내 *1997 Ph.D. 블로그 항목으로 논문 *.
μ \mu μ M M M dim = n \dim=n dim = n g g g ∇ \nabla ∇ Δ M \Delta_M Δ M k − t Δ / 2 ( x , y ) k_{-t\Delta/2}(x,y) k − t Δ/2 ( x , y ) M M M
따라서 k − t Δ / 2 ( x , x ) = d D M ∗ μ / g d x k_{-t\Delta/2}(x,x) = dDM_*\mu/\sqrt{g}dx k − t Δ/2 ( x , x ) = d D M ∗ μ / g d x μ \mu μ Ω t ( M ) ∣ x \Omega_t(M)\vert_x Ω t ( M ) ∣ x D M DM D M 참고: D M − 1 Ω t ∣ x DM^{-1}\Omega_t\vert_x D M − 1 Ω t ∣ x
Ω t 0 \Omega_t^0 Ω t 0 M M M
Ω t [ γ ] \Omega_t[\gamma] Ω t [ γ ] M M M γ \gamma γ γ 0 \gamma_0 γ 0
D M − 1 Ω t 0 [ γ ] DM^{-1}\Omega_t^0[\gamma] D M − 1 Ω t 0 [ γ ] M M M γ ( s ) = D M ( s ℓ ( γ ) t e ⃗ 1 ) , 0 ≤ s ≤ t \gamma(s) = DM(\frac{s\ell(\gamma)}{t}\vec{e}^1), 0\leq s \leq t γ ( s ) = D M ( t s ℓ ( γ ) e 1 ) , 0 ≤ s ≤ t S γ 0 ( s ) n − 1 ( k ℓ ( γ 0 ) ) , 0 ≤ s ≤ t , k → ∞ S_{\gamma_0(s)}^{n-1}(k\ell(\gamma_0)), 0\leq s \leq t, k\rightarrow\infty S γ 0 ( s ) n − 1 ( k ℓ ( γ 0 )) , 0 ≤ s ≤ t , k → ∞ γ 0 \gamma_0 γ 0 γ 0 \gamma_0 γ 0 D M DM D M Ω t 0 [ γ ] \Omega_t^0[\gamma] Ω t 0 [ γ ] det g ( x ⃗ ) = 1 \det{g(\vec{x})} = 1 det g ( x ) = 1
d s 2 = d x ⊙ d x + h ( x , y ) d x ⊙ d y + ( 1 + h 2 ( x , y ) ) d y ⊙ d y g ( x , 0 ) = 0 , σ 1 = d x + h ( x , y ) d y σ 2 = h ( x , y ) d x + ( 1 + h 2 ( x , y ) ) d y σ 1 ∧ σ 2 = d x ∧ d y d σ 1 = ∂ h ∂ x d x ∧ d y d σ 2 = ( − ∂ h ∂ y + 2 h ∂ h ∂ x ) d x ∧ d y \begin{aligned}
ds^2 &= dx\odot dx + h(x,y) dx\odot dy + (1+h^2(x,y)) dy\odot dy \\
g(x,0) &= 0,\\
\sigma_1 &= dx + h(x,y)dy \\
\sigma_2 &= h(x,y)dx + (1+h^2(x,y))dy \\
\sigma_1 \wedge \sigma_2 &= dx \wedge dy\\
d\sigma_1 &= \frac{\partial h}{\partial x}dx\wedge dy \\
d\sigma_2 &= (-\frac{\partial h}{\partial y}+2h\frac{\partial h}{\partial x}) dx\wedge dy \\
\end{aligned} d s 2 g ( x , 0 ) σ 1 σ 2 σ 1 ∧ σ 2 d σ 1 d σ 2 = d x ⊙ d x + h ( x , y ) d x ⊙ d y + ( 1 + h 2 ( x , y )) d y ⊙ d y = 0 , = d x + h ( x , y ) d y = h ( x , y ) d x + ( 1 + h 2 ( x , y )) d y = d x ∧ d y = ∂ x ∂ h d x ∧ d y = ( − ∂ y ∂ h + 2 h ∂ x ∂ h ) d x ∧ d y
따라서 연결 1-form α : = A d x + B d y \alpha := Adx + Bdy α := A d x + B d y
