3.The Einstein-Hilbert functional The gradient flow for E# If ∂ ∂ s g i j = v i j \frac{\partial}{\partial s} g_{i j}=v_{i j} ∂ s ∂ g ij = v ij
∂ ∂ s d μ = 1 2 V d μ \begin{equation*} \frac{\partial}{\partial s} d \mu=\frac{1}{2} V d \mu \tag{2.11} \end{equation*} ∂ s ∂ d μ = 2 1 V d μ ( 2.11 ) Proof:
Ex1 :(Variation of the inverse of g g g g i j g j k = δ k i g^{i j} g_{j k}=\delta_{k}^{i} g ij g jk = δ k i
∂ ∂ s g i j = − g i k g j ℓ ∂ ∂ s g k ℓ \begin{equation*} \frac{\partial}{\partial s} g^{i j}=-g^{i k} g^{j \ell} \frac{\partial}{\partial s} g_{k \ell} \tag{2.14} \end{equation*} ∂ s ∂ g ij = − g ik g j ℓ ∂ s ∂ g k ℓ ( 2.14 ) Proof:
If ∂ ∂ s g i j = v i j \frac{\partial}{\partial s} g_{i j}=v_{i j} ∂ s ∂ g ij = v ij
d d s E = ∫ M ( − Δ V + ∇ p ∇ q v p q − ⟨ v , R c ⟩ + 1 2 R V ) d μ = ∫ M ⟨ v , 1 2 R g − R c ⟩ d μ \begin{align*} \frac{d}{d s} E & =\int_{M}\left(-\Delta V+\nabla_{p} \nabla_{q} v_{p q}-\langle v, \mathrm{Rc}\rangle+\frac{1}{2} R V\right) d \mu \\ & =\int_{M}\left\langle v, \frac{1}{2} R g-\mathrm{Rc}\right\rangle d \mu \tag{2.15} \end{align*} d s d E = ∫ M ( − Δ V + ∇ p ∇ q v pq − ⟨ v , Rc ⟩ + 2 1 R V ) d μ = ∫ M ⟨ v , 2 1 R g − Rc ⟩ d μ ( 2.15 ) where
E ( g ) ≑ ∫ M R d μ E(g) \doteqdot \int_{M} R d \mu E ( g ) ≑ ∫ M R d μ Proof:
Remark:Note that (twice) the gradient flow of E E E
∂ ∂ t g i j = 2 ( ∇ E ( g ) ) i j = R g i j − 2 R i j \begin{equation*} \frac{\partial}{\partial t} g_{i j}=2(\nabla E(g))_{i j}=R g_{i j}-2 R_{i j} \tag{2.16} \end{equation*} ∂ t ∂ g ij = 2 ( ∇ E ( g ) ) ij = R g ij − 2 R ij ( 2.16 ) Ex2 :. Given a metric g 0 g_{0} g 0
C ≑ { u g 0 : u > 0 and Vol ( u g 0 ) = 1 } \mathcal{C} \doteqdot\left\{u g_{0}: u>0 \text { and } \operatorname{Vol}\left(u g_{0}\right)=1\right\} C ≑ { u g 0 : u > 0 and Vol ( u g 0 ) = 1 } be the space of unit volume metrics conformal to g 0 g_{0} g 0 C \mathcal{C} C E E E
Proof:
Ex3 . Show that when n ≥ 3 n \geq 3 n ≥ 3 E E E [ g 0 ] ≑ { g : g = e u g 0 \left[g_{0}\right] \doteqdot\left\{g: g=e^{u} g_{0}\right. [ g 0 ] ≑ { g : g = e u g 0 u ∈ C ∞ ( M ) } \left.u \in C^{\infty}(M)\right\} u ∈ C ∞ ( M ) } ∂ ∂ t g = − R g \frac{\partial}{\partial t} g=-R g ∂ t ∂ g = − R g
Proof:
In other words, we consider the more general class of flows
∂ ∂ t g i j = − 2 ( R i j + ∇ i ∇ j f ) \frac{\partial}{\partial t} g_{i j}=-2\left(R_{i j}+\nabla_{i} \nabla_{j} f\right) ∂ t ∂ g ij = − 2 ( R ij + ∇ i ∇ j f ) where f f f ∇ i ∇ j f = ( L ∇ f g ) i j \nabla_{i} \nabla_{j} f=\left(\mathcal{L}_{\nabla f} g\right)_{i j} ∇ i ∇ j f = ( L ∇ f g ) ij
Ex4. Define a 1-parameter family of diffeomorphisms Ψ ( t ) \Psi(t) Ψ ( t ) M → M M \rightarrow M M → M
d d t Ψ ( t ) = ∇ g ( t ) f ( t ) Ψ ( 0 ) = id M \begin{aligned} \frac{d}{d t} \Psi(t) & =\nabla_{g(t)} f(t) \\ \Psi(0) & =\operatorname{id}_{M} \end{aligned} d t d Ψ ( t ) Ψ ( 0 ) = ∇ g ( t ) f ( t ) = id M Show that g ˉ ( t ) ≑ Ψ ( t ) ∗ g ( t ) \bar{g}(t) \doteqdot \Psi(t)^{*} g(t) g ˉ ( t ) ≑ Ψ ( t ) ∗ g ( t ) f ˉ ( t ) ≑ f ∘ Ψ ( t ) \bar{f}(t) \doteqdot f \circ \Psi(t) f ˉ ( t ) ≑ f ∘ Ψ ( t )
