4.Evolution of geometric quantities Variation of the Christoffel symbols# Lemma 1 (Variation of Christoffel symbols). If g ( s ) g(s) g ( s ) ∂ ∂ s g i j = v i j \frac{\partial}{\partial s} g_{i j}=v_{i j} ∂ s ∂ g ij = v ij
∂ ∂ s Γ i j k = 1 2 g k ℓ ( ∇ i v j ℓ + ∇ j v i ℓ − ∇ ℓ v i j ) \begin{equation*} \frac{\partial}{\partial s} \Gamma_{i j}^{k}=\frac{1}{2} g^{k \ell}\left(\nabla_{i} v_{j \ell}+\nabla_{j} v_{i \ell}-\nabla_{\ell} v_{i j}\right) \tag{2.23} \end{equation*} ∂ s ∂ Γ ij k = 2 1 g k ℓ ( ∇ i v j ℓ + ∇ j v i ℓ − ∇ ℓ v ij ) ( 2.23 ) Proof:
**Remark:**In coordinate-free notation, (2.23) is
⟨ ( ∂ ∂ s ∇ ) ( X , Y ) , Z ⟩ = 1 2 ( ( ∇ X v ) ( Y , Z ) + ( ∇ Y v ) ( X , Z ) − ( ∇ Z v ) ( X , Y ) ) \begin{equation*} \left\langle\left(\frac{\partial}{\partial s} \nabla\right)(X, Y), Z\right\rangle=\frac{1}{2}\left(\left(\nabla_{X} v\right)(Y, Z)+\left(\nabla_{Y} v\right)(X, Z)-\left(\nabla_{Z} v\right)(X, Y)\right) \tag{2.24} \end{equation*} ⟨ ( ∂ s ∂ ∇ ) ( X , Y ) , Z ⟩ = 2 1 ( ( ∇ X v ) ( Y , Z ) + ( ∇ Y v ) ( X , Z ) − ( ∇ Z v ) ( X , Y ) ) ( 2.24 ) This formula may be derived directly from differentiating (1.4).
Corollary: (Evolution of Christoffel symbols under RF). Under the Ricci flow ∂ ∂ t g i j = − 2 R i j \frac{\partial}{\partial t} g_{i j}=-2 R_{i j} ∂ t ∂ g ij = − 2 R ij
∂ ∂ t Γ i j k = − g k ℓ ( ∇ i R j ℓ + ∇ j R i ℓ − ∇ ℓ R i j ) \begin{equation*} \frac{\partial}{\partial t} \Gamma_{i j}^{k}=-g^{k \ell}\left(\nabla_{i} R_{j \ell}+\nabla_{j} R_{i \ell}-\nabla_{\ell} R_{i j}\right) \tag{2.25} \end{equation*} ∂ t ∂ Γ ij k = − g k ℓ ( ∇ i R j ℓ + ∇ j R i ℓ − ∇ ℓ R ij ) ( 2.25 ) Evolution of Laplacian under RF# Lemma: (Evolution of Laplacian under RF). If ( M n , g ( t ) ) \left(M^{n}, g(t)\right) ( M n , g ( t ) )
∂ ∂ t ( Δ g ( t ) ) = 2 R i j ⋅ ∇ i ∇ j \frac{\partial}{\partial t}\left(\Delta_{g(t)}\right)=2 R_{i j} \cdot \nabla_{i} \nabla_{j} ∂ t ∂ ( Δ g ( t ) ) = 2 R ij ⋅ ∇ i ∇ j where Δ g ( t ) \Delta_{g(t)} Δ g ( t ) n = 2 n=2 n = 2 ∂ ∂ t ( Δ ) = R Δ \frac{\partial}{\partial t}(\Delta)=R \Delta ∂ t ∂ ( Δ ) = R Δ
