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Conformal transformations and gluing formulas Klaus Kirsten Baylor - - PowerPoint PPT Presentation

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Conformal transformations and gluing formulas Klaus Kirsten Baylor University Microlocal and Global Analysis, Interactions with Geometry University of Potsdam, March 4, 2019 Joint work with Yoonweon Lee (Inha University, Korea) J. Math. Phys.

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Conformal transformations and gluing formulas

Klaus Kirsten

Baylor University

Microlocal and Global Analysis, Interactions with Geometry University of Potsdam, March 4, 2019

Joint work with Yoonweon Lee (Inha University, Korea)

  • J. Math. Phys. 56 (2015) 123501 (19pp)
  • J. Geom. Phys. 117 (2017) 197-213

Symmetry 10 (2018) 31 (16pp)

  • J. Geom. Anal. 28 (2018) 3856-3891

The BFK-gluing formula and the curvature tensors on a two-dimensional compact manifold; submitted The gluing formula, conformal transformations, and geometry; in preparation Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 1 / 20

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Analytical surgery M = M1 ∪N M2

M1 N M2

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 2 / 20

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Analytical surgery M = M1 ∪N M2

M1 N M2

Eigenvalue problem for the Laplacian: −∆i ϕ(i)

k

= λ(i)

k

ϕ(i)

k ,

ϕ(i)

k

  • N = 0,

i = 1, 2,

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 2 / 20

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Analytical surgery M = M1 ∪N M2

M1 N M2

Eigenvalue problem for the Laplacian: −∆i ϕ(i)

k

= λ(i)

k

ϕ(i)

k ,

ϕ(i)

k

  • N = 0,

i = 1, 2, − ∆ ϕk = λk ϕk.

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 2 / 20

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Outline

1 Introduction

Heat kernel Zeta function BFK-gluing formula

2 Polynomial q(λ) and geometry 3 Applications: Casimir forces in pistons 4 Outlook

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 3 / 20

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Introduction

What are spectral functions?

Eigenvalue problem for a suitable differential operator P def. on Y : Puℓ(x) = λℓuℓ(x), Buℓ|x∈∂Y = 0 0 < λ1 ≤ λ2..., λℓ → ∞ as ℓ → ∞.

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 4 / 20

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Introduction

What are spectral functions?

Eigenvalue problem for a suitable differential operator P def. on Y : Puℓ(x) = λℓuℓ(x), Buℓ|x∈∂Y = 0 0 < λ1 ≤ λ2..., λℓ → ∞ as ℓ → ∞. Heat kernel (m + 1 = dim(Y )): K(t) =

∞

  • k=1

e−tλk

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 4 / 20

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Introduction

What are spectral functions?

Eigenvalue problem for a suitable differential operator P def. on Y : Puℓ(x) = λℓuℓ(x), Buℓ|x∈∂Y = 0 0 < λ1 ≤ λ2..., λℓ → ∞ as ℓ → ∞. Heat kernel (m + 1 = dim(Y )): K(t) =

∞

  • k=1

e−tλk ∼ t−(m+1)/2

∞

  • n=0,1/2,1,...

cn tn

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 4 / 20

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Introduction

What are spectral functions?

Eigenvalue problem for a suitable differential operator P def. on Y : Puℓ(x) = λℓuℓ(x), Buℓ|x∈∂Y = 0 0 < λ1 ≤ λ2..., λℓ → ∞ as ℓ → ∞. Heat kernel (m + 1 = dim(Y )): K(t) =

∞

  • k=1

e−tλk ∼ t−(m+1)/2

∞

  • n=0,1/2,1,...

cn tn cn =

  • Y

dx an(x) +

  • ∂Y

dy bn(y)

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 4 / 20

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Gluing formula for the heat kernel: K1(t) + K2(t) − K(t)

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 5 / 20

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Gluing formula for the heat kernel: K1(t) + K2(t) − K(t)

t→0

∼ t−(m+1)/2

∞

  • n=1/2,1,...

tn

  • N

dy ˜ bn(y)

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 5 / 20

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Gluing formula for the heat kernel: K1(t) + K2(t) − K(t)

t→0

∼ t−(m+1)/2

∞

  • n=1/2,1,...

tn

  • N

dy ˜ bn(y)

The combination of the heat kernels has a small-t asymptotics only depending on a density on N.

