# Ngô Quốc Anh

## November 1, 2013

### The Yamabe problem: The work by Thierry Aubin

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 22:05

Following the previous note about the work of Trudinger, today we talk about the work of Aubin regarding to the Yamabe problem, that is the following simple PDE

$\displaystyle -\Delta \varphi + R\varphi = C_0 \varphi^\frac{n+2}{n-2}.$

In his elegant paper entitlde “Équations différentielles non linéaires et problème de Yamabe concernant la courbure scalaire” published in J. Math. Pures Appl. in 1976, Aubin proved the existence for almost all manifolds for $n\geqslant 6$.

By using the notations used in the note about the work of Yamabe, they are

$\displaystyle {F_q}(u) = \frac{{\displaystyle\int_M {\left( {\frac{{4(n - 1)}}{{n - 2}}|\nabla u{|^2} + R{u^2}} \right)d{v_g}} }}{{{{\left( {\displaystyle\int_M {|u{|^q}d{v_g}} } \right)}^{2/q}}}}$

where $q \leqslant \frac{2n}{n-2}$ and

$\displaystyle {\mu _q} = \mathop {\inf }\limits_{u \in {H^1}(M)} {F_q}(u),$

Aubin proved that

Theorem 1. If $\mu_{2n/(n-2)}$ satisfies

$\displaystyle\mu_{2n/(n-2)}

then the Yamabe problem is solvable where $\omega_n$ is the volume of the unit sphere in $\mathbb R^n$.

In fact, he proved a stronger result saying that in any case, there holds

$\displaystyle\mu_{2n/(n-2)} \leqslant n(n-1)\omega_n^{2/n},$

and the equality occurs if and only if $M$ is conformally equivalent to the sphere with standard metric. Having this result, to solve the Yamabe problem, we have only to exhibit a test function $\psi$ such that $F_{2n/(n-2)}(\psi).

To prove the above estimate, he made use of his breakthrough on the best constant for the Sobolev embedding, actually, it says that

Theorem 2 (Best constants for the Sobolev embedding). The Sobolev embedding theorem holds for any complete manifold $M$ of dimension $n$ with bounded curvature and positive injectivity radius. Moreover, for any $\varepsilon>0$, there exists a constant $A_q(\varepsilon)$ such that every $\varphi \in W^{1,q}(M)$

$\displaystyle \|\varphi\|_{L^p} \leqslant \big(K(n,q)+\varepsilon \big) \|\nabla \varphi\|_{L^q} + A_q(\varepsilon) \|\varphi\|_q$

with $1/p=1/q-1/n>0$.

When $q=2$, there holds $K(n,2)=2(\omega_n)^{-1/n}(n(n-2))^{-1/2}$. A standard argument shows that there is a sequence of smooth positive functions $\psi_i$ (after normalizing) such that

$\displaystyle \|\psi_i\|_{2n/(n-2)} = 1, \quad \|\psi_i\|_{2} \to 0, \quad \|\nabla \psi_i\|_2 \to K(n,2)^{-1}.$

Consequently,

$\displaystyle F_{2n/(n-2)}(\psi_i)=\frac{4(n-1)}{n-2}\|\nabla \psi_i\|_2^2 + \int_M R\psi_i^2 dv_g \to n(n-1)\omega_n^{2/n}.$

From the definition of $\mu_{2n/(n-2}$ we immediately have $\displaystyle\mu_{2n/(n-2)} \leqslant n(n-1)\omega_n^{2/n}$ as claimed.

Now, to prove Theorem 1, Aubin used variational methods. Assuming $M$ has unit volume. Indeed, by normalizing, we can use

$\displaystyle {\mu _q} = \inf \left\{ {\frac{{4(n - 1)}}{{n - 2}}\int_M {|\nabla u{|^2}d{v_g}} + \int_M {R{u^2}d{v_g}} :\int_M {|u{|^q}d{v_g}} = 1} \right\}.$

When $q<\frac{2n}{n-2}$, it is clear that $\mu_q$ is finite since we can use $L^q$-norm to control $L^2$-norm. Then it is a standard routine to obtain existence result for $q<\frac{2n}{n-2}$. Once we have this, we can send $q \to \frac{2n}{n-2}$ to obtain a sequence of minimizing solutions, say $\{u_i\}_i$. To obtain some sort of precompactness property for the sequence $\{u_i\}$ we observe by using constant functions that $\mu_q \leqslant \int_M R dv_g$. Moreover, $\|u_i\|_2 \leqslant 1$ for any $i$. Finally, we can control $\|\nabla u_i\|_2$ as follows

$\displaystyle\frac{{4(n - 1)}}{{n - 2}}\left\| {\nabla {u_i}} \right\|_2^2 \leqslant {\mu _q} + \mathop {\sup }\limits_M |R| \leqslant \int_M {Rd{v_g}} + \mathop {\sup }\limits_M |R|.$

In other words, the sequence $\{u_i\}_i$ is bounded in $H^1(M)$. Then, up to subsequences, we have a weak limit $u \in H^1(M)$. Since $u_i$ solves the subcritical equation, that is to say

$\displaystyle \frac{{4(n - 1)}}{{n - 2}}\int_M {\nabla {u_i}\nabla \varphi d{v_g}} + \int_M {Ru_i\varphi d{v_g}} = {\mu _q}\int_M {u_i^{q - 1}\varphi d{v_g}}$

we can pass to the limit as $q\to \frac{2n}{n-2}$ (for a detailed explanation, check my paