Periods of Calabi-Yau Manifolds
in
Physics and Number Theory
TIFR 15 March 2011
Tuesday, 22 March 2011
Periods of Calabi-Yau Manifolds
in
Physics and Number Theory
With X. de la Ossa and F. Rodriguez Villegas
TIFR 15 March 2011
Tuesday, 22 March 2011
Aims
• To explain why the periods of CY manifolds are
important to string theorists.
• These periods are also important to number theorists
because they encode important arithmetic information
about the manifold.
• I want to speculate about the role of `quantum
corrections’ and mirror symmetry for the zeta-function.
Tuesday, 22 March 2011
CY manifolds arise in the “compactification” of string
theory from D=10 down to D=4. A desirable consequence of
string theory is spacetime supersymmetry and the question
arises as to whether this process preserves supersymmetry.
δb
δf
= f�
= b � + ∇�
Very quickly we come to the condition
∇� = 0 .
Which we take as a condition on the manifold
[∇, ∇]� = 0 .
Given � we construct
Jm n = −i�† γm n �
Tuesday, 22 March 2011
Ωmnr = �Tγmnr �
h
pq
=
1
0
0
0
1
0
h11
h
21
0
h
h11
1
0
21
0
0
1
An example is the quintic threefold:
M : P (x, ψ) =
5
�
i=1
x5i − 5ψ x1 x2 x3 x4 x5 .
This has h11 = 1 and h21 = 101 .
Tuesday, 22 March 2011
h
pq
=
1
0
0
0
1
0
h11
h
21
0
h
h11
1
0
21
0
0
1
An example is the quintic threefold:
M : P (x, ψ) =
5
�
i=1
x5i − 5ψ x1 x2 x3 x4 x5 .
This has h11 = 1 and h21 = 101 .
Tuesday, 22 March 2011
The existence of the holomorphic form is a very useful
fact since the form can be constructed explicitly
Ω
=
=
Tuesday, 22 March 2011
1
2πi
�
�abcde
a
b
c
d
e
�abcde x dx dx dx dx
P
xa dxb dxc dxd
∂P
∂xe
The manifold depends on parameters
and these may be thought of as the
coefficients in the defining polynomials.
There is more canonical way to give
coordinates on the space of complex
structures. The holomorphic form is
defined up to scale but is otherwise
3
H
(M, Z).
unique so defines a line in
The coordinates of this line are
�Γ =
�
Ω
Γ
taken over a basis of cycles Γ ∈ H3 (M, Z) and these
are the periods.
Tuesday, 22 March 2011
We may think of the following polynomial as defining a
one-parameter subfamily of the 101 parameter space of
quintics
M : P (x, ψ) =
5
�
5
xi
i=1
− 5ψ x1 x2 x3 x4 x5 .
In this case there is a simple relation between this family
and the one-parameter mirror family
�
W = M/Γ
Γ : (x1 , x2 , x3 , x4 , x5 ) �→ (ζ n1 x1 , ζ n2 x2 , ζ n3 x3 , ζ n4 x4 , ζ n5 x5 )
where ζ 5 = 1 and
�
i
Tuesday, 22 March 2011
ni ≡ 0 mod 5
Periods for the mirror quintic
Consider Ω for the mirror quintic family, which has just
one complex structure parameter so
b3 = 1 + h21 + h21 + 1 = 4
The quantities
Ω, Ω , Ω , Ω , Ω
�
��
���
����
are all 3-forms, and there are five of them, so there is
a linear relation
LΩ = 0
Tuesday, 22 March 2011
�
5ψ
−
(2πi)3
�
5ψ
−
(2πi)4
�
=
5ψ
−
(2πi)5
�
=
1
(2πi)5
�
1
(2πi)5
�
=
=
=
=
Tuesday, 22 March 2011
x1 dx2 dx3 dx4
∂P
∂x5
γ2 ×γ3 ×γ4
x1 dx2 dx3 dx4 dx5
P
γ2 ×γ3 ×γ4 ×γ5
dx1 dx2 dx3 dx4 dx5
P
γ1 ×γ2 ×γ3 ×γ4 ×γ5
Γ
Γ
dx1 dx2 dx3 dx4 dx5
�
�
P
i
5
i (x )
1
2
3
4
5
x x x x x 1 − 5ψ x1 x2 x3 x4 x5
�
∞ �
1 5
5 5 m
�
dx dx dx dx dx
(x ) + . . . + (x )
1 x2 x3 x4 x5
x1 x2 x3 x4 x5
5ψ
x
m=0
1
2
∞
�
(5n)! 1
5 (5ψ)5n
(n!)
