Winding of simple walks on the square lattice

Timothy Budd∗

5th February 2020

Abstract

A method is described to count simple diagonal walks on Z2with a �xed starting point and

endpoint on one of the axes and a �xed winding angle around the origin. The method involves the

decomposition of such walks into smaller pieces, the generating functions of which are encoded in

a commuting set of Hilbert space operators. The general enumeration problem is then solved by

obtaining an explicit eigenvalue decomposition of these operators involving elliptic functions. By

further restricting the intermediate winding angles of the walks to some open interval, the method

can be used to count various classes of walks restricted to cones in Z2of opening angles that are

integer multiples of π/4.

We present three applications of this main result. First we �nd an explicit generating function for

the walks in such cones that start and end at the origin. In the particular case of a cone of angle 3π/4these walks are directly related to Gessel’s walks in the quadrant, and we provide a new proof of

their enumeration. Next we study the distribution of the winding angle of a simple random walk on

Z2around a point in the close vicinity of its starting point, for which we identify discrete analogues

of the known hyperbolic secant laws and a probabilistic interpretation of the Jacobi elliptic functions.

Finally we relate the spectrum of one of the Hilbert space operators to the enumeration of closed

loops in Z2with �xed winding number around the origin.

Keywords: Lattice walks; Random walks; Winding angle; Generating functions; Elliptic functions

1 Introduction

Counting of lattice paths has been a major topic in combinatorics (and probability and physics) for

many decades. Especially the enumeration of various types of lattice walks con�ned to convex cones

in Z2, like the positive quadrant, has attracted much attention in recent years, due mainly to the rich

algebraic structure of the generating functions involved (see e.g [15, 5] and references therein) and the

relations with other combinatorial structures (e.g. [4, 31]). The study of lattice walks in non-convex

cones has received much less attention. Notable exception are walks on the slit plane [12, 16] and the

three-quarter plane [14]. When describing the plane in polar coordinates, the con�nement of walks to

cones of di�erent opening angles (with the tip positioned at the origin) may equally be phrased as a

restriction on the angular coordinates of the sites visited by the walk. One may generalize this concept

by replacing the angular coordinate by a notion of winding angle of the walk around the origin, in

which case one can even make sense of cones of angles larger than 2π . It stands to reason that a �ne

control over the winding angle in the enumeration of lattice walks brings us a long way in the study of

walks in (especially non-convex) cones.

Although the winding angle of lattice walks seems to have received little attention in the com-

binatorics literature, probabilistic aspects of the winding of long random walks have been studied in

∗Institute for Mathematics, Astrophysics and Particle Physics, Radboud University Nijmegen. T.Budd@science.ru.nl

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Figure 1: The winding angle of a simple diagonal walk w ∈ W of length |w | = 19 from (2, 0)to (−1, 3). The (full) winding angle θw is indicated in green, and the winding angle increment

θwi − θwi−1in blue (which is negative in this example).

considerable detail [21, 22, 36, 38]. In particular, it is known that under suitable conditions on the steps of

the random walk the winding angle after n steps is typically of order logn, and that the angle normalized

by 2/logn converges in distribution to a standard hyperbolic secant distribution. The methods used all

rely on coupling to Brownian motion, for which the winding angle problem is easily studied with the

help of its conformal properties. Although quite generally applicable in the asymptotic regime, these

techniques tell us little about the underlying combinatorics.

In this paper we initiate the combinatorial study of lattice walks with control on the winding angle, by

looking at various classes of simple (rectilinear or diagonal) walks onZ2. As we will see, the combinatorial

tools described in this paper are strong enough to bridge the gap between the combinatorial study of

walks in cones and the asymptotic winding of random walks. Before describing the main results of the

paper, we should start with some de�nitions.

We let W be the set of simple diagonal walks w in Z2of length |w | ≥ 0 avoiding the origin, i.e. w

is a sequence (wi ) |w |i=0in Z2 \ {(0, 0)} with wi −wi−1 ∈ {(1, 1), (1,−1), (−1,−1), (−1, 1)} for 1 ≤ i ≤ |w |.

We de�ne the winding angle θwi ∈ R ofw up to time i to be the di�erence in angular coordinates of wi

and w0 including a contribution 2π (resp. −2π ) for each full counterclockwise (resp. clockwise) turn of

w around the origin up to time i . Equivalently, (θwi )|w |i=0

is the unique sequence in R such that θw0= 0

and θwi − θwi−1is the (counterclockwise) angle between the segments ((0, 0),wi−1) and ((0, 0),wi ) for

1 ≤ i ≤ |w |. The (full) winding angle of w is then θw B θw|w | . See Figure 1 for an example.

Main result The Dirichlet space D is the Hilbert space of complex analytic functions f on the unit

disc D = {z ∈ C : |z | < 1} that vanish at 0 and have �nite norm ‖ f ‖2D

with respect to the Dirichlet inner

product

〈f ,д〉D =∫Df ′(z)д′(z)dA(z) =

∞∑n=1

n [zn]f (z) [zn]д(z),

where the measure dA(z1 + iz2) B 1

π dz1dz2 is chosen such that

∫D

dA(z) = 1. See [3] for a review

of its properties. We denote by (en)∞n=1the standard orthogonal basis de�ned by en(z) B zn , which

is unnormalized since ‖en ‖2D = n. With this notation 〈en , f 〉D = n[zn]f (z) for any analytic function

f ∈ D.

For k ∈ (0, 1) we let vk : C \ {z ∈ R : z2 ≥ k} → C be the analytic function de�ned by the elliptic

integral

vk (z) B1

4K(k)

∫ z

0

dx√(k − x2)(1 − kx2)

(1)

along the simplest path from 0 to z, where K(k) is the complete elliptic integral of the �rst kind with

elliptic modulus k (see Appendix A for de�nitions and notation). The appearance of this elliptic integral

in lattice walks enumeration is a natural one since2

π K(4t) is precisely the generating function for

2

excursions of the simple diagonal walk from the origin (see (58) in Appendix A),

2

πK(4t) =

∞∑n=0

(2n

n

)2

t2n . (2)

The incomplete elliptic integral vk does not have a comparably simple combinatorial interpretation, but

provides an important conformal mapping from a slit disk onto a rectangle in the complex plane, as

detailed in Section 2.1.

For �xed k we use the conventional notation

k ′ =√

1 − k2and k1 =

1 − k ′1 + k ′

for the complementary modulus k ′ and the descending Landen transformation k1 of k , which both take

values in (0, 1) again (see Appendix A). Using these we introduce a family (fm)∞m=1of analytic functions

by setting (notice the k1 in vk1(z)!)

fm(z) B cos(2πm(vk1(z) + 1/4)) − cos(πm/2), (3)

which satis�es fm(0) = 0. Even though vk1(z) has branch cuts at z = ±

√k1, we will see (Lemma 6, 7, 8)

that fm(z) has radius of convergence around 0 equal to 1/√k1 > 1 and has �nite norm with respect to

the Dirichlet inner product, hence fm ∈ D. According to Proposition 9 the norm of fm is given explicitly

by

‖ fm ‖2D =m(q−mk − qmk )

4

, (4)

where qk ∈ (0, 1) is the (elliptic) nome of modulus k (see (61) in Appendix A), which is analytic for k in

the unit disk. Once properly normalized the family of functions ( ˆfm)∞m=1provides an orthonormal basis

of D, i.e.

ˆfm(z) Bfm(z)‖ fm ‖D

=cos(2πm(vk1

(z) + 1/4)) − cos(πm/2)√m4(q−mk − qmk )

,⟨

ˆfn , ˆfm⟩D= 1n=m .

The main technical result of this paper is the following.

Theorem 1. For `,p ≥ 1 and α ∈ π2Z, let W(α )

`,p be the set of (possibly empty) simple diagonal walksw onZ2 \ {(0, 0)} that start at (p, 0), end on one of the axes at distance ` from the origin, and have full windingangle θw = α .

(i) LetW (α )`,p (t) B

∑w ∈W(α )

`,pt |w | be the generating function of W(α )

`,p . For k = 4t ∈ (0, 1) �xed, there

exists a compact self-adjoint operator Y(α )k on D with eigenvectors (fm)∞m=1such that

W (α )`,p (t) = 〈e`,Y

(α )k ep〉D, Y(α )k fm =

2K(k)π

1

mqm |α |/πk fm . (5)

(ii) Let W(α, I )`,p ⊂ W

(α )`,p be the subset of the aforementioned walks that have intermediate winding angles

in an interval I ⊂ R, i.e. θwi ∈ I for i = 1, 2, . . . , |w | − 1, and letW (α, I )`,p (t) be the corresponding

generating function. If I = (β−, β+) with β± ∈ π2Z ∪ {±∞}, α ∈ [β−, β+] ∩ π

2Z and α , 0 or α , β±,

then the generating functionW (α, I )`,p (t) is related to a matrix element of a compact self-adjoint operator

3

on D with the same eigenvectors (fm)∞m=1, as described in the table below.

α β− β+ W (α, I )`,p (t) Eigenvalues

> 0 0 α1

`p〈e`,A(α )k ep〉 A(α )k fm =

π2K (k )

mq−mα /πk −qmα /π

k

fm

> 0 < 0 α1

`〈e`, J(α,β−)k ep〉 J(α,β−)k fm =

q2mβ−/πk −1

q2mβ−/πk −q2mα /π

k

qmα/πk fm

≥ 0 < 0 > α 〈e`,B(α,β−,β+)k ep〉 B(α,β−,β+)k fm =2K (k )π

q2mβ−/πk −1

mqmα /πk

q2mα /πk −q2mβ+/π

k

q2mβ−/πk −q2mβ+/π

k

fm

The remaining cases follow from the symmetries (α , β−, β+) → (−α ,−β+,−β−) and (α , β−, β+) →(α ,α − β+,α − β−), and the cases β± = ±∞ agree with the corresponding limits β± → ±∞ (usingthat qk ∈ (0, 1)).

(iii) The statement of (ii) remains valid for β± ∈ π4Z∪ {±∞} and α ∈ [β−, β+] ∩ π

2Z and α , 0 or α , β±

as long as ` and p are even.

We emphasize that the convention for the index placement in W(α, I )`,p , that is used also for other

sets of walks throughout this work, is that the �rst index corresponds to the endpoint of the walk

and the second index to the starting point. The reason for this choice is precisely the relation of the

generating functionW (α )`,p to the matrix element at position (`,p) of a Hilbert space operator. As we will

see, composition of particular families of walks often corresponds to the composition of the respective

operators (or multiplication of the respective in�nite matrices). The index placement re�ects the natural

right-to-left ordering in the composition of these operators.

(a) W(3π /2)4,2 (b) W

(π ,(0,3π /2))3,1 (c) W

(π /2,(−π /4,π ))4,4

Figure 2: Examples of walks enumerated by Theorem 1.

Theorem 1 is stated for �xed real values of k = 4t ∈ (0, 1), while from a combinatorial point of

view it may be preferable to think of the generating functionsW (α, I )`,p (t) as formal power series in the

variable t . This raises the question to what extent the eigenvalue decomposition can be understood

on the level of formal power series. To this end we prove in Proposition 10 that for any m,p ≥ 1 the

coe�cient [zp ] ˆfm(z) = 1

p

⟨ep , ˆfm

⟩D

is an analytic function in k around 0, such that we may interpret

ˆfm(z) = ˆfm(z; t) as taking values in the ring R[[z, t]] of formal power series in the variables z and t , see

Section 2.2 for an explicit expansion form = 1, . . . , 6. With this observation Theorem 1 can be largely

recast as a set of identities on formal power series. To this end we denote the multivariate generating

function of W(α, I )`,p by

W (α, I )(x ,y; t) =∑`,p≥1

x `ypW (α, I )`,p (t) ∈ R[[x ,y, t]],

4

(a) Diagonal excursion. (b) Rectilinear excursion.

Figure 3: Example of an excursion in diagonal and rectilinear form together with its winding

angle.

and the eigenvalues of A(α )k , J(α,β−)k , B(α,β−,β+)k byA(α )m (t), J(α,β−)m (t), B(α,β−,β+)m (t) respectively (this includes

Y(α )k = B(α,−∞,∞)k as a special case). These eigenvalues, as given in Theorem 1(ii), are all analytic functions

of t in the unit disk (see (70) and (2) for the power series representations of qk and K(k)). Provided

α , 0, themth eigenvalue converges to 0 asm →∞ in the formal topology of R[[t]], i.e. for any n ≥ 0

the coe�cient [tn]A(α )m (t) vanishes for all m large enough. Theorem 1 implies the convergent series

identities in R[[x ,y, t]] for α , 0,

W (α,(0,α ))(x ,y; t) =∞∑

m=1

A(α )m (t) ˆfm(x ; t) ˆfm(y; t),

W (α,(β−,α ))(x ,y; t) = y ∂∂y

∞∑m=1

J(α,β−)m (t) ˆfm(x ; t) ˆfm(y; t),

W (α,(β−,β+))(x ,y; t) = xy∂

∂x

∂

∂y

∞∑m=1

B(α,β−,β+)m (t) ˆfm(x ; t) ˆfm(y; t),

with α , β− and β+ satisfying the respective assumptions described in Theorem 1(ii) and (iii). We leave

it as an open problem whether these identities and an expression forˆfm(x ; t) can be derived using

exclusively formal power series.

As an example of how to compute explicit generating functions with the help of Theorem 1, let us

look at the set W(π )3,3 of simple diagonal walks from (3, 0) to (−3, 0) that have winding angle π around

the origin. According to Theorem 1(i) its generating function satis�es

W (π )3,3 (t) =

⟨e3,Y

(π )k e3

⟩D=

2K(k)π

∞∑m=1

1

mqmk

⟨e3, ˆfm

⟩2

D= 10 t6 + 280 t8 + 5661 t10 + · · · ,

where we used the series expansions of qk and K(k) (respectively (70) and (58) in Appendix A) and that

of

⟨e3, ˆfm

⟩D

in Section 2.2.

Application: Excursions Theorem 1 can be used to count many specialized classes of walks involving

winding angles. The �rst quite natural counting problem we address is that of the (diagonal) excursionsE from the origin, i.e. E is the set of (non-empty) simple diagonal walks starting and ending at the

origin with no intermediate returns (Figure 3a). Actually, in this case we may equally well consider

simple rectilinear walks on Z2, thanks to the obvious linear mapping between the two types of walks

(Figure 3b). Even though walksw ∈ E do not completely avoid the origin, we may still naturally assign a

winding angle sequence to them by imposing that the �rst and last step do not contribute to the winding

5

(a)

(b)

Figure 4: Excursions (a) of length 2n+2 staying in the angular interval (−π/4,π/2) are in bijection

with Gessel walks (b) in the quadrant starting and ending at the origin with 2n steps.

angle, i.e. θw1= θw

0= 0 and θw = θw|w | = θ

w|w |−1

. In Proposition 18 we prove that the generating function

for excursions with winding angle α ∈ π2Z is given (for k = 4t ∈ (0, 1) �xed) by

F (α )(t) B∑w ∈E

t |w |1{θw=α } =2π

K(k)

∞∑n=1

(1 − qnk )2

1 − q4nk

qn( 2

π |α |+1)k .

Since the summand is analytic in t around 0 and O(t2n) for any n, the relation implies an identity of

formal power series in R[[t]] for any α ∈ π2Z.

Similarly to Theorem 1(ii) one may further restrict the full winding angle sequence of w to lie in an

open interval I = (β−, β+) with β− ∈ π4Z<0, β+ ∈ π

4Z>0 and α ∈ I ∩ π

2Z. In this case it is more natural to

also �x the starting direction, say w1 = (1, 1), and we denote the corresponding generating function by

F (α, I )(t). Observe in particular that F (α,R)(t) = F (α )(t)/4. We prove in Theorem 23 that the generating

function F (α, I )(t) is given by the �nite sum

F (α, I )(t) = π

8δ

∑σ ∈(0,δ )∩ π

2Z

(cos

(4σα

δ

)− cos

(4σ (2β+ − α)

δ

))F

(t ,

4σ

δ

), δ B 2(β+ − β−), (6)

where F (t ,b) B ∑α ∈ π

2Z F(α )(t) eibα is the “characteristic function” associated to F (α )(t). Since F (t ,b) =

F (t , 4 − b), the terms in the right-hand side of (6) corresponding to σ and δ − σ are actually identical,

therefore leaving only b δπ c distinct terms. For non-integer values of b (see Proposition 18 for the full

expression) F (t ,b) can be expressed in closed form as

F (t ,b) = 1

cos

(πb2

) 1 −π tan

(πb4

)2K(k)

θ ′1

(πb4,√qk

)θ1

(πb4,√qk

) , (b ∈ R \ Z) (7)

where θ1(z,q) is the �rst Jacobi theta function (see (64) for a de�nition). Again (6) and (7) imply the

equivalent identities on the level of formal power series in R[[t]] or C[[t]]. In Proposition 20 we prove

that the power series F (t ,b) is algebraic in t for any b ∈ Q \ Z but that is it transcendental for b ∈ Z. By

inspecting the terms appearing in (6) we �nd that F (α, I )(t) is transcendental if and only if β+, β− ∈ π2Z+ π

4

or both β+, β− ∈ πZ + π2

and α ∈ πZ (see Theorem 23).

A special case of excursions for which the generating function is algebraic is α = 0 and I =

(−π/4,π/2), see Figure 4a. After removal of the �rst and last step of the walk and a linear transformation,

these correspond precisely to walks in the quadrant starting and ending at the origin with steps in

{(−1, 0), (1, 0), (−1,−1), (1, 1)}, also known as Gessel walks. Around 2000 Ira Gessel conjectured that

6

Figure 5: The winding angles θ�j+ 1

2

and θ•j+ 1

2

of a simple random walk at time j + 1

2around,

respectively, a dual lattice point adjacent to the origin and the origin itself.

the generating function for such walks is given (in our notation) by

1

t2F (0,(−π /4,π /2))(t) =

∞∑n=0

t2n16

n (5/6)n(1/2)n(2)n(5/3)n

= 1 + 2t2 + 11t4 + 85t6 + · · · ,

where (a)n = Γ(a+n)/Γ(a) is the descending Pochhammer symbol (see also Sloane’s Online Encyclopediaof Integer sequences (OEIS) sequence A135404). The �rst computer-aided proof of this conjecture appeared

in [30], and it was followed by several “human” proofs in [11, 13, 5]. Here we provide an alternative

proof using Theorem 23. Indeed, we have explicitly

F (0,(−π /4,π /2))(t) = 1

8

F

(t ,

4

3

)+

1

8

F

(t ,

8

3

)=

1

4

F

(t ,

4

3

)=

1

2

[ √3π

2K(4t)θ ′

1

( π3,√qk

)θ1

( π3,√qk

) − 1

],

where used that F (t ,b) = F (t , 4 − b). According to our discussion above this is an algebraic power

series in t , a fact about F (0,(−π /4,π /2))(t) that was �rst observed in [10]. In Corollary 24 we deduce an

explicit algebraic equation for F (0,(−π /4,π /2))(t), and check that it agrees with a known equation for∑∞n=0

t2n+216

n (5/6)n (1/2)n(2)n (5/3)n .

Application: Unconstrained random walks Let (Wi )i≥0 be a simple random walk on Z2started

at the origin. A natural question is to ask for the (approximate) distribution of the winding angle θzjof the random walk around some point z ∈ R2

up to time j. As mentioned before, this question has

been addressed successfully in the literature in the limit j → ∞ using coupling to Brownian motion.

If z ∈ (Z + 1/2)2, then 2θzj /log j is known [36, 21, 22, 38] to converge in distribution to the hyperbolic

secant law with density1

2sech(πx/2)dx (recall that sechy = 1/coshy = 2/(ey + e−y )). If instead z ∈ Z2

and one conditions the random walk not to hit z before time j , then 2θzj /log j converges to a “hyperbolic

secant squared law” with densityπ4

sech2(πx/2)dx [36].

In Section 4 we complement these results by deriving exact statistics at �nite j with the help of

Theorem 1. To this end we look at the winding angles around two points in the vicinity of the starting

point, namely (−1/2,−1/2) and the origin itself. To be precise, let θ�j+1/2 be the winding angle of (Wi )iaround (−1/2,−1/2) up to time j + 1/2, i.e. halfway its step fromWj toWj+1 (see Figure 5). Similarly,

let θ•j+1/2 be the winding angle around the origin, ignoring the �rst step and with the convention that

θ•j+1/2 = ∞ if (Wi )i has returned to the origin strictly before time j + 1.

If ζk is a geometric random variable with parameter k ∈ (0, 1), i.e. P[ζk = j] = k j (1 − k) for j ≥ 0,

then the following “discrete hyperbolic secant laws” hold (Theorem 25):

P[θ�ζk+1/2 ∈ (α −

π2,α + π

2)]= ck sech (4Tkα) + k−1

k 1{α=0} for α ∈ π2Z,

P[θ•ζk+1/2 ∈ (α −

π2,α + π

2)]= Ck sech

(4Tk (α − π

4))

sech

(4Tk (α + π

4))

for α ∈ π2Z + π

4,

7

(a) Lodd

2(b) Leven

−1

Figure 6: Two examples of rooted loops with di�erent index and parity.

where ck , Ck and Tk are explicit functions of k (see Theorem 25).

As a consequence we �nd probabilistic interpretations of the Jacobi elliptic functions cn and dn (see

Appendix A) as characteristic functions of winding angles,

E exp

(ib{θ�ζk+1/2}πZ+ π2

)= cn(K(k)b,k), E exp

(ib{θ�ζk−1/2}πZ

)= dn(K(k)b,k).

Here {·}A denotes rounding to the nearest element of A ⊂ R and we set θ�−1/2 = 0 by convention.

Since cn(y, 1) = dn(y, 1) = sech(y) = Eeiyη is the characteristic function of the aforementioned

hyperbolic secant distribution, we may directly conclude the convergence in distribution as k → 1 of

the winding angle θ�ζk+1/2/K(k) at geometric time ζk . A more delicate singularity analysis, which is

beyond the scope of this paper, would yield the same distributional limit for 2θ�j+1/2/log(j) as j →∞, in

accordance with previous work.

Application: Loops The last application we discuss utilizes the fact that the eigenvalues of the oper-

ators in Theorem 1 have much simpler expressions than the components

⟨ep , fm

⟩D

of the eigenvectors.

It is therefore worthwhile to seek combinatorial interpretations of traces of (combinations of) operators,

the values of which only depend on the eigenvalues. When writing out the trace in terms of the basis

(ep )∞p=1it is clear that such an interpretation must involve walks that start and end at arbitrary but equal

distance from the origin. If the full winding angle is taken to be a multiple of 2π then such a walk forms

a loop, i.e. it starts and ends at the same point.

A natural combinatorial set-up is described in Section 5. There we consider the set Ln of rooted loopsof index n, n ∈ Z \ {0}, which are simple diagonal walks avoiding the origin that start and end at an

arbitrary but equal point in Z2and have winding angle 2πn around the origin. The set Ln = Leven

n ∪Lodd

n

naturally partitions into loops that visit only sites of even respectively odd parity ((x ,y) ∈ Z2with x +y

even respectively odd), see Figure 6. We introduce the generating functions

Ln(t) B∑w ∈Ln

t |w |

|w | Lodd

n (t) =∑

w ∈Lodd

n

t |w |

|w | , Leven

n (t) =∑

w ∈Leven

n

t |w |

|w | ,

where we have included a factor 1/|w | for convenience, such that the generating functions of the set

Ln is actually tL′n(t). Observe that L1(t) is precisely the generating function of unrooted loops of index

1, i.e. rooted loops modulo rerooting (but preserving orientation), because the equivalence class of a

rooted loop w of index 1 contains precisely |w | elements. This is not true for n ≥ 2, since Ln contains

rooted loops w that cover themselves multiple times and thus have equivalence classes with less than

|w | elements.

8

Theorem 28 states that these generating functions for n > 0 are given by

Ln(t) =1

ntr J(2πn,−∞)k =

1

n

q2nk

1 − q2nk

, Lodd

n (t) =1

n

q2nk

1 − q4nk

, Leven

n (t) = 1

n

q4nk

1 − q4nk

.

Here J(2πn,−∞)k is the operator J(α,β−)k with α = 2πn and β− = −∞ appearing in Theorem 1(ii) whose mth

eigenvalue q2nmk can be obtained as the β− → −∞ limit of the displayed expression (see the �nal remark

of Theorem 1(ii) ). Note as before that these imply the equivalent identities on the level of power series

in R[[t]].A simple probabilistic consequence is the following. Let (Wi )2`i=0

be a simple random walk on Z2

started at the origin and conditioned to return after 2` steps. For each point z ∈ R2we let the index Iz

be the signed number of times (Wi )2`i=0winds around z in counterclockwise direction, i.e. 2π Iz is the

winding angle of (Wi )2`i=0around z. If z lies on the trajectory of (Wi )2`i=0

, then we set Iz = ∞. We let theclusters Cn of index n be the set of connected components of {z ∈ R2

: Iz = n}, and for c ∈ Cn we let |c |and |∂c | respectively be the area and boundary length of component c . See Figure 7 for an example. The

expectation values for the total area and boundary length of the clusters of index n then satisfy

E

[ ∑c ∈Cn|c |

]=

42`(

2``

)2

2`

n[k2`]

q2nk

1 − q4nk

∼ `

2πn2,

E

[ ∑c ∈Cn(|∂c | − 2)

]=

42`(

2``

)2

4`

n[k2`]

q2nk

1 + q2nk

∼ 2π 3`

log2 `,

with the asymptotics as ` →∞ indicated. The �rst result should be compared to the analogous result

for Brownian motion: Garban and Ferreras proved in [28] using Yor’s work [40] that the expected area

of the set of points with index n with respect to a unit time Brownian bridge in R2is equal to 1/(2πn2).

Perhaps surprisingly, we �nd that the expected boundary length all the components of index n (minus

twice the expected number of such components) grows asymptotically at a rate that is independent of n,

contrary to the total area.

Open question 1. Does E[∑

c ∈C0 �nite |c |], i.e. the total area of the �nite clusters of index 0, have a

similarly explicit expression? Based on the results of [28] we expect it to be asymptotic to π`/30 as ` →∞.

Finally we mention one more potential application of the enumeration of loops in Theorem 28 in the

context of random walk loop soups [33], which are certain Poisson processes of loops on Z2. A natural

Figure 7: A simple walk on Z2together with its clusters of index 1 (light blue) and 2 (dark blue).

The largest cluster c has area |c | = 9 and boundary length |∂c | = 20 (notice that both sides of the

“slit” contribute to the length).

9

quantity to consider in such a system is the winding �eld which roughly assigns to any point z ∈ R2

the total index of all the loops in the process [24, 23]. Theorem 28 may be used to compute explicit

expectation values (one-point functions of the corresponding vertex operators to be precise) in the

massive version of the loop soups. We will pursue this direction elsewhere.

Discussion The connection between the enumeration of walks and the explicitly diagonalizable

operators on Dirichlet space may seem a bit magical to the reader. So perhaps some comments are in

order on how we arrived at this result, which originates in the combinatorial study of planar maps.

A planar map is a connected multigraph (a graph with multiple edges and loops allowed) that is

properly embedded in the 2-sphere (edges are only allowed to meet at their extremities), viewed up to

orientation-preserving homeomorphisms of the sphere. The connected components of the complement

of a planar map are called the faces, which have a degree equal to the number of bounding edges. There

exists a relatively simple multivariate generating function for bipartite planar maps, i.e. maps with all

faces of even degree, that have two distinguished faces of degree p and ` and a �xed number of faces

of each degree (see e.g. [25]). The surprising fact is that this generating function has a form that is

very similar to that of the generating functionW (π ,(−π ,π ))`,p (t) of diagonal walks from (p, 0) to (−`, 0) that

avoid the slit {(x , 0) : x ≤ 0} until the end. A combinatorial explanation (in the particular case of critical

planar maps) appears in [20] using a peeling exploration [18, 19],

If one further decorates the planar maps by a rigid O(n) loop model [8], then the combinatorial

relation extends to one between walks of �xed winding angle W(α )`,p with α ∈ πZ and planar maps with

two distinguished faces and certain collections of non-intersecting loops separating the two faces. The

combinatorics of the latter has been studied in considerable detail in [8, 7, 6, 20], which has inspired our

treatment of the simple walks on Z2in this paper. Further details on the connection and an extension

to more general lattice walks with small steps (i.e. steps in {−1, 0, 1}2) will be provided in forthcoming

work.

Finally, we point out that these methods can be used to determine Green’s functions (and more

general resolvents) for the Laplacian on regular lattices in the presence of a conical defect or branched

covering, which are relevant to the study of various two-dimensional statistical systems. As an example,

in recent work [32] Kenyon and Wilson computed the Green’s function on the branched double cover

of the square lattice, which has applications in local statistics of the uniform spanning tree on Z2as well

as dimer systems.

Acknowledgements

The author would like to thank Kilian Raschel, Alin Bostan and Gaëtan Borot for their suggestions on how

to prove Corollary 24, and Thomas Prellberg for suggesting to extend Theorem 25 to absorbing boundary

conditions. The author is indebted to an anonymous referee for numerous corrections and suggestions

to improve the exposition. This work was supported by a public grant as part of the Investissement

d’avenir project, reference ANR-11-LABX-0056-LMH, LabEx LMH, as well as ERC-Advanced grant

291092, “Exploring the Quantum Universe” (EQU). Part of this work was done while the author was at

the Niels Bohr Institute, University of Copenhagen and Institut de Physique Théorique, CEA, Université

Paris-Saclay.

10

(a) J`,p , J`,p (t) = 1

`

⟨e`, Jkep

⟩D

(b) B`,p , B`,p (t) =⟨e`,Bkep

⟩D

(c) A`,p , A`,p (t) = 1

`p

⟨e`,Akep

⟩D

Figure 8: Examples of walks contributing to the three types of building blocks.

2 Winding angle of walks starting and ending on an axis

Our strategy towards proving Theorem 1 will be to �rst prove part (ii) for three special cases (see Figure

8),

J`,p B W( π

2,(− π

2, π

2))

`,p , B`,p B W(0,(− π

2, π

2))

`,p , A`,p B W( π

2,(0, π

2))

`,p .

We de�ne three linear operators Jk , Bk and Ak on D by specifying their matrix elements with respect to

the standard basis (ei )i≥1 in terms of the corresponding generating functions J`,p (t), B`,p (t) and A`,p (t)(with k = 4t ) as

1

`

⟨e`, Jkep

⟩D= J`,p (t),

⟨e`,Bkep

⟩D= B`,p (t),

1

`p

⟨e`,Akep

⟩D= A`,p (t).

The motivation to de�ne the operators in this way is given in Proposition 2.

The results in this section will require some Hilbert space terminology (see [35, Chapter IV] or [41,

Chapter 1] for introductions). A linear operator T on a Hilbert space H is bounded if there exists a

constant C > 0 such that ‖Ty‖H ≤ C‖y‖H for all y ∈ H. A bounded operator T is said to be compactif T({y ∈ H : ‖y‖H ≤ 1}) is compact in the norm topology on H, i.e. the topology on H induced by

the distance d(x ,y) = ‖x − y‖H. Finally, a linear operator T is self-adjoint if 〈x ,Ty〉H = 〈Tx ,y〉H for all

x ,y ∈ H. For our purposes, the most useful characterization is that T is self-adjoint and compact if and

only if there exists an orthonormal basis (an)∞n=1of H and a sequence of real numbers (λn)∞n=1

satisfying

λn → 0 as n →∞ such that Ty =∑∞

n=1λn 〈y,an〉Han for all y ∈ H.

Proposition 2. The linear operators Jk , Bk , Ak on D are bounded and satisfy Jk = AkBk . Moreover Jkand Ak are compact.

Proof. First we verify that for �xed t ∈ (0, 1/4) we have an exponential bound A`,p (t) < Ce−`−p for

some C > 0. To this end observe that A`,p can be identi�ed with a subset of (unconstrained) diagonal

walks starting at 0 and ending on the line {(x ,y) ∈ Z2: y − x = p + `}. Hence

A`,p (t) ≤∞∑

m,n=0

tm+n 2n(m + n

m

) (m

m+p+`2

)1{m+p+`

2∈Z} =

1

√1 − 4t

(1 − 2t −

√1 − 4t

2t

)`+p, (8)

which falls o� exponentially in ` + p. It follows that Ak has �nite Hilbert-Schmidt norm,

∞∑p=1

‖Akep ‖2D‖ep ‖2D

=

∞∑p, `=1

p `A2

`,p (t) < ∞.

In particular, Ak is bounded and compact, see e.g. [35, Theorem VI.22(e)].

11

On the other hand, the matrix elements of Bk with respect to the orthonormal basis (ep/√p)p≥1

are given by B`,p (t)/√`p ≤ B`,p (t) ≤ b`−p , where bn is the generating function of (unconstrained)

diagonal walks from the origin to (n, 0). In particular bn is smaller than the right-hand side of (8) when

` + p = n and therefore falls o� exponentially, ensuring that (bn)n∈Z are the Fourier coe�cients of

some continuous real 2π -periodic function ϕ : R→ R. By a classical result [29, Appendix] the Toeplitz

operator associated to ϕ, i.e. an operator with matrix elements (b`−p )`,p≥1, is bounded because ϕ is

bounded. Since the operator norm of this Toeplitz operator bounds that of Bk , Bk is also bounded.

For `,p,m ≥ 1, composition of walks determines a bijection between pairs of walks in A`,m ×Bm,p

and walks in J`,p for which the last intersection with the horizontal axis occurs at (m, 0). Hence

1

`

⟨e`, Jkep

⟩D= J`,p (t) =

∞∑m=1

A`,m(t)Bm,p (t) =∞∑

m=1

1

`m〈e`,Akem〉D

⟨em ,Bkep

⟩D=

1

`

⟨e`,AkBkep

⟩D,

implying that Jk = AkBk . Since Ak is compact and Bk is bounded, their composition Jk is bounded and

compact [41, Theorem 1.15]. �

2.1 The operator Jk

We wish to enumerate the walksw ∈ J`,p , p, ` ≥ 1, that start at (p, 0) and end at (0, `) while maintaining

strictly positive �rst coordinate until the end. By looking at both coordinates of the walks separately, we

easily see that these walks are in bijection with pairs of simple walks of equal length on Z, the �rst of

which starts at p and ends at 0 while staying positive until the end, while the second starts at 0 and ends

at ` without further restrictions. For �xed length n, such walks only exist if both n − p and n − ` are

non-negative even integers, in which case the ballot theorem tells us that the former walks are counted

bypn

( n(n+p)/2

)and the latter by

( n(n+`)/2

). Therefore the generating function J`,p (t) is given explicitly by

J`,p (t) =∞∑n=1

p

n

(n

(n + `)/2

) (n

(n + p)/2

)1{n−p and n−` non-negative and even}t

n . (9)

It is non-trivial only when p + ` is even, in which case it has radius of convergence equal to 1/4.

For �xed k = 4t ∈ (0, 1) we denote byψk : D→ C the analytic function given by

ψk (x) =1 −√

1 − kx2

√k x

,

which maps the unit disk D biholomorphically onto a strict subsetψk (D) ⊂ D, its inverse being given

by 2/(√k(z + 1/z)) for z ∈ ψk (D). It induces a linear operator Rk on the Dirichlet space D of analytic

functions f on D that vanish at the origin by setting Rk f B f ◦ψk .

Lemma 3. The linear operator Rk is bounded and Jk = R†kRk . In particular, Jk is self-adjoint and injective.

Proof. Since the Dirichlet inner product is preserved under conformal mapping, in the sense that

〈f ◦ψ ,д ◦ψ 〉D(D) = 〈f ,д〉D(ψ (D)) for any biholomorphic functionψ on D [3, Section 2.1], we have

‖Rk f ‖2D = ‖ f ◦ψk ‖2

D =

∫ψk (D)

| f ′(z)|2dA(z) ≤ ‖ f ‖2D,

implying that Rk is bounded.

For n − p non-negative and even we may compute

[xn]ψk (x)p =(k

4

)n/2[zn]

(1 −√

1 − 4z2

2z

)p=

(k

4

)n/2 pn[un−p ]

(1 + u2

)n=

(k

4

)n/2 pn

(n

(n + p)/2

), (10)

12

where in the second equality we applied Lagrange inversion to the Catalan series (1 −√

1 − 4z2)/(2z).Therefore⟨

e`,R†kRkep⟩D=

⟨Rke`,Rkep

⟩D=

∞∑n=1

n [xn]ψk (x)` [xn]ψk (x)p

=

∞∑n=1

(k

4

)nn`

n

(n

(n + `)/2

)p

n

(n

(n + p)/2

)1{n−p and n−` non-negative and even}

= ` J`,p (t) =⟨e`, Jkep

⟩D,

which shows that Jk = R†kRk . It follows that Jk is self-adjoint and injective, since 〈f , Jk f 〉D =∫ψk (D)

| f ′(z)|2dA(z) = 0 i� f is the zero function. �

The injectivity and self-adjointness of Jk imply the following very useful property for the triple of

operators Jk , Ak and Bk .

Lemma 4. The operators Jk , Ak and Bk are self-adjoint, commuting, and simultaneously diagonalizable.

Proof. It is clear from the de�nitions that Bk and Ak are self-adjoint, and according to Lemma 3 the

same is true for Jk . The relation Jk = AkBk from Proposition 2 then implies that all three operators

(mutually) commute. Since both Jk and Ak are compact, they are simultaneously diagonalizable [42,

Corollary 3.2.10], in the sense that there exists an orthonormal basis (bm)∞m=1of D such that each bm is

an eigenvector of both Jk and Ak . According to Lemma 3, Jk is injective and therefore Ak must be too.

Observe that, for any m ≥ 1, Jkbm = BkAkbm with Akbm , 0, and therefore bm is an eigenvector of Bktoo. �

The remainder of the subsection is devoted to diagonalizing Jk , while in Section 2.3 we will determine

the corresponding eigenvalues for the remaining two operators Ak and Bk . Finally, in Section 2.4 we

will prove Theorem 1 by taking suitable compositions of the operators Jk , Bk , and Ak .

In order to diagonalize Jk it su�ces to �nd an orthogonal basis (fm)∞m=1of D consisting of analytic

functions that are also orthogonal with respect to the Dirichlet inner-product onψk (D),

〈f ,д〉D(ψk (D)) B∫ψk (D)

f ′(z)д′(z) dA(z).

Indeed, if 〈fn , fm〉D = 〈fn , fm〉D(ψk (D)) = 0 form , n then

〈fn , Jk fm〉D =⟨fn ,R†kRk fm

⟩D= 〈fn , fm〉D(ψk (D)) =

〈fm , fm〉D(ψk (D))〈fm , fm〉D

〈fn , fm〉D for allm,n ≥ 1,

implying that fm is an eigenvector of Jk with eigenvalue 〈fm , fm〉D(ψk (D))/〈fm , fm〉D.

To obtain such a basis we seek an injective holomorphic mapping that takes both D andψk (D) to

su�ciently simple domains. As we will see the elliptic integral vk1(z) introduced in (1) does this job.

First we notice (using (72) in Appendix A) that vk (z) can be expressed in terms of the inverse function

arcsn(·,k) (in a suitable neighbourhood of the origin) of the Jacobi elliptic function sn(·,k)with modulus

k ,

vk (z) =1

4K(k)

∫ z

0

dx√(k − x2)(1 − kx2)

=arcsn

(z√k,k

)4K(k) . (11)

Denote by zk the analytic function

zk (v) =√k sn(4K(k)v,k), (12)

13

Figure 9: The analytic function vk1maps the disk with two slits onto a rectangle of width

1

2and

height 2Tk = K(k ′)/(2K(k)). The shaded domains representψk (D) and its image Hk under vk1.

which will provide the inverse to vk on a suitable domain (see Lemma 5 below). As depicted in Figure 9

and proved in the next lemma, after the removal of two slits vk1maps both D andψk (D) to rectangles,

with the same width but di�erent heights.

Lemma 5. The analytic function vk1maps D \ {z ∈ R : z2 ≥ k1}, respectively ψk (D) \ {z ∈ R :

z2 ≥ k1}, biholomorphically onto the open rectangle Hk1B (−1/4, 1/4) + i(−Tk1

/2,Tk1/2), respectively

Hk B (−1/4, 1/4)+ i(−Tk/2,Tk/2), whereTk = K(k ′)/(4K(k)) satis�esTk1= 2Tk (see (78) in Appendix A).

The inverse mapping zk1(v) extends to a surjective holomorphic map from the in�nite strip R + i(−Tk ,Tk )

to D. Moreover, zk1(−v) = zk1

(v + 1/2) = −zk1(v) and

(i) zk1maps [− 1

4, 1

4] bijectively to [−

√k1,√k1] with zk1

(0) = 0 and zk1(± 1

4) = ±

√k1;

(ii) zk1maps [− 1

4, 1

4] + iTk bijectively to {v ∈ C : |v | = 1, Im(v) ≥ 0} with zk1

(± 1

4+ iTk ) = ±1 and

zk1(iTk ) = i ;

(iii) zk1maps ± 1

4+ i[0,Tk ] bijectively to ±[

√k1, 1] with zk1

(± 1

4) = ±

√k1 and zk1

(± 1

4+ iTk ) = ±1.

Proof. It is well-known that sn(·,k)maps the open rectangle (−K(k),K(k))+i(0,K(k ′)) biholomorphically

onto the upper half plane (see e.g. [2, §47]) with the boundary [−K(k),K(k)]mapped bijectively to [−1, 1]and the boundaries ±K(k) + i[0,K(k ′)] to ±[1,k]. Hence zk1

as de�ned in (12) maps (− 1

4, 1

4) + i(0,Tk1

)biholomorphically onto the upper half plane with the boundary [− 1

4, 1

4]mapped bijectively to [−

√k1,√k1]

and the boundaries ± 1

4+ i[0,Tk1

] to ±[√k1, 1/

√k1]. Moreover the pseudo-periodicity and oddness of

sn(·,k) (see Appendix A) imply that zk1(−v) = zk1

(v + 1/2) = −zk1(v) as well as zk1

(0) = 0, showing

property (i).

According to (74) and (59) we have the identity sn(u,k) sn(u+iK(k ′),k) = 1/k , from which it follows

by setting u = x − iK(k ′)/2 with x ∈ R that | sn(x + iK(k ′)/2,k)| = 1/√k . Hence, |zk1

(x + iTk )| = 1

and therefore zk1must map [− 1

4, 1

4] + iTk bijectively to {v ∈ C : |v | = 1, Im(v) ≥ 0}. The part of

the rectangle (− 1

4, 1

4) + i(0,Tk1

) lying below the line [− 1

4, 1

4] + iTk is then mapped biholomorphically

by zk1to the open upper half disk. Together with property (i) and zk1

(−v) = −zk1(v) this shows that

zk1maps Hk1

biholomorphically onto D \ {z ∈ R : z2 ≥ k1}, providing the inverse of vk1restricted

to the latter set. Thanks to the pseudo-periodicity zk1(v + 1/2) = −zk1

(v), the function zk1maps the

in�nite strip R+ i(−Tk1/2,Tk1

/2) surjectively to D. Furthermore, sn(u,k) sn(u + iK(k ′),k) = 1/k implies

that sn(iK(k ′)/2,k)2 = −1/k and therefore zk1(iTk ) = i , proving property (ii). Property (iii) is a direct

consequence of the �rst two and the way zk1maps the boundaries ± 1

4+ i[0,Tk1

].It remains to show that zk1

maps Hk to ψk (D) \ {z ∈ R : z2 ≥ k1}. The descending Landen

transformation (79) relates the Jacobi elliptic functions sn(·,k) and sn(·,k1) through

sn(u,k) = (1 + k1) sn(u/(1 + k1),k1)1 + k1 sn

2(u/(1 + k1),k1).

14

From the arguments above we know that if u ∈ (−K(k),K(k)) + i(−K(k ′)/2,K(k ′)/2) then | sn(u,k)| <1/√k , in which case we may invert the relation to√

k1 sn(u/(1 + k1),k1) =1 −

√1 − k2

sn2(u,k)

k sn(u,k) = ψk (√k sn(u,k)). (13)

The sign of the square root is determined by noticing, with the help of (78), thatu

1+k1

∈ (−K(k1),K(k1))+i(−K ′(k1)/4,K ′(k1)/4) and therefore | sn(u/(1 + k1),k1)| < 1/

√k1 whereas | sn(u,k)| < 1/

√k . Setting

u = 4K(k)v and using (78) the relation (13) becomes zk1(v) = ψk (zk (v)) valid for v ∈ Hk1

. From our

previous considerations, with k1 replaced by k , we know that zk (Hk ) = D \ {z ∈ R : z2 ≥ k}. Since

ψk (√k) =

√k1, we conclude that

zk (Hk ) = ψk (D \ {z ∈ R : z2 ≥ k}) = ψk (D) \ {z ∈ R : z2 ≥ k1},

which �nishes the proof. �

Let Hk1be the Hilbert space of analytic functions д on the in�nite strip R + i(−Tk ,Tk ) that satisfy

д(0) = 0 and д(v + 1) = д(v) = д(1/2 −v) and that have �nite norm with respect to the inner product

de�ned via an integral over the rectangle Hk1,

〈д,h〉Hk1

=

∫Hk

1

д′(v)h′(v)dA(v). (14)

Lemma 6. The pullback f 7→ f ◦ zk1determines an isomorphism D→ Hk1

of Hilbert spaces.

Proof. According to Lemma 5, zk1maps the in�nite strip R + i(−Tk ,Tk ) surjectively to the disk D and

satis�es zk1(v + 1) = zk1

(v) = zk1(1/2 −v). Any function f ∈ D is thus mapped to an analytic function

д = f ◦ zk1on the strip R + i(−Tk ,Tk ) satisfying д(0) = 0 and д(v + 1) = д(v) = д(1/2 − v). Moreover,

using the conformal invariance of the Dirichlet inner product, we have that

‖ f ‖2D =∫D| f ′(z)|2dA(z) =

∫D\{z∈R:z2>k1 }

| f ′(z)|2dA(z) =∫Hk

1

|д′(v)|2dA(v) = ‖д‖2Hk1

.

Hence д = f ◦ zk1∈ Hk1

, and more generally 〈f1 ◦ zk1, f2 ◦ zk1

〉Hk1

= 〈f1, f2〉D for f1, f2 ∈ D.

Finally we need to show that f 7→ f ◦ zk1is surjective. Observe that д ∈ Hk1

is determined (via

its periodicities) by its values in Hk1. Therefore д ◦ vk1

determines an analytic function in D \ {z ∈R : z2 > k1}. Pairs of points just above and below the slits correspond (under the mapping vk1

) to

complex conjugate pairs of points of real part ±1/4 (see Lemma 5(iii)). Since д(1/4 + iy) = д(1/4 − iy)and д(−1/4 + iy) = д(−1/4 − iy) for y ∈ (0,Tk ), the function д ◦ vk1

is continuous across the slits and

therefore extends to an analytic function f ∈ D, which clearly satis�es f ◦ zk1= д. �

Since Hk1deals with periodic functions, it is natural to seek an orthogonal basis of Hk1

consisting

of trigonometric functions.

Lemma 7. The sequence (дm)∞m=1de�ned by дm(v) = cos(2πm(v + 1/4)) − cos(πm/2) is an orthogonal

basis ofHk1and

‖дm ‖2Hk1

=m

4

(q−mk − qmk ). (15)

Proof. It should be clear from the de�nition that дm ∈ Hk1for anym ≥ 1. Computing the inner product

of дm and дn with the help of Stokes’ theorem, we �nd

〈дm ,дn〉Hk1

=

∫Hk

1

д′m(v)д′n(v)dA(v) =1

2πi

∫∂Hk

1

дm(v)д′n(v)dz

15

where we used that dA(v) = dA(x + iy) is the measure associated to the 2-form1

π dx ∧ dy = 1

2π i dv ∧ dv

and the second integral traces the boundary ∂Hk1in counterclockwise direction. With the help of the

periodicity of the functions дm and дn this reduces to

〈дm ,дn〉Hk1

=1

2πi

∫1/4

−1/4(дm(x + iTk )д′n(x − iTk ) − дm(x − iTk )д′n(x + iTk ))dx

=1

2πi

∫1/2

−1/2дm(x + iTk )д′n(x − iTk )dx

= n i

∫1/2

−1/2cos(2πm(x + iTk + 1

4)) sin(2πn(x − iTk + 1

4)).

This integral vanishes form , n, while form = n the integral is straightforwardly evaluated to

〈дm ,дm〉Hk1

=m

2

sinh(4πmTk ) =m

4

(q−mk − qmk ). (16)

It remains to show that the linear span of (дm)∞m=1is dense in Hk1

. To this end observe that any analytic

function д ∈ Hk1admits a Fourier series representation

д(v) =∞∑

m=−∞ame

2πmi(v+1/4), (17)

that is absolutely convergent on the strip R + i(−Tk ,Tk ) (and the same is true for its v-derivative). The

identities д(v) = д(1/2−v) and д(0) = 0 imply that a−m = am and a0+2

∑∞m=1

am cos(πm/2) = 0. Hence

we have the absolutely convergent sum

д(v) = a0 + 2

∞∑m=1

am cos(2πm(v + 1/4)) =∞∑

m=1

2am дm(v).

From here one may check that 〈дm ,д〉Hk1

= 2am ‖дm ‖2Hk1

and therefore if д , 0 then 〈дm ,д〉Hk1

, 0 for

somem ≥ 1, implying (дm)∞m=1is dense. �

Transferring the basis (дm)∞m=1of Hk1

to D using the isomorphism of Lemma 6 gives precisely the

basis (fm)∞m=1announced in (3) in the introduction. In the following it will be useful to know that the

functions fm can be analytically continued beyond the unit disc D:

Lemma 8. For eachm ≥ 1, fm(z) = дm ◦vk1(z) has radius of convergence equal to 1/

√k1.

Proof. From the initial remarks in the proof of Lemma 5 it follows that zk1maps (−1/4, 1/4)+i(−2Tk , 2Tk )

biholomorphically to the double-slit plane C\ {z ∈ R : z2 ≥√k1}. Since дm is analytic on C, the function

fm can be analytically continued to the double-slit plane. It was also shown in Lemma 5 that the

boundary segments 1/4± i[0, 2Tk ] of the rectangle are both mapped by zk1to the slit [

√k1, 1/

√k1]. Since

дm(1/4 + iy) = дm(1/4 − iy) for y ∈ R, we �nd that fm is continuous across the slit [√k1, 1/

√k1], and

similar arguments apply to the other slit [−1/√k1,−√k1]. Hence, fm can be analytically continued

to C \ {z ∈ R : z2 ≥ 1/√k1}. To �nish the proof we observe that fm(z) necessarily has a branch cut

at z = 1/√k1 because zk1

(v + 2iTk ) = 1/zk1(v) and z ′k1

(1/4) = 0 imply that z ′k1

(1/4 + 2iTk ) = 0 while

д′m(1/4 + 2iTk ) , 0. �

We are now ready to perform the diagonalization of Jk .

Proposition 9. The family (fm)∞m=1forms an orthogonal basis of D consisting of eigenvectors of Jk

satisfying

Jk fm =1

qm/2k + q−m/2k

fm , ‖ fm ‖2D =m

4

(q−mk − qmk

).

16

Proof. The functions (дm)∞m=1of Lemma 7 satisfy дm = fm ◦ zk1

. Therefore Lemma 6 and Lemma 7

together imply that (fm)∞m=1is an orthogonal basis of D and that

〈fm , fn〉D = 〈дm ,дn〉Hk1

=1

2

m sinh(2mπTk1)1{m=n } =

m

4

(q−mk − qmk

)1{m=n } .

By Lemma 5, 〈fm , fn〉D(ψk (D)) = 〈дm ,дn〉Hk , so by replacing k1 → k in the previous expression we �nd

〈fm , Jk fn〉D = 〈fm , fn〉D(ψk (D)) =1

2

m sinh(2mπTk )1{m=n } .

Hence fm is an eigenvector of Jk with eigenvalue given by

〈fm , Jk fm〉D〈fm , fm〉D

=sinh(2mπTk )sinh(4mπTk )

=1

2 cosh(2mπTk )=

1

qm/2k + q−m/2k

,

where we made use of (78) and (61). �

Notice that this veri�es Theorem 1(ii) for α = π/2 and I = (−π/2,π/2).

2.2 Analytic structure of the eigenvectors

Proposition 10. For any p,m ≥ 1, the coe�cient [zp ] ˆfm(z) = 1

p

⟨ep , ˆfm

⟩Dof the normalized eigenvector

ˆfm = fm/‖ fm ‖D is analytic in k around 0.

Proof. From the de�nition (1) of vk and the fact that K(k) = π2+ O(k2) is analytic in k at 0, it follows

that [zp ]vk (√k z) is analytic in k at k = 0 for any p ≥ 1. Since k1 =

1

4k2 + O(k4), the same is true for

[zp ]vk1(√k1 z) as well as [zp ]fm(

√k1z) = [zp ]дm(vk1

(√k1 z)) for any p,m ≥ 1. Since

√k1 is analytic in k

around k = 0, [zp ]fm(z) has a Laurent series expansion around k = 0 with a pole of order at most p at

k = 0. The inverse norm 1/‖ fm ‖D = 1/√

m4(q−mk − qmk ) as computed in Proposition 9 is analytic in k for

any m (see (70)), implying that [zp ] ˆfm(z) has a Laurent series expansion around k = 0 too. Recalling

from Lemma 8 that fm(z) has radius of convergence larger than 1, we deduce from Cauchy’s inequality

and Lemma 11 below that���[zp ] ˆfm(z)��� ≤ sup

|z |=1

| ˆfm(z)| ≤ 2 for any p,m ≥ 1 and k ∈ (0, 1/2).

We conclude that [zp ] ˆfm(z) has no pole at k = 0, thus �nishing the proof. �

Lemma 11. We have the bound | ˆfm(z)| < 2 for all k ∈ (0, 1/2), |z | ≤ 1 andm ≥ 1.

Proof. The function дm(v) = cos(2πm(v +1/4))−cos(πm/2) of Lemma 7 evaluated atv = x + iy satis�es

|дm(v)|2 =(cos( 1

2mπ ) − cos(2mπ (x + 1/4)) cosh(2mπy)

)2

+ sin2(2mπ (x + 1/4)) sinh

2(2mπy)= cos

2( 12mπ ) − 2 cos( 1

2mπ ) cos(2mπ (x + 1/4)) cosh(2mπy)

+1

2

cos(4mπ (x + 1/4)) + 1

2

cosh(4mπy)

≤ 3

2

+ 2 cosh(2mπy) + 1

2

cosh(4mπy) = 4 cosh4(mπy).

For v ∈ Hk1= (−1/4, 1/4) + i(−Tk ,Tk ) we therefore have

|дm(v)| ≤ 2 cosh2(πmTk ) =

1

2

(qm/4k + q−m/4k )2, (18)

17

Figure 10: The re�ection principle in the vertical axis yields a bijection between walks from (p, 0)to (−`, 0) and walks (p, 0) to (`, 0) that visit the vertical axis at least once.

which translates into the bound

| fm(z)| ≤1

2

(qm/4k + q−m/4k )2 for all |z | ≤ 1,m ≥ 1 and k ∈ (0, 1).

According to (63), we have qk ∈ (0, 1

20) when k ∈ (0, 1/2) and therefore

1

2(qm/4k + q−m/4k )2 ≤

√q−mk − qmk .

The claim then follows from | ˆfm(z)| = | fm(z)|/√

m4(q−mk − qmk ). �

It follows from the proof of Proposition 10 that one can compute the series expansion of

⟨ep , ˆfm

⟩D=

p[zp ] ˆfm(z) to arbitrary order in k , given the series expansions of vk1(√k1z) (which follows from the

de�nition (1)), K(k) and qk (which are described in Appendix A). The table below shows the �rst few

terms of1√m

⟨ep , ˆfm

⟩D

in t = k/4 for p ≤ 3 andm ≤ 6.

1√m

⟨ep , ˆfm

⟩D

p = 1 p = 2 p = 3

m = 1 −1 + 3t 4

2+ 24t6 + 2765t 8

8+ · · · 0 −3t2 − 24t4 − 477t 6

2− 2652t8 + · · ·

m = 2 0 1 − 4t4 − 64t6 − 1835t 8

2+ · · · 0

m = 3 t2 + 8t4 + 82t6 + 944t8 + · · · 0 −1 + 9t4 + 144t6 + 2052t8 + · · ·m = 4 0 −2t2 − 16t4 − 160t6 − 1792t8 + · · · 0

m = 5 −t4 − 16t6 − 230t8 + · · · 0 3t2 + 24t4 + 231t6 + 2472t8 + · · ·m = 6 0 3t4 + 48t6 + 684t8 + · · · 0

2.3 The operators Bk and Ak

Recall that B`,p (t) =⟨e`,Bkep

⟩D

is the generating function for the set B`,p of simple diagonal walks

from (p, 0) to (`, 0) that maintain strictly positive �rst coordinate. A simple re�ection principle (see

Figure 10) teaches us that B`,p (t) is given by

B`,p (t) = B`−p (t) − B−`−p (t), (19)

where Bm(t), m ∈ Z, is the generating function of simple diagonal walks from (0, 0) to (m, 0) without

further restrictions.

Lemma 12. For �xed t ∈ (0, 1/4), Bm(t) can be expressed as a contour integral as

Bm(t) =1

2πi

∫γ

dz

zm+1

√1 − 4t2(z + 1/z)2

, (20)

where γ traces the unit circle in counterclockwise direction.

18

Proof. The contribution of walks from (0, 0) to (m, 0) of length 2n andm even is(2n

n

) (2n

n +m/2

)t2n =

1

2πi

∫γ

dz

zm+1

(2n

n

)(z + 1/z)2nt2n .

The result then follows from summing over n ≥ 0 and relying on absolute convergence to interchange

the summation and integral. �

Based on the similarity between the integrand of (20) and the one in the de�nition (1) of vk1, we

�nd the following useful representation of Bk .

Lemma 13. If f ,д ∈ D are analytic in a neighbourhood of the closed unit disk, then

〈f ,Bkд〉D =2K(k)π

∫γ ′(f (z) − f (z−1))(д(z) − д(z−1))v ′k1

(z)dz, (21)

where γ ′ traces the upper half of the unit circle starting at 1 and ending at −1.

Proof. From the de�nition (1) we see that the integrand of (21) is continuous and bounded on the upper-

half circle, and therefore the right-hand side of (21) converges absolutely. Hence, it su�ces to check the

identity for f = e` and д = ep , p, ` ≥ 1. Combining (19) and Lemma 12 we �nd

〈e`,Bep〉D = B`−p (t) − B−`−p (t) =1

2

(B`−p (t) − B−`−p (t) + Bp−`(t) − B`+p (t)

)=−1

4πi

∫γ

(z` − z−`)(zp − z−p )dzz√

1 − 4t2(z + 1/z)2=−1

2πi

∫γ ′

(z` − z−`)(zp − z−p )dzz√

1 − 4t2(z + 1/z)2, (22)

where in the last equality we used that both sides vanish for p + ` odd, while for p + ` even the upper

and lower half circles contribute equally. Note that

(k1 − z2)(1 − k1z2) = −z2((1 + k1)2 − k1(z + 1/z)2) = −z2(1 + k1)2

(1 − k2

4

(z + 1/z)2).

Hence, for z on the upper-half circle (1) implies that

v ′k1

(z) = 1

4K(k1)1√

(k1 − z2)(1 − k1z2)=

i

4K(k)1

z√

1 − k2

4(z + 1/z)2

,

where we used (78) and the sign on the right-hand is determined by the fact that v ′k1

(z) has positive

imaginary part when z is on the upper-half circle close to 1. Combining with (22) we indeed reproduce

the right-hand side of (21). �

With the contour integral representation in hand it is now straightforward to evaluate Bk (and Ak

subsequently) with respect to the basis (fm)∞m=1.

Proposition 14. The linear operators Bk and Ak are compact and have the same eigenvectors (fm)∞m=1as

Jk satisfying

Bk fm =2K(k)π

1

m

1 − qmk1 + qmk

fm ,

Ak fm =π

2K(k)m

q−m/2k − qm/2k

fm .

19

Proof. According to Lemma 8 the functions fm(z) are analytic for |z | < 1/√k1, and therefore we may

apply Lemma 13 to obtain

〈fn ,Bk fm〉D =2K(k)π

∫γ ′(fn(z) − fn(z−1))(fm(z) − fm(z−1))v ′k1

(z)dz

= −2K(k)π

∫1/4

−1/4(дn(v + iTk ) − дn(v − iTk ))(дm(v + iTk ) − дm(v − iTk ))dv,

where we changed integration variables z = zk1(v), using that zk1

(v − iTk ) = 1/zk1(v + iTk ) and that zk1

maps (−1/4, 1/4) + iTk to γ ′ but with opposite orientation. Substituting the expression for дm yields

〈fn ,Bk fm〉D = −2K(k)π

∫1/2

0

(cos(2πn(v + iTk )) − cos(2πn(v − iTk )))

× (cos(2πm(v + iTk )) − cos(2πm(v − iTk )))dv

=8K(k)π

sinh(2πnTk ) sinh(2πmTk )∫

1/2

0

sin(2πnv) sin(2πmv)dv

=2K(k)π

sinh2(2πmTk )1{m=n } .

Together with (4) we conclude that fm is an eigenvector of Bk (which we already knew from Lemma 4

and Proposition 9) with eigenvalue

〈fm ,Bk fm〉D〈fm , fm〉D

=4K(k)π

sinh2(2mπTk )

m sinh(4mπTk )=

2K(k)π

tanh(2mπTk )m

=2K(k)π

1

m

1 − qmk1 + qmk

.

Since Jk = AkBk , we �nd using Proposition 9 that

Ak fm =π

2K(k)m1 + qmk1 − qmk

1

qm/2k + q−m/2k

fm =π

2K(k)m

q−m/2k − qm/2k

fm .

The eigenvalues ofBk andAk both approach 0 asm →∞, implying compactness (see e.g. [41, Proposition

1.19]). Note that we already established compactness of Ak in Proposition 2. �

2.4 Proof of Theorem 1

We start with part (i) with α ∈ π2Z. Given a walk w ∈ W(α )

`,p , let (si )Ni=1be the sequence of times at which

w alternates between the axes, i.e. s0 = 0 and for each j ≥ 0 we set sj+1 = inf{s > sj : |θws − θwsj | = π/2}provided it exists (otherwise sj = sN is the last entry in the sequence). Let (α j )Nj=0

and (`j )Nj=0be the

sequences of winding angles and distances to the origin de�ned by α j = θwsj respectively `j = |wsj | for

0 ≤ j ≤ N . It is now easy to see that for 0 ≤ j < N the part of the walk between time sj and sj+1 is (up

to a unique rotation around the origin and/or re�ection in the horizontal axis) of the form of a walk

w (j) ∈ J`j+1, `j . Similarly, the last part of the walk between time sN and |w | is (up to rotation) of the form

of a walk w (N ) ∈ B`,`N . See Figure 11 for an example.

In fact, this construction is seen to yield a bijection between W(α )`,p and the set of tuples(

N , (`j )Nj=0, (α j )Nj=0

, (w (j))Nj=0

)where N ≥ 0, `0 = p, `j ≥ 1, (α j )Nj=0

is a simple walk onπ2Z from α0 = 0 to αN = α , w (j) ∈ J`j+1, `j for

0 ≤ j < N and w (N ) ∈ B`,`N . If we denote by

a(α )N =

(N

N− 2απ

2

)1{N− 2α

π even and non-negative}

20

(a) (b) (c)

Figure 11: (a) A walk w ∈ W(−π /2)5,3 with N = 3 axis alternations; (b) its decomposition into

w (0) ∈ J1,3, w (1),w (2) ∈ J1,1, w (3) ∈ B5,1; (c) its winding angle sequence (θwi ).

the number of simple walks onπ2Z from 0 to α of length N ≥ 0, then we may identify the generating

function of W(α )`,p as

W (α )`,p (t) =

∞∑N=0

a(α )N

∞∑`1, ..., `N =1

B`,`N (t)N−1∏i=0

J`i+1, `i (t)

=

∞∑N=0

a(α )N

∞∑`1, ..., `N =1

⟨e`,Bke`N

⟩D

N−1∏i=0

1

`i+1

⟨e`i+1, Jke`i

⟩D

=

∞∑N=0

a(α )N

⟨e`,Bk JNk ep

⟩D.

Since the eigenvalues of Jk are all strictly smaller than 1/2 and a(α )N ≤ 2N

, the operator

∑MN=0

a(α )N Bk JNkis compact for any M ≥ 0 and converges as M → ∞ (in the operator norm topology) to a compact

self-adjoint operator Y(α )k satisfying

W (α )`,p (t) =

⟨e`,Y

(α )k ep

⟩D. (23)

With a little help of (10), we �nd the (formal) generating function

a(α )(x) =∞∑

N=0

a(α )N xN =1

√1 − 4x2

(1 −√

1 − 4x2

2x

)2 |α |/π

.

Then one may deduce after some simpli�cation that

Y(α )k fm = Bk a(α ) (Jk ) fm =2K(k)π

1

m

1 − qmk1 + qmk

a(α )(

1

qm/2k + q−m/2k

)fm

=2K(k)π

1

mq |α |m/πk fm ,

in agreement with part (i) of Theorem 1.

We could easily extend this result to the case I = (β−, β+) with β± ∈ π2Z ∪ {±∞} such that 0 ∈ I

and α ∈ I ∩ π2Z, by replacing a(α )(x) by the generating function of simple walks on

π2Z con�ned to an

interval. Instead, we choose to discuss a re�ection principle at the level of the simple diagonal walks,

which allows us to directly generalize to the case of β± ∈ π4Z ∪ {±∞} of part (iii).

21

(a) (b)

Figure 12: Example with α = 0, I = (−π/4,π/2). (a) A walk w ∈ W(−π /2)4,6 is depicted in black, and

its re�ection w ′ ∈ W(0)4,6 in gray. (b) The corresponding winding angle sequences (θwi ) and (θw ′i ).

Lemma 15. Suppose ` and p are both odd and β± ∈ π2Z, or ` and p are both even and β± ∈ π

4Z, such that

0 ∈ I . If in addition α ∈ π2Z ∩ I , then the generating function forW(α, I )

`,p is given by

W (α, I )`,p (t) =

∞∑n=−∞

(W (α+nδ )

`,p (t) −W (2β+−α+nδ )`,p (t)

), δ B 2(β+ − β−).

Proof. Consider any walk w ∈ ⋃∞n=−∞W

(2β+−α+nδ )`,p and let s = inf{j ≥ 0 : θwj < I } be the �rst time w

leaves I , which is well-de�ned since θw < I . It is not hard to see that under the stated conditions on

`,p, β± the winding angle sequence (θwi )|w |i=0

cannot cross β± without visiting β±, and therefore θws = β±and ws lies on an axis or a diagonal of Z2

. Then we let w ′ be the walk obtained from w by re�ecting the

portion of w after time s in this axis or diagonal (see Figure 12). Then θw′= 2β+ − θw or θw

′= 2β− − θw .

Hence w ′ ∈ ⋃∞n=−∞W

(α+nδ )`,p . It is not hard to see that this mapping w 7→ w ′ is injective (the inverse

w ′ 7→ w is given by the exact same re�ection operation). Moreover, any walk w ′ ∈ ⋃∞n=−∞W

(α+nδ )`,p is

obtained in such way provided (θw ′i )|w ′ |i=0

visits β± at least once. Clearly, the only walks w ′ not satisfying

the latter condition are the ones in W(α, I )`,p . The claimed result for the generating function W (α, I )

`,p (t)readily follows (absolute convergence is granted because

∑α ′∈ π

2ZW

(α ′)`,p (t) < ∞). �

Inspired by this result let us introduce for β± ∈ π4Z such that 0 ∈ I and α ∈ π

2Z ∩ I , the operator

B(α,β−,β+)k on D de�ned by

B(α,β−,β+)k B∞∑

n=−∞

(Y(α+nδ )k − Y(2β+−α+nδ )k

), δ B 2(β+ − β−), (24)

which by construction satis�es B(α,β−,β+)k = B(−α,−β+,−β−)k . By Theorem 1(i) it is well-de�ned, compact

and self-adjoint and has eigenvalues

2K(k)π

1

m

∞∑n=−∞

(q |α+nδ |m/πk − q |2β+−α+nδ |m/πk

)=

2K(k)π

q2mβ−/πk − 1

mqmα/πk

q2mα/πk − q2mβ+/π

k

q2mβ−/πk − q2mβ+/π

k

(25)

for α ≥ 0, while the eigenvalues for α < 0 are obtained by the substitution (α , β−, β+) → (−α ,−β+,−β−).Lemma 15 then tells us that

W (α, I )`,p (t) =

⟨e`,B

(α,β−,β+)k ep

⟩D

(26)

holds under the conditions stated in the lemma, which exactly veri�es part (ii) and (iii) for 0,α ∈ I and

β± �nite.

22

Similarly when β− = −∞ or β+ = ∞ one may introduce the operators B(α,−∞,β+)k = Y(α )k − Y(2β+−α )k

and B(α,β−,∞)k = Y(α )k − Y(2β−−α )k . It is straightforward to check that then (26) still holds, and that the

eigenvalues are given by (25) in the appropriate limit β− → −∞ or β+ →∞.

Next, let us consider the case 0 < α ∈ π2Z and I = (0,α). The case α = π/2 with the corresponding

operator A(π /2)k = Ak has already been settled in Proposition 14, so let us assume α ≥ π . Any such

walk w ∈ W(α,(0,α ))`,p is naturally encoded in a triple w (1),w (2),w (3) of walks with w (1) ∈ A`1,p , w (2) ∈

W(α−π ,(−π /2,α−π /2))`2, `1

and w (3) ∈ A`,`2for some `1, `2 ≥ 1. Hence

W (α,(0,α ))`,p (t) =

∞∑`1, `2=1

1

``2

⟨e`,Ake`2

⟩D

⟨e`2,B(α−π ,−π /2,α−π /2)k e`1

⟩D

1

`1p

⟨e`1,Akep

⟩D

=1

`p

⟨e`,AkB

(α−π ,−π /2,α−π /2)k Akep

⟩D.

One may easily verify the claimed eigenvalues of A(α )k = AkB(α−π ,−π /2,α−π /2)k Ak and its compactness,

thus settling Theorem 1(ii) for the operator A(α ).Finally, a similar argument shows that for 0 < α ∈ π

2Z, 0 > β− ∈ π

4Z,

W(α,(β−,α ))`,p (t) =

∞∑1=1

1

``1

⟨e`,Ake`1

⟩D

⟨e`1,B(α−π /2,β−,α )k ep

⟩D=

1

`

⟨e`,AkB

(α−π /2,β−,α )k ep

⟩D,

and once again one may directly verify the eigenvalues of J(α,β−)k = AkB(α−π /2,β−,α )k . This veri�es

Theorem 1(ii) and (iii) for the operator J(α,β−)k , thereby �nishing the proof of Theorem 1.

3 Excursions

3.1 Counting excursions with �xed winding angle

Recall from the introduction the set of excursions E consisting of (non-empty) simple diagonal walks

starting and ending at the origin with no intermediate returns. For such an excursion w ∈ E we have a

well-de�ned winding angle sequence (θwi )|w |i=0

with θw1= θw

0= 0 and θw = θw|w | = θ

w|w |−1

. Our �rst goal

is to compute the generating function of excursions with winding angle equal to α ∈ π2Z,

F (α )(t) B∑w ∈E

t |w |1{θw=α } . (27)

To this end we cannot directly apply Theorem 1(i) because the excursions start and end at the origin.

Nevertheless, a combinatorial trick allows us to relate F (α )(t) to the generating functionsW (α′)

`,p with

α ′ > |α |. In order for this trick to work we �rst have to establish a bound onW (α )`,p as α gets large.

Lemma 16. For t ∈ (0, 1/4), there exists aC > 0 such that∑

`,p≥1W (α )

`,p (t) < Cq|α |π −

1

4

k for all α ∈ π2Z \ {0}.

Proof. Let r ∈ (1, 1/√k1). Since

ˆfm has radius of convergence larger than r (Lemma 8), Cauchy’s

inequality implies that

���[zp ] ˆfm(z)��� ≤ r−p sup |z |=r | ˆfm(z)|. Therefore

∞∑p=1

���⟨ep , ˆfm⟩D

��� = ∞∑p=1

p���[zp ] ˆfm(z)

��� ≤ r

(r − 1)2 sup

|z |=r| ˆfm(z)| =

r

(r − 1)21

‖ fm ‖Dsup

|z |=r| fm(z)|. (28)

23

Given that vk1maps the double-slit disc D \ {z ∈ R : z2 ≥ k1} onto Hk1

(Lemma 5), we may choose

r ∈ (1, 1/√k1) such that vk1

maps {z ∈ C : |z | ≤ r } \ {z ∈ R : z2 ≥ k1} into the slightly larger rectangle

(− 1

4, 1

4) + i(− 5

4Tk ,

5

4Tk ). The proof of Lemma 11 shows that within this triangle |дm(v)| satis�es

|дm(v)| ≤ 2 cosh2( 5

4mπTk ) =

1

2

(q− 5

16m

k + q5

16m

k

)2

≤ 2q− 5

8m

k . (29)

The maximum of |дm(v)| on the latter is attained in the corners ±1/4 ± i 5

4Tk of the rectangle (see also

the proof of Lemma 11), where it is bounded by

cosh(2πm 5

4Tk ) ≤ q−5m/8

k .

By (4), there exists a c ′ > 0 (depending on t ) such that ‖ fm ‖D ≥ c ′q−m/2k for allm ≥ 1. It then follows

from (28) that we can �nd a c > 0 such that

∞∑p=1

���⟨ep , ˆfm⟩D

��� ≤ cq−5m/8k

‖ fm ‖D≤ c

c ′q−m/8k for allm ≥ 1. (30)

With the help of Theorem 1(i) we may evaluate for α ∈ π2Z \ {0},

∞∑`,p=1

W (α )`,p (t) =

∞∑`,p=1

⟨e`,Y

(α )k ep

⟩D=

∑`,p≥1

∞∑m=1

2K(k)π

1

mqm |α |/πk

⟨e`, ˆfm

⟩D

⟨ep , ˆfm

⟩D,

which according to our bound (30) is seen to be absolutely convergent. Interchanging the summations

and performing the sum over ` and p, we �nd that there exists a c ′′ > 0 such that

∞∑`,p=1

W (α )`,p (t) ≤ c ′′

∞∑m=1

1

mqm( |α |π −

1

4)

k = c ′′q|α |π −

1

4

k

∞∑m=1

1

mq(m−1)( |α |π −

1

4)

k ≤ c ′′q|α |π −

1

4

k

∞∑m=1

1

mq

1

4(m−1)

k .

Since the latter sum converges, we have obtained the desired estimate. �

Lemma 17. For α ∈ π2Z, the series F (α )(t) may be expressed as the absolutely convergent sum

F (α )(t) = 4

∞∑m, `,p=1

(−1)`+p+m+1mW ( |α |+mπ /2)2`,2p (t). (31)

Proof. By Lemma 16,

∑∞`,p=1

W ( |α |+mπ /2)2`,2p (t) < Cq

m2+|α |π −

1

4

k for some C > 0 and all α ∈ π2Z and m ≥ 1,

from which it follows that the right-hand side of (31) is absolutely convergent.

Next we use that for α ′ ∈ π2Z>0, the sets

⋃p∈4Z>0

W(α ′)`,p and

⋃p∈4Z>0−2

W(α ′)`,p are nearly in bijection.

Indeed, a walk in the former is mapped to a unique walk in the latter by moving its starting point by

(±2, 0) depending on the direction of the �rst step (see Figure 13), while keeping the other sites �xed.

It is not hard to see that any walk

⋃p∈4Z>0−2

W(α ′)`,p is obtained in such way except for those walks in

W(α ′)`,2

that have w1 = (1,±1). The generating function of such walks is therefore given by∑w ∈W(α

′)`,2

t |w |1{ |w1 |=√

2} =∞∑p=1

(−1)p+1W (α′)

`,2p (t).

An analogous argument for the endpoint then yields

S (α′)(t) B

∑w ∈W(α

′)2,2

t |w |1{ |w1 |= |w |w |−1|=√

2} =∞∑

p, `=1

(−1)p+`W (α′)

2`,2p (t). (32)

24

Figure 13: By changing the starting point the dark blue (solid) walk in W(π )2,4 is mapped to a walk in

W(π )2,2 . The dark green (dashed) walk w ∈ W(π )

2,2 satis�es |w1 | = |w |w |−1| =√

2, such that moving

its starting point and endpoint to the origin gives an excursion.

LetX (α′)

be the set of four possible tuples (w1,w |w |−1,θw|w |−1

−θw1)whenw ∈ W(α

′)2,2 and |w1 | = |w |w |−1

| =√

2. For instance, if α ′ = 3π/2 then

X (3π /2) = {((1, 1), (−1,−1),π ), ((1, 1), (1,−1), 3π/2), ((1,−1), (−1,−1), 3π/2), ((1,−1), (1,−1), 2π )} .

Then we may write

S (α′)(t) =

∑(x,y,α )∈X (α ′)

∑w ∈W(α

′)2,2

t |w |1{w1=x,w |w |−1=y } =

∑(x,y,α )∈X (α ′)

∑w ∈E

t |w |1{w1=x,w |w |−1=y }1{θw=α },

where we used the obvious mapping to excursions by merely moving the starting point and endpoint

to the origin (see Figure 13). By the rotational symmetry of the set of excursions we may replace the

1{w1=x,w |w |−1=y } by 1/4 in the last sum. By observing that α takes exactly the four values α ′ − π/2, α ′,

α ′, α ′ + π/2, one �nds that

S (α′)(t) = 1

4

∑(x,y,α )∈X (α ′)

∑w ∈E

t |w |1{θw=α } =1

4

F (α′−π /2)(t) + 1

2

F (α′)(t) + 1

4

F (α′+π /2)(t).

From here it is an easy check by substitution that for α ∈ π2Z≥0,

4

∞∑m=1

(−1)m+1mS (α+mπ /2)(t) = F (α )(t) = F (−α )(t),

which together with (32) yields the claimed identity. �

We are now in the position to apply Theorem 1(i) and explicitly evaluate F (α )(t), as well as its

“characteristic function”

F (t ,b) B∑w ∈E

t |w |eibθw=

∑α ∈ π

2Z

F (α )(t) eibα . (33)

Notice that b 7→ F (t ,b) is periodic with period 4, and that by symmetry F (α )(t) = F (−α )(t) and therefore

F (t ,b) is real and F (t ,−b) = F (t ,b).

25

Proposition 18. The excursion generating functions are given by

F (α )(t) = 2π

K(k)

∞∑n=1

(1 − qnk )2

1 − q4nk

qn( 2

π |α |+1)k , (34)

F (t ,b) =

1

cos

(πb2

) 1 −π tan

(πb4

)2K(k)

θ ′1

(πb4,√qk

)θ1

(πb4,√qk

) for b ∈ R \ Z,

1 − π2K (k ) for b ∈ 4Z,

1 − 2E(k )π for b ∈ 2Z + 1,

−1 +4E(k )π − (1 − k2) 2K (k )π for b ∈ 4Z + 2,

(35)

where K(k) and E(k) are the complete elliptic integrals of the �rst and second kind and θ1(z,q) is a Jacobitheta function (see Appendix A for de�nitions).

Proof. Combining Lemma 17 with Theorem 1(i) we �nd

F (α )(t) = 4

∞∑m, `,p=1

(−1)`+p+m+1m⟨e2`,Y

( |α |+mπ /2)k e2p

⟩D

=4K(k)π

∞∑m,n=1

(−1)m+1m

nqn( 2

π |α |+m)k

1

‖ f2n ‖2

( ∞∑p=1

(−1)p⟨f2n , e2p

⟩D

)2

=4K(k)π

∞∑n=1

1

n

qn( 2

π |α |+1)k

(1 + qnk )21

‖ f2n ‖2

( ∞∑p=1

(−1)p⟨f2n , e2p

⟩D

)2

, (36)

where we used that

⟨fn , e2p

⟩D= 0 for n odd. Since f2m(z) has radius of convergence larger than one

(Lemma 8) and is even, we have

∞∑p=1

(−1)p⟨f2n , e2p

⟩D=

∞∑p=1

(−1)p 2p [z2p ]f2n(z) = i f ′2n(i).

From the de�nition (1) of vk1one may read o� that v ′k1

(i) = 1/(4K(k1)(1 + k1)) = 1/(4K(k)) using (78),

while vk1(i) = iTk using Lemma 5(ii). Together with (3) we then �nd

∞∑p=1

(−1)p 〈e2p , f2n〉 = iд′

2n(iTk )4K(k) = (−1)n π

K(k)n sinh(4πnTk ) = (−1)n π

2K(k)n(q−nk − q

nk ). (37)

Combining with (36) and (4) we arrive at

F (α )(t) = 2π

K(k)

∞∑n=1

(q−nk − qnk )

2

(1 + qnk )2qn( 2

π |α |+1)k

q−2nk − q2n

k

=2π

K(k)

∞∑n=1

(1 − qnk )2

1 − q4nk

qn( 2

π |α |+1)k ,

establishing the �rst identity (34) of the Proposition.

Using the identity ∑α ∈ π

2Z

x2

π |α |eibα =x−1 − x

x + x−1 − 2 cos(πb/2) for 0 < x < 1

26

we obtain with some algebraic manipulation

F (t ,b) = 2π

K(k)

∞∑n=1

(1 − qnk )2

1 − q4nk

1 − q2nk

qnk + q−nk − 2 cos(πb/2) (38)

=2π

K(k)

∞∑n=1

(1 − 2

qnk + q−nk

)1

qnk + q−nk − 2 cos(πb/2) (39)

=2π

K(k)

∞∑n=1

[(1 − 1

cos(πb/2)

)1

qnk + q−nk − 2 cos(πb/2) +

1

cos(πb/2)1

qnk + q−nk

], (40)

where the last equality only holds for b ∈ R \ (2Z + 1). The �rst term in the sum can be handled for

b ∈ R \ Z by recognizing that

∞∑n=1

1

qnk + q−nk − 2 cos(πb/2) =

∞∑n=1

∞∑m=1

qmnk

sin(πbm/2)sin(πb/2) =

∞∑m=1

qmk1 − qmk

sin(πbm/2)sin(πb/2)

=1

4 sin(πb/2)

[θ ′

1(πb/4,√qk )

θ1(πb/4,√qk )− cot(πb/4)

],

where in the last equality we used [1, 16.29.1]. The second term follows from [1, 17.3.22],

∞∑n=1

1

qnk + q−nk= −1

4

+K(k)2π.

Hence, for b ∈ R \ Z we have

F (t ,b) = 1

cos

(πb2

) 1 −π tan

(πb4

)2K(k)

θ ′1

(πb4,√qk

)θ1

(πb4,√qk

) ,as claimed.

It remains to check the value of F (t ,b) at b = 0, 1, 2, since F (t ,b + 4) = F (t ,b) and F (t ,−b) = F (t ,b)as observed below (33). To this end we require two identities that follow from [17, (23) and (26)], namely

∞∑n=1

1

(qnk + q−nk )2

= −1

8

+1

2

∞∑n=−∞

1

(qnk + q−nk )2

= −1

8

+1

2π 2K(k)E(k),

∞∑n=1

1

(qn−1/2k + q1/2−n

k )2=

1

2π 2(K(k)E(k) − (1 − k2)K2(k)).

Starting from (39) or (40) and using these identities we obtain

F (t , 0) = 2π

K(k)

∞∑n=1

1

qnk + q−nk= 1 − π

2K(k) ,

F (t , 1) = 4π

K(k)

[1

2

∞∑n=1

1

qnk + q−nk−∞∑n=1

1

(qnk + q−nk )2

]=

4π

K(k)

[K(k)4π− E(k)K(k)

2π 2

],

F (t , 2) = 2π

K(k)

[2

∞∑n=1

1

(qn/2k + q−n/2k )2−∞∑n=1

1

qnk + q−nk

]=

2π

K(k)

[2

∞∑n=1

1

(qnk + q−nk )2

+ 2

∞∑n=1

1

(qn−1/2k + q1/2−n

k )2−∞∑n=1

1

qnk + q−nk

]=

2π

K(k)

[2

E(k)K(k)π 2

− (1 − k2)K2(k)π 2− K(k)

2π

],

which reduce to formulas in (35). �

27

There are more elementary ways to compute the quantity F (t ,b) at integer values of b, by directly

considering the contributions of the individual walks to (33). Indeed, F (t , 0) just counts all excursions

from the origin without intermediate returns, and therefore is related to (2) via

2

πK(4t) =

∞∑n=0

F (t , 0)n = 1

1 − F (t , 0) .

Similarly, F (t , 1) precisely counts excursions that stay strictly within a single diagonal half-plane, while

F (t , 2) counts excursions that stay strictly within a single quadrant. This follows from a re�ection

principle that is a special case of Theorem 23 below, and indeed one may check that they are related to

the cone generating functions F (0,(−π /2,π /2))(t) = F (t , 1)/4 and F (0,(−π /4,π /4))(t) = F (t , 2)/4 of Theorem

23. From elementary considerations we could therefore have determined that

F (t , 1) = 4

∞∑n=0

t2n+2

n∑p=0

1

p + 1

(2p

p

) (2(n − p)n − p

) (2n

2p

)= 1 − 2F1(− 1

2, 1

2; 1; (4t)2), (41)

F (t , 2) = 4

∞∑n=0

t2n+2

(1

n + 1

(2n

n

))2

= −1 + 2F1(− 1

2,− 1

2; 1; (4t)2), (42)

in terms of the hypergeometric function 2F1(a,b; c; z) = ∑∞n=0

(a)n (b)n(c)n

znn!

. With the help of [1, 17.3.10]

and by noting that (2) implies1

16t∂∂t t

∂∂t F (t , 2) =

2

π K(4t) these can be shown to reproduce the explicit

expressions in Proposition 18.

By (41) and (42) and the fact [1, 17.3.9] that2

π K(k) = 2F1( 12, 1

2; 1;k2), all of the functions F (t ,b) at in-

teger values b can be expressed as (a rational function of) a hypergeometric function 2F1(± 1

2,± 1

2; 1; (4t)2)

with appropriate signs. It is a classical result [37] that all of these, hence all of F (t ,b) for b ∈ Z, are

transcendental in t . Actually something stronger is true:2

π K(k) and2

π E(k) are algebraically independent

in k . Although we are certain this is a classical result, we were unable to �nd a reference.

Lemma 19. The series 2

π K(k) and2

π E(k) are algebraically independent in k .

Proof. At k = 1/√

2 the complete elliptic integrals K(k) and E(k) take the value [9, Theorem 1.7]

2K(1/√

2)π

=Γ2( 1

4)

2π 3/2 ,2E(1/

√2)

π=

8π 2 + Γ4( 14)

4π 3/2Γ2( 14).

However, π and Γ( 14) (and q

1/2 = eπ ) are famously [34] known to be algebraically independent over Q.

Hence, k ,2

π K(k) and2

π E(k) cannot satisfy a nontrivial polynomial equation with rational coe�cients. �

The situation is surprisingly di�erent at other rational values of b:

Corollary 20. The power series t 7→ F (t ,b) is algebraic if b ∈ Q \ Z and transcendental if b ∈ Z.

Proof. We have already dealt with the case b ∈ Z, so let us assume b ∈ Q \ Z. Using the Landen

transformation θ ′1(u,q)/θ1(u,q) + θ ′4(u,q)/θ4(u,q) = θ ′1(u,

√q)/θ1(u,

√q), which follows e.g. from the

series representations [1, 16.29.1 & 16.29.4], we may rewrite F (t ,b) as

F (t ,b) = 1

cos

(πb2

) 1 −π tan

(πb4

)2K(k)

©­­«θ ′

1

(πb4,qk

)θ1

(πb4,qk

) + θ ′4(πb4,qk

)θ4

(πb4,qk

) ª®®¬ , (43)

which in turn can be expressed using Jacobi’s zeta function Z (u,k) (see (75) in Appendix A) as1

F (t ,b) = 1

cos

(πb2

) [1 − tan

(πb

4

) (2Z (u,k) + cn (u,k) dn (u,k)

sn (u,k)

)], u = K(k)b/2. (44)

1Thanks to Kilian Raschel for pointing out this relation!

28

The numbers cos(πb/2), tan(πb/4) are algebraic because cos(πb/2) and sin(πb/2) occur as roots of

Chebyshev polynomials. To show that F (t ,b) is algebraic, we thus have to check that sn(u,k), cn(u,k),dn(u,k), and Z (u,k) are algebraic in k .

The Jacobi elliptic functions sn, cn, dn satisfy addition formulas [1, 16.17.1-3], e.g.

sn(x + y) = snx cny dny + sny cnx dnx

1 − k2sn

2 x sn2y

,

where all elliptic functions are understood to have modulus k . These addition formulas allow one to

express sn(nx ,k) as a rational function of k , sn(x ,k), cn(x ,k) and dn(x ,k). Setting x = K(k)b/2 such

that sn(nx ,k) = 0 and eliminating cn(x ,k) and dn(x ,k) using the quadratic relations (73), one obtains a

polynomial equation for sn(x ,k)with coe�cients that are polynomials in k , implying that sn(K(k)b/2,k)is algebraic in k and similarly for cn(K(k)b/2,k) and dn(K(k)b/2,k).

The Jacobi zeta function Z also satis�es an addition formula (76), that can be used together with the

addition formulas for sn, cn and dn to express Z (nx ,k) − nZ (x ,k) as a rational function of k , sn(x ,k),cn(x ,k) and dn(x ,k). Setting x = K(k)b/2, we have that Z (nx ,k) = 0 [1, 17.4.29-30] and sn(x ,k), cn(x ,k)and dn(x ,k) are algebraic in k , so the same is true for Z (K(k)b/2,k). �

3.2 Asymptotic behaviour

As an intermezzo let us look at a probabilistic consequence of Proposition 18 in the limit t → 1/4. For a

simple random walk on Z2started at the origin we consider the total winding angle around the origin

up to its �rst return to the origin, ignoring the contributions from the �rst and last step (as we have

been doing all along for excursions).

Corollary 21. Consider a simple random walk on Z2 started at the origin. The probability that its windingangle around the origin upon its �rst return equals πm/2 form ∈ Z is

1

π

(−ψ

(|m | + 1

4

)+ 2ψ

(|m | + 2

4

)−ψ

(|m | + 3

4

)),

whereψ (x) = Γ′(x)/Γ(x) is the digamma function.

Proof. Letm ∈ Z be �xed. First we rewrite the sum in (34) with α =mπ/2 as

∞∑n=1

(1 − qnk )2

1 − q4nk

qn( |m |+1)k =

∞∑n=1

∞∑p=0

qn( |m |+4p+1)k (1 − qnk )

2

=

∞∑p=0

[1

1 − q |m |+4p+1

k

− 2

1 − q |m |+4p+2

k

+1

1 − q |m |+4p+3

k

]As k → 1, i.e. qk → 1, the latter is asymptotically equal to

∞∑p=0

[1

|m | + 4p + 1

− 2

|m | + 4p + 2

+1

|m | + 4p + 3

]1

1 − qk(1 + o(1))

=1

4

(−ψ

(|m | + 1

4

)+ 2ψ

(|m | + 2

4

)−ψ

(|m | + 3

4

))1

1 − qk(1 + o(1)),

where we used the series representationψ (x+1)−ψ (1) = ∑∞p=1

(1

p −1

x+p

)of the digamma function. From

the de�nition (61) of the nome it follows that K(k) logqk = −πK(k ′), so K(k)(1 − qk ) → πK(0) = π 2/2as k → 1. Hence, we �nd from (34) that as t → 1/4,

F (mπ /2)(t) → 1

π

(−ψ

(|m | + 1

4

)+ 2ψ

(|m | + 2

4

)−ψ

(|m | + 3

4

)).

From the de�nition (27) it is easily seen that this is precisely the desired probability. �

29

Next we look at the growth of the coe�cients [t2`]F (t ,b) as ` →∞.

Lemma 22. For �xed b ∈ [0, 2] the coe�cients of t 7→ F (t ,b) satisfy the estimate

[t2`]F (t ,b) ∼

sin

2

(πb4

)Γ(1+b)π

42(`+1−b)

`b+1for b ∈ (0, 2)

π` log

2 `4

2` for b = 0

1

4π `34

2` for b = 2.

The analogous results for other b ∈ R follow from F (t ,b) = F (t ,−b) = F (t ,b + 4).

Proof. We �rst consider the case b ∈ (0, 1) ∪ (1, 2). For �xed z ∈ R \ πZ, by (62) and (64) the functions

q−1/4k θ1(z,qk ) and q−1/4

k θ ′1(z,qk ) are analytic in k ∈ C \ {w ∈ R : w2 ≥ 1} and both have only a single

simple zero at k = 0 [1, 16.36.2]. Similarly θ4(z,qk ) and θ ′4(z,qk ) are analytic in k ∈ C \ {w ∈ R : w2 ≥ 1}

and θ4(z,qk ) has no zeroes [1, 16.36.2]. It follows that

θ ′1(z,qk )

θ1 (z,qk )+θ ′

4(z,qk )

θ4 (z,qk )is analytic in k ∈ C \ {w ∈ R : w2 ≥ 1} for z ∈ R \ πZ. Since the same is true for K(k) and K(k) , 0

(see (62)), we deduce from the expression (43) that k 7→ F (k/4,b) with b ∈ (0, 1) ∪ (1, 2) is analytic in

k ∈ C \ {z ∈ R : z2 ≥ 1}. Hence, we should focus on the behaviour of F (k/4,b) as k2 → 1, which

corresponds to qk → 1, k ′→ 0 and qk ′ → 0.

For b �xed we may use Jacobi’s identity (see e.g. [2, §22 (6)])

θ1

(πb

4

,√qk

)=

i√

2Tke− πb2

32Tk θ1

(−i πb

8Tk,q2

k ′

).

Taking logarithmic derivatives in b on both sides, we �nd

π

4K(k)

θ ′1

(πb4,√qk

)θ1

(πb4,√qk

) = − 1

K(k)©­­«iπ

8Tk

θ ′1

(−i πb

8Tk,q2

k ′

)θ1

(−i πb

8Tk,q2

k ′

) + πb

16Tk

ª®®¬= − π

2K(k ′)©­­«iθ ′

1

(−2iπbTk ′,q

2

k ′

)θ1

(−2iπbTk ′,q

2

k ′

) + b

2

ª®®¬ ,where we used that Tk = K(k ′)/(4K(k)) = 1/(16Tk ′), see (61).

From the de�nition (64) with sin((2n + 1)(−2iπbTk ′)) = i2(q(n+1/2)b

k ′ − q−(n+1/2)bk ′ ) we �nd the series

representations

i θ1

(−2iπbTk ′,q

2

k ′)=

∞∑n=0

(−1)n(q(n+1/2)(2n−b+1)k ′ − q(n+1/2)(2n+b+1)

k ′

)= q

1

2(1−b)

k ′ − q1

2(1+b)

k ′ +O(q

3

2(3−b)

k ′

),

θ ′1

(−2iπbTk ′,q

2

k ′)=

∞∑n=0

(−1)n(2n + 1)(q(n+1/2)(2n−b+1)k ′ + q(n+1/2)(2n+b+1)

k ′

)= q

1

2(1−b)

k ′ + q1

2(1+b)

k ′ +O(q

3

2(3−b)

k ′

),

where the asymptotics as qk ′ → 0 are valid because 0 < b < 2. Together with K(k ′) = π2+2πqk ′ +O(q2

k ′)(see (69)), we obtain

π

4K(k)

θ ′1

(πb4,√qk

)θ1

(πb4,√qk

) = (1 − b

2

)(1 − 4qk ′) + 2qbk ′ +O

(q2bk ′ + q

2

k ′

).

30

Hence, for b ∈ (0, 1) ∪ (1, 2) as k → ±1 we �nd from (35) that2

F (t ,b) =1 + (b − 2) tan

(πb4

)cos

(πb2

) − 4

tan

(πb4

)cos

(πb2

) (qbk ′ + (b − 2)qk ′

)+O

(q2bk ′ + q

2

k ′

).

Since qk ′ = (1 − k2)/16 +O((1 − k2)2), the dominant contribution at the singularity is proportional to

(1 − k2)b and then a transfer theorem [27, Theorem 3A] tells us that

[t2`]F (t ,b) ∼ −41−2b

tan

(πb4

)cos

(πb2

)Γ(−b)

42`

`b+1

= sin2

(πb

4

)Γ(1 + b)

π

42(`+1−b)

`b+1

, (45)

where we used the re�ection formula Γ(1 − z)Γ(z) = π/sin(πz). This proves the claimed asymptotics for

b ∈ (0, 1) ∪ (1, 2).Next we look at F (t ,b) for b = 0, 1, 2. Using (61) and Legendre’s relation (60) each can be expressed

in terms of K ′(k), E ′(k) and qk ′ , which in turn are analytic in 1 − k2,

F (t , 0) = 1 − π

2K(k) = 1 +π 2

2K ′(k) logqk ′= 1 +

π

log1−k2

16

+O(1 − k2

),

F (t , 1) = 1 − 2

E(k)π= 1 +

2

π 2(K ′(k) − E ′(k)) logqk ′ −

1

K ′(k)

= 1 − 2

π+

1

2π(1 − k2) log

1 − k2

16

+O(1 − k2

),

F (t , 2) = −1 +2

K ′(k) +4

π 2

(E ′(k) − 1 + k2

2

K ′(k))

logqk ′

=4

π− 1 − 1

π(1 − k2) − 1

8π(1 − k2)2 log

1 − k2

16

+O((1 − k2)2

).

In each case the term containing log1−k2

16produces the main contribution to the asymptotics [t2`]F (t ,b).

Using transfer theorems [27, Theorem 3A & remarks at the end of Section 3] we establish that

[t2`]F (t , 0) ∼ π

` log2 `

42`, [t2`]F (t , 1) ∼ 1

2π`24

2`, [t2`]F (t , 2) ∼ 1

4π`34

2` .

Finally, observe that the formula for b = 1 agrees with (45), while this is not the case for b = 0 and

b = 2. �

3.3 Excursions restricted to an angular interval

We turn to the problem of enumerating excursions w with winding angle sequence (θwi )|w |i=0

restricted

to lie fully within an interval I ⊂ R. For convenience we let E′ = {w ∈ E : w1 = (1, 1)} be the set of

excursions that leave the origin in a �xed direction. Then we let F (α, I )(t) be the generating function

F (α, I )(t) =∑w ∈E′

t |w |1{θw=α and θwi ∈I for 0≤i≤ |w | } .

Notice that with this de�nition F (α,R)(t) = F (α )(t)/4.

2The constant term is exactly the characteristic function of the probability distribution in Corollary 21.

31

Theorem 23. For I = (β−, β+), β− ∈ π4Z<0 and β+ ∈ π

4Z>0, and α ∈ I ∩ π

2Z the generating function

F (α, I )(t) is given by the �nite sum

F (α, I )(t) = π

8δ

∑σ ∈(0,δ )∩ π

2Z

(cos

(4σα

δ

)− cos

(4σ (2β+ − α)

δ

))F

(t ,

4σ

δ

), δ B 2(β+ − β−). (46)

It is transcendental in t if one of the following conditions is satis�ed:

• β+, β− ∈ π2Z + π

4,

• β+, β− ∈ πZ + π2and α ∈ πZ.

Otherwise it is algebraic in t . Moreover, its coe�cients satisfy the asymptotic estimate

[t2`]F (α, I )(t) ∼(cos

(2πα

δ

)− cos

(2π (2β+ − α)

δ

))sin

2

(π 2

2δ

)Γ

(1 + 2π

δ

)4δ

42(`+1−2π /δ )

`1+2π /δ .

Proof. By a re�ection principle that is completely analogous to that used in Lemma 15 we observe that

F (α, I )(t) is given by the sum

F (α, I )(t) = 1

4

∞∑n=−∞

(F (α+nδ )(t) − F (2β+−α+nδ )(t)

), δ B 2(β+ − β−),

where the factor 1/4 is due to the fourfold di�erence between E and E′.

Then the discrete Fourier transform

π

2δ

∑σ ∈(0,δ )∩ π

2Z

(e−4iσα/δ − e−4iσ (2β+−α )/δ

)e4iσα ′/δ = 1{α−α ′∈δZ} − 1{2β+−α−α ′∈δZ}

with α ′ ∈ π2Z leads to

F (α, I )(t) = π

8δ

∑σ ∈(0,δ )∩ π

2Z

(e−4iσα/δ − e−4iσ (2β+−α )/δ

) ∑α ′∈ π

2Z

F (α′)(t)e4iσα ′/δ .

Using that the rightmost sum, by de�nition (33), equals F (t , 4σ/δ ) = F (t , 4(δ − σ )/δ ), the latter is seen

to agree exactly with (46).

Since F (α, I )(t) is expressed in terms of trigonometric functions at rational angles (meaning rational

multiples of π ) and F (t ,b) at rational values of b, it follows from Corollary 20 that F (α, I )(t) is algebraic

unless the sum includes one or more F (t ,b) at integer values of b. Note that F (t , 0) and F (t , 4) cannot

occur, while F (t , 3) = F (t , 1) and both occur with the same prefactor, because the summand in (46) is

invariant under the replacement σ → δ − σ . By Lemma 19 and (35), the series F (t , 1) and F (t , 2) are

algebraically independent, so F (α, I )(t) is transcendental if and only if at least one of F (t , 1) and F (t , 2) is

present with a nonzero prefactor. We check this case by case:

• β+, β− ∈ π2Z + π

4: F (α, I )(t) is transcendental because it contains a term with σ = δ/2 equal to

(cos(2α) − cos(4β+ − 2α))F (t , 2) = 2 cos(2α)F (t , 2) = ±2F (t , 2);

• β+ − β− ∈ π2Z + π

4: F (α, I )(t) is algebraic because it contains no term with σ = δ/2 or σ = δ/4.

If β+, β− ∈ π2Z then the term with σ = δ/2 equals (cos(2α) − cos(4β+ − 2α))F (t , 2) = 0, so we should

focus on σ = δ/4. If it exists, its prefactor equals cos(α) − cos(2β+ − α) = cos(α)(1 − cos(2β+)).

• β+, β− ∈ π2Z, β+ − β− ∈ πZ + π

2: F (α, I )(t) is algebraic, because it contains no term with σ = δ/4;

32

• β+, β− ∈ πZ: F (α, I )(t) is algebraic, because 1 − cos(2β+) = 0;

• β+, β− ∈ πZ + π2

, α ∈ πZ + π2

: F (α, I )(t) is algebraic, because cos(α) = 0;

• β+, β− ∈ πZ + π2

, α ∈ πZ: F (α, I )(t) is transcendental, because cos(α)(1 − cos(2β+)) = ±2.

Since these six cases exhaust all choices of β−, β+, and α , we have proven the criterion in the Theorem

statement.

According to Lemma 22 the asymptotics of F (α, I )(t) is determined by the terms σ = π/2 and

σ = δ − π/2. If δ > π then π/2 , δ − π/2 and the corresponding terms are equal and involve F (t ,b) at

b = 2π/δ ∈ (0, 2). If δ = π , then there is a single term σ = π/2 that involves F (t , 2), but its asymptotic

growth is twice larger than what one gets for F (t ,b) with b ∈ (0, 2) as b → 2 (see Lemma 22). Hence, in

either case Lemma 22 implies that as ` →∞,

[t2`]F (α, I )(t) ∼ π

4δ

(cos

(2πα

δ

)− cos

(2π (2β+ − α)

δ

))sin

2

(π 2

2δ

)Γ

(1 + 2π

δ

)π

42(`+1−2π /δ )

`1+2π /δ ,

in accordance with the claimed result. �

As a special case we look at excursions that stay in the angular interval (−π/4,π/2), see Figure 4.

Corollary 24 (Gessel’s lattice path conjecture). The generating function of excursions that stay in theangular interval (−π/4,π/2) with winding angle α = 0 is

F (0,(−π /4,π /2))(t) = 1

4

F

(t ,

4

3

)=

1

2

[ √3π

2K(4t)θ ′

1

( π3,√qk

)θ1

( π3,√qk

) − 1

]=

∞∑n=0

t2n+216

n (5/6)n(1/2)n(2)n(5/3)n

. (47)

Proof. The �rst two equalities follow from Theorem 23 and Proposition 18 respectively (using that

F (t , 4 − b) = F (t ,b)). It remains to show that our generating function reproduces the known formula

[30, 10]

∞∑n=0

t2n+216

n (5/6)n(1/2)n(2)n(5/3)n

=1

2

[2F1

(−1

2

,−1

6

;

2

3

; (4t)2)− 1

],

which we will do by showing they both solve the same algebraic equation (and checking that the �rst

few terms in the expansion agree).

Denoting y = 1

2F (t , 4/3) + 1 and using (44) we get (with k = 4t )

y =√

3

(2Z (u,k) + cn (u,k) dn (u,k)

sn (u,k)

), u =

2K(k)3

.

Applying the addition formula (76) to Z (u + u + u,k) twice one �nds

0 = Z (3u,k) = 3Z (u,k) − k2(sn(2u,k) sn

2(u,k) + sn(2u,k) sn(3u,k) sn(u,k))

= 3Z (u,k) − k2sn(2u,k) sn

2(u,k),

where we used that Z (2K(k),k) = sn(2K(k),k) = 0. Using the double argument formula [1, 16.18.1]

sn(2u,k) = 2 sn(u,k) cn(u,k) dn(u,k)1 − k2

sn4(u,k)

as well as the quadratic relations (73), one may express y in terms of x B dn(u,k) as

y = −x (1 − x2)2 + 3k2

(1 − x2)2 − k2

√x2 + k2 − 1

3(1 − x2) . (48)

33

Using the various addition theorems applied to sn(u+u+u,k) = 0 and rewriting in terms of x = dn(u,k)one �nds after a slightly tedious calculation that x solves

k2 =(1 − x)(1 + x)3

1 + 2x.

Eliminating k from (48), y is then seen to be given by

y =3 − (1 − x)2√

3(1 + 2x).

Finally one may check that the last two displays (with k2 = 16t2) parametrize the curve

27y8 −(4608t4 + 4032t2 + 18

)y4 +

(−32768t6 + 67584t4 + 4224t2 − 8

)y2

− 65536t8 − 114688t6 − 50688t4 − 448t2 − 1 = 0,

which is equivalent to the polynomial in [10, Corollary 2] (after substituting t → t2, T → (y −

1)/(2t2)). �

4 Winding angle distribution of an unconstrained random walk

Let (Wi )i be the simple random walk on Z2started at the origin. In this section we use the results of

Theorem 1 to determine statistics of the winding angle θzj of (Wi )i up to time j around a lattice point

z ∈ Z2or a dual lattice point z ∈ (Z+ 1/2)2. In the case that (Wi )i hits the lattice point z at time i , we set

θzi = θzi−1

and θzj = ∞ for j > i . From a physicist’s point of view, we may regard the z ∈ Z2as an obstacle

with absorbing boundary conditions and z ∈ (Z + 1/2)2 as one with re�ecting boundary conditions.

As alluded to in the introduction, the asymptotics of θzj in the limit j →∞ has been well-studied in

the literature. If z ∈ (Z + 1/2)2 then one has the convergence in distribution to the hyperbolic secantdistribution [36, 21, 22, 38]

P

[2θzj

log j∈ (a,b)

]j→∞−−−−→

∫ b

a

1

2

sech

(πx2

)dx .

Otherwise if z ∈ Z2, then conditionally on θzj < ∞ one has the convergence [36] (see also [26])

P

[2θzj

log j∈ (a,b)

�����θzj < ∞]

j→∞−−−−→

∫ b

a

π

4

sech2

(πx2

)dx .

As we will see now, these asymptotic laws have discrete counterparts if the combinatorics is setup

appropriately. First of all we restrict to winding angles around points that are close to the origin. To be

precise, we consider the lattice point z• B (0, 0) and the dual lattice point z� B (−1/2,−1/2) (or any of

the other dual lattice points at distance 1/√

2 from the origin) and we denote the corresponding winding

angles by θ•j and θ�j respectively. Moreover, instead of considering the winding angle at integer time

j, we look at angles θ•j+1/2 and θ�j+1/2 at half-integer time j + 1/2 (see Figure 5), which prevents them

from taking certain unwanted exact multiples of π/4. Finally, we replace the �xed time by a geometric

random variable.

Theorem 25 (Hyperbolic secant laws). If ζk is a geometric random variable with parameter k ∈ (0, 1)independent of the random walk, i.e. P[ζk = j] = k j (1 − k) for j ≥ 0, then

P[θ�ζk+1/2 ∈ (α −

π2,α + π

2)]= ck sech (4Tkα) + k−1

k 1{α=0} for α ∈ π2Z,

P[θ•ζk+1/2 ∈ (α −

π2,α + π

2)]= Ck sech

(4Tk (α − π

4))

sech

(4Tk (α + π

4))

for α ∈ π2Z + π

4,

where ck = π/(2kK(k)), Ck = π sinh (2Tkπ ) /(2K(k)) and as usual Tk = K(k ′)/(4K(k)).

34

A consequence of these hyperbolic secant laws is the following probabilistic interpretation of the

Jacobi elliptic functions cn and dn.

Corollary 26 (Jacobi elliptic functions as characteristic functions). Let θ�j+1/2 be the winding anglearound (−1/2,−1/2) as in Theorem 25 with the convention θ�−1/2 = 0. If we denote by {·}A : R → A

rounding to the closest element of A ⊂ R (con�icts do not arise here) then we have the characteristicfunctions

E exp

[ib{θ�ζk−1/2}πZ

]= dn(K(k)b,k), (49)

E exp

[ib{θ�ζk+1/2}πZ+ π2

]= cn(K(k)b,k), (50)

where cn(·,k) and dn(·,k) are Jacobi elliptic functions with modulus k .

Even though Theorem 25 and Corollary 26 are stated for simple walks on the square lattice, the

proofs that follow deal exclusively with simple diagonal walks. In order to approach the problem we

require a new building block, in the sense of Section 2, that involves walks that start on an axis but end

at a general point. In particular we will consider the set Cn of non-empty diagonal walks w starting at

(n, 0) and staying strictly inside the positive quadrant, i.e. wi ∈ Z2

>0for 1 ≤ i ≤ |w |, with generating

function Cn(t).

Lemma 27. For anym > 0, Cn(t) satis�es∞∑n=1

(1

2

+Cn(t))〈en , fm〉D = (−1)m π

8(1 − k)K(k)m(q−mk − qmk )

(qm/4k + q−m/4k )2.

Proof. Let us consider the set of all (possibly empty) diagonal walks starting at (p, 0) for p > 0, which has

generating function 1/(1−k). Among these the walks that visit the vertical axis for the �rst time at (±`, 0)for ` > 0 have generating function

2

1−k J`,p (see Figure 14a). On the other hand, the walks that �rst visit

the origin have generating function that can be expressed as the alternating sum2

1−k∑∞

`=1(−1)`+1 J2`,p

using an argument analogous to the one in the proof of Lemma 17. Hence, the remaining walks, those

starting at (p, 0) and avoiding the vertical axis, have generating function

1

1 − k −2

1 − k

∞∑=1

J`,p (t) −2

1 − k

∞∑=1

(−1)`+1 J2`,p =1

1 − k

∞∑=1

1

`

⟨e`,

[I − 2

(1 − cos

π`

2

)Jk

]ep

⟩D

.

By further decomposing such walks at their last intersection with the horizontal axis (see Figure 14b),

we �nd that the latter generating function equals

∞∑n=1

(1 + 2Cn(k))⟨en ,Bkep

⟩D,

where the addition of 1 in parentheses takes into account the walks ending on the horizontal axis. The

equality of the last two displays together with Propositions 14 and 9 implies that

∞∑n=1

(1

2

+Cn(k))〈en , fm〉D =

π

2K(k)m(1 + qmk )

1 − qmk

∞∑n=1

(1

2

+Cn(k))〈en ,Bk fm〉D

=π

4(1 − k)K(k)m(1 + qmk )

1 − qmk

∞∑=1

1

`

⟨e`,

[I − 2

(1 − cos

π`

2

)Jk

]fm

⟩D

=π

4(1 − k)K(k)m(1 + qmk )

1 − qmk

∞∑=1

[1 − 2

1 − cosπ `2

qm/2k + q−m/2k

]1

`〈e`, fm〉D .

35

(a) (b)

Figure 14: (a) An unconstrained walk starting at (1, 0) that visits the vertical axis decomposes

into a walk in J3,1 (in black) and another unconstrained walk (in blue). (b) A walk starting at

(1, 0) and avoiding the vertical axis decomposes into a walk in B5,1 (in green) and a walk in C5

(in black).

Since fm has a radius of convergence larger than one (Lemma 8) and zk1( 1

4+ iTk ) = 1 and zk1

(±iTk ) = ±i(Lemma 5(ii)), we have the special values fm(±i) = дm(±iTk ) and fm(1) = дm( 1

4+iTk ), with дm as de�ned

in Lemma 7. Therefore we may evaluate

∞∑=1

1

`〈e`, fm〉D =

∞∑=1

[z`]fm(z) = fm(1) = (−1)m cosh(2πmTk ) − cos

πm

2

,

∞∑=1

cosπ `2

`〈e`, fm〉D =

∞∑=1

i` + (−i)`2

[z`]fm(z) =fm(i) + fm(−i)

2

= cos

πm

2

(cosh(2πmTk ) − 1) .

Combining the last two displays yields

∞∑=1

[1 − 2

1 − cosπ `2

qm/2k + q−m/2k

]1

`〈e`, fm〉D = (−1)m(cosh(2πmTk ) − 1) = (−1)m

2

(qm/4k − q−m/4k )2,

which leads to the stated result. �

Proof of Theorem 25. For α ∈ π2Z+ π

4and p ≥ 1 we introduce the set G

(α )p of non-empty simple diagonal

walks w starting at (p, 0) and ending at an arbitrary location that have no intermediate visits to the

origin, i.e. wi , (0, 0) for i ≤ |w | − 1, and that have winding angle θw|w |−1/2 ∈ (α −π4,α + π

4). See Figure

15a for an example with p = 1 and α = −5π/4.

It is not hard to see that such a walk can be uniquely encoded in a triple (w ′,w ′′, s), where w ′ ∈W(α−π /4)n,p ∪W(α+π /4)n,p for some n ≥ 1, w ′′ is empty or w ′′ ∈ Cn and s ∈ {(1, 1), (1,−1), (−1,−1), (−1, 1)}

a single step (see Figure 15b). Any such triple gives rise to a walk in G(α )p unless w ′′ is empty, in which

case s is only allowed to take two of the four possible steps. Consequently the generating function

G(α )p (t) of G(α )p satis�es the relation

G(α )p (t) = 4t∞∑n=1

(1

2

+Cn(t)) (W (α−π /4)n,p (t) +W (α+π /4)n,p (t)

).

36

(a) (b)

Figure 15: (a) An unconstrained walk w with θw|w |−1/2 ∈ (−3π/2,−π ). (b) Its decomposition into

w ′ ∈ W(−3π /2)3,1 (blue), w ′′ ∈ C3 (black) and s = (1,−1) (green).

Figure 16: The probability that the winding angle θ�j+ 1

2

around (− 1

2,− 1

2) of the simple random

walk (Wi )i up to time j + 1

2takes values in (α − π

4,α + π

4) is related to the counting of walks

w ∈ G(α )1

.

With the help of Theorem 1(i) and Lemma 27 this evaluates to

G(α )p (t) = k∞∑n=1

(1

2

+Cn(t)) ⟨

en , (Y(α−π /4)k + Y(α+π /4)k )ep⟩D

= k∞∑

m,n=1

(1

2

+Cn(t))〈en , fm〉D

⟨ep , (Y(α−π /4)k + Y(α+π /4)k )fm

⟩D

‖ fm ‖2D

= k∞∑

m,n=1

(1

2

+Cn(t))〈en , fm〉D

8K(k)π

qm |α |/πk (qm/4k + q−m/4k )m2(q−mk − qmk )

⟨ep , fm

⟩D

=k

1 − k

∞∑m=1

(−1)mm

qm |α |/πk

qm/4k + q−m/4k

⟨ep , fm

⟩D. (51)

We turn to the distribution of the winding angle θ�ζk+1/2 around (−1/2,−1/2). By a suitable rotation

and translation of the walk (see Figure 16) we may relate the generating function G(α )p (t) at p = 1 with

the probability that θ�ζk+1/2 takes values in (α − π4,α + π

4) with α ∈ π

2Z + π

4,

P[θ�ζk+1/2 ∈ (α −

π4,α + π

4)]=

1 − kk

G(α )1(k/4).

Note that 〈e1, fm〉D = 0 form even, while

〈e1, fm〉D = f ′m(0) = (−1)(m+1)/22πmv ′k1

(0) = (−1)(m+1)/2πm

2

√k1K(k1)

=(−1)(m+1)/2πm

kK(k)

37

for m odd, where we used that 2

√k1K(k1) = 2/(

√k1 + 1/

√k1)K(k) = kK(k) according to (78) and (77).

With this we can evaluate (51) at p = 1 to

G(α )1(t) = 1

1 − kπ

K(k)

∞∑=0

(−1)`q(2`+1) |α |/πk

q(2`+1)/4k + q−(2`+1)/4

k

.

In particular, for α ∈ π2Z and α , 0 we have

P[θ�ζk+1/2 ∈ (α −

π2,α + π

2)]=

1 − kk

(G(α− π

4)

1(k/4) +G(α+

π4)

1(k/4)

)=

π

kK(k)

∞∑=0

(−1)`q(2`+1) |α |/πk =

π

kK(k)1

qα/πk + q−α/πk

=π

2kK(k) sech (4Tkα) .

Note that the second equality does not hold for α = 0, but using that these probabilities with α ∈ πZmust sum to one, we deduce with the help of

∑α ∈πZ sech(4Tkα) = 2K(k)/π [17, Equation (19)] that

P[θ�ζk+1/2 ∈ (−

π2, π

2)]= 1 −

∑α ∈πZα,0

P[θ�ζk+1/2 ∈ (α −

π2,α + π

2)]= 1 − 1

k+

π

2kK(k) .

In the case of the winding angle around the origin and α ∈ π2Z + π

4, we may identify

P[θ•ζk+1/2 ∈ (α −

π2,α + π

2)]=

1 − kk

(G(α−

π4)(k/4) + G(α+ π4 )(k/4)

), (52)

where G(α′)(t), α ′ ∈ π

2Z, is the generating function for the set

ˆG(α′)

of non-empty simple diagonal

walks w starting at (0, 0) and ending at arbitrary location with no intermediate visits to the origin, i.e.

wi , (0, 0) for 1 ≤ i ≤ |w | −1, and θw|w |−1/2 ∈ (α′− π

4,α ′+ π

4). We claim that (52) for α ∈ (π

2Z+ π

4)\ {±π

4}

can be computed via the absolutely convergent sum

G(α−π4)(t) + G(α+ π4 )(t) = 4

∞∑p=1

(−1)p−1G(α )2p (t). (53)

Note that this sum fails to be absolutely convergence for α = ±π/4, sinceG(±π /4)2p (t) = 2t +O(t2) for any

p ≥ 1. By an argument similar to that used in Lemma 17, the alternating sum counts walks w ′ ∈ G(α )2

starting at (2, 0) and having w ′1= (1,±1). By moving the starting point of w ′ to the origin we obtain a

walk w with winding angle θw|w |−1/2 = θw ′|w ′ |−1/2 ∓ π/4 if w1 = (1,±1). Since θw

′

|w ′ |−1/2 ∈ (α −π4,α + π

4),

these walks correspond precisely to {w ∈ ˆG(α−π4)

: w1 = (1, 1)} and {w ∈ ˆG(α+π4)

: w1 = (1,−1)}, and

include therefore precisely 1/4 of the desired walksˆG(α−

π4) ∪ ˆG(α+

π4). This veri�es the expression (53).