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 5 / 20

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Laplacian under conformal scaling: ∆f = |g|−1/2∂i

  • |g|1/2gij∂jf
  • gℓ = ℓ2g =

⇒ ∆ℓf = 1 ℓ2 ∆f = ⇒ λℓ

k = 1

ℓ2 λk

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 6 / 20

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Laplacian under conformal scaling: ∆f = |g|−1/2∂i

  • |g|1/2gij∂jf
  • gℓ = ℓ2g =

⇒ ∆ℓf = 1 ℓ2 ∆f = ⇒ λℓ

k = 1

ℓ2 λk Heat kernel under conformal scaling: K ℓ

i (t)

=

∞

  • k=1

e−tλℓ

i,k

t→0

∼ t−(m+1)/2

  • n=0,1/2,1,...

cℓ

i,ntn

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 6 / 20

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Laplacian under conformal scaling: ∆f = |g|−1/2∂i

  • |g|1/2gij∂jf
  • gℓ = ℓ2g =

⇒ ∆ℓf = 1 ℓ2 ∆f = ⇒ λℓ

k = 1

ℓ2 λk Heat kernel under conformal scaling: K ℓ

i (t)

=

∞

  • k=1

e−tλℓ

i,k

t→0

∼ t−(m+1)/2

  • n=0,1/2,1,...

cℓ

i,ntn t→0

∼ t ℓ2 −(m+1)/2

  • n=0,1/2,1,...

ci,n t ℓ2 n

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 6 / 20

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Laplacian under conformal scaling: ∆f = |g|−1/2∂i

  • |g|1/2gij∂jf
  • gℓ = ℓ2g =

⇒ ∆ℓf = 1 ℓ2 ∆f = ⇒ λℓ

k = 1

ℓ2 λk Heat kernel under conformal scaling: K ℓ

i (t)

=

∞

  • k=1

e−tλℓ

i,k

t→0

∼ t−(m+1)/2

  • n=0,1/2,1,...

cℓ

i,ntn t→0

∼ t ℓ2 −(m+1)/2

  • n=0,1/2,1,...

ci,n t ℓ2 n = ⇒ cℓ

i,n

= ℓm+1−2nci,n

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 6 / 20

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Heat kernel coefficients under conformal scaling: cℓ

i,n

= ℓm+1−2nci,n =

  • Mi

(ℓm+1dx) ℓ−2nan(x) +

  • N

(ℓmdy) ℓ1−2nbn(y)

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 7 / 20

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Heat kernel coefficients under conformal scaling: cℓ

i,n

= ℓm+1−2nci,n =

  • Mi

(ℓm+1dx) ℓ−2nan(x) +

  • N

(ℓmdy) ℓ1−2nbn(y) Example: m = 2, n = 3/2: cℓ

i,3/2 = ci,3/2 =

  • N

dy

  • b1R + b2Rrr + b3K 2 + b4KabKab
  • Klaus Kirsten (Baylor University)

Gluing formula Potsdam, March 4, 2019 7 / 20

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Zeta function: ζ(s) =

∞

  • ℓ=1

λ−s

ℓ ,

ℜs > m + 1 2

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 8 / 20

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Zeta function: ζ(s) =

∞

  • ℓ=1

λ−s

ℓ ,

ℜs > m + 1 2 Functional determinant: ” ln det P = ln

∞

  • ℓ=1

λℓ =

∞

  • ℓ=1

ln λℓ = − d ds

∞

  • ℓ=1

λ−s

ℓ

  • s=0

= −ζ′(0) ”

D.B. Ray and I.M. Singer, Adv. Math. 7 (1971) 145-210 Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 8 / 20

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Zeta function: ζ(s) =

∞

  • ℓ=1

λ−s

ℓ ,

ℜs > m + 1 2 Functional determinant: ” ln det P = ln

∞

  • ℓ=1

λℓ =

∞

  • ℓ=1

ln λℓ = − d ds

∞

  • ℓ=1

λ−s

ℓ

  • s=0

= −ζ′(0) ”