d σ 1 = α ∧ σ 1 d σ 2 = σ 2 ∧ α ⟹ A = ∂ h ∂ y − 3 h ∂ h ∂ x B = h ∂ h ∂ y − ( 1 + 3 h 2 ) ∂ h ∂ x K = ∂ B ∂ x − ∂ A ∂ y = ∂ h ∂ x ∂ h ∂ y − 6 h ∂ h ∂ x 2 − ( 1 + 3 h 2 ) ∂ 2 h ∂ x ∂ x − ∂ 2 h ∂ y ∂ y + 2 h ∂ 2 h ∂ x ∂ y , h = h ( y ) ⟹ α = ∂ ∂ y ( h d x + h 2 2 d y ) K ( y ) = − ∂ 2 h ∂ y ∂ y has Galilean Symmetry: h ↦ h ( y , β ) = h ( y ) + β y . \begin{aligned}
d\sigma_1 &= \alpha \wedge \sigma_1 \\
d\sigma_2 &= \sigma_2 \wedge \alpha \\
\implies \\
A &= \frac{\partial h}{\partial y} - 3h\frac{\partial h}{\partial x} \\
B &= h\frac{\partial h}{\partial y} - (1+3h^2)\frac{\partial h}{\partial x} \\
\\
K &= \frac{\partial B}{\partial x} - \frac{\partial A}{\partial y} \\
&= \frac{\partial h}{\partial x}\frac{\partial h}{\partial y} - 6h\frac{\partial h}{\partial x}^2 - (1+3h^2)\frac{\partial^2 h}{\partial x \partial x}
- \frac{\partial^2 h}{\partial y \partial y} + 2h \frac{\partial^2 h}{\partial x \partial y} ,\\
h &= h(y) \implies \\
\alpha &= \frac{\partial }{\partial y} (hdx +\frac{h^2}{2} dy)\\
K(y) &= -\frac{\partial^2 h}{\partial y \partial y}\ \\
\text{has Galilean Symmetry:} \\
h \mapsto h(y,\beta) &= h(y) + \beta y \ .\\
\end{aligned} d σ 1 d σ 2 ⟹ A B K h α K ( y ) has Galilean Symmetry: h ↦ h ( y , β ) = α ∧ σ 1 = σ 2 ∧ α = ∂ y ∂ h − 3 h ∂ x ∂ h = h ∂ y ∂ h − ( 1 + 3 h 2 ) ∂ x ∂ h = ∂ x ∂ B − ∂ y ∂ A = ∂ x ∂ h ∂ y ∂ h − 6 h ∂ x ∂ h 2 − ( 1 + 3 h 2 ) ∂ x ∂ x ∂ 2 h − ∂ y ∂ y ∂ 2 h + 2 h ∂ x ∂ y ∂ 2 h , = h ( y ) ⟹ = ∂ y ∂ ( h d x + 2 h 2 d y ) = − ∂ y ∂ y ∂ 2 h = h ( y ) + β y .
중요 따라서, h = h ( y ) h=h(y) h = h ( y ) c ⃗ ˙ ( t ) = − α ( γ ˙ ( t ) ) c ⃗ ( t ) = − ∂ y ( h d x / d t + h 2 / 2 d y / d t ) c ⃗ ( t ) \dot{\vec{c}}(t) = -\alpha(\dot{\gamma}(t))\vec{c}(t) = -\partial_y(h\ dx/dt + h^2/2\ dy/dt)\vec{c}(t) c ˙ ( t ) = − α ( γ ˙ ( t )) c ( t ) = − ∂ y ( h d x / d t + h 2 /2 d y / d t ) c ( t )
c ⃗ ( t ) = e x p ( − ( ( h + β y ) d x / d t + ( h 2 / 2 + ( β y ) 2 / 2 + β y h ) d y / d t ) ∣ x 0 , y 0 x t , y t ) c ⃗ ( 0 ) \vec{c}(t) = exp(-((h+\beta y)\ dx/dt + (h^2/2 + (\beta y)^2/2 + \beta y h) \ dy/dt)|_{x_0,y_0}^{x_t,y_t})\vec{c}(0) c ( t ) = e x p ( − (( h + β y ) d x / d t + ( h 2 /2 + ( β y ) 2 /2 + β y h ) d y / d t ) ∣ x 0 , y 0 x t , y t ) c ( 0 )
전송 곡선의 기능입니다. γ \gamma γ 독립 입니다. 이 사실은 개발 맵이 루프를 보존한다는 것을 의미합니다.
