The Gursky–Streets equations

Abstract

Gursky–Streets introduced a formal Riemannian metric on the space of conformal metrics in a fixed conformal class of a compact Riemannian four-manifold in the context of the $\sigma _2$-Yamabe problem. The geodesic equation of Gursky–Streets’ metric is a fully nonlinear degenerate elliptic equation. Using this geometric structure and the geodesic equation, Gursky–Streets proved an important result in conformal geometry, that the solution of the $\sigma _2$-Yamabe problem is unique (the existence of such a solution was known more than a decade ago). A key ingredient is the convexity of Chang–Yang’s ${{\mathcal {F}}}$-functional along the (smooth) geodesic. However Gursky–Streets have only proved uniform $C^{0, 1}$ regularity for a perturbed equation and it turns out that the uniform $C^{1, 1}$ regularity is very delicate. Without such a uniform $C^{1, 1}$ regularity, Gursky–Streets arguments of the uniqueness theorem are very complicated. In this paper we establish the uniform $C^{1, 1}$ regularity of the Gursky–Streets equation. In the course of deriving regularity, we also obtain very interesting and new convexity regarding matrices in $\Gamma _2^+$. As an application, we can establish strictly the convexity of ${{\mathcal {F}}}$-functional along $C^{1, 1}$ geodesic. This gives a straightforward proof of the uniqueness of solutions of $\sigma _2$-Yamabe problem.

Appendix: The metric structure and the uniqueness of \(\sigma _2\)-Yamabe problem

In this section we verify the geodesic convexity of the functional ${{\mathcal {F}}}$ of Chang–Yang and give a much more straightforward proof of uniqueness of $\sigma _2$-Yamabe problem. Such a proof of uniqueness follows the formal metric picture set up in Gursky–Streets [23] and our $C^{1, 1}$ regularity is the key to carry out the approach technically. With $C^{1, 1}$ regularity, one can prove further that ${{\mathcal {C}}}^+$ is a metric space with Gursky–Streets metric which has nonpositive curvature in the sense of Alexanderov, following [9, Sect. 5] and [11, Sect. 5]. Since the proof is very long and the results are not needed in the rest of the paper in any sense, we skip the details.

We fix some notations. Consider the approximating geodesic equation, given two fixed boundary datum $u_0, u_1$,

$$\begin{aligned} u_{tt}\sigma _2(A_u)-\langle T_1(A_u), \nabla u_t\otimes \nabla u_t\rangle =sf. \end{aligned}$$

We have obtained uniform $C^{1, 1}$ estimates for any smooth $f>0$. We take $f\equiv 1$ in particular to get an approximating geodesic $u^s$ and denote u to be its limit. We refer u as the geodesic connecting $u_0, u_1$. We will need the following curvature weighted Poincare-inequalities, due to B. Andrews [1].

Lemma 5.1

(Andrews) Let $(M^n, g)$ be a compact Riemannian manifold with positive Ricci curvature. Given a Lipschitz function $\phi$ with $\int _M \phi dv=0$, then

$$\begin{aligned} \frac{n}{n-1}\int _M \phi ^2 dv\le \int _M (Ric^{-1}) (\nabla \phi , \nabla \phi )dv, \end{aligned}$$

with the equality if and only if $\phi \equiv 0$ or $(M^n, g)$ is isometric to the round sphere.

Gursky–Streets obtained a weaker form of this inequality for $n=4$,

Lemma 5.2

(Gursky–Streets [23]) Let $(M^4, g)$ be a closed Riemannian manifold such that $A_g\in \Gamma ^+_2$. Given a Lipschitz function $\phi$, then

$$\begin{aligned} \int _M \frac{1}{\sigma _2(A_g)} T_1(A_g)(\nabla \phi , \nabla \phi ) dv\ge 4\int _M \phi ^2 dv-\frac{4}{\int _M dv}\left( \int _M \phi dv\right) ^2. \end{aligned}$$

The equality holds if and only if $\phi$ is a constant or $(M^4, g)$ is isometric to the round sphere.

The main result in this section is the convexity of the functional ${{\mathcal {F}}}$ along the $C^{1, 1}$ geodesic.

Theorem 5.1

Given $u_0, u_1\in {{\mathcal {C}}}^+$, let $u^s$ be the approximating geodesic satisfying

$$\begin{aligned} u_{tt}\sigma _2(A_u)-\langle T_1(A_u), \nabla u_t\otimes \nabla u_t\rangle =s. \end{aligned}$$

Then ${{\mathcal {F}}}$ is convex along the $C^{1, 1}$ geodesic u. In particular ${{\mathcal {F}}}$ achieves its minimum energy at any smooth critical point.