D.B. Ray and I.M. Singer, Adv. Math. 7 (1971) 145-210

BFK-gluing formula:

ln det (−∆ + λ) − ln det (−∆1 + λ) − ln det (−∆2 + λ) = ln det R(λ) + q(λ)

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 8 / 20

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Zeta function: ζ(s) =

∞

  • ℓ=1

λ−s

ℓ ,

ℜs > m + 1 2 Functional determinant: ” ln det P = ln

∞

  • ℓ=1

λℓ =

∞

  • ℓ=1

ln λℓ = − d ds

∞

  • ℓ=1

λ−s

ℓ

  • s=0

= −ζ′(0) ”

D.B. Ray and I.M. Singer, Adv. Math. 7 (1971) 145-210

BFK-gluing formula:

ln det (−∆ + λ) − ln det (−∆1 + λ) − ln det (−∆2 + λ) = ln det R(λ) + q(λ)

q(λ) =

[m/2]

  • j=0

qjλj :integral of some local density over N.

  • D. Burghelea, L. Friedlander and T. Kappeler, J. Funct. Anal. 107 (1992) 34-65
  • G. Carron, Am. J. Math. 124 (2002) 307-352

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 8 / 20

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Let h ∈ C ∞(N), φi ∈ C ∞(Mi), such that (−∆1 + λ)φ1 = 0, (−∆2 + λ)φ2 = 0, φ1 |N = φ2|N = h,

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 9 / 20

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Let h ∈ C ∞(N), φi ∈ C ∞(Mi), such that (−∆1 + λ)φ1 = 0, (−∆2 + λ)φ2 = 0, φ1 |N = φ2|N = h, then the Dirichlet-to-Neumann map is defined as follows: R(λ) : C ∞(N) → C ∞(N), R(λ)(h) = (∂rφ1 − ∂rφ2|N .

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 9 / 20

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Polynomial q(λ) and geometry

BFK-gluing formula:

ln det (−∆ + λ) − ln det (−∆1 + λ) − ln det (−∆2 + λ) = ln det R(λ) + q(λ)

q: integral of some local density over N.

  • D. Burghelea, L. Friedlander and T. Kappeler, J. Funct. Anal. 107 (1992) 34-65
  • G. Carron, Am. J. Math. 124 (2002) 307-352

M1 N M2

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 10 / 20

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Polynomial q(λ) and geometry

BFK-gluing formula:

ln det (−∆ + λ) − ln det (−∆1 + λ) − ln det (−∆2 + λ) = ln det R(λ) + q(λ)

q: integral of some local density over N.

  • D. Burghelea, L. Friedlander and T. Kappeler, J. Funct. Anal. 107 (1992) 34-65
  • G. Carron, Am. J. Math. 124 (2002) 307-352

M1 N M2

Product structure near N: ds2 = dr2 + dN2 q(λ) = − ln 2 (ζ∆N+λ(0) + d(0))

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 10 / 20

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Behavior under conformal scaling g → ℓ2g:

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 11 / 20

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Behavior under conformal scaling g → ℓ2g: Zeta determinant of Laplacians: ∆ → 1 ℓ2 ∆

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 11 / 20

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Behavior under conformal scaling g → ℓ2g: Zeta determinant of Laplacians: ∆ → 1 ℓ2 ∆ = ⇒ λn → 1 ℓ2 λn

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 11 / 20

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Behavior under conformal scaling g → ℓ2g: Zeta determinant of Laplacians: ∆ → 1 ℓ2 ∆ = ⇒ λn → 1 ℓ2 λn ζℓ(s) =

∞

  • n=1

λn ℓ2 −s = ℓ2sζ(s)

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 11 / 20

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Behavior under conformal scaling g → ℓ2g: Zeta determinant of Laplacians: ∆ → 1 ℓ2 ∆ = ⇒ λn → 1 ℓ2 λn ζℓ(s) =

∞

  • n=1

λn ℓ2 −s = ℓ2sζ(s) = ⇒ ζ′

ℓ(0) = 2 ln ℓ ζ(0) + ζ′(0)

