Mirzakhani’s “Simple Geodesics and Weil-Petersson volumes”

Mirzakhani's "Simple Geodesics and Weil-Petersson volumes"

Lewis Bowen October 21, 2020

Lewis Bowen (UT Austin)

Mirzakhani

October 21, 2020 1 / 22

Goal:

Present a recursive formula for the volume of Mg,n(L1, . . . , Ln) the moduli space of genus g hyperbolic surfaces with n geodesic boundary components of lengths Li [0, ) when the Euler characteristic is negative.

Lewis Bowen (UT Austin)

Mirzakhani

October 21, 2020 2 / 22

vol(M1,1) Theorem (McShane's identity)

Let X be a hyperbolic once-punctured torus. Then

(1 + exp( (X )))-1 = 1/2

where the sum is over all simple closed geodesics (s.c.g.) on X .

Proposition

vol(M1,1) = 2/6.

Proof.

M1,1 = T1,1/Mod1,1. T1,1 has nice description in terms of Fenchel-Nielsen coordinates:

T1,1 = {( , ) : 0 < , R}

depending on a s.c.c. on S = S1,1.

If we had a fundamental domain for the action Mod1,1 T1,1 then vol(M1,1) equals the volume of the fund. domain and we're done.

We won't use a fund. domain. Instead we find a finite measure on an intermediate cover

T1,1/ Stab() that projects to the volume form.

Lewis Bowen (UT Austin)

Mirzakhani

October 21, 2020 3 / 22

vol(M1,1)

Proof.

Fix a s.c.c. on S = S1,1. Then

M1,1 := T1,1/ Stab() = {(X , ) : X M1,1 : a s.c.g. on X } = {( , ) : 0 < , R}/( , ) ( , + ).

Let

f (x)

=

1 1+ex

.

So

McShane's

identity

states

f ( (X )) = 1/2.

vol(M1,1) = 2 vol(M1,1) f ( (X ))

= 2 M1,1 f ( (X )) d volM1,1 (X , )

x

x

2

=2

0

f (x) dy dx = 2

0

0

1 + ex

dx =

. 6

Lewis Bowen (UT Austin)

Mirzakhani

October 21, 2020 4 / 22

vol(M1,1)(L) Theorem (Generalized McShane's identity)

Let X be a hyperbolic one holed-torus with geodesic boundary of length L. Then

D(L, (), ()) = L

where the sum is over all simple closed geodesics (s.c.g.) on X , not including the boundary and

ex /2

+

e

y +z 2

D(x, y , z) = 2 log

e-x /2

+

e

y +z 2

.

Proposition

vol(M1,1(L))

=

L2 24

+ 2/6.

Proof.

M1,1(L) = T1,1(L)/Mod1,1. T1,1(L) has nice description in terms of Fenchel-Nielsen coordinates:

T1,1(L) = {( , ) : 0 < , R}

depending on a s.c.c. on S = S1,1.

Lewis Bowen (UT Austin)

Mirzakhani

October 21, 2020 5 / 22

vol(M1,1)(L) Proof.

Fix a s.c.c. on S = S1,1. Then

M1,1(L) := T1,1(L)/ Stab() = {(X , ) : X M1,1(L) : a s.c.g. on X } = {( , ) : 0 < , R}/( , ) ( , + ).

L vol(M1,1(L)) = vol(M1,1(L)) D(L, (), ())

=

D(L,

M1,1 (L)

(),

()) d volM1,1(L)(X , )

x

=

D(L, y , y ) dy dx = xD(L, x, x) dx.

00

0

D(x, y , z) = H(y + z, x)

x

where H(x, y ) =

1

1+exp

x +y 2

+

1

1+exp

x -y 2

.

Lewis Bowen (UT Austin)

Mirzakhani

October 21, 2020 6 / 22

Proof.

Set V (L) = vol(M1,1(L)).

LV (L) =

xD(L, x, x) dx.

0

Take

L

of

both

sides:

(LV (L)) =

L

=

x D(L, x, x) dx = xH(2x, L) dx

0 L

0

x

x

0

1

+

exp

2x +L 2

+

1

+

exp

2x -L 2

dx

Set y1 = x + L/2, y2 = x - L/2 to get

(LV (L)) =

L

L/2

y1 - L/2 1 + ey1

dy1

+

-L/2

y2 + L/2 1 + ey2

dy2

=

0

y1 - L/2 1 + ey1

dy1

-

L/2 0

y1 - L/2 1 + ey1

dy1

+

0

y2 + L/2 1 + ey2

dy2

+

0 -L/2

y2 + L/2 1 + ey2

d

=

2

x 0 1 + ex dx -

L/2 0

y1 - L/2 1 + ey1

dy1

+

0 -L/2

y2 + L/2 1 + ey2

dy2

=

L/2

2/6 -

(y - L/2)

0

1

1

1 + ey + 1 + e-y

dy = 2/6 + L2/8.

Lewis Bowen (UT Austin)

Mirzakhani

October 21, 2020 7 / 22

Proof.

L

(LV (L))

=

2/6

+

L2/8.

LV (L) =

L 0

2/6

+

x 2/8

dx

=

L2/6

+

L3/24.

V (L) = vol(M1,1(L)) = 2/6 + L2/24.

Lewis Bowen (UT Austin)

Mirzakhani

October 21, 2020 8 / 22

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