Hearing the size of a triangle

Zhiqin Lu

Graduate colloquium of the University of California at Irvine

December 5th, 2008

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Fourier expansion

We assume that f (x) is a smooth period function on [0,2π].Then we have the well-known Fourier expansion

f (x) =a0

2+∞∑

k=1

(ak cos kx + bk sin kx)

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Good thing of the Fourier expansion1 Split a function into simplier pieces;2 Geometry interpretation;3 Connecting differential operators to linear

algebra=functional analysis.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Good thing of the Fourier expansion1 Split a function into simplier pieces;2 Geometry interpretation;3 Connecting differential operators to linear

algebra=functional analysis.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Good thing of the Fourier expansion1 Split a function into simplier pieces;2 Geometry interpretation;3 Connecting differential operators to linear

algebra=functional analysis.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Geometric intepretation

Pythagorean theorem or Parseval equality (Bessel inequality)

a20 +

∞∑k=1

(a2k + b2

k ) =1π

∫ 2π

0f (x)2dx

Euclidean geometry on the space spanned by1, cos kx , sin kx.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Geometric intepretation

Pythagorean theorem or Parseval equality (Bessel inequality)

a20 +

∞∑k=1

(a2k + b2

k ) =1π

∫ 2π

0f (x)2dx

Euclidean geometry on the space spanned by1, cos kx , sin kx.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Geometric intepretation

Pythagorean theorem or Parseval equality (Bessel inequality)

a20 +

∞∑k=1

(a2k + b2

k ) =1π

∫ 2π

0f (x)2dx

Euclidean geometry on the space spanned by1, cos kx , sin kx.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

We consider the differential operator

∆ =∂2

∂t2

We have

∆ 1 = 0

∆ cos kx = −k2 cos kx

∆ sin kx = −k2 sin kx

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

We consider the differential operator

∆ =∂2

∂t2

We have

∆ 1 = 0

∆ cos kx = −k2 cos kx

∆ sin kx = −k2 sin kx

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Under the basis 1, cos kx , sin kx, ∆ is an infinite dimensionalmatrix

∆ =

0−1

−1−4

−4. . .

. . .

Consider ∆ as a differential operator AND an linear operator.Warning: Infinite dimensional space is quiet different from finitedimensional space. If we don’t have estimate, strange thingscould happen.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Under the basis 1, cos kx , sin kx, ∆ is an infinite dimensionalmatrix

∆ =

0−1

−1−4

−4. . .

. . .

Consider ∆ as a differential operator AND an linear operator.Warning: Infinite dimensional space is quiet different from finitedimensional space. If we don’t have estimate, strange thingscould happen.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Under the basis 1, cos kx , sin kx, ∆ is an infinite dimensionalmatrix

∆ =

0−1

−1−4

−4. . .

. . .

Consider ∆ as a differential operator AND an linear operator.Warning: Infinite dimensional space is quiet different from finitedimensional space. If we don’t have estimate, strange thingscould happen.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

How to work on high dimensions?

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Let Ω be a bounded domain in Rn with smooth boundary. Whatdo we mean by period functions on Ω?

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

For a general domain, there is no such things as period function.

Boundary conditions!

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

For a general domain, there is no such things as period function.

Boundary conditions!

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Tow kinds of boundary conditions:1 Dirichlet boundary condition

f |∂Ω = 0

2 Neumann boundary condition

∂f∂n|∂Ω = 0

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Tow kinds of boundary conditions:1 Dirichlet boundary condition

f |∂Ω = 0

2 Neumann boundary condition

∂f∂n|∂Ω = 0

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Can we find building blocks of functions on Ω satisfying theDirichlet boundary condition?Answer: Eigenvalues and Eigenfuntions!

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Can we find building blocks of functions on Ω satisfying theDirichlet boundary condition?Answer: Eigenvalues and Eigenfuntions!