Proof:
**Ex1:**Given ∂ ∂ s g i j = v i j \frac{\partial}{\partial s} g_{i j}=v_{i j} ∂ s ∂ g ij = v ij ∂ ∂ s ( Δ g ( s ) ) \frac{\partial}{\partial s}\left(\Delta_{g(s)}\right) ∂ s ∂ ( Δ g ( s ) )
Proof:
if ∂ ∂ s g i j = v i j \frac{\partial}{\partial s} g_{i j}=v_{i j} ∂ s ∂ g ij = v ij ∂ ∂ s R i j = 1 2 ∇ ℓ ( ∇ i v j ℓ + ∇ j v i ℓ − ∇ ℓ v i j ) − 1 2 ∇ i ∇ j V \begin{equation*} \frac{\partial}{\partial s} R_{i j}=\frac{1}{2} \nabla_{\ell}\left(\nabla_{i} v_{j \ell}+\nabla_{j} v_{i \ell}-\nabla_{\ell} v_{i j}\right)-\frac{1}{2} \nabla_{i} \nabla_{j} V \tag{2.29} \end{equation*} ∂ s ∂ R ij = 2 1 ∇ ℓ ( ∇ i v j ℓ + ∇ j v i ℓ − ∇ ℓ v ij ) − 2 1 ∇ i ∇ j V ( 2.29 ) Commuting derivatives in (2.29) yields the variation of Ricci formula: ∂ ∂ s R i j = − 1 2 ( Δ L v i j + ∇ i ∇ j V − ∇ i ( div v ) j − ∇ j ( div v ) i ) \begin{equation*} \frac{\partial}{\partial s} R_{i j}=-\frac{1}{2}\left(\Delta_{L} v_{i j}+\nabla_{i} \nabla_{j} V-\nabla_{i}(\operatorname{div} v)_{j}-\nabla_{j}(\operatorname{div} v)_{i}\right) \tag{2.31} \end{equation*} ∂ s ∂ R ij = − 2 1 ( Δ L v ij + ∇ i ∇ j V − ∇ i ( div v ) j − ∇ j ( div v ) i ) ( 2.31 ) Here Δ L \Delta_{L} Δ L
Δ L v i j ≑ Δ v i j + 2 R k i j ℓ v k ℓ − R i k v j k − R j k v i k \begin{equation*} \Delta_{L} v_{i j} \doteqdot \Delta v_{i j}+2 R_{k i j \ell} v_{k \ell}-R_{i k} v_{j k}-R_{j k} v_{i k} \tag{2.32} \end{equation*} Δ L v ij ≑ Δ v ij + 2 R kij ℓ v k ℓ − R ik v jk − R jk v ik ( 2.32 ) Note that (2.31) may be rewritten as
∂ ∂ s ( − 2 R i j ) = Δ L v i j + ∇ i X j + ∇ j X i \frac{\partial}{\partial s}\left(-2 R_{i j}\right)=\Delta_{L} v_{i j}+\nabla_{i} X_{j}+\nabla_{j} X_{i} ∂ s ∂ ( − 2 R ij ) = Δ L v ij + ∇ i X j + ∇ j X i where X = 1 2 ∇ V − div v X=\frac{1}{2} \nabla V-\operatorname{div} v X = 2 1 ∇ V − div v
Remark: By (1.50) we see that the Hodge Laplacian acts on 2-forms formally in the same way that the Lichnerowicz Laplacian acts on symmetric 2 -tensors.