Proof

Let $u^s$ be the unique smooth solution of the equation,

$$\begin{aligned} u_{tt}\sigma _2(A_u)-\langle T_1(A_u), \nabla u_t\otimes \nabla u_t\rangle =s. \end{aligned}$$

(5.1)

Denote u to be “the geodesic”, which is the limit of $u^s$ when $s\rightarrow 0$. Consider the functional ${{\mathcal {F}}}(u)$ and ${{\mathcal {F}}}(u^s)$ for $t\in [0, 1]$. By the uniform estimate, we know that $u^s$ converges to u in $C^{1, \alpha }([0, 1]\times M)$ for any $\alpha \in [0, 1)$. Moreover, we compute

$$\begin{aligned} \begin{aligned}&\int _M \Delta u |\nabla u|^2dV-\int _M \Delta u^s |\nabla u^s|^2 dV\\&\quad =\int _M \Delta u (|\nabla u|^2-|\nabla u^s|^2)dV+\int _M |\nabla u^s|^2 \Delta (u-u^s) dV\\&\quad =\int _M \Delta u (|\nabla u|^2-|\nabla u^s|^2)dV+\int _M \nabla (|\nabla u^s|^2) \nabla (u-u^s) dV \end{aligned} \end{aligned}$$

It follows that, $\int _M \Delta u^s |\nabla u^s|^2dV$ converges to $\int _M \Delta u |\nabla u|^2dV$ (uniformly with respect to t) when $s\rightarrow 0$. Using the formula (2), it implies that ${{\mathcal {F}}}(u^s)$ converges to ${{\mathcal {F}}}(u)$ uniformly w.r.t t. In particular ${{\mathcal {F}}}(u)$ is continuous w.r.t $t\in [0, 1]$. A similar argument shows that $\int _M \Delta u |\nabla u|^2 dV$ is Lipschitz in t and hence ${{\mathcal {F}}}(u)$ is Lipschitz in t. Denote the conformal invariant total $\sigma _2$ curvature as

$$\begin{aligned} \sigma =\int _M \sigma _2(g_u^{-1}A_u)dV_u\;\text {and}\; {{\bar{\sigma }}}=\sigma V_u^{-1} \end{aligned}$$

where $V_u$ is the total volume of $g_u$. Along the path $u^s$, using the variational structure of ${{\mathcal {F}}}$ [3] (see the computation as in [23]), we have

$$\begin{aligned} \frac{d{{\mathcal {F}}}(u^s)}{dt}=\int _M u^s_t(-\sigma _2(g^{-1}_{u^s}A_{u^s})+{{\bar{\sigma }}} ) dV_{u^s} \end{aligned}$$

To compute the second derivative we need to be careful about the conformal factor. We compute the second derivative [using (2.1), Lemma 2.1 and the Eq. (5.1)],

$$\begin{aligned} \begin{aligned} \frac{d^2{{\mathcal {F}}}(u^s)}{dt^2}&=\int _M \left( -u^s_{tt}\sigma _2({g_u^s}^{-1}A_{u^s})-u_t^s\langle T_1({g_{u^s}}^{-1}A_{u^s}), \nabla ^2 u^s_t\rangle _{g_{u^s}}\right) dV_{u^s}\\&\quad +{{\bar{\sigma }}} \int _M\left[ u^s_{tt}-4\left( u^s_t-\underline{u^s_t}\right) ^2 \right] dV_{u^s}\\&=\int _M \left( -u_{tt}\sigma _2(A_{u^s})+\langle T_1(A_{u^s}), \nabla u^s_t\otimes \nabla u^s_t\rangle \right) dV\\&\quad +{{\bar{\sigma }}} \int _M\left[ u^s_{tt}-4\left( u^s_t-\underline{u^s_t}\right) ^2 \right] dV_{u^s}\\&=-s\int _M dV+{{\bar{\sigma }}} \int _M\left[ u^s_{tt}-4\left( u^s_t-\underline{u^s_t}\right) ^2 \right] dV_{u^s}, \end{aligned} \end{aligned}$$

(5.2)

where we use the notation of average,

$$\begin{aligned} \underline{u^s_t}={V_{u^s}}^{-1}\int _M u^s_t dV_{u^s}. \end{aligned}$$

We compute, using the equation (5.1),

$$\begin{aligned} \int _M u^s_{tt} dV_{u^s}= & {} \int _M \frac{1}{\sigma _2(g^{-1}_{u^s}A_{u^s})} \langle T_1(g_{u^s}^{-1}A_{u^s}), \nabla u^s_t\otimes \nabla u^s_t\rangle _{g_{u^s}} dV_{u^s}\\&+s\int _M \frac{1}{\sigma _2(g^{-1}_{u^s}A_{u^s})}dV \end{aligned}$$