n=0
3
4
5
�
5ψ
−
(2πi)3
�
5ψ
−
(2πi)4
�
=
5ψ
−
(2πi)5
�
=
1
(2πi)5
�
1
(2πi)5
�
=
=
=
=
Tuesday, 22 March 2011
x1 dx2 dx3 dx4
∂P
∂x5
γ2 ×γ3 ×γ4
x1 dx2 dx3 dx4 dx5
P
γ2 ×γ3 ×γ4 ×γ5
1
2πi
�
dx1
x1
dx1 dx2 dx3 dx4 dx5
P
γ1 ×γ2 ×γ3 ×γ4 ×γ5
Γ
Γ
dx1 dx2 dx3 dx4 dx5
�
�
P
i
5
i (x )
1
2
3
4
5
x x x x x 1 − 5ψ x1 x2 x3 x4 x5
�
∞ �
1 5
5 5 m
�
dx dx dx dx dx
(x ) + . . . + (x )
1 x2 x3 x4 x5
x1 x2 x3 x4 x5
5ψ
x
m=0
1
2
∞
�
(5n)! 1
5 (5ψ)5n
(n!)
n=0
3
4
5
�
5ψ
−
(2πi)3
�
5ψ
−
(2πi)4
�
=
5ψ
−
(2πi)5
�
=
1
(2πi)5
�
1
(2πi)5
�
=
=
=
=
Tuesday, 22 March 2011
x1 dx2 dx3 dx4
∂P
∂x5
γ2 ×γ3 ×γ4
x1 dx2 dx3 dx4 dx5
P
γ2 ×γ3 ×γ4 ×γ5
1
2πi
�
dx1
x1
dx1 dx2 dx3 dx4 dx5
P
γ1 ×γ2 ×γ3 ×γ4 ×γ5
Γ
Γ
dx1 dx2 dx3 dx4 dx5
�
�
P
i
5
i (x )
1
2
3
4
5
x x x x x 1 − 5ψ x1 x2 x3 x4 x5
�
∞ �
1 5
5 5 m
�
dx dx dx dx dx
(x ) + . . . + (x )
1 x2 x3 x4 x5
x1 x2 x3 x4 x5
5ψ
x
m=0
1
2
∞
�
(5n)! 1
5 (5ψ)5n
(n!)
n=0
3
4
5
5n
We expect the period to satisfy a fourth order differential
equation, and indeed it does
L� = 0 ;
∞
�
(5n)! n
�(λ) =
λ
,
5
(n!)
n=0
1
λ=
(5ψ)5
where
L = ϑ − 5λ
4
4
�
d
ϑ=λ
.
dλ
(5ϑ + i) ;
i=1
The point λ = 0 is a regular singular point with all four
indices equal to zero. Thus the solutions near the origin
are asymptotic to
2
3
1, log λ, log λ, log λ
Tuesday, 22 March 2011
More precisely the solutions are of the form
�0 (λ) = f0 (λ)
�1 (λ) = f0 (λ) log λ + f1 (λ)
2
�2 (λ) = f0 (λ) log λ + 2f1 (λ) log λ + f2 (λ)
3
2
�3 (λ) = f0 (λ) log λ + 3f1 (λ) log λ + 3f2 (λ) log λ + f3 (λ)
where the fj (λ) are power series. These series will enter
into the Yukawa couplings and into the calculation of the
numbers of Fp -rational points of M . Recall that these
solutions may be found by the method of Fröbenius. That
is by seeking solutions
�(λ, ε) =
∞
�
m=0
Tuesday, 22 March 2011
am (ε) λ
m+ε
to the eqn. L �(λ, ε) = ε4 λε .
Yukawa couplings and
integral series
This is the expression for the Yukawa coupling
yttt = 5
�
2πi
5
�3
2
ψ
�0 (ψ)2 (1 − ψ 5 )
�
dψ
dt
�3
= 5+
where, in these expressions
1 �1 (λ)
t=
2πi �0 (λ)
Tuesday, 22 March 2011
and q = exp(2πit)
∞
�
nk k 3 q k
k=0
1 − qk
,
1 �1 (λ)
t=
2πi �0 (λ)
q = exp(2πit)
These relations give the mirror map. Note that integers
arise here also
λ
= q + 154 q + 179139 q + 313195944 q
+657313805125 q 5 + 1531113959577750 q 6
7
+3815672803541261385 q
+9970002717955633142112 q 8 + . . . .
Tuesday, 22 March 2011
2
3
4
Arithmetic of the Quintic
Now ask what is, for a physicist, a very strange question:
For the quintic
5
�
i=1
x5i − 5ψ x1 x2 x3 x4 x5 = 0
How many solutions to this equation are there with integer
k
x and how does this number vary with ψ ?
k
x
Since the
are coordinates in a projective space we are
free to multiply by a common scale, so there is no
difference between a solution in integers and a solution in
rational numbers. This formulation is better because Q is a
field but is still hard.
Easier: How many solutions are there over a finite field?
Tuesday, 22 March 2011
Field Theory for Physicists
A field F is a set on which + and × are defined and have
the usual associative and distributive properties.
F is an abelian group with respect to and +
∗
F = F \ {0} is an abelian group with respect to × .
Finite fields are uniquely classified by the number of
N
elements, which is p for some prime p and integer N.
F7
Tuesday, 22 March 2011
x
0
1
2
3
4
5
6
x−1
*
1
4
5
2
3
6
An old result, going back to Fermat, is a = a, write this as
p
p−1
a(a
− 1) = 0
it follows that
ap−1 =

1, if a �= 0

0, if a = 0 .