Z − Δ / 2 ( t ) : = ∫ M k − t Δ / 2 ( x , x ) g d x = ∑ j = 0 ∞ e − λ i t / 2 Z_{-\Delta/2}(t) := \int_M k_{-t\Delta/2}(x,x) \sqrt{g}dx = \sum_{j=0}^\infty e^{-\lambda_i t/2} Z − Δ/2 ( t ) := ∫ M k − t Δ/2 ( x , x ) g d x = ∑ j = 0 ∞ e − λ i t /2
마지막으로 Radon-Nicodym 파생 상품에서 다음을 정의하십시오.
D M ∗ μ ( Ω t ) : = ∫ M D M ∗ μ ( Ω t ∣ x g d x ) D M ∗ μ ( Ω t 0 ) : = ∫ M D M ∗ μ ( Ω t 0 ∣ x g d x ) D M ∗ μ ( Ω t [ γ ] ) : = ∫ M D M ∗ μ ( Ω t [ γ ] ∣ x g d x ) \begin{aligned}
DM_*\mu(\Omega_t) &:= \int_M DM_*\mu(\Omega_t\vert_x \sqrt{g}dx)\\
DM_*\mu(\Omega^0_t) &:= \int_M DM_*\mu(\Omega^0_t\vert_x \sqrt{g}dx)\\
DM_*\mu(\Omega_t[\gamma]) &:= \int_M DM_*\mu(\Omega_t[\gamma]\vert_x \sqrt{g}dx) \\
\end{aligned} D M ∗ μ ( Ω t ) D M ∗ μ ( Ω t 0 ) D M ∗ μ ( Ω t [ γ ]) := ∫ M D M ∗ μ ( Ω t ∣ x g d x ) := ∫ M D M ∗ μ ( Ω t 0 ∣ x g d x ) := ∫ M D M ∗ μ ( Ω t [ γ ] ∣ x g d x )
Z − Δ / 2 ( t ) = D M ∗ μ ( Ω t ) = D M ∗ μ ( Ω t 0 ) + ∑ { γ } D M ∗ μ ( Ω t [ γ ] ) D M ∗ μ ( Ω t 0 ) ≈ t → 0 ( 2 π t ) − n / 2 ( v o l ( M ) + t / 6 ∫ M K ( x ) g d x + O ( t 2 ) ) by McKean-Singer D M ∗ μ ( Ω t [ γ ] ) = e − ℓ ( γ ) 2 / 2 t ∫ M D M ∗ μ ( e t < J B B t | B t > Ω t 0 [ γ ] ∣ x g d x ) by Cameron-Martin = e − ℓ ( γ ) 2 / 2 t ∫ T γ 0 M E ( e t J B ∣ Ω t 0 [ γ ] ∣ x ( τ ) ) d x 1 ( τ ) … d x n ( τ ) d τ \begin{aligned}
Z_{-\Delta/2}(t) = DM_*\mu(\Omega_t) &= DM_*\mu(\Omega^0_t) + \sum_{\set{\gamma}} DM_*\mu(\Omega_t[\gamma]) \\
DM_*\mu(\Omega_t^0) &\approx_{t\rightarrow 0} (2\pi t)^{-n/2}(vol(M) + t/6\int_M K(x)\sqrt{g} dx + O(t^2))\space \small\text{by McKean-Singer}\\
DM_*\mu(\Omega_t[\gamma]) &= e^{-\ell(\gamma)^2/2t}\int_M DM_*\mu(e^{\bra{J_BB_t}\ket{B_t}} _t \Omega_t^0[\gamma]\vert_x\sqrt{g}dx)\space\small \text{ by Cameron-Martin}\\
&= e^{-\ell(\gamma)^2/2t}\int_{T_{\gamma_0}M} E(e^{J_B}_{t} | \Omega_t^0[\gamma]\vert_{x(\tau)})dx^1(\tau)\dots dx^n(\tau) d\tau\\