∂ g ˉ i j ∂ t = − 2 R ˉ i j ∂ f ˉ ∂ t = − Δ ˉ f ˉ + ∣ ∇ ˉ f ˉ ∣ 2 − R ˉ \begin{align*} \frac{\partial \bar{g}_{i j}}{\partial t} & =-2 \bar{R}_{i j} \tag{2.17}\\ \frac{\partial \bar{f}}{\partial t} & =-\bar{\Delta} \bar{f}+|\bar{\nabla} \bar{f}|^{2}-\bar{R} \tag{2.18} \end{align*} ∂ t ∂ g ˉ ij ∂ t ∂ f ˉ = − 2 R ˉ ij = − Δ ˉ f ˉ + ∣ ∇ ˉ f ˉ ∣ 2 − R ˉ ( 2.17 ) ( 2.18 ) Proof:
the gradient flow for F \mathcal{F} F # If we define
F ( g , f ) ≑ ∫ M ( R + ∣ ∇ f ∣ 2 ) e − f d μ \begin{equation*} \mathcal{F}(g, f) \doteqdot \int_{M}\left(R+|\nabla f|^{2}\right) e^{-f} d \mu \tag{2.21} \end{equation*} F ( g , f ) ≑ ∫ M ( R + ∣∇ f ∣ 2 ) e − f d μ ( 2.21 ) then the gradient flow for F \mathcal{F} F e − f d μ e^{-f} d \mu e − f d μ
∂ ∂ t g i j = − 2 ( R i j + ∇ i ∇ j f ) ∂ ∂ t f = − R − Δ f \begin{aligned} \frac{\partial}{\partial t} g_{i j} & =-2\left(R_{i j}+\nabla_{i} \nabla_{j} f\right) \\ \frac{\partial}{\partial t} f & =-R-\Delta f \end{aligned} ∂ t ∂ g ij ∂ t ∂ f = − 2 ( R ij + ∇ i ∇ j f ) = − R − Δ f Proof:
∂ ∂ s ( e − f d μ ) = 0 \frac{\partial}{\partial s}\left(e^{-f} d \mu\right)=0 ∂ s ∂ ( e − f d μ ) = 0 E ( g ) ≑ ∫ M R e − f d μ \mathcal{E}(g) \doteqdot \int_M R e^{-f} d \mu E ( g ) ≑ ∫ M R e − f d μ ∫ M ∣ ∇ f ∣ 2 e − f d μ = 4 ∫ M ∣ ∇ e − f / 2 ∣ 2 d μ \int_M|\nabla f|^2 e^{-f} d \mu=4 \int_M\left|\nabla e^{-f / 2}\right|^2 d \mu ∫ M ∣∇ f ∣ 2 e − f d μ = 4 ∫ M ∇ e − f /2 2 d μ
d d s E = ∫ M ∂ R ∂ s e − f d μ = − ∫ M ⟨ v , R c ⟩ e − f d μ + ∫ M ( − Δ V + ∇ i ∇ j v i j ) e − f d μ \frac{d}{d s} \mathcal{E} =\int_{M} \frac{\partial R}{\partial s} e^{-f} d \mu =-\int_{M}\langle v, \mathrm{Rc}\rangle e^{-f} d \mu+\int_{M}\left(-\Delta V+\nabla_{i} \nabla_{j} v_{i j}\right) e^{-f} d \mu d s d E = ∫ M ∂ s ∂ R e − f d μ = − ∫ M ⟨ v , Rc ⟩ e − f d μ + ∫ M ( − Δ V + ∇ i ∇ j v ij ) e − f d μ
∫ M ( ∂ ∂ s ∣ ∇ f ∣ 2 ) e − f d μ = ∫ M ( − v i j ∇ i f ∇ j f + 2 ∇ f ⋅ ∇ ( ∂ f ∂ s ) ) e − f d μ \int_{M}\left(\frac{\partial}{\partial s}|\nabla f|^{2}\right) e^{-f} d \mu =\int_{M}\left(-v_{i j} \nabla_{i} f \nabla_{j} f+2 \nabla f \cdot \nabla\left(\frac{\partial f}{\partial s}\right)\right) e^{-f} d \mu ∫ M ( ∂ s ∂ ∣∇ f ∣ 2 ) e − f d μ = ∫ M ( − v ij ∇ i f ∇ j f + 2∇ f ⋅ ∇ ( ∂ s ∂ f ) ) e − f d μ
Ex 6. Show that
∫ M ∣ ∇ f ∣ 2 e − f d μ = ∫ M ( Δ f ) e − f d μ \int_{M}|\nabla f|^{2} e^{-f} d \mu=\int_{M}(\Delta f) e^{-f} d \mu ∫ M ∣∇ f ∣ 2 e − f d μ = ∫ M ( Δ f ) e − f d μ This shows that we may rewrite F \mathcal{F} F
F ( g , f ) = ∫ M ( R + 2 Δ f − ∣ ∇ f ∣ 2 ) e − f d μ \mathcal{F}(g, f)=\int_{M}\left(R+2 \Delta f-|\nabla f|^{2}\right) e^{-f} d \mu F ( g , f ) = ∫ M ( R + 2Δ f − ∣∇ f ∣ 2 ) e − f d μ Motivations for doing this are given by (1.84) and (5.17).