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 11 / 20

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Behavior under conformal scaling g → ℓ2g: Zeta determinant of Laplacians: ∆ → 1 ℓ2 ∆ = ⇒ λn → 1 ℓ2 λn ζℓ(s) =

∞

  • n=1

λn ℓ2 −s = ℓ2sζ(s) = ⇒ ζ′

ℓ(0) = 2 ln ℓ ζ(0) + ζ′(0)

Zeta determinant of the Dirichlet-to-Neumann map: (−∆1 + λ)φ1 = 0, (−∆2 + λ)φ2 = 0, φ1 |N = φ2|N = h, R(h) := (∂rφ1 − ∂rφ2|N

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 11 / 20

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Behavior under conformal scaling g → ℓ2g: Zeta determinant of Laplacians: ∆ → 1 ℓ2 ∆ = ⇒ λn → 1 ℓ2 λn ζℓ(s) =

∞

  • n=1

λn ℓ2 −s = ℓ2sζ(s) = ⇒ ζ′

ℓ(0) = 2 ln ℓ ζ(0) + ζ′(0)

Zeta determinant of the Dirichlet-to-Neumann map: (−∆1 + λ)φ1 = 0, (−∆2 + λ)φ2 = 0, φ1 |N = φ2|N = h, R(h) := (∂rφ1 − ∂rφ2|N = ⇒ Rℓ = 1 ℓ R = ⇒

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 11 / 20

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Behavior under conformal scaling g → ℓ2g: Zeta determinant of Laplacians: ∆ → 1 ℓ2 ∆ = ⇒ λn → 1 ℓ2 λn ζℓ(s) =

∞

  • n=1

λn ℓ2 −s = ℓ2sζ(s) = ⇒ ζ′

ℓ(0) = 2 ln ℓ ζ(0) + ζ′(0)

Zeta determinant of the Dirichlet-to-Neumann map: (−∆1 + λ)φ1 = 0, (−∆2 + λ)φ2 = 0, φ1 |N = φ2|N = h, R(h) := (∂rφ1 − ∂rφ2|N = ⇒ Rℓ = 1 ℓ R = ⇒ ζ′

ℓ,DN(0) = ln ℓ ζDN(0) + ζ′ DN(0).

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 11 / 20

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Behavior under conformal scaling g → ℓ2g: Zeta determinant of Laplacians: ∆ → 1 ℓ2 ∆ = ⇒ λn → 1 ℓ2 λn ζℓ(s) =

∞

  • n=1

λn ℓ2 −s = ℓ2sζ(s) = ⇒ ζ′

ℓ(0) = 2 ln ℓ ζ(0) + ζ′(0)

Zeta determinant of the Dirichlet-to-Neumann map: (−∆1 + λ)φ1 = 0, (−∆2 + λ)φ2 = 0, φ1 |N = φ2|N = h, R(h) := (∂rφ1 − ∂rφ2|N = ⇒ Rℓ = 1 ℓ R = ⇒ ζ′

ℓ,DN(0) = ln ℓ ζDN(0) + ζ′ DN(0).

Transformation of the BFK-gluing formula: ln ℓ

  • (2 [ζ(0) − ζ1(0) − ζ2(0)] − ζDN(0)
  • = q(λ) − qℓ(λ)

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 11 / 20

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Once we show ζDN(0) = 2 [ζ(0) − ζ1(0) − ζ2(0)]

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 12 / 20

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Once we show ζDN(0) = 2 [ζ(0) − ζ1(0) − ζ2(0)] this implies qℓ(λ) = q(λ)

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 12 / 20

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Once we show ζDN(0) = 2 [ζ(0) − ζ1(0) − ζ2(0)] this implies qℓ(λ) = q(λ) Large-λ asymptotic of ζ′

DN(0):

−ζ′

DN(0) λ→∞

∼

∞

  • j=0

πjλ

m−j 2

+

m

  • j=0

sjλ

m−j 2 ln λ Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 12 / 20

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Once we show ζDN(0) = 2 [ζ(0) − ζ1(0) − ζ2(0)] this implies qℓ(λ) = q(λ) Large-λ asymptotic of ζ′

DN(0):

−ζ′

DN(0) λ→∞

∼

∞

  • j=0

πjλ

m−j 2

+

m

  • j=0

sjλ

m−j 2 ln λ

− ζℓ′

DN(0)