\end{aligned} Z − Δ/2 ( t ) = D M ∗ μ ( Ω t ) D M ∗ μ ( Ω t 0 ) D M ∗ μ ( Ω t [ γ ]) = D M ∗ μ ( Ω t 0 ) + { γ } ∑ D M ∗ μ ( Ω t [ γ ]) ≈ t → 0 ( 2 π t ) − n /2 ( v o l ( M ) + t /6 ∫ M K ( x ) g d x + O ( t 2 )) by McKean-Singer = e − ℓ ( γ ) 2 /2 t ∫ M D M ∗ μ ( e t ⟨ J B B t ∣ B t ⟩ Ω t 0 [ γ ] ∣ x g d x ) by Cameron-Martin = e − ℓ ( γ ) 2 /2 t ∫ T γ 0 M E ( e t J B ∣ Ω t 0 [ γ ] ∣ x ( τ ) ) d x 1 ( τ ) … d x n ( τ ) d τ
Horocyclic coordinates : h = h ( y ) ⟹ = ℓ ( γ 0 ) 2 π t ∫ R e − ( 1 + h 2 ( y ) ) ℓ ( γ ) 2 / 2 t 2 sinh − K ( y ) ℓ ( γ ) / 2 ℓ ( γ ) d y , h ( y ) = y ⟹ = e − ℓ ( γ ) 2 / 2 t ℓ ( γ 0 ) 2 2 π t sinh − K ℓ ( γ ) / 2 ∫ R e − y 2 ℓ ( γ ) 2 / 2 t ℓ ( γ ) d y 2 π t \begin{aligned}
\text{Horocyclic coordinates}: \\
h&= h(y) \implies \\
&=\frac{\ell(\gamma_0)}{2\pi t}\int_{\Reals}\frac{e^{- (1+h^2(y))\ell(\gamma)^2/2t}}{2\sinh \sqrt{-K(y)}\ell(\gamma)/2}\ell(\gamma) dy\ ,\\
h(y) = y \implies \\
&= \frac{e^{-\ell(\gamma)^2/2t}\ell(\gamma_0)}{2\sqrt{2\pi t}\sinh \sqrt {-K}\ell(\gamma)/2}\int_{\Reals}e^{-y^2\ell(\gamma)^2/2t}{\frac{\ell(\gamma)dy}{\sqrt{2\pi t}}}
\end{aligned} Horocyclic coordinates : h h ( y ) = y ⟹ = h ( y ) ⟹ = 2 π t ℓ ( γ 0 ) ∫ R 2 sinh − K ( y ) ℓ ( γ ) /2 e − ( 1 + h 2 ( y )) ℓ ( γ ) 2 /2 t ℓ ( γ ) d y , = 2 2 π t sinh − K ℓ ( γ ) /2 e − ℓ ( γ ) 2 /2 t ℓ ( γ 0 ) ∫ R e − y 2 ℓ ( γ ) 2 /2 t 2 π t ℓ ( γ ) d y
다음에서 dim = 2 \dim = 2 dim = 2 K = − κ 2 K = -\kappa^2 K = − κ 2
det ∣ I − J γ ∣ = ( e κ ℓ ( γ ) − 1 ) ( 1 − e − κ ℓ ( γ ) ) = 2 sinh κ ℓ ( γ ) / 2 γ ( t ) = γ 0 ( k t ) ⟹ D M ∗ μ ( Ω t [ γ ] ) = e − k 2 ℓ ( γ 0 ) 2 / 2 t ℓ ( γ 0 ) 2 2 π t sinh k κ ℓ ( γ 0 ) / 2 \begin{aligned}
\det |I-J_\gamma| &= (e^{\kappa\ell(\gamma)} - 1)(1 - e^{-\kappa\ell(\gamma)}) = 2 \sinh \kappa\ell(\gamma)/2\\
\gamma(t) = \gamma_0(kt)\implies \\
DM_*\mu(\Omega_t[\gamma])&=\frac{e^{-k^2\ell(\gamma_0)^2/2t}\ell(\gamma_0)}{2\sqrt{2\pi t}\sinh k\kappa\ell(\gamma_0)/2}\\