Commutator formula for Hessian and Lichnerowicz heat operator# Lemma: (Commutator formula for the Hessian and the Lichnerowicz heat operator). Under the Ricci flow, the Hessian and the Lichnerowicz Laplacian heat operator commute. That is, for any function f f f
∇ i ∇ j ( ∂ f ∂ t − Δ f ) = ( ∂ ∂ t − Δ L ) ∇ i ∇ j f \begin{equation*} \nabla_{i} \nabla_{j}\left(\frac{\partial f}{\partial t}-\Delta f\right)=\left(\frac{\partial}{\partial t}-\Delta_{L}\right) \nabla_{i} \nabla_{j} f \tag{2.33} \end{equation*} ∇ i ∇ j ( ∂ t ∂ f − Δ f ) = ( ∂ t ∂ − Δ L ) ∇ i ∇ j f ( 2.33 ) Proof:
In other words,
[ ∇ i ∇ j , ∂ ∂ t − Δ L ] = 0 \left[\nabla_{i} \nabla_{j}, \frac{\partial}{\partial t}-\Delta_{L}\right]=0 [ ∇ i ∇ j , ∂ t ∂ − Δ L ] = 0 where Δ L ≑ Δ \Delta_{L} \doteqdot \Delta Δ L ≑ Δ
**Corollary:**If g ( t ) g(t) g ( t ) f ( t ) f(t) f ( t ) ∂ f ∂ t = Δ f \frac{\partial f}{\partial t}=\Delta f ∂ t ∂ f = Δ f
∂ ∂ t ( ∇ ∇ f ) = Δ L ( ∇ ∇ f ) \begin{equation*} \frac{\partial}{\partial t}(\nabla \nabla f)=\Delta_{L}(\nabla \nabla f) \tag{2.36} \end{equation*} ∂ t ∂ ( ∇∇ f ) = Δ L ( ∇∇ f ) ( 2.36 ) Ex2. (Commutator of ∂ ∂ t + Δ L \frac{\partial}{\partial t}+\Delta_{L} ∂ t ∂ + Δ L ∇ ∇ \nabla \nabla ∇∇ ∇ i ∇ j ( ∂ f ∂ t + Δ f ) = ( ∂ ∂ t + Δ L ) ∇ i ∇ j f − 2 ( ∇ i R j ℓ + ∇ j R i ℓ − ∇ ℓ R i j ) ∇ ℓ f \nabla_{i} \nabla_{j}\left(\frac{\partial f}{\partial t}+\Delta f\right)=\left(\frac{\partial}{\partial t}+\Delta_{L}\right) \nabla_{i} \nabla_{j} f-2\left(\nabla_{i} R_{j \ell}+\nabla_{j} R_{i \ell}-\nabla_{\ell} R_{i j}\right) \nabla_{\ell} f ∇ i ∇ j ( ∂ t ∂ f + Δ f ) = ( ∂ t ∂ + Δ L ) ∇ i ∇ j f − 2 ( ∇ i R j ℓ + ∇ j R i ℓ − ∇ ℓ R ij ) ∇ ℓ f
Proof:
**Ex3.**Show that under the Ricci flow, for any 1-form X X X
( ∂ ∂ t − Δ L ) ( L X ♮ g ) = L [ ( ∂ ∂ t − Δ d ) X ] ♮ g \left(\frac{\partial}{\partial t}-\Delta_{L}\right)\left(\mathcal{L}_{X^{\natural}} g\right)=\mathcal{L}_{\left[\left(\frac{\partial}{\partial t}-\Delta_{d}\right) X\right]^{\natural} g} ( ∂ t ∂ − Δ L ) ( L X ♮ g ) = L [ ( ∂ t ∂ − Δ d ) X ] ♮ g that is,
( ∂ ∂ t − Δ L ) ( ∇ i X j + ∇ j X i ) = ∇ i Y j + ∇ j Y i \begin{equation*} \left(\frac{\partial}{\partial t}-\Delta_{L}\right)\left(\nabla_{i} X_{j}+\nabla_{j} X_{i}\right)=\nabla_{i} Y_{j}+\nabla_{j} Y_{i} \tag{2.38} \end{equation*} ( ∂ t ∂ − Δ L ) ( ∇ i X j + ∇ j X i ) = ∇ i Y j + ∇ j Y i ( 2.38 ) where Y ≑ ( ∂ ∂ t − Δ d ) X Y \doteqdot\left(\frac{\partial}{\partial t}-\Delta_{d}\right) X Y ≑ ( ∂ t ∂ − Δ d ) X
Lemma: If ( M n , g ( t ) ) \left(M^{n}, g(t)\right) ( M n , g ( t ) ) X X X