= − ln ℓ ζDN(0) − ζ′

DN(0)

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 12 / 20

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Once we show ζDN(0) = 2 [ζ(0) − ζ1(0) − ζ2(0)] this implies qℓ(λ) = q(λ) Large-λ asymptotic of ζ′

DN(0):

−ζ′

DN(0) λ→∞

∼

∞

  • j=0

πjλ

m−j 2

+

m

  • j=0

sjλ

m−j 2 ln λ

− ζℓ′

DN(0)

= − ln ℓ ζDN(0) − ζ′

DN(0) λ→∞

∼

∞

  • j=0

πj λ ℓ2 m−j

2

+

m

  • j=0

sj λ ℓ2 m−j

2

ln λ ℓ2

  • Klaus Kirsten (Baylor University)

Gluing formula Potsdam, March 4, 2019 12 / 20

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Comparing ln ℓ terms: ζDN(0) = 2

m

  • j=0

sℓ

j

λ ℓ2 m−j

2 Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 13 / 20

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Comparing ln ℓ terms: ζDN(0) = 2

m

  • j=0

sℓ

j

λ ℓ2 m−j

2

= 2

m

  • j=0

sjλ

m−j 2 Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 13 / 20

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Comparing ln ℓ terms: ζDN(0) = 2

m

  • j=0

sℓ

j

λ ℓ2 m−j

2

= 2

m

  • j=0

sjλ

m−j 2

Compare the large-λ asymptotics of the gluing formula: ζ′(0) − ζ′

1(0) − ζ′ 2(0) = q(λ) + ζ′ DN(0)

to see ζDN(0) = 2 [ζ(0) − ζ1(0) − ζ2(0)]

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 13 / 20

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Comparing ln ℓ terms: ζDN(0) = 2

m

  • j=0

sℓ

j

λ ℓ2 m−j

2

= 2

m

  • j=0

sjλ

m−j 2

Compare the large-λ asymptotics of the gluing formula: ζ′(0) − ζ′

1(0) − ζ′ 2(0) = q(λ) + ζ′ DN(0)

to see ζDN(0) = 2 [ζ(0) − ζ1(0) − ζ2(0)] and thus qℓ(λ) = q(λ)

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 13 / 20

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So far from conformal scaling: q(λ) = 1 4π

  • N

dy

  • b0(−λ) + b1R + b2Rrr + b3K 2 + b4KabKab
  • Klaus Kirsten (Baylor University)

Gluing formula Potsdam, March 4, 2019 14 / 20

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So far from conformal scaling: q(λ) = 1 4π

  • N

dy

  • b0(−λ) + b1R + b2Rrr + b3K 2 + b4KabKab
  • From product metric case near N:

q(λ) = 1 4π(− ln 2)

  • N

dy

  • −λ + 1

6R

  • =

⇒ b0 = − ln 2, b1 = −1 6 ln 2.

What else can be found by using general conformal transformations?

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 14 / 20

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More general conformal transformations: P = −∆ − E

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 15 / 20

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More general conformal transformations: P = −∆ − E Define one-parameter family of operators as follows: g(ǫ) = e2ǫF(x)g, E(ǫ) = e−2ǫF(x)

  • E + 1

2(m − 1)ǫ∆F + 1 4(m − 1)2ǫ2F;rF;r

  • Klaus Kirsten (Baylor University)

Gluing formula Potsdam, March 4, 2019 15 / 20

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More general conformal transformations: P = −∆ − E Define one-parameter family of operators as follows: g(ǫ) = e2ǫF(x)g, E(ǫ) = e−2ǫF(x)

  • E + 1

2(m − 1)ǫ∆F + 1 4(m − 1)2ǫ2F;rF;r

  • P(ǫ)

= −∆ǫ − E(ǫ)

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 15 / 20

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More general conformal transformations: P = −∆ − E Define one-parameter family of operators as follows: g(ǫ) = e2ǫF(x)g, E(ǫ) = e−2ǫF(x)

  • E + 1

2(m − 1)ǫ∆F + 1 4(m − 1)2ǫ2F;rF;r

  • P(ǫ)

= −∆ǫ − E(ǫ) = e−2ǫF(x)P.