Hence it follows that

$$\begin{aligned} \begin{aligned} \frac{d^2{{\mathcal {F}}}(u^s)}{dt^2}&= -s\int _M dV+s{{\bar{\sigma }}} \int _M \frac{1}{\sigma _2(g^{-1}_{u^s}A_{u^s})}dV\\&\quad + {{\bar{\sigma }}} \int _M \left[ \frac{1}{\sigma _2(g^{-1}_{u^s}A_{u^s})} \langle T_1(g_{u^s}^{-1}A_{u^s}), \nabla u^s_t\otimes \nabla u^s_t\rangle _{g_{u^s}} -4\left( u^s_t-\underline{u^s_t}\right) ^2 \right] dV_{u^s}. \end{aligned} \end{aligned}$$

(5.3)

By Lemma 5.2 we know that

$$\begin{aligned} \frac{d^2{{\mathcal {F}}}(u^s)}{dt^2}> -s \int _M dV. \end{aligned}$$

This shows the convexity of ${{\mathcal {F}}}(u^s)+s t^2 \int _M dV$. Taking $s\rightarrow 0$, this implies the convexity of ${{\mathcal {F}}}$ along the geodesic u. The second part of the statement follows directly. Note that the second part of the statement was verified by Gursky–Streets [Lemma 6.1] [23]. $\square$

If we denote ${{\bar{\Gamma }}}_2^+$ to be the closure of $\Gamma _2^+$ with respect to $C^{1, 1}$ topology, then all geodesic segments lie in ${{\bar{\Gamma }}}_2^+$ and ${{\mathcal {F}}}$ is continuous. Now we suppose $u_0, u_1\in {{\mathcal {C}}}^+$ are two smooth critical points of ${{\mathcal {F}}}$ (hence $u_0, u_1$ are both minimizers of ${{\mathcal {F}}}$); let u(t) be the geodesic connecting $u_0, u_1$, then by the convexity of ${{\mathcal {F}}}$ along geodesic, we know that u(t) are all minimizers of ${{\mathcal {F}}}$ (over ${{\bar{\Gamma }}}_2^+$). Next we prove the following,

Corollary 5.3

Let u be the $C^{1, 1}$ geodesic connecting $u_0, u_1$. Then either $u_1=u_0+const$, or $(M, g_{u^i})$ is isometric to the round sphere.

Proof

Since ${{\mathcal {F}}}$ achieves its minimum at $u_0$ and $u_1$, by the convexity of ${{\mathcal {F}}}$ we know that ${{\mathcal {F}}}$ remains constant along the geodesic u. In other words, u(t) minimizes ${{\mathcal {F}}}$ for any $t\in [0, 1]$ over ${{\bar{\Gamma }}}_2^+$. First we note that at $u_0$, $f(s)=F(u_0+sv)$ has the property that $f^{''}(0)\ge 0$ by direct computations, for any given function v. For simplicity we can assume that v satisfies $\int _M v dV_{u_0}=0$ and we write $u(s)=u_0+sv$. We compute (with the notation $u=u(s)$)

$$\begin{aligned} f^{'}(s)=-\int _M v (\sigma _2(g_{u}^{-1}A_u)-{{\bar{\sigma }}}(s)) dV_u \end{aligned}$$

and

$$\begin{aligned} f^{''}(s)=\int _M \langle T_1(g_{u}^{-1}A_u), \nabla v\otimes \nabla v\rangle dV_u-4{{\bar{\sigma }}}(s) \int _M v^2 dV_u, \end{aligned}$$

where ${{\bar{\sigma }}}(s)=\sigma V_{u}^{-1}$. Since $u_0$ satisfies that $\sigma _2(g_{u}^{-1}A_u)={{\bar{\sigma }}}$, hence at $s=0$, we have

$$\begin{aligned} f^{''}(0)={{\bar{\sigma }}}\left( \int _M \sigma _2(g_u^{-1}A_u)\langle T_1(g_{u}^{-1}A_u, \nabla v\otimes \nabla v\rangle dV_u -4\int _M v^2dV_u\right) \end{aligned}$$