∂ ∂ t X i = Δ X i + R k i X k \begin{equation*} \frac{\partial}{\partial t} X^{i}=\Delta X^{i}+R_{k}^{i} X^{k} \tag{2.39} \end{equation*} ∂ t ∂ X i = Δ X i + R k i X k ( 2.39 ) then h i j ≑ ∇ i X j + ∇ j X i = ( L X g ) i j h_{i j} \doteqdot \nabla_{i} X_{j}+\nabla_{j} X_{i}=\left(\mathcal{L}_{X} g\right)_{i j} h ij ≑ ∇ i X j + ∇ j X i = ( L X g ) ij
∂ ∂ t h i j = Δ L h i j ≑ Δ h i j + 2 R k i j ℓ h k ℓ − R i k h k j − R j k h i k \begin{equation*} \frac{\partial}{\partial t} h_{i j}=\Delta_{L} h_{i j} \doteqdot \Delta h_{i j}+2 R_{k i j \ell} h_{k \ell}-R_{i k} h_{k j}-R_{j k} h_{i k} \tag{2.40} \end{equation*} ∂ t ∂ h ij = Δ L h ij ≑ Δ h ij + 2 R kij ℓ h k ℓ − R ik h kj − R jk h ik ( 2.40 ) Proof:
**Remark:**A special case of (2.38) is formula (2.33). We may see this as follows. Since ( ∂ ∂ t − Δ d ) d f = d ( ∂ ∂ t − Δ ) f \left(\frac{\partial}{\partial t}-\Delta_{d}\right) d f=d\left(\frac{\partial}{\partial t}-\Delta\right) f ( ∂ t ∂ − Δ d ) df = d ( ∂ t ∂ − Δ ) f X = d f X=d f X = df
( ∂ ∂ t − Δ L ) ( 2 ∇ i ∇ j f ) = ∇ i ( ( ∂ ∂ t − Δ d ) ( d f ) ) j + ∇ j ( ( ∂ ∂ t − Δ d ) ( d f ) ) i = 2 ∇ i ∇ j ( ( ∂ ∂ t − Δ ) f ) \begin{aligned} \left(\frac{\partial}{\partial t}-\Delta_{L}\right)\left(2 \nabla_{i} \nabla_{j} f\right) & =\nabla_{i}\left(\left(\frac{\partial}{\partial t}-\Delta_{d}\right)(d f)\right)_{j}+\nabla_{j}\left(\left(\frac{\partial}{\partial t}-\Delta_{d}\right)(d f)\right)_{i} \\ & =2 \nabla_{i} \nabla_{j}\left(\left(\frac{\partial}{\partial t}-\Delta\right) f\right) \end{aligned} ( ∂ t ∂ − Δ L ) ( 2 ∇ i ∇ j f ) = ∇ i ( ( ∂ t ∂ − Δ d ) ( df ) ) j + ∇ j ( ( ∂ t ∂ − Δ d ) ( df ) ) i = 2 ∇ i ∇ j ( ( ∂ t ∂ − Δ ) f ) which is (2.33).
Ex4 Show that if X X X
∇ k ∇ j X i + R ℓ k j i X ℓ = 0 \begin{equation*} \nabla_{k} \nabla_{j} X_{i}+R_{\ell k j i} X_{\ell}=0 \tag{2.41} \end{equation*} ∇ k ∇ j X i + R ℓ kji X ℓ = 0 ( 2.41 ) Tracing (2.41), we have
Δ X i + R ℓ i X ℓ = 0 \begin{equation*} \Delta X_{i}+R_{\ell i} X_{\ell}=0 \tag{2.42} \end{equation*} Δ X i + R ℓ i X ℓ = 0 ( 2.42 ) Hence, if M n M^{n} M n
Proof:
Lemma (Evolution of the Ricci tensor under RF). Under the Ricci flow,
∂ ∂ t R i j = Δ L R i j = Δ R i j + 2 R k i j ℓ R k ℓ − 2 R i k R j k \begin{equation*} \frac{\partial}{\partial t} R_{i j}=\Delta_{L} R_{i j}=\Delta R_{i j}+2 R_{k i j \ell} R_{k \ell}-2 R_{i k} R_{j k} \tag{2.43} \end{equation*} ∂ t ∂ R ij = Δ L R ij = Δ R ij + 2 R kij ℓ R k ℓ − 2 R ik R jk ( 2.43 ) Proof:
Ex6 . Calculate the evolution equation for R i j − α R g i j R_{i j}-\alpha R g_{i j} R ij − α R g ij α ∈ R \alpha \in \mathbb{R} α ∈ R
Proof:
