(ℂ) ⟶ ℂ
(X, Y) ⟼ tr(XY)

is nondegenerate, each of the three equations in (23) describes an affine hyperplane in M_{2}(ℂ). We first suppose that κ(t_{1}, t_{2}, t_{12}) ≠ 2, that is,

(24)  4 − t_{1}^{2} − t_{2}^{2} − t_{12}^{2} + t_{1}t_{2}t_{12} ≠ 0.

Lemma 7. There exists A_{1}, A_{2} ∈ SL(2) satisfying (21) such that W is nonempty.

Proof. Since κ(t_{1}, t_{2}, t_{12}) ≠ 2. the pair A_{1}, A_{2} generates an irreducible representation. In particular {𝕀, A_{1}, A_{2}} is a linearly independent subset of the 4-dimensional vector space M_{2}(ℂ). The three conditions of (23) are independent, so W ⊂ M_{2}(ℂ) is an affine line. □

Let W_{0}, W_{1} ∈ W be distinct elements in this line. Then the function function

ℂ ⟶ ℂ
s ⟼ det(sW_{1} + (1 − s)W_{0})

is polynomial of degree ≤ 2, and is thus onto unless it is constant.

Lemma 8. Let W_{0}, W_{1} ∈ M_{2}(ℂ). Then

det(W_{0} + s(W_{1} − W_{0})) =
det(W_{0}) + s (tr(W_{0})tr(W_{1} − W_{0}) − tr(W_{0}(W_{1} − W_{0}))
+ s^{2} det(W_{1} − W_{0})

Proof. Clearly

(25)  tr(W_{0} + s (W_{1} − W_{0})) = tr(W_{0}) + s tr(W_{1} − W_{0}).

Now

(26)  det(W) = \frac{tr(W)^{2} − tr(W^{2})}{2}

whenever W ∈ M_{2}(ℂ). Now apply (26) to (25) taking

W = W_{0} + s(W_{1} − W_{0})
□

Thus the restriction det|_{W} is constant only if det(W_{1} − W_{0}) = 0. Work in the slice

A_{1} = [\begin{matrix}t_{1} & −1 \\ 1 & 0 \end{matrix}] , A_{2} = [\begin{matrix}0 & ξ \\ −ξ^{−1} & t_{2} \end{matrix}] ,

where ξ + ξ^{−1} = t_{12}.

The matrix W_{0} ∈ SL(2) defined by:

W_{0} = [\begin{matrix}t_{3} & ((t_{13} − t_{1}t_{3})ξ + t_{23})ξ/(ξ^{2} − 1) \\ ((t_{13} − t_{1}t_{3}) + t_{23}ξ)/(ξ^{2} − 1) & 0 \end{matrix}] ,

satisfies (23). Any other W ∈ W must satisfy

tr(W − W_{0}) = 0,
(27)  tr(A_{2}(W − W_{0})) = 0,
tr(A_{1}(W − W_{0})) = 0.

Lemma 9. Any solution W − W_{0} of (27) is a multiple of

[A_{1}, A_{2}] = [\begin{matrix}ξ^{−1} − ξ & −t_{2} + t_{1}ξ \\ −t_{2} + ξ^{−1}t_{1} & ξ − ξ^{−1} \end{matrix}] .

Proof. The first equation in (27) asserts that W − W_{0} lies in the subspace SL(2, ℂ), upon which the trace form is nondegenerate. The second and third equations assert that W −W_{0} is orthogonal to A_{1} and A_{2}. By (24), A_{1}, A_{2} and 𝕀 are linearly independent in M_{2}(ℂ), so the solutions of (27) form a one-dimensional linear subspace. The Lie product

[A_{1}, A_{2}] = A_{1}A_{2} − A_{2}A_{1}

is nonzero, lies in SL(2, ℂ), and since

tr(A_{i}A_{1}A_{2}) = tr(A_{i}A_{2}A_{1})

for i = 1, 2, orthogonal to A_{1} and A_{2}. The lemma follows. □

Parametrize W explicitly as

W = W_{0} + s[A_{1}, A_{2}].

Since

det([A_{1}, A_{2}) = 4 − (t_{1}^{2} + t_{2}^{2} + t_{12}^{2} − t_{1}t_{2}t_{12}) = 2 − κ(t_{1}, t_{2}, t_{12}) ≠ 0,

Thus the polynomial

W ⟶^{det} ℂ

is nonconstant, and hence onto. Taking W_{1} ∈ det_{W}^{−1}(1), the proof is complete assuming (24).

Now consider the case when 4 − t_{1}^{2} − t_{2}^{2} − t_{12}^{2} + t_{1}t_{2}t_{12} = 0. In that case there exist a_{1}, a_{2} ∈ ℂ^{∗} such that

t_{i} = a_{i} + (a_{i})^{−1}

for i = 1, 2. Then either

(28)  t_{12} = a_{1}a_{2} + (a_{1}a_{2})^{−1}

or

(29)  t_{12} = a_{1}(a_{2})^{−1} + (a_{1})^{−1}a_{2}.

In the first case (28), set

A_{1} := [\begin{matrix}a_{1} & t_{13} − a_{1}t_{3} \\ 0 & (a_{1})^{−1} \end{matrix}] ,
A_{2} := [\begin{matrix}a_{2} & t_{23} − a_{2}t_{3} \\ 0 & (a_{2})^{−1} \end{matrix}] ,
A_{3} := [\begin{matrix}t_{3} & −1 \\ 1 & 0 \end{matrix}]

and in the second case (28), set

A_{1} := [\begin{matrix}(a_{1})^{−1} & t_{13} − (a_{1})^{−1}t_{3} \\ 0 & a_{1} \end{matrix}] ,

A_{2} := [\begin{matrix}a_{2} & t_{23} − a_{2}t_{3} \\ 0 & (a_{2})^{−1} \end{matrix}] ,

A_{3} := [\begin{matrix}t_{3} & −1 \\ 1 & 0 \end{matrix}]

obtaining (A_{1}, A_{2}, A_{3}) ∈ SL(2) explicitly solving (21) and (22). This completes the proof of Proposition 6.. □

Writing

W = a_{0}I + a_{1}A_{1} + a_{2}A_{2}

Acknowlegement. I am grateful to Hyman Bass, Richard Brown, Carlos Florentino, Linda Keen and Sean Lawton for their interest in this manuscript, and for providing valuable suggestions.

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Mathematics Department, University of Maryland, College Park, MD 20742 USA

  E-mail address: wmg@math.umd.edu