Tag Archives: sigma2

Bach flat four dimensional manifold and sigma2 functional

We want to find the necessary condition of being the critical points of {\int\sigma_2} on four dimensional manifold.

1. Preliminary

Suppose {(M^n,g)} is a Riemannian manifold with {n=4}. {P_g} is the Schouten tensor

\displaystyle P_g=\frac{1}{n-2}\left(Ric-\frac{R}{2(n-1)}g\right)

and denote {J=\text{\,Tr\,} P_g}. Define

\displaystyle \sigma_2(g)=\frac{1}{2}[(\text{\,Tr\,} P_g)^2-|P_g|_g^2]

\displaystyle I_2(g)=\int_M \sigma_2(g)d\mu_g

where {|P|_g^2=\langle P,P\rangle_g}. It is well known that {I_2(g)} is conformally invariant.

Suppose {g(t)=g+th} where {h} is a symmetric 2-tensor. We want to calculate the first derivative of {I_2(g(t))} at {t=0}. To that end, let us list some basic facts (see the book of Toppings). Firstly denote {(\delta h)_j=-\nabla^i{h_{ij}}} the divergence operator and

\displaystyle G(h)=h-\frac{1}{2}(\text{\,Tr\,} h) g

\displaystyle (\Delta_L h)_{ij}=(\Delta h)_{ij}-h_{ik}Ric_{jl}g^{kl}-h_{jk}Ric_{il}g^{kl}+2R_{ikjl}h^{kl}

where {\Delta_L} is the Lichnerowicz Laplacian. Then the first variation of Ricci curvature and scalar curvature are

\displaystyle \dot{R}=\delta^2h-\Delta(\text{\,Tr\,} h)-\langle h,Ric\rangle \ \ \ \ \ (1)

\displaystyle \dot{Ric}=-\frac{1}{2}\Delta_Lh-\frac{1}{2}L_{(\delta G(h))^\sharp}g=-\frac{1}{2}\Delta_Lh-\frac{1}{2}L_{(\delta h)^\sharp}g-\frac{1}{2}Hess(\text{\,Tr\,} h)

\displaystyle =-\frac{1}{2}\Delta_Lh-d(\delta h)-\frac{1}{2}Hess(\text{\,Tr\,} h)

where we were using upper dot to denote the derivative with respect to {t}.

2. First variation of the sigma2 functional

Lemma 1 {(M^4,g)} is a critical point of {I_2(g)} if and only if

\displaystyle \Delta P-Hess(J)+2\mathring{Rm}(P)-2JP-|P|_g^2g=0 \ \ \ \ \ (2)

where {(\mathring{Rm}(P))_{ij}=R_{ikjl}P^{kl}}.

Proof:

\displaystyle \frac{d}{dt}\big|_{t=0} I_2(g(t))=\int_M J\dot J-\langle\dot P,P\rangle+\langle h,P\wedge P\rangle+\frac{1}{2}\sigma_2\text{\,Tr\,} h \,d\mu_g

where {(P\wedge P)_{ij}=P_{ik}P_{jl}g^{kl}}. Since we have

\displaystyle \int_M\langle P,\dot P\rangle\\ =\frac{1}{n-2}\int_M\langle P, \dot Ric-\dot Jg-Jh\rangle =\frac{1}{n-2}[\langle P, \dot Ric\rangle-\dot J J-J\langle h,P\rangle]

\displaystyle =\frac{1}{n-2}[-\frac{1}{2}\langle h,\Delta_L P\rangle+\langle h,Hess(J)\rangle-\frac{1}{2}\Delta J \text{\,Tr\,} h-\dot J J-J\langle h,P\rangle]

Plugging this into the derivative of {I_2} to get

\displaystyle (n-2)\frac{d}{dt}\big|_{t=0} I_2(g(t))\\ =\int_M\frac{1}{2}\langle h,\Delta_L P\rangle-\langle h,Hess(J)\rangle+\frac{1}{2}\Delta J \text{\,Tr\,} h\\