If $v\ne 0$ is fixed, then $f^{''}(0)=c>0$ if $(M, g_{u_0})$ is not isometric to the round sphere, by Lemma 5.2. Now we assume that $(M, g_{u_i})$ is not isometric to the round sphere, for $i=1, 2$. Hence we have $f^{''}(0)=c>0$, where c is a fixed constant depending only on v and $u_0$. It follows that, for $s\in [-\delta , \delta ]$ with $\delta$ sufficiently small, $f(0)\le f(s)-c s^2/2$. In other words,

$$\begin{aligned} F(u_0)\le F(u_0+sv)-cs^2/2. \end{aligned}$$

Clearly $F(u(t)+sv)$ is a continuous function in t, similar to the argument in Theorem 5.1. Hence for t sufficiently small, we have

$$\begin{aligned} F(u(t))=F(u_0)\le F(u(t)+sv). \end{aligned}$$

Now take $u=u(t)$. We need to compute the first variation of ${{\mathcal {F}}}$ at u. We need the following, at $s=0$,

$$\begin{aligned} \frac{\partial {{\mathcal {F}}}(u+sv)}{\partial s}=-\int _M v (\sigma _2(g_u^{-1}A_u)-{{\bar{\sigma }}}) dV_u \end{aligned}$$

(5.4)

If u is smooth, then (5.4) follows directly [3]. A main point is that (5.4) holds using the fact $T_1$ is divergence free (when $n=4$). When $u\in C^{1, 1}$, then $T_1(g_u^{-1}A_u)$ is divergence free in the following sense: for any smooth vector $X=(X^i)$, we have

$$\begin{aligned} \int _M \sum _j T_1(g_u^{-1}A_u) ^{ij} \nabla _j X^i dV_u=0 \end{aligned}$$

(5.5)

We can choose a sequence of smooth function $u_n$ such that $u_n$ converges to u in $W^{2, p}$ and $u_n$ has uniform $C^{1, 1}$ bound. A direct approximation argument gives (5.5). Given (5.5), (5.4) follows directly as in [3]; the point is that the following one-form $\alpha$ is still closed for $u\in C^{1, 1}$ and it gives the first variation of ${{\mathcal {F}}}$, by the computation as in [3] together with (5.5), where

$$\begin{aligned} \alpha (v)=-\int _M v (\sigma _2(g_u^{-1}A_u)-{{\bar{\sigma }}}) dV_u. \end{aligned}$$

Since ${{\mathcal {F}}}(u)\le {{\mathcal {F}}}(u+sv)$, we have at $s=0$, for any v,

$$\begin{aligned} \frac{\partial {{\mathcal {F}}}(u+sv)}{\partial s}=-\int _M v (\sigma _2(g_u^{-1}A_u)-{{\bar{\sigma }}}) dV_u= 0 \end{aligned}$$

(5.6)

Since v can be any smooth function (by adding a constant to v does not change the integral), it follows that we can choose v to be any $L^1$ measurable function. In particular we have $\sigma _2(g_u^{-1}A_u)-{{\bar{\sigma }}}=0$, where ${{\bar{\sigma }}}=\sigma V_u^{-1}$. It follows that $\sigma _2(g_u^{-1}A_u)>0$. Hence $u\in C^{1, 1}$ is a strong solution of the uniform elliptic equation

$$\begin{aligned} \sigma _2(A_u)=\sigma V_u^{-1} e^{-4u} \end{aligned}$$

(5.7)

and the standard elliptic regularity then gives the smoothness of u (in space direction). Hence $u(t): M\rightarrow {\mathbb {R}}$ is smooth for small t and it solves the Eq. (5.7). Taking derivative with respect to t, the elliptic regularity then implies that $u_t$ is smooth in space direction. (One can make this rigorous by taking difference quotient in time direction and get elliptic regularity, which implies that $u_t$ is smooth in space direction). Note that we do not assert at the moment that u is smooth in space time, even though we know this holds a posteriori. Nevertheless we can directly compute, similar as in (5.3),

$$\begin{aligned} \frac{d^2 {{\mathcal {F}}}(u(t))}{dt^2}={{\bar{\sigma }}} \int _M \left[ \frac{1}{\sigma _2(g^{-1}_{u}A_{u})} \langle T_1(g_{u}^{-1}A_{u}), \nabla u^s_t\otimes \nabla u_t\rangle _{g_{u}} -4\left( u_t-\underline{u_t}\right) ^2 \right] dV_{u}=0. \end{aligned}$$

This implies that $u_t=\text {const}$ or $(M^4, g_u)$ is isometric to the round sphere $S^4$, by Lemma 5.2. $\square$

This gives a direct proof of the uniqueness of $\sigma _2$-Yamabe problem.

Corollary 5.4

Let $(M^4, g)$ be a compact four manifold with ${{\mathcal {C}}}^+\ne \emptyset$.

• (1) There exists a unique solution to the $\sigma _2$-Yamabe problem in [g] if $(M^4, g)$ is not conformally equivalent to the round $S^4$.

• (2) In $[g_{S^4}]$, all solutions to the $\sigma _2$-problem are round metrics.