8 VIN DE SILVA, JOEL W. ROBBIN AND DIETMAR A. SALAMON

Definition 2.1 (Traces). Fix two (not necessarily distinct) intersection points

x, y ∈ α ∩ β.

(i) Let w : Σ \ (α ∪ β) → Z be a two-chain. The triple Λ = (x, y, w) is called an

(α, β)-trace if there exists an element u ∈ D(x, y) such that w is given by (1).

In this case Λ =: Λu is also called the (α, β)-trace of u and we sometimes write

wu := w.

(ii) Let Λ = (x, y, w) be an (α, β)-trace. The triple ∂Λ := (x, y, ∂w) is called the

boundary of Λ.

(iii) A one-chain ν : (α \ β) ∪ (β \ α) → Z is called an (x, y)-trace if there exist

smooth curves γα : [0, 1] → α and γβ : [0, 1] → β such that γα(0) = γβ(0) = x,

γα(1) = γβ(1) = y, γα and γβ are homotopic in Σ with fixed endpoints, and

(3) ν(z) =

deg(γα,z), for z ∈ α \ β,

− deg(γβ,z), for z ∈ β \ α.

Remark 2.2. Assume Σ is simply connected. Then the condition on γα and

γβ to be homotopic with fixed endpoints is redundant. Moreover, if x = y then a

one-chain ν is an (x, y)-trace if and only if the restrictions

να := ν|α\β, νβ := −ν|β\α

are constant. If x = y and α, β are embedded circles and A, B denote the positively

oriented arcs from x to y in α, β, then a one-chain ν is an (x, y)-trace if and only if

να|α\(A∪β) = να|A\β − 1

and

νβ|β\(B∪α) = νβ|B\α − 1.

In particular, when walking along α or β, the function ν only changes its value at

x and y.

Lemma 2.3. Let x, y ∈ α∩β and u ∈ D(x, y). Then the boundary of the (α, β)-

trace Λu of u is the triple ∂Λu = (x, y, ν), where ν is given by (2). In other words,

if w is given by (1) and ν is given by (2) then ν = ∂w.

Proof. Choose an embedding γ : [−1, 1] → Σ such that u is transverse to

γ, γ(t) ∈ Σ \ (α ∪ β) for t = 0, γ(−1), γ(1) are regular values of u, γ(0) ∈ α \ β

is a regular value of u|D∩R, and γ intersects α transversally at t = 0 such that

orientations match in

Tγ(0)Σ = Tγ(0)α ⊕ R ˙ γ (0).

Denote Γ := γ([−1, 1]). Then

u−1(Γ)

⊂ D is a 1-dimensional submanifold with

boundary

∂u−1(Γ)

=

u−1(γ(−1))

∪

u−1(γ(1))

∪

(

u−1(γ(0))

∩ R)

)

.

If z ∈ u−1(Γ) then

im du(z) + Tu(z)Γ = Tu(z)Σ,

Tzu−1(Γ)

=

du(z)−1Tu(z)Γ.

We orient

u−1(Γ)

such that the orientations match in

Tu(z)Σ = Tu(z)Γ ⊕

du(z)iTzu−1(Γ).

In other words, if z ∈

u−1(Γ)

and u(z) = γ(t), then a nonzero tangent vector

ζ ∈

Tzu−1(Γ)

is positive if and only if the pair (˙ γ (t),du(z)iζ) is a positive basis

of Tγ(t)Σ. Then the boundary orientation of

u−1(Γ)

at the elements of

u−1(γ(1))