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

We will only consider the Dirichlet eigenvalues in this talk.Let the Laplace operator be defined

∆ =∂2

∂x21

+ · · ·+ ∂2

∂x2n

In order to be able to write the operator into an infinite matrix,we have to do some preparations.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

We will only consider the Dirichlet eigenvalues in this talk.Let the Laplace operator be defined

∆ =∂2

∂x21

+ · · ·+ ∂2

∂x2n

In order to be able to write the operator into an infinite matrix,we have to do some preparations.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

We will only consider the Dirichlet eigenvalues in this talk.Let the Laplace operator be defined

∆ =∂2

∂x21

+ · · ·+ ∂2

∂x2n

In order to be able to write the operator into an infinite matrix,we have to do some preparations.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

We need to make the operator “symmetric”, or self-adjoint. Letf ,g be functions vanished on ∂Ω. Then by the second Green’stheorem, we have ∫

Ωf ∆g =

∫Ω

∆f g.

That is, in the abstract notation, we have

(∆f ,g) = (f ,∆g).

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

We need to make the operator “symmetric”, or self-adjoint. Letf ,g be functions vanished on ∂Ω. Then by the second Green’stheorem, we have ∫

Ωf ∆g =

∫Ω

∆f g.

That is, in the abstract notation, we have

(∆f ,g) = (f ,∆g).

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

We need to make the operator “symmetric”, or self-adjoint. Letf ,g be functions vanished on ∂Ω. Then by the second Green’stheorem, we have ∫

Ωf ∆g =

∫Ω

∆f g.

That is, in the abstract notation, we have

(∆f ,g) = (f ,∆g).

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

But that is not enough...The linear space of smooth function is not a complete innerproduct space.This is different from finite dimensional case, where all thespaces are complete.The good space is the L2 space.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

But that is not enough...The linear space of smooth function is not a complete innerproduct space.This is different from finite dimensional case, where all thespaces are complete.The good space is the L2 space.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

But that is not enough...The linear space of smooth function is not a complete innerproduct space.This is different from finite dimensional case, where all thespaces are complete.The good space is the L2 space.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

But that is not enough...The linear space of smooth function is not a complete innerproduct space.This is different from finite dimensional case, where all thespaces are complete.The good space is the L2 space.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

If a function f is not smooth, then ∆f may not make sense.We have to be happy (or familiar with?) a linear operator thatonly defined on a subspace (densely defined).Question: ∆ is an operator defined on C∞(Ω), which is densein L2(Ω). Is it possible to extend the operator to the wholespace?Answer: not possible. Because ∆ is a closed graph operator. Ifpossible to extend, then the operator is in fact bounded. (closedgraph theorem)

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

If a function f is not smooth, then ∆f may not make sense.We have to be happy (or familiar with?) a linear operator thatonly defined on a subspace (densely defined).Question: ∆ is an operator defined on C∞(Ω), which is densein L2(Ω). Is it possible to extend the operator to the wholespace?Answer: not possible. Because ∆ is a closed graph operator. Ifpossible to extend, then the operator is in fact bounded. (closedgraph theorem)

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

If a function f is not smooth, then ∆f may not make sense.We have to be happy (or familiar with?) a linear operator thatonly defined on a subspace (densely defined).Question: ∆ is an operator defined on C∞(Ω), which is densein L2(Ω). Is it possible to extend the operator to the wholespace?Answer: not possible. Because ∆ is a closed graph operator. Ifpossible to extend, then the operator is in fact bounded. (closedgraph theorem)

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

If a function f is not smooth, then ∆f may not make sense.We have to be happy (or familiar with?) a linear operator thatonly defined on a subspace (densely defined).Question: ∆ is an operator defined on C∞(Ω), which is densein L2(Ω). Is it possible to extend the operator to the wholespace?Answer: not possible. Because ∆ is a closed graph operator. Ifpossible to extend, then the operator is in fact bounded. (closedgraph theorem)

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

DefinitionA non-zero function f is called an eigenfunction, if there is areal number λ, such that

∆f = −λf .

Fact: There is a basis of L2(Ω) made from eigenfunctions!This is the higher dimensional analog of Fourier expansion.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

DefinitionA non-zero function f is called an eigenfunction, if there is areal number λ, such that

∆f = −λf .