With the help of (51) and (37) (and absolute convergence) this evaluates to

G(α−π4)(t) + G(α+ π4 )(t) = 4k

1 − k

∞∑p,m=1

(−1)m+p−1

m

qm |α |/πk

qm/4k + q−m/4k

⟨e2p , fm

⟩D

=k

1 − kπ

K(k)

∞∑n=1

(−1)n−1(q−n/2k − qn/2k )q2n |α |/πk

=k

1 − kπ

K(k)

(1

1 + q1

2+ 2α

π− 1

1 + q−1

2+ 2α

π

), (54)

where we dropped the absolute value | · | around α because the expression is invariant under α → −α .

Combining with (52) this leads to the claimed probability for α , ±π4

.

38

Finally we show that (54) is valid also for α = ±π4

. To this end it is su�cient to verify that the

resulting expression for

∑α ′∈ π

2Z G(α ′)(t) corresponds to the generating function of walks from the

origin with no intermediate returns. The latter is related to the excursion generating function F (t ,b) of

Proposition 18 by ∑α ′∈ π

2Z

G(α′)(t) = k

1 − k (1 − F (t , 0)) =k

1 − kπ

2K(k) .

The sum of (54) over α ∈ π2Z + π

4, thus double counting each G(α

′)(t) for α ′ ∈ π2Z, indeed gives twice

this expression. �

Proof of Corollary 26. First we note that for α ∈ πZ,

P[θ�ζk−1/2 ∈ (α −

π2,α + π

2)]= (1 − k)1{α=0} + k P

[θ�ζk+1/2 ∈ (α −

π2,α + π

2)]

=π

2K(k) sech

(αK(k ′)K(k)

).

The desired characteristic functions then follow with the help of [1, 16.23.2 & 16.23.3],

E exp

(ib{θ�ζk−1/2}πZ

)=

π

K(k)

∞∑n=−∞

eibnπ

qnk + q−nk= dn(K(k)b,k),

E exp

(ib{θ�ζk+1/2}πZ+ π2

)=

π

kK(k)

∞∑n=−∞

eib(n+1/2)π

qn+1/2k + q−n−1/2

k

= cn(K(k)b,k).

�

5 Winding angle of loops

Another, rather interesting application of Theorem 1 is the counting of loops on Z2. To be precise,

for integer n , 0, let the set Ln of rooted loops of index n be the set of simple diagonal walks w

on Z2 \ {(0, 0)} that start and end at the same (arbitrary) point and have winding angle θw = 2πn.

The set Ln = Leven

n ∪ Lodd

n naturally partitions into the even loops Leven

n supported on {(x ,y) ∈ Z2:

x + y even, (x ,y) , (0, 0)} and the odd loops Lodd

n on {(x ,y) ∈ Z2: x + y odd}.

Theorem 28. The (“inverse-size biased”) generating functions for Levenn and Lodd

n are given by

Levenn (t) B∑

w ∈Levenn

t |w |

|w | =1

|n | trD PevenJ(2π |n |,−∞)k =1

|n |q4 |n |k

1 − q4 |n |k

,

Loddn (t) B∑

w ∈Loddn

t |w |

|w | =1

|n | trD PoddJ(2π |n |,−∞)k =1

|n |q2 |n |k

1 − q4 |n |k

,

where Peven (respectively Podd) is the projection operator onto the even (respectively odd) functions in D.

Proof. Without loss of generality we will take n > 0, since the case of negative n then follows from

symmetry. Let us consider the subset

ˆLn B {w ∈ Ln : w0 ∈ {(x , 0) : x > 0} and θwi < θw

for 0 ≤ i < |w |}

of rooted loops of index n > 0 that start on the positive x-axis and that attain the winding angle

2nπ only at the very end. ClearlyˆLn =

⋃p≥1

W(2πn,(−∞,2πn))p,p in the notation of Theorem 1. Similarly

39

ˆLeven/odd

n B ˆLn ∩ Leven/odd =⋃

p even/oddW(2πn,(−∞,2πn))p,p . The generating functions of

ˆLeven

n andˆLodd

n

are therefore given by∑w ∈ ˆLeven/odd

n

t |w | =∑

p even/odd

1

p

⟨ep , J

(2πn,−∞)k ep

⟩D= trD Peven/oddJ(2nπ ,−∞)k =

∑m even/odd

q2mnk ,

where we used that (according to Theorem 1(ii)) J(2nπ ,−∞)k has eigenvalues (q2nmk )m≥1 and that the even

and odd subspaces of D are spanned by the even respectively odd elements of the basis (fm)m≥1. Hence,∑w ∈ ˆLeven

n

t |w | =q4nk

1 − q4nk

and

∑w ∈ ˆLodd

n

t |w | =q2nk

1 − q4nk

. (55)

Now suppose we take a general loop w ∈ Ln . We denote by w (j) ∈ Ln , 1 ≤ j ≤ |w |, the cyclic

permutation of w given by the walk w (j) B (w j ,w j+1, . . . ,w |w |,w1, . . . ,w j ). We claim that among these

|w | cyclic permutations are exactly n elements ofˆLn .

To see this, let (i`)m`=1be the sequence of increasing times (in {1, 2, . . . , |w |}) at which w intersects

the positive x-axis. Then w (i` ), 1 ≤ ` ≤ m, are potential candidates for walks inˆLn , since they start on

the positive x-axis. For each such walkw ′ = w (i` ) we may consider the winding angle sequence (θw ′i )|w ′ |i=0

as well as the subsequence (α (`)j )mj=0of (θw ′i )

|w ′ |i=0

containing just those angles in 2πZ. Then (α (`)j )mj=0

describes a walk on 2πZ from 0 to 2πn with steps in {−2π , 0, 2π }, and w (i` ) ∈ ˆLn precisely when this

walk stays strictly below 2πn until the very end. Since the walks (α (`)j )mj=0, 1 ≤ ` ≤ m, correspond

precisely to an equivalence class under cyclic permutation of the increments, a well-known cycle lemma

(or ballot theorem) tells us that the latter condition, hence w (i` ) ∈ ˆLn , is satis�ed for exactly n values of

`.

The claim implies that∑w ∈Ln

t |w |

|w | =1

n

∑w ∈Ln

t |w |

|w |

|w |∑j=1

1{w (j )∈ ˆLn } =1

n

∑w ∈ ˆLn

t |w |

|w |

|w |∑j=1

1{w (j )∈Ln } =1

n

∑w ∈ ˆLn

t |w | .

This identity restricted to the even and odd subspaces together with (55) then gives the desired expres-

sions. �

For the rest of this section we switch to simple rectilinear rooted loops on Z2. For such a loop w

we let the index Iw : R2 → Z be de�ned by setting Iw (z) = 0 when z lies on the trajectory of w and

otherwise 2π Iw (z) is the winding angle of w around the point z. By a suitable a�ne transformation we

may now equally think of Lodd

n , n , 0, as the set of simple rooted loops w on Z2with index Iw (z) = n

with respect to some �xed o�-lattice point, say, z = (1/2, 1/2). Similarly, Leven

n , n , 0, is in 1-to-1

correspondence with such loops that have index n with respect to a �xed lattice point, say, the origin.

The following probabilistic result takes advantage of this point of view.

Corollary 29. For ` ≥ 1, letW = (Wi )2`i=0be a simple random walk on Z2 conditioned to return to the

origin after 2` steps. For n , 0, let Cn be the set of connected components of (IW )−1(n). Then

E

[ ∑c ∈Cn|c |

]=

42`(

2``

)2

2`

n[k2`]

q2nk

1 − q4nk

∼ `

2πn2,

E

[ ∑c ∈Cn(|∂c | − 2)

]=

42`(

2``

)2

4`

n[k2`]

q2nk

1 + q2nk

∼ 2π 3`

log2 `,

where |c | is the area of c ∈ Cn and |∂c | the boundary length of c (see Figure 7). The asymptotic formulashold as ` →∞ and n �xed.

40

Proof. The left and right sides of the identities are all seen to be invariant under n → −n, so we may

restrict to the case n > 0. A simple counting exercise shows that for any c ∈ Cn the area and boundary

length of c can be expressed in terms of the cardinalities |c ∩ Z2 | and |c ∩ (Z + 1/2)2 | as

|c | = |c ∩ (Z + 1/2)2 | and |∂c | = 2|c ∩ (Z + 1/2)2 | − 2|c ∩ Z2 | + 2.

Hence, we have that ∑c ∈Cn|c | =

∑z∈(Z+1/2)2

1{IW (z)=n } =∑w

1{Iw (1/2,1/2)=n },∑c ∈Cn(|c | + 1 − |∂c |/2) =

∑z∈Z2

1{IW (z)=n } =∑w

1{Iw (0,0)=n },

where the last sum on both lines is over all possible translationsw ofW by a vector in Z2. The collection

of translations that have index Iw (1/2, 1/2) = n (respectively Iw (0, 0) = n) indexed by all possible

W precisely determines a partition of the loops of length 2` in Lodd

n (respectively Leven

n ). Since the

probability of any particular walkW is

(2``

)−2

we �nd with the help of Theorem 28 that

E

[ ∑c ∈Cn|c |

]=

1(2``

)2[t2`]

∑w ∈Lodd

n

t |w | =1(

2``

)22` [t2`]Lodd

n (t) =4

2`(2``

)2

2`

n[k2`]

q2nk

1 − q4nk

,

E

[ ∑c ∈Cn(|c | + 1 − |∂c |/2)

]=

1(2``

)2[t2`]

∑w ∈Leven

n

t |w | =1(

2``

)22` [t2`]Leven

n (t) = 42`(

2``

)2

2`

n[k2`]

q4nk

1 − q4nk

.

The �rst line and the di�erence between the two lines agree with the claimed formulas.

Since k 7→ qk is analytic in C \ {k ∈ R : |k | ≥ 1} and |qk | < 1 (see (62) in Appendix A), it follows

that both k 7→ q2nk /(1 − q

4nk ) and k 7→ q2n

k /(1 + q2nk ) are ∆-analytic [27] with singularities at k = ±1.

Since qk = exp

(π 2

logqk′

)= 1 + π 2

log(1−k2) +O(log−2(1 − k2)) as k → ±1, we �nd

q2nk

1 − q4nk

= − 1

4n

log(1 − k2)π 2

+O(1) and

q2nk

1 + q2nk

=1

2

+n

2

π 2

log(1 − k2) +O(log−2(1 − k2)).

Standard transfer theorems (see [27]) then imply

[k2`]q2nk

1 − q4nk

∼ 1

4nπ 2`and [k2`]

q2nk

1 + q2nk

∼ π 2n

2` log2 `

as ` →∞.

Together with 42`

(2``

)−2 ∼ π` these give rise to the stated asymptotics. �

A Elliptic functions

This work depends heavily on elliptic functions and their properties. Given that the required properties

are scattered in the literature and that di�erent sources often use inconsistent notation, we provide here

a summary of the de�nitions and required material. We refer to [1, 2, 9] for background and proofs.

The complete elliptic integral of the �rst kind with elliptic modulus k in the open unit disc D is de�ned

by

K(k) =∫

1

0

dy√(1 − y2)(1 − k2y2)

, (56)

41

while the complete elliptic integral of the second kind is

E(k) =∫

1

0

√1 − k2y2

1 − y2dy. (57)

Both are analytic in k ∈ D with series expansions around the origin given by [1, 17.3.11-12]

K(k) = π

2

∞∑n=0

(2n

n

)2(k

4

)2n

, E(k) = π

2

∞∑n=0

1

1 − 2n

(2n

n

)2(k

4

)2n

. (58)

For k ∈ D, the complementary modulus k ′ ∈ D and the associated elliptic integrals K ′ and E ′ are de�ned

as

k ′ =√

1 − k2, K ′(k) = K(k ′), E ′(k) = E(k ′), (59)

where the primes should not be confused with the derivatives. They satisfy Legendre’s relation [1,

17.3.13]

E(k)K ′(k) + E ′(k)K(k) − K(k)K ′(k) = π2. (60)

The elliptic nome qk ∈ D and the quarter-period ratio Tk ∈ (0,∞) + iR are

qk = e−πK′(k )/K (k ) = e−4πTk , Tk =

K ′(k)4K(k) . (61)

The functions K(k), qk and Tk can be analytically continued to the double-slit plane k ∈ C \ {z ∈ R :

z2 ≥ 1}, where they satisfy [39, Section 2]

|qk | < 1, Re(Tk ) > 0, K(k) , 0. (62)

Note that (61) and (59) imply that q1/√

2= e−π , therefore

|qk | ≤ e−π < 1

20for |k | ≤ 1√

2

. (63)

For q ∈ D and z ∈ C, the Jacobi theta functions are de�ned via the absolutely convergent series

θ1(z,q) = 2

∞∑n=0

(−1)nq(n+1/2)2sin((2n + 1)z), (64)

θ2(z,q) = 2

∞∑n=0

q(n+1/2)2cos((2n + 1)z), (65)

θ3(z,q) = 1 + 2

∞∑n=1

qn2

cos(2nz), (66)

θ4(z,q) = 1 + 2

∞∑n=1

(−1)nqn2

cos(2nz). (67)

By (62) the theta functions k 7→ θi (z,qk ) evaluated at q = qk and z �xed are analytic in the slit plane

k ∈ C \ {z ∈ R : z2 ≥ 1}. The squared elliptic modulus and the corresponding complete elliptic integral

can be expressed in terms of q = qk via [9, Theorem 2.1 & (2.1.13)]

k2 =θ2(0,q)4θ3(0,q)4

= 16q − 128q2 + 704q4 + · · · , (68)

K(k) = π

2

θ3(0,q)2 =π

2

+ 2πq + 2πq2 + 2πq4 + · · · . (69)

42

By reversion of the �rst of these series one obtains the series expansion of qk around k = 0 (with radius

of convergence equal to 1),

qk =

(k

4

)2

+ 8

(k

4

)4

+ 84

(k

4

)6

+ 992

(k

4

)8

+ · · · , (70)

which can also be understood on the level of formal power series as the reversion of the series k2 =

θ2(0,q)4/θ3(0,q)4 ∈ R[[q]].The Jacobi elliptic functions sn(·,k), cn(·,k) and dn(·,k) are de�ned in terms of the Jacobi theta

functions θi (z,q) with argument z = π2K (k )u and nome q = qk via

sn(u,k) = θ3(0,qk )θ2(0,qk )

θ1(z,qk )θ4(z,qk )

, cn(u,k) = θ4(0,qk )θ2(0,qk )

θ2(z,qk )θ4(z,qk )

, dn(u,k) = θ4(0,qk )θ3(0,qk )

θ3(z,qk )θ4(z,qk )

. (71)

Alternatively, they are the unique biperiodic meromorphic functions periodic under u → u + 4K(k) and

u → u + 4iK ′(k) satisfying [2, §24]

u =

∫sn(u,k )

0

dy√(1 − y2)(1 − k2y2)

=

∫1

cn(u,k )

dy√(1 − y2)(k ′2 + k2y2)

=

∫1

dn(u,k )

dy√(1 − y2)(y2 − k ′2)

, (72)

in a neighbourhood of u = 0. They satisfy the quadratic relations [1, 16.9.1]

sn2(u,k) + cn

2(u,k) = k2sn

2(u,k) + dn2(u,k) = 1 (73)

and various argument shift relations [1, 16.8], e.g.

sn(u + iK ′(k),k) = 1

k sn(u,k) , sn(u + 2iK ′(k),k) = sn(u,k). (74)

Series representations of the Jacobi elliptic functions can be obtained from (72) by expanding the

integrand in y, followed by integration and series reversion, [1, 16.22]

sn(u,k) = u − 1

6

(1 + k2)u3 +1

120

(1 + 14k2 + k4)u5 + · · · ,

cn(u,k) = 1 − 1

2

u2 +1

24

(1 + 4k2)u4 − 1

720

(1 + 44k2 + 16k4)u6 + · · · ,

dn(u,k) = 1 − 1

2

k2u2 +1

24

(4k2 + k4)u4 − 1

720

(16k2 + 44k4 + k6)u6 + · · · .

The Jacobi zeta function is given in terms of the Jacobi theta functions with argument z = π2K (k )u by

[1, 16.34.1 & 16.34.4]

Z (u,k) = π

2K(k)θ ′

4(z,k)

θ4(z,k)=

π

2K(k)θ ′

1(z,k)

θ1(z,k)− cn(u,k) dn(u,k)

sn(u,k) , (75)

where θ ′i (z,k) = ∂∂zθi (z,k). It is periodic in u with period 2K(k) [1, 17.4.30] and satis�es the addition

formula [1, 17.4.35]

Z (u +v,k) = Z (u,k) + Z (v,k) − k2sn(u,k) sn(v,k) sn(u +v,k). (76)

We often encounter elliptic functions in which the modulus k ∈ (0, 1) is replaced by its descendingLanden transformation k1 ∈ (0, 1) de�ned as

k1 =1 − k ′1 + k ′

=1 −√

1 − k2

1 +√

1 − k2

, k =2

√k1 + 1/

√k1

. (77)

43

The elliptic integrals, quarter-period ratio and nome transform as [9, Theorem 1.2]

K(k) = (1 + k1)K(k1), K ′(k) = 1

2

(1 + k1)K ′(k1), Tk1= 2Tk , qk1

= q2

k . (78)

The Jacobi elliptic and theta functions also transform relatively nicely under Landen transformations,

e.g. [1, 16.12.2]

sn(u,k) = (1 + k1) sn(u/(1 + k1),k1)1 + k1 sn

2(u/(1 + k1),k1). (79)

See [1, Section 16.12], [2, §38] [9, Section 2.7] for more such identities.

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