\displaystyle \quad +(n-1)\dot J J+J\langle h,P\rangle+(n-2)\langle h,P\wedge P\rangle+\frac{n-2}{2}\sigma_2\text{\,Tr\,} h d\mu_g

In order to simplify the above equation, we recall the definition of Lichnerowicz Laplacian {\Delta_L}

\displaystyle (\Delta_LP)_{ij}=(\Delta P)_{ij}-2P_{ik}Ric_{jl}g^{kl}+2R_{ikjl}P^{kl}

\displaystyle =(\Delta P)_{ij}-2(n-2)P_{ik}P_{jl}g^{kl}-2JP_{ij}+2R_{ikjl}P^{kl}

Apply (1) to get

(n-1)\dot J J=\frac12J\dot R=\frac12J[\delta^2h-\Delta(\text{\,Tr\,} h)-\langle h,Ric\rangle]
=\frac{1}{2}[\langle h, Hess(J)\rangle-\text{\,Tr\,} h\Delta J-(n-2)J\langle h, P\rangle-J^2\text{\,Tr\,} h]

Therefore we can simplify it to be

\displaystyle (n-2)\frac{d}{dt}\big|_{t=0} I_2(g(t)) =\int_M\frac{1}{2}\langle h,\Delta P\rangle-\frac{1}{2}\langle h, Hess(J)\rangle+h^{ij}R_{ikjl}P^{kl}

\displaystyle -\frac{n-2}{2}J\langle h,P\rangle-\frac{1}{2} J^2 \text{\,Tr\,} h+\frac{n-2}{2}\sigma_2\text{\,Tr\,} h d\mu_g

Let us denote {(\mathring{Rm}(P))_{ij}=R_{ikjl}P^{kl}}. Using the fact {n=4} and the definition of {\sigma_2},

\displaystyle \frac{d}{dt}\big|_{t=0} I_2(g(t))=\frac{1}{4}\int_M\langle h,Q\rangle d\mu_g

where

\displaystyle Q=\Delta P-Hess(J)+2\mathring{Rm}(P)-2JP-|P|_g^2g

\Box

Remark 1 It is easy to verify {\text{\,Tr\,} Q=0}, this is equivalent to say {I_2} is invariant under conformal change. More precisely, letting {h=2ug}, then

\displaystyle \frac{d}{dt}\big|_{t=0} I_2(g(t))=\frac{1}{4}\int_M\langle h,Q\rangle d\mu_g=\frac{1}{2}\int_M u\text{\,Tr\,} Q d\mu_g=0.

Remark 2 If {g} is an Einstein metric with {Ric=2(n-1)\lambda g}, then {P=\lambda g}, {J=n\lambda} and

\displaystyle \mathring{Rm}(P)=\lambda Ric=2(n-1)\lambda^2 g

It is easy to verify that {Q=0}. In other words, Einstein metrics are critical points of {I_2}.

Are there any non Einstein metric which are critical points of {I_2}?

Here is one example. Suppose {M=\mathbb{S}^2\times N}, where {\mathbb{S}^2} is the sphere with standard round metric and {(N,g_N)} is a two dimensional compact manifolds with sectional curvature {-1}. {M} is endowed with the product metric. We can prove {Ric=g_{S^2}-g_N}, {P=\frac{1}{2}g_{S^2}-\frac{1}{2}g_N}, {J=0}, {\mathring{Rm}(P)=g_{prod}} and consequently {Q=0}.

Note that the above example is a locally conformally flat manifold. For this type of manifold, we have the following lemma which can say

Lemma 2 Suppose {g} is locally conformally flat and {Q=0}, then

Proof: When {g} is locally conformally flat,

\displaystyle \mathring{Rm}(g)=JP+|P|_g^2g-2P\wedge P

{Q=0} is equivalent to

\displaystyle \Delta P-Hess(J)+|P|_g^2g-4P\wedge P=0

Actually this is equivalent to the Bach tensor {B} is zero. \Box

3. Another point of view

We have the Euler Characteristic formula for four dimensional manifolds

\displaystyle 8\pi^2\chi(M)=\int_M (|W|_g^2+\sigma_2) d\mu_g

therefore the critical points for {\int_M \sigma_2d\mu_g} will be the same as the critical points of {\int_M |W|_g^2d\mu_g}. However, the functional

\displaystyle g\rightarrow \int_M |W|_g^2d\mu_g

is well studied by Bach. The critical points of this functional satisfy Bach tensor equal to 0.