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 15 / 20

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More general conformal transformations: P = −∆ − E Define one-parameter family of operators as follows: g(ǫ) = e2ǫF(x)g, E(ǫ) = e−2ǫF(x)

  • E + 1

2(m − 1)ǫ∆F + 1 4(m − 1)2ǫ2F;rF;r

  • P(ǫ)

= −∆ǫ − E(ǫ) = e−2ǫF(x)P. Well known transformation behavior: d dǫ

  • ǫ=0 ζǫ′

i (0)

= 2c m+1

2 (F) = 2ζi(0, F)

d dǫ

  • ǫ=0 ζǫ′

DN(0)

= ζDN(0, F)

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 15 / 20

slide-52
SLIDE 52

d dǫ|ǫ=0 of gluing formula:

2 [ζ1(0, F) + ζ2(0, F) − ζ(0, F)] = d dǫ |ǫ=0qǫ(E) − ζDN(0, F)

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 16 / 20

slide-53
SLIDE 53

d dǫ|ǫ=0 of gluing formula:

2 [ζ1(0, F) + ζ2(0, F) − ζ(0, F)] = d dǫ |ǫ=0qǫ(E) − ζDN(0, F) Compare both sides: F;rr : −1 4 = 1 2b0 − 4b1 − 2b2

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 16 / 20

slide-54
SLIDE 54

d dǫ|ǫ=0 of gluing formula:

2 [ζ1(0, F) + ζ2(0, F) − ζ(0, F)] = d dǫ |ǫ=0qǫ(E) − ζDN(0, F) Compare both sides: F;rr : −1 4 = 1 2b0 − 4b1 − 2b2 = ⇒ b2 = 1 12 ln 2 + 1 8

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 16 / 20

slide-55
SLIDE 55

d dǫ|ǫ=0 of gluing formula:

2 [ζ1(0, F) + ζ2(0, F) − ζ(0, F)] = d dǫ |ǫ=0qǫ(E) − ζDN(0, F) Compare both sides: F;rr : −1 4 = 1 2b0 − 4b1 − 2b2 = ⇒ b2 = 1 12 ln 2 + 1 8 KF;r : − 5 16 = 1 2b0 − 4b1 − b2 + 4b3 + 2b4

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 16 / 20

slide-56
SLIDE 56

d dǫ|ǫ=0 of gluing formula:

2 [ζ1(0, F) + ζ2(0, F) − ζ(0, F)] = d dǫ |ǫ=0qǫ(E) − ζDN(0, F) Compare both sides: F;rr : −1 4 = 1 2b0 − 4b1 − 2b2 = ⇒ b2 = 1 12 ln 2 + 1 8 KF;r : − 5 16 = 1 2b0 − 4b1 − b2 + 4b3 + 2b4 = ⇒ 2b3 + b4 = − 1 24 ln 2 − 3 32.

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 16 / 20

slide-57
SLIDE 57

d dǫ|ǫ=0 of gluing formula:

2 [ζ1(0, F) + ζ2(0, F) − ζ(0, F)] = d dǫ |ǫ=0qǫ(E) − ζDN(0, F) Compare both sides: F;rr : −1 4 = 1 2b0 − 4b1 − 2b2 = ⇒ b2 = 1 12 ln 2 + 1 8 KF;r : − 5 16 = 1 2b0 − 4b1 − b2 + 4b3 + 2b4 = ⇒ 2b3 + b4 = − 1 24 ln 2 − 3 32. Final information found from example M = B2 × S1: b3 + b4 = 1 32 ln 2 + 3 128 = ⇒ b3 = − 7 96 ln 2 − 15 128, b4 = 5 48 ln 2 + 9 64.

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 16 / 20

slide-58
SLIDE 58

Applications: Casimir forces in pistons

Piston configuration: Mi = Mi × S1

M1 N M2

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 17 / 20

slide-59
SLIDE 59

Applications: Casimir forces in pistons

Piston configuration: Mi = Mi × S1

M1 N M2

Massless scalar field with Dirichlet boundary condition −∆i ϕ(i)

k

= λ(i)

k

ϕ(i)

k ,

ϕ(i)

k

  • N = 0,

i = 1, 2.

