## Tag Archives: sigma2

### Some calculation of sigma2 II

We want find the critical metric of the following functional restricted to the space of conformal metrics of unit volume

$\displaystyle I(g)=\frac{n-4}{2}t\int_M J^2_gd\mu_g+4\int_M\sigma_2(A_g)d\mu_g$

Here ${(M^n,g)}$ is a Riemannian metric with ${n\geq 5}$, ${A}$ is the Schouten tensor and ${J}$ is ${tr_gA}$.

Now let us differentiate the two terms respectively, suppose ${g(s)=(1+s\psi)^{\frac{4}{n-4}}g=u(s)^{\frac{4}{n-2}}g}$. From conformal change property of scalar curvature, we get

$\displaystyle J_{g(s)}=u(s)^{-\frac{n+2}{n-2}}\left(-\frac{2}{n-2}\Delta_{g}u(s)+Ju(s)\right)$

$\displaystyle \dot J=-\frac{2}{n-2}\Delta_g\dot u-\frac{4}{n-2}J\dot u$

Then

$\displaystyle \frac{d}{ds}|_{t=0}\int_{M}J_{g(s)}^2d\mu_{g(s)}=\int_{M}2J\dot{J}+\frac{2n}{n-2}J^2\dot ud\mu$

$\displaystyle =\int_M -\frac{4}{n-2}J\Delta \dot u+\frac{2(n-4)}{n-2}J^2\dot ud\mu$

$\displaystyle =\int_M\left(-\frac{4}{n-4}\Delta J+2J^2\right)\psi d\mu$

where we have used the fact ${\dot u(0)=\frac{n-2}{n-4}\psi}$. Next, it is easy to know

$\displaystyle \frac{d}{ds}|_{t=0}\int_M \sigma_2(A_{g(s)})d\mu_{g(s)}=2\int_{M}\sigma_2(A)\psi d\mu$So

$\displaystyle I'(g)\psi=2t\int_{M}\left(-\Delta J+\frac{n-4}{2}J\right)\psi d\mu_g+8\int_{M}\sigma_2(A)\psi d\mu_g$

Now assume ${g(s)}$ keep the unit volume infinitesimally, that is ${\int_M \psi d\mu=0}$, then ${I'(g)=0}$ under this constraint means

$\displaystyle t\left(-\Delta J+\frac{n-4}{2}J\right)+4\sigma_2(A)=const.$

Remark: M.J. Gursky, F. Hang, Y.-J. Lin, Riemannian Manifolds with Positive Yamabe Invariant and Paneitz Operator, Int. Math. Res. Not.

### 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.

### 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

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

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