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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