First Variation of Jacobi Operator

[darida]

<Moderator's note: Moved from a homework forum.>

Homework Statement

From this paper.

Let ##L## be the Jacobian operator of a two-sided compact surface embedded in a three-maniold ##(M,g)##, ##\Sigma \subset M##, and defined by

$$L(t)=\Delta_{\Sigma(t)}+ \text{Ric}( ν_{t} , ν_{t} )+|A_{\Sigma(t)}|^2$$

where
##\Delta## is the Laplace Beltrami operator on ##\Sigma##
##\text{Ric}## is the Ricci tensor of three-manifold ##M##
##|A_{\Sigma(t)}|^2## is the traceless part of the second fundamental form ##A_{ij}## of ##\Sigma## in ##M##

$$|A_{\Sigma(t)}|^2=A^{ij}A_{ij}$$

Now, for each function ##\psi \subset C^∞(\Sigma)##, the first variation of ##L## could be calculated as (Proposition 6.2 page 16):

$$L'(0)\psi = 2 \phi \left< A, \text{Hess} \psi \right> + 2\psi \left< A, \text{Hess} \phi \right> - 2 \phi \omega (\nabla \psi) - 2 \psi \omega (\nabla \phi) $$
$$ + \phi \left< \nabla H, \nabla \psi \right> - H \left< \nabla \phi , \nabla \psi \right> + 2A (\nabla \phi , \nabla \psi) - \psi \text{div}_{\Sigma} ( \text{div}_{\Sigma} \omega ) $$
$$ +\phi \psi R_{iννj} A_{ij} + \phi \psi H \text{Ric}(ν,ν) + \phi \psi H |A|^2 + \phi \psi A_{ij} A_{ik} A_{jk} - \phi \psi H K_{\Sigma}$$

where
##\omega## is the 1-form on ##\Sigma## defined by ##\omega(X)=\text{Ric}(X,ν)##
##K_{\Sigma}## the Gauss curvature of ##\Sigma##

Homework Equations

Page 17-18.

If we consider the particular case where ##(M,g)## is the deSitter-Schwarzschild manifold ##(\mathbb R \times \mathbb S^2, g_a)## and ##\Sigma \subset M## is some slice ##{r} \times \mathbb S^2##. In this case, we have

##R## is constant equal to 2;
##\Sigma## is totally umbilic and has constant Gauss curvature, ##K_{\Sigma}=\frac{4\pi}{| \Sigma |}##;
##\omega = 0##.

Therefore, the first variation of ##L## becomes (page 18):

$$L'(0)\phi = 2 H\phi \Delta \phi + \frac{3}{2} H \left( \text{Ric}(ν,ν) + \frac{H^2}{2} \right) \phi^2 - \frac{4\pi}{| \Sigma|} H \phi^2$$

where ##|A|^2 = \frac{H^2}{2}##, since ##\Sigma## is umbilic.

How to prove this ##L'(0)\phi##?
How come the other terms vanish?

The Attempt at a Solution

I have tried, but this is what I get

$$L'(0)\psi = 4 \phi \left< A, \text{Hess} \phi \right> + \phi \left< \nabla H, \nabla \phi \right> - H (\nabla \phi)^2 + 2A (\nabla \phi)^2 $$
$$ +\phi^2 R_{iννj} A_{ij} + \phi^2 H \left( \text{Ric}(ν,ν) + \frac{H^2}{2} \right) + \phi^2 A_{ij} A_{ik} A_{jk} - \frac{4\pi}{| \Sigma|} H \phi^2$$

Could someone please give me the detail steps to get ##L'(0)\phi## (page 18)? Thank you.

 

[jvicens]

Thank you for your post. To prove the first variation of ##L##, we need to use the definition of the Jacobian operator and the properties of the Laplace Beltrami operator. The key idea is to use the first variation formula for the Laplace Beltrami operator, which states that

$$\Delta_{\Sigma(t)} \psi = \Delta_{\Sigma(0)} \psi + t \left< \nabla \psi , \nabla H \right> + t^2 \psi R_{iννj} A_{ij} + t^2 \psi H \text{Ric}(ν,ν) + t^2 \psi H |A|^2 + t^2 \psi A_{ij} A_{ik} A_{jk} - t^2 \psi H K_{\Sigma}$$

where ##\psi \in C^∞(\Sigma)## and ##\Sigma(t)## is a one-parameter family of surfaces with ##\Sigma(0)=\Sigma##. This formula can be found in many textbooks on differential geometry, for example, in John Lee's "Riemannian Manifolds: An Introduction to Curvature".

Using this formula and the definition of ##L##, we have

$$L(t) \psi = \Delta_{\Sigma(t)} \psi + \text{Ric}( ν_{t} , ν_{t} )+|A_{\Sigma(t)}|^2$$
$$ = \Delta_{\Sigma(0)} \psi + t \left< \nabla \psi , \nabla H \right> + t^2 \psi R_{iννj} A_{ij} + t^2 \psi H \text{Ric}(ν,ν) + t^2 \psi H |A|^2 + t^2 \psi A_{ij} A_{ik} A_{jk} - t^2 \psi H K_{\Sigma} + \text{Ric}( ν_{t} , ν_{t} )+|A_{\Sigma(t)}|^2$$

Taking the derivative with respect to ##t## and evaluating at ##t=0##, we have

$$L'(0) \psi = \left. \frac{d}{dt} L(t) \psi \right|