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I'm reading a paper on the Min-Oo Conjecture (http://arxiv.org/abs/1004.3088), and I'm stuck on the following step in a proposition:

Given a metric $g_0(t)$ on the upper hemisphere $\mathbb{S}^n_+$, and the standard metric $\bar{g}$ on the sphere $\mathbb{S}^n$ restricted to the hemisphere, we define another metric $g(t)$ on the upper hemisphere by:

$g(t) = g_0(t) + \frac{1}{2(n-1)}t^2 u \bar{g}$

In the paper they say this implies:

$R_{g(t)}=R_{g_0(t)} - \frac{1}{2} t^2 (\Delta u + nu) + O(t^3)$

I'm not sure if I should break down and calculate like mad, or if there is a better way to see this. The conditions we have on $u$ are simply

$u|_{\partial \mathbb{S}^n_+}= 0$

I appreciate all help. Cheers!

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You forgot to include the definition of $g_0(t) = \bar{g} + t \mathcal{L}_X \bar{g}$, and that the Laplacian is the one relative to the standard metric $\bar{g}$. Both facts are (I think) somewhat important. –  Willie Wong Apr 23 '14 at 15:40
You're right, thanks for pointing that out! –  Michael Pinkard Apr 24 '14 at 14:13

2 Answers 2

up vote 6 down vote accepted

Willie Wong gives the correct computational method of computing the desired formula. If you happen to believe the general formula for the derivative of the scalar curvature $R$, you can save yourself the trouble of going all the way back to the definition of curvature. This formula can be found in Besse's book "Einstein Manifolds" (p. 63) (NB: I strongly suggest that you derive this formula for yourself if you have never done it, the computation is essentially sketched in Willie Wong's answer. In particular, my answer is not "shorter/easier" unless you already have done the general computation. However, the general formula is quite well known, so perhaps you're comfortable with it already).

The formula for the derivative of scalar curvature of $g$ in the direction of a symmetric $2$-tensor $h$ reads $$ DR|_g(h) = \sum_{i,j=1}^n ((D_g)_{e_i,e_j}^2h)(e_i,e_j) - \Delta_g(tr_g h) - g(Ric_g,h) $$ From this, we see that for the $g(t)$ considered in Proposition 13 $$ g(t) = g_0(t) + t^2\left(\frac{1}{2(n-1)} u \overline g\right) := g_0(t) + t^2h, $$ if we consider the $t^2h$ term as a perturbation of $g_0(t)$ of the form $sh$, then \begin{align*} R_{g(t)} & = R_{g_0(t)} + t^2 DR|_{g_0(t)}h +O(t^4)\\ & = R_{g_0(t)} + t^2\left( \sum_{i,j=1}^n ((D_{g_0(t)})_{e_i,e_j}^2h)(e_i,e_j) - \Delta_{g_0(t)}(tr_{g_0(t)} h) - g_0(t)(Ric_{g_0(t)}, h) \right)+O(t^4)\\ & = R_{g_0(t)} + t^2\left( \sum_{i,j=1}^n ((D_{\overline g})_{e_i,e_j}^2h)(e_i,e_j) - \Delta_{\overline g}(tr_{\overline g} h) - (n-1)tr_{\overline g} h + O(t)\right)+O(t^4)\\ & = R_{g_0(t)} + t^2\left( \frac{1}{2(n-1)}\Delta_{\overline g} u - \frac{n}{2(n-1)}\Delta_{\overline g} u - \frac{n}{2(n-1)} u + O(t)\right)+O(t^4)\\ & = R_{g_0(t)} - \frac 12 t^2(\Delta_{\overline g}u +nu) +O(t^3) \end{align*} Here, I've used the fact that $\Gamma_{g_0(t)} = \Gamma_{\overline g} + O(t)$.

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Thanks! That makes complete sense! –  Michael Pinkard Apr 23 '14 at 20:16
You're welcome! –  Otis Chodosh Apr 24 '14 at 4:00
Two comments: (a) the first variation formula that you quoted only works for $DR|_{\bar{g}}$, that is, the variation near the metric of the round sphere, or Einstein manifolds with the correct $\lambda$. For general metrics $g_0(t)$ the last term should be replaced by a Ricci term. (b) Terry Tao gave a short derivation of the general variation formula for the scalar curvature (equation 15 in his blog entry). –  Willie Wong Apr 24 '14 at 7:48
@WillieWong, Whoops you're right! Its fixed now, thanks! –  Otis Chodosh Apr 24 '14 at 15:12

One needs to use that $g_0(t) = \bar{g} + t \mathcal{L}_X \bar{g}$ (a linear in $t$ perturbation of $\bar{g}$) as specified in the paper.

Write the Christoffel symbol $\bar{\Gamma}$ for those of $\bar{g}$, then the Christoffel symbol for $g(t)$ can be written as

$$ \Gamma(t) = \bar{\Gamma} + t \Gamma' + t^2 \Gamma'' + O(t^3)$$

Since the Riemann curvature is schematically

$$ \mathrm{d} \Gamma + \Gamma^2 $$

we have

$$ \mathrm{Riem} = \underbrace{\overline{\mathrm{Riem}} + t\mathrm{d}\Gamma' + t\bar{\Gamma}\Gamma' + t^2 (\Gamma')^2 }_{\mathrm{Riem} \text{ of } g_0} + t^2 (\mathrm{d} \Gamma'' + \bar{\Gamma}\Gamma'') + O(t^3) \tag{*} $$

The Ricci curvature decomposes similarly. The Scalar curvature is the metric trace of the Ricci. We can write the inverse metric

$$ g(t)^{-1} = (g_0(t))^{-1} - \frac{1}{2(n-1)}t^2 u \bar{g}^{-1} + O(t^3) \tag{**} $$


$$ (g_0(t))^{-1} = \bar{g}^{-1} + O(t) $$

From this it follows that

$$ R_{g(t)} - R_{g_0(t)} = \text{Double trace of quadratic term of *} + \text{Quadratic term of **}\cdot \overline{\mathrm{Ricci}} $$

and a simple computation gives the desired result.

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