Fact: There is a basis of L2(Ω) made from eigenfunctions!This is the higher dimensional analog of Fourier expansion.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

DefinitionA non-zero function f is called an eigenfunction, if there is areal number λ, such that

∆f = −λf .

Fact: There is a basis of L2(Ω) made from eigenfunctions!This is the higher dimensional analog of Fourier expansion.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Under the above basis, we get

∆ =

−λ1

−λ2−λ3

. . .

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

The inverse of the operator is

∆−1 =

−λ−1

1−λ−1

2−λ−1

3. . .

Thus estimating the eigenvalues from above or from below arevery interesting, because they tell the size of the Laplacianoperator.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

There are a lot of questions in the field that can be solved usingthe knowledge at the first-year graduate level.For example, there is a very old theorem of Karen Uhlenbeck,stating that a generic domain in R2 is simple. That is, theLaplacian, as a infinite matrix, whose eigenvalues are all ofmultiplicity one.More recently, Hillairet and Judge proved that generic polygonsof n > 3 are simple:

L. Hillairet and C. JudgeGeneric spectral simplitcity of polygonsarXiv/0703616

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

There are a lot of questions in the field that can be solved usingthe knowledge at the first-year graduate level.For example, there is a very old theorem of Karen Uhlenbeck,stating that a generic domain in R2 is simple. That is, theLaplacian, as a infinite matrix, whose eigenvalues are all ofmultiplicity one.More recently, Hillairet and Judge proved that generic polygonsof n > 3 are simple:

L. Hillairet and C. JudgeGeneric spectral simplitcity of polygonsarXiv/0703616

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

There are a lot of questions in the field that can be solved usingthe knowledge at the first-year graduate level.For example, there is a very old theorem of Karen Uhlenbeck,stating that a generic domain in R2 is simple. That is, theLaplacian, as a infinite matrix, whose eigenvalues are all ofmultiplicity one.More recently, Hillairet and Judge proved that generic polygonsof n > 3 are simple:

L. Hillairet and C. JudgeGeneric spectral simplitcity of polygonsarXiv/0703616

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

There are a lot of questions in the field that can be solved usingthe knowledge at the first-year graduate level.For example, there is a very old theorem of Karen Uhlenbeck,stating that a generic domain in R2 is simple. That is, theLaplacian, as a infinite matrix, whose eigenvalues are all ofmultiplicity one.More recently, Hillairet and Judge proved that generic polygonsof n > 3 are simple:

L. Hillairet and C. JudgeGeneric spectral simplitcity of polygonsarXiv/0703616

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

They left the following conjecture: Generic triangle is simple.Using Math 210, we can solve the problem.(joint with Rowlett)

TheoremGeneric triangle is simple!

Proof. The parameter space (In the terminology of geometry,the moduli space is a 2-d orbifold.) of a triangle of fixed area isa 2-dimensional object.The set Fm such that the first m eigenvalues are distinct is adense open set on the parameter space.Thus by the Baire category theorem,⋂

Fm

is dense and open, thus generic.Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

They left the following conjecture: Generic triangle is simple.Using Math 210, we can solve the problem.(joint with Rowlett)

TheoremGeneric triangle is simple!

Proof. The parameter space (In the terminology of geometry,the moduli space is a 2-d orbifold.) of a triangle of fixed area isa 2-dimensional object.The set Fm such that the first m eigenvalues are distinct is adense open set on the parameter space.Thus by the Baire category theorem,⋂

Fm

is dense and open, thus generic.Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

They left the following conjecture: Generic triangle is simple.Using Math 210, we can solve the problem.(joint with Rowlett)

TheoremGeneric triangle is simple!

Proof. The parameter space (In the terminology of geometry,the moduli space is a 2-d orbifold.) of a triangle of fixed area isa 2-dimensional object.The set Fm such that the first m eigenvalues are distinct is adense open set on the parameter space.Thus by the Baire category theorem,⋂

Fm

is dense and open, thus generic.Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

They left the following conjecture: Generic triangle is simple.Using Math 210, we can solve the problem.(joint with Rowlett)

TheoremGeneric triangle is simple!