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 17 / 20

slide-60
SLIDE 60

Applications: Casimir forces in pistons

Piston configuration: Mi = Mi × S1

M1 N M2

Massless scalar field with Dirichlet boundary condition −∆i ϕ(i)

k

= λ(i)

k

ϕ(i)

k ,

ϕ(i)

k

  • N = 0,

i = 1, 2. Finite temperature quantum field theory: relevant object ζ′

1(0) + ζ′ 2(0)

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 17 / 20

slide-61
SLIDE 61

Applications: Casimir forces in pistons

Piston configuration: Mi = Mi × S1

M1 N M2

Massless scalar field with Dirichlet boundary condition −∆i ϕ(i)

k

= λ(i)

k

ϕ(i)

k ,

ϕ(i)

k

  • N = 0,

i = 1, 2. Finite temperature quantum field theory: relevant object ζ′

1(0) + ζ′ 2(0) − ζ′(0)

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 17 / 20

slide-62
SLIDE 62

Applications: Casimir forces in pistons

Piston configuration: Mi = Mi × S1

M1 N M2

Massless scalar field with Dirichlet boundary condition −∆i ϕ(i)

k

= λ(i)

k

ϕ(i)

k ,

ϕ(i)

k

  • N = 0,

i = 1, 2. Finite temperature quantum field theory: relevant object

E glue

Cas (β) = −1

2 ∂ ∂β (ζ′

1(0) + ζ′ 2(0) − ζ′(0))

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 18 / 20

slide-63
SLIDE 63

Applications: Casimir forces in pistons

Piston configuration: Mi = Mi × S1

M1 N M2

Massless scalar field with Dirichlet boundary condition −∆i ϕ(i)

k

= λ(i)

k

ϕ(i)

k ,

ϕ(i)

k

  • N = 0,

i = 1, 2. Finite temperature quantum field theory: relevant object

E glue

Cas (β) = −1

2 ∂ ∂β (ζ′

1(0) + ζ′ 2(0) − ζ′(0)) = 1

2 ∂ ∂β (ζ′

DN(0) − q(λ))

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 18 / 20

slide-64
SLIDE 64

Applications: Casimir forces in pistons

Piston configuration: Mi = Mi × S1

M1 N M2

Massless scalar field with Dirichlet boundary condition: −∆i ϕ(i)

k

= λ(i)

k

ϕ(i)

k ,

ϕ(i)

k

  • N = 0,

i = 1, 2. Finite temperature quantum field theory: relevant object FCas(β) = 1 2 ∂ ∂a ∂ ∂β

  • ζ′

1(0) + ζ′ 2(0) − ζ′(0)

  • Klaus Kirsten (Baylor University)

Gluing formula Potsdam, March 4, 2019 19 / 20

slide-65
SLIDE 65

Applications: Casimir forces in pistons

Piston configuration: Mi = Mi × S1

M1 N M2

Massless scalar field with Dirichlet boundary condition: −∆i ϕ(i)

k

= λ(i)

k

ϕ(i)

k ,

ϕ(i)

k

  • N = 0,

i = 1, 2. Finite temperature quantum field theory: relevant object FCas(β) = 1 2 ∂ ∂a ∂ ∂β

  • ζ′

1(0) + ζ′ 2(0) − ζ′(0)

  • = −1

2 ∂ ∂a ∂ ∂β ζ′

DN(0)

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 19 / 20

slide-66
SLIDE 66

Outlook

Understand the quantity ∂ ∂a ∂ ∂β ζ′

DN(0)

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 20 / 20

slide-67
SLIDE 67

Outlook

Understand the quantity ∂ ∂a ∂ ∂β ζ′

DN(0)

Show a gluing formula for the case where N has a boundary!

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 20 / 20

slide-68
SLIDE 68

Outlook

Understand the quantity ∂ ∂a ∂ ∂β ζ′

DN(0)

Show a gluing formula for the case where N has a boundary! How to incorporate other choices of boundary conditions?

Klaus Kirsten (Baylor University) Gluing formula Potsdam, March 4, 2019 20 / 20