There is another elementary fact that is very useful
�
n
a
=
a∈Fp
�
a∈Fp
(ba)
n
= b
n
�
n
a .
a∈Fp
It follows that, for a �= 0 ,
�
a∈Fp
Tuesday, 22 March 2011
an =

0, if p − 1 does not divide n

−1, if p − 1 divides n .
A Zero’th Order Result
Take now x ∈
νλ
5
Fp
and 5ψ ∈ Fp , p �= 5 and let
� �
�
= # x � P (x, ψ) = 0 , λ =
1
(5ψ)5
This number can be computed mod p with relative ease.
νλ ≡
��
x∈F5p
1 − P (x, ψ)
p−1
[p/5]
νλ
Tuesday, 22 March 2011
�
� (5m)!
[p/5]
m
≡
�0 (λ) =
λ
.
5
(m!)
m=0
p-Adic Numbers
νλ is a definite number so we may seek to compute it
exactly.
νλ =
(0)
νλ
+
(1)
νλ
p+
(2)
νλ
p +
2
(3)
νλ
(j)
νλ
p +
3
(4)
νλ
p
4
with 0 ≤
≤ p − 1 and evaluate mod p , mod p and so
on. This leads naturally to p-adic analysis. Given r ∈ Q we
write
2
m
m0 α
r =
=
p
n
n0
The p-adic norm of r is defined by
�r�p = p−α ,
Tuesday, 22 March 2011
�0�p = 0
3
Counting the Points Exactly
The result is most simply stated for the case 5� | p − 1
νλ
=
p
f0 (Λ) +
�
1
+
3!
p
1−p
�
�
p
1−p
p �
f1 (Λ)
�3
1
+
2!
p ���
f3 (Λ)
�
p
1−p
1
+
4!
�
�2
p ��
f2 (Λ)
p
1−p
�4
p ����
f4 (Λ)
+ O(p5 )
In this expression
Λ = Teich(λ) = lim λ
n→∞
Tuesday, 22 March 2011
pn
p−1
�
(5m)! m
p
and f0 (Λ) =
Λ
5
(m!)
m=0
We can also perform the sum in this expression for the
number of points
νλ =
p−1
�
βm Λ
m
m=0
with coefficients
βm
Tuesday, 22 March 2011
am(1+p+p2 +...+pn+1 )
m
5
= lim
= (−1) G5m G−m
n→∞ am(1+p+p2 +...+pn )
Mirror Symmetry and the
Zeta Function
We work now over Fpr and denote by Nr (ψ) the number
of projective solutions to the equation P (x, ψ) = 0 .
ζ(T, ψ) = exp
�∞
�
� Nr (ψ)T r
r=1
r
Numerator of deg. 2h21 + 2 dep. on the cpx. structure of M
=
.
11
Denominator of deg. 2h + 2
Tuesday, 22 March 2011
Explicitly for the quintic we have
20
ρ
30
ρ
R0 (T, ψ) RA (p T , ψ) RB (p T , ψ)
ζM (T, ψ) =
(1 − T )(1 − pT )(1 − p2 T )(1 − p3 T )
ρ
ρ
ρ
ρ
R0 (T, ψ)
ζW (T, ψ) =
(1 − T )(1 − pT )101 (1 − p2 T )101 (1 − p3 T )
where ρ = 1, 2, 4 is the smallest integer such that 5|(p − 1)
ρ
Tuesday, 22 March 2011
Explicitly for the quintic we have
20
ρ
30
ρ
R0 (T, ψ) RA (p T , ψ) RB (p T , ψ)
ζM (T, ψ) =
(1 − T )(1 − pT )(1 − p2 T )(1 − p3 T )
ρ
ρ
ρ
ρ
R0 (T, ψ)
ζW (T, ψ) =
(1 − T )(1 − pT )101 (1 − p2 T )101 (1 − p3 T )
where ρ = 1, 2, 4 is the smallest integer such that 5|(p − 1)
ρ
Tuesday, 22 March 2011
Explicitly for the quintic we have
20
ρ
30
ρ
R0 (T, ψ) RA (p T , ψ) RB (p T , ψ)
ζM (T, ψ) =
(1 − T )(1 − pT )(1 − p2 T )(1 − p3 T )
ρ
ρ
ρ
ρ
R0 (T, ψ)
ζW (T, ψ) =
(1 − T )(1 − pT )101 (1 − p2 T )101 (1 − p3 T )
where ρ = 1, 2, 4 is the smallest integer such that 5|(p − 1)
ρ
Tuesday, 22 March 2011
The 5-adic Limit
The desired relations are true in the 5-adic limit. For all
p and ψ
R0 (T, ψ) = (1 − T )(1 − p T )(1 − p2 T )(1 − p3 T ) + O(52 )
RA (T, ψ)20 RB (T, ψ)30 = (1 − p T )100 (1 − p2 T )100 + O(52 )
Compare this with the quantum corrections to the
classical Yukawa coupling
yttt
(0)
yttt
= 1+
∞
�
1
nk k 3 q k
5
k=0
1 − qk
� 2�
= 1+O 5
since Lian and Yau have shown that 5 |nk k for each k .
3
Tuesday, 22 March 2011
3