\end{aligned} det ∣ I − J γ ∣ γ ( t ) = γ 0 ( k t ) ⟹ D M ∗ μ ( Ω t [ γ ]) = ( e κ ℓ ( γ ) − 1 ) ( 1 − e − κ ℓ ( γ ) ) = 2 sinh κ ℓ ( γ ) /2 = 2 2 π t sinh kκ ℓ ( γ 0 ) /2 e − k 2 ℓ ( γ 0 ) 2 /2 t ℓ ( γ 0 )
다음에서 dim = 3 \dim=3 dim = 3 ( z , z ˉ ) (z,\bar{z}) ( z , z ˉ )
J D M ( x ⃗ + ( τ + ℓ ( γ ) ) e ⃗ 1 ) = ( e κ ℓ ( γ ) 0 0 0 e − κ ℓ ( γ ) + i θ ( γ ) 0 0 0 e − κ ℓ ( γ ) − i θ ( γ ) ) ⟹ det I − ⊥ γ 0 k = ∣ 1 − e − k ( κ ℓ ( γ 0 ) − i θ ( γ 0 ) ) ∣ 2 \begin{aligned}
J_{DM(\vec{x}+(\tau+\ell(\gamma))\vec{e}^1)} &=
\begin{pmatrix}
e^{\kappa\ell(\gamma)} && 0 && 0\\
0 && e^{-\kappa\ell(\gamma)+i\theta(\gamma)} && 0 \\
0 && 0 && e^{-\kappa\ell(\gamma)-i\theta(\gamma)} \\
\end{pmatrix}\\
\implies& \\
\det I-{\perp_{\gamma_0}}^k &= |1-e^{-k(\kappa\ell(\gamma_0)-i\theta(\gamma_0))}|^2
\end{aligned} J D M ( x + ( τ + ℓ ( γ )) e 1 ) ⟹ det I − ⊥ γ 0 k = e κ ℓ ( γ ) 0 0 0 e − κ ℓ ( γ ) + i θ ( γ ) 0 0 0 e − κ ℓ ( γ ) − i θ ( γ ) = ∣1 − e − k ( κ ℓ ( γ 0 ) − i θ ( γ 0 )) ∣ 2
및 이후 z = x 2 + i x 3 ⟹ d z ˉ ∧ d z = ( d x 2 − i d x 3 ) ∧ ( d x 2 + i d x 3 ) = 2 i d x 2 ∧ d x 3 z=x^2+ix^3 \implies d\bar{z}\wedge dz= (dx^2-idx^3)\wedge(dx^2+idx^3) = 2idx^2\wedge dx^3 z = x 2 + i x 3 ⟹ d z ˉ ∧ d z = ( d x 2 − i d x 3 ) ∧ ( d x 2 + i d x 3 ) = 2 i d x 2 ∧ d x 3
κ = 1 ⟹ D M ∗ μ ( Ω t [ γ ] ) = e − k 2 ℓ ( γ 0 ) 2 / 2 t ℓ ( γ 0 ) 2 2 π t ( 1 − e − k ℓ ( γ 0 ) ) ∣ e k ℓ ( γ 0 ) / 2 − e − k ( ℓ ( γ 0 ) / 2 − i θ ( γ 0 ) ) ∣ \begin{aligned}
\kappa &= 1 \implies \\
DM_*\mu(\Omega_t[\gamma])
&=\frac{e^{-k^2\ell(\gamma_0)^2/2t}\ell(\gamma_0)}{2\sqrt{2\pi t (1-e^{-k\ell(\gamma_0)})}|e^{k\ell(\gamma_0)/2}-e^{-k(\ell(\gamma_0)/2-i\theta(\gamma_0))}|}\\
\end{aligned} κ D M ∗ μ ( Ω t [ γ ]) = 1 ⟹ = 2 2 π t ( 1 − e − k ℓ ( γ 0 ) ) ∣ e k ℓ ( γ 0 ) /2 − e − k ( ℓ ( γ 0 ) /2 − i θ ( γ 0 )) ∣ e − k 2 ℓ ( γ 0 ) 2 /2 t ℓ ( γ 0 )