\displaystyle B_{ij}=\nabla^k\nabla^l W_{likj}+\frac{1}{2}Ric^{kl}W_{likj}

Obviously, {B=0} for Einstein metric, but not all Bach flat metrics are Einstein. For example {B=0} for any locally conformally flat manifolds.

Advertisements

Some calculations of sigma_2

On four-manifold {(M^4,g_0)}, we define Shouten tensor

\displaystyle A = Ric-\frac 16 Rg

and Einstein tensor and gravitational tensor

\displaystyle E=Ric - \frac 14 Rg\quad S=-Ric+\frac{1}{2}Rg

Suppose {\sigma_2} is the elemantary symmetric function

\displaystyle \sigma_2(\lambda)=\sum_{i<j}\lambda_i\lambda_j

Thinking of {A} as a tensor of type {(1,1)}. {\sigma_2(A)} is defined as {\sigma_2} applied to eigenvalues of {A}. Then

\displaystyle \sigma_2(A)= \frac{1}{2}[(tr_g A)^2-\langle A, A\rangle_g] \ \ \ \ \ (1)

Notice {A=E+\frac{1}{12}Rg}. Easy calculation reveals that

\displaystyle \sigma_2(A)=-\frac{1}{2}|E|^2+\frac{1}{24}R^2 \ \ \ \ \ (2)

Under conformal change of metric {g=e^{2w}g_0}, we have

\displaystyle R= e^{-2w}(R_0-6\Delta_0 w-6|\nabla_0 w|^2) \ \ \ \ \ (3)

\displaystyle A=A_0-2\nabla^2_0 w+2dw\otimes dw-|\nabla_0w|^2g_0 \ \ \ \ \ (4)

\displaystyle S=S_0+2\nabla_0^2w-2\Delta_0wg_0-2dw\otimes dw-|\nabla_0 w|^2g_0 \ \ \ \ \ (5)

We want to solve the equation {\sigma_2(A)=f>0}, which is equivalent to solve

\displaystyle \sigma_2(A_0-2\nabla^2_0w+2dw\otimes dw-|\nabla_0w|^2g_0)=f

This is an fully nonlinear equation of Monge-Ampere type. Under local coordinates, the above equation can be treated as

\displaystyle F(\partial_i\partial_j w,\partial_kw,w,x)=f

where {F(p_{ij},v_k,s,x):\mathbb{R}^{n\times n}\times\mathbb{R}^n\times\mathbb{R}\times\mathbb{R}^n\rightarrow \mathbb{R}}. This equation is elliptic if the matrix {\left(\frac{\partial F}{\partial p_{ij}}\right)} is positive definite. In order to find that matrix, we need the linearized operator

\displaystyle L[\phi]=\frac{\partial F}{\partial p_{ij}}(\nabla_0^2\phi)_{ij}=\frac{d}{dt}|_{t=0}F(\partial_i\partial_j w+t\partial_i\partial_j\phi,\partial_kw,w,x) \ \ \ \ \ (6)

Using the elementary identity

\displaystyle \frac{d}{dt}\rvert_{t=0}\sigma_2(H+tG)=tr_gH\cdot tr_gG-\langle H, G\rangle_g. \ \ \ \ \ (7)

for any fixed matrix {H} and {G}. Now plug in {H=A} is Schouten tensor and {B=-2\nabla_0^2\phi}. One can calculate them as

\displaystyle tr_g H\cdot tr_g G=\langle \frac{1}{3}Rg, G\rangle_g \ \ \ \ \ (8)

Then we get

\displaystyle L[\phi]=\langle S,G\rangle_g=-2\langle S,\nabla^2_0\phi\rangle_g