Proof. The parameter space (In the terminology of geometry,the moduli space is a 2-d orbifold.) of a triangle of fixed area isa 2-dimensional object.The set Fm such that the first m eigenvalues are distinct is adense open set on the parameter space.Thus by the Baire category theorem,⋂

Fm

is dense and open, thus generic.Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

They left the following conjecture: Generic triangle is simple.Using Math 210, we can solve the problem.(joint with Rowlett)

TheoremGeneric triangle is simple!

Proof. The parameter space (In the terminology of geometry,the moduli space is a 2-d orbifold.) of a triangle of fixed area isa 2-dimensional object.The set Fm such that the first m eigenvalues are distinct is adense open set on the parameter space.Thus by the Baire category theorem,⋂

Fm

is dense and open, thus generic.Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

They left the following conjecture: Generic triangle is simple.Using Math 210, we can solve the problem.(joint with Rowlett)

TheoremGeneric triangle is simple!

Proof. The parameter space (In the terminology of geometry,the moduli space is a 2-d orbifold.) of a triangle of fixed area isa 2-dimensional object.The set Fm such that the first m eigenvalues are distinct is adense open set on the parameter space.Thus by the Baire category theorem,⋂

Fm

is dense and open, thus generic.Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

They left the following conjecture: Generic triangle is simple.Using Math 210, we can solve the problem.(joint with Rowlett)

TheoremGeneric triangle is simple!

Proof. The parameter space (In the terminology of geometry,the moduli space is a 2-d orbifold.) of a triangle of fixed area isa 2-dimensional object.The set Fm such that the first m eigenvalues are distinct is adense open set on the parameter space.Thus by the Baire category theorem,⋂

Fm

is dense and open, thus generic.Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Question: If we know all the eigenvalues, can we tell the shapeof the domain?Very famous paper

Mark KacCan one hear the shape of a drum?Amer. Math Monthly, 1966

In general the answer is No. But this might be the starting pointof the spectrum geometry: to what extent can be tell about thegeometry of the domain drum from the information ofeigenvalues?

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Question: If we know all the eigenvalues, can we tell the shapeof the domain?Very famous paper

Mark KacCan one hear the shape of a drum?Amer. Math Monthly, 1966

In general the answer is No. But this might be the starting pointof the spectrum geometry: to what extent can be tell about thegeometry of the domain drum from the information ofeigenvalues?

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Question: If we know all the eigenvalues, can we tell the shapeof the domain?Very famous paper

Mark KacCan one hear the shape of a drum?Amer. Math Monthly, 1966

In general the answer is No. But this might be the starting pointof the spectrum geometry: to what extent can be tell about thegeometry of the domain drum from the information ofeigenvalues?

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Question: If we know all the eigenvalues, can we tell the shapeof the domain?Very famous paper

Mark KacCan one hear the shape of a drum?Amer. Math Monthly, 1966

In general the answer is No. But this might be the starting pointof the spectrum geometry: to what extent can be tell about thegeometry of the domain drum from the information ofeigenvalues?

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Can we hear the shape of a triangle?The answer is yes!And the finite number of eigenvalue will do the job!

P-K Chang and D. DeTurckOn hearing the shape of a triangle.Proc. Amer. Math. Society, 105(4), 1989, pp 1033-1038

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Can we hear the shape of a triangle?The answer is yes!And the finite number of eigenvalue will do the job!

P-K Chang and D. DeTurckOn hearing the shape of a triangle.Proc. Amer. Math. Society, 105(4), 1989, pp 1033-1038

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Can we hear the shape of a triangle?The answer is yes!And the finite number of eigenvalue will do the job!

P-K Chang and D. DeTurckOn hearing the shape of a triangle.Proc. Amer. Math. Society, 105(4), 1989, pp 1033-1038

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

What they proved was, for any triangle, there is a number Nsuch that there are no other triangle such that the first Neigenvalues match with the given triangle.A pd. D thesis problem: how many eigenvalues is enough? Is100 enough?

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

What they proved was, for any triangle, there is a number Nsuch that there are no other triangle such that the first Neigenvalues match with the given triangle.A pd. D thesis problem: how many eigenvalues is enough? Is100 enough?

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

The determinant of the Laplacian plays an important role inMirror symmetry.Formally,

det ∆ =∏λi 6=0

λi

How to well define?

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

The determinant of the Laplacian plays an important role inMirror symmetry.Formally,

det ∆ =∏λi 6=0

λi

How to well define?

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

ζ function regularization (Math 220 required).Let

ζ(s) =∑λi 6=0

1λs

Then we can prove that the function is holomorphic, if Re(s) isvery negative.By Math 220, we know that we can extend the function as ameromorphic function on C. In particular, at the original pointthe function ζ(s) is holomorphic.Formally, we have

ζ ′(s) =∑ 1

λs logλ−1i

Thus formally, we have

ζ ′(0) = − log det ∆

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

ζ function regularization (Math 220 required).Let

ζ(s) =∑λi 6=0

1λs

Then we can prove that the function is holomorphic, if Re(s) isvery negative.By Math 220, we know that we can extend the function as ameromorphic function on C. In particular, at the original pointthe function ζ(s) is holomorphic.Formally, we have

ζ ′(s) =∑ 1

λs logλ−1i

Thus formally, we have

ζ ′(0) = − log det ∆

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

ζ function regularization (Math 220 required).Let

ζ(s) =∑λi 6=0

1λs

Then we can prove that the function is holomorphic, if Re(s) isvery negative.By Math 220, we know that we can extend the function as ameromorphic function on C. In particular, at the original pointthe function ζ(s) is holomorphic.Formally, we have

ζ ′(s) =∑ 1

λs logλ−1i

Thus formally, we have

ζ ′(0) = − log det ∆

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

ζ function regularization (Math 220 required).Let

ζ(s) =∑λi 6=0

1λs

Then we can prove that the function is holomorphic, if Re(s) isvery negative.By Math 220, we know that we can extend the function as ameromorphic function on C. In particular, at the original pointthe function ζ(s) is holomorphic.Formally, we have

ζ ′(s) =∑ 1

λs logλ−1i

Thus formally, we have

ζ ′(0) = − log det ∆

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Riemann ζ function

Defineζ(z) = 1 +

12z + · · ·+ 1

nz + · · ·

For Re(z) > 1, the above series converges absolutely. Thus

ζ(z) defines a holomorphic function on Re(z) > 1.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Riemann ζ function

Defineζ(z) = 1 +

12z + · · ·+ 1

nz + · · ·

For Re(z) > 1, the above series converges absolutely. Thus

ζ(z) defines a holomorphic function on Re(z) > 1.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

ζ function regularization

We have the identity∫ ∞0

tse−tndt =Γ(s + 1)

ns+1 .

sn

∫ ∞0

ts−1e−tndt =

∫ ∞0

tse−tndt

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

ζ function regularization

We have the identity∫ ∞0

tse−tndt =Γ(s + 1)

ns+1 .

sn

∫ ∞0

ts−1e−tndt =

∫ ∞0

tse−tndt

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

The same method is true for the ζ function of eigenvalues. Wecan define

ζ(z) =∑λi 6=0

1λz

i.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

By Mirror Symmetry, we know that for a compact CY 3 fold X ,there is the mirror pair X ′ such that the quantum field theorieson both manifolds are identical.We introduce the so-called BCOV conjecture.The BOCV conjecture is related to the quintic mirror pair.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

By Mirror Symmetry, we know that for a compact CY 3 fold X ,there is the mirror pair X ′ such that the quantum field theorieson both manifolds are identical.We introduce the so-called BCOV conjecture.The BOCV conjecture is related to the quintic mirror pair.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

By Mirror Symmetry, we know that for a compact CY 3 fold X ,there is the mirror pair X ′ such that the quantum field theorieson both manifolds are identical.We introduce the so-called BCOV conjecture.The BOCV conjecture is related to the quintic mirror pair.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

We consider the quintics in CP4

Z 50 + · · ·+ Z 5

4 − 5λZ0 · · ·Z4 = 0.

For general complex number λ, it is a smooth hypersurface. Infact, it is a CY 3-fold.The CY manifold and its mirror pair is the most studied.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

We consider the quintics in CP4

Z 50 + · · ·+ Z 5

4 − 5λZ0 · · ·Z4 = 0.

For general complex number λ, it is a smooth hypersurface. Infact, it is a CY 3-fold.The CY manifold and its mirror pair is the most studied.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

We consider the quintics in CP4

Z 50 + · · ·+ Z 5

4 − 5λZ0 · · ·Z4 = 0.

For general complex number λ, it is a smooth hypersurface. Infact, it is a CY 3-fold.The CY manifold and its mirror pair is the most studied.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

What is the BCOV Conjecture?

BCOV refers to M. Bershadsky, S. Cecotti, H. Ooguri, C.Vafa.Very important paper

Bershadsky, M. and Cecotti, S. and Ooguri, H. andVafa, C.Kodaira-Spencer theory of gravity and exact results forquantum string amplitudesComm. Math. Phys, 165(2):311–427, 1994.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

What is the BCOV Conjecture?

BCOV refers to M. Bershadsky, S. Cecotti, H. Ooguri, C.Vafa.Very important paper

Bershadsky, M. and Cecotti, S. and Ooguri, H. andVafa, C.Kodaira-Spencer theory of gravity and exact results forquantum string amplitudesComm. Math. Phys, 165(2):311–427, 1994.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

What is the BCOV Conjecture?

BCOV refers to M. Bershadsky, S. Cecotti, H. Ooguri, C.Vafa.Very important paper

Bershadsky, M. and Cecotti, S. and Ooguri, H. andVafa, C.Kodaira-Spencer theory of gravity and exact results forquantum string amplitudesComm. Math. Phys, 165(2):311–427, 1994.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

The BCOV Conjecture is related to the g = 1 Mirror symmetry.Based on physics observation, they conjecture a relationbetween two (presumably unrelated) Calabi-Yau threefolds.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

For the general quintics

Z 50 + · · ·+ Z 5

4 − 5λZ0 · · ·Z4 = 0,

its mirror pair was explicitly constructed by Greene and Plesser.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

For the general quintics

Z 50 + · · ·+ Z 5

4 − 5λZ0 · · ·Z4 = 0,

its mirror pair was explicitly constructed by Greene and Plesser.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

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DefinitionThe mirror map is the holomorphic map from a neighborhood of∞ ∈ P1 to a neighborhood of 0 ∈ ∆ defined by the followingformula

q := (5ψ)−5 exp

5y0(ψ)

∞∑n=1

(5n)!

(n!)5

5n∑

j=n+1

1j

1(5ψ)5n

,

where |ψ| 1, and

y0(ψ) :=∞∑

n=0

(5n)!

(n!)5(5ψ)5n , |ψ| > 1.

The inverse of the mirror map is denoted by ψ(q).

Zhiqin Lu Graduate colloquium of the University of California at Irvine

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Define the multi-valued function F top1,B(ψ) as

F top1,B(ψ) :=

(ψ

y0(ψ)

) 623

(ψ5 − 1)−16 q

dψdq

,

andF top

1,A(q) := F top1,B(ψ(q)).

Zhiqin Lu Graduate colloquium of the University of California at Irvine

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Conjecture

(A) Let ng(d) be the genus-g degree-d instanton number of aquintic in CP4 for g = 0,1. Then the following identity holds:

− qddq

log F top1,A(q) =

5012−

∞∑n,d=1

n1(d)2nd qnd

1− qnd −∞∑

d=1

n0(d)2d qd

12(1− qd ).

Zhiqin Lu Graduate colloquium of the University of California at Irvine

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Conjecture (A) was proved by Aleksey Zinger.

Aleksey ZingerThe Reduced Genus-One Gromov-Witten Invariants ofCalabi-Yau HypersurfacesArXiv: 0705.2397v2, 2007.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

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Setup of Conjecture (B)

Let X be a compact Kähler manifold.

Let ∆ = ∆p,q be the Laplacian on (p,q) forms;By compactness, the spectrum of ∆ are eigenvalues:

0 ≤ λ0 ≤ λ1 ≤ · · · ≤ λn → +∞.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

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Setup of Conjecture (B)

Let X be a compact Kähler manifold.

Let ∆ = ∆p,q be the Laplacian on (p,q) forms;By compactness, the spectrum of ∆ are eigenvalues:

0 ≤ λ0 ≤ λ1 ≤ · · · ≤ λn → +∞.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

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Setup of Conjecture (B)

Let X be a compact Kähler manifold.

Let ∆ = ∆p,q be the Laplacian on (p,q) forms;By compactness, the spectrum of ∆ are eigenvalues:

0 ≤ λ0 ≤ λ1 ≤ · · · ≤ λn → +∞.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

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Setup of Conjecture B

Bershadsky-Ceccotti-Ooguri-Vafa defined

T def=∏p,q

(det ∆p,q)(−1)p+qpq.

Why define such a strange quantity?Answer: Riemann-Roch-Grothendieck theorem

Zhiqin Lu Graduate colloquium of the University of California at Irvine

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Setup of Conjecture B

Bershadsky-Ceccotti-Ooguri-Vafa defined

T def=∏p,q

(det ∆p,q)(−1)p+qpq.

Why define such a strange quantity?Answer: Riemann-Roch-Grothendieck theorem

Zhiqin Lu Graduate colloquium of the University of California at Irvine

Hearing triangle

Setup of Conjecture B

Bershadsky-Ceccotti-Ooguri-Vafa defined

T def=∏p,q

(det ∆p,q)(−1)p+qpq.

Why define such a strange quantity?Answer: Riemann-Roch-Grothendieck theorem

Zhiqin Lu Graduate colloquium of the University of California at Irvine

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Conjecture

(B) Let ‖ · ‖ be the Hermitian metric on the line bundle

(π∗KW/CP1)⊗62 ⊗ (T (CP1))⊗3|CP1\D

induced from the L2-metric on π∗KW/CP1 and from the

Weil-Petersson metric on T (CP1). Then the following identityholds:

τBCOV(Wψ) = Const.

∥∥∥∥∥ 1F top

1,B(ψ)3

(Ωψ

y0(ψ)

)62

⊗(

qddq

)3∥∥∥∥∥

23

,

where Ω is the local holomorphic section of the (3,0) forms.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

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Conjecture B was proved by Fang-L-Yoshikawa.

H. Fang, Z. Lu, and K-I, YoshikawaAsymptotic behavior of the BCOV torsion of Calabi-YaumoduliArXiv: math/0601411, 2006, Journal of Diff. Geom. 2008.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

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Combining Conjecture A and B, we verified the MirrorSymmetry prediction to the case g = 1. For higher genus, theB-side of the conjectures have not been set up.On the other hand, the g = 0 Mirror Symmetry Conjecture wasproved by Lian-Liu-Yau and Givental.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

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Combining Conjecture A and B, we verified the MirrorSymmetry prediction to the case g = 1. For higher genus, theB-side of the conjectures have not been set up.On the other hand, the g = 0 Mirror Symmetry Conjecture wasproved by Lian-Liu-Yau and Givental.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

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A life time question: Can we hear the shape of our Universe?

Zhiqin Lu Graduate colloquium of the University of California at Irvine

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Not known, but joint with physicists Mike Douglas, we provedthe number of Universes is finite, if the string theory is true.

Zhiqin Lu Graduate colloquium of the University of California at Irvine

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Thank you!

Zhiqin Lu Graduate colloquium of the University of California at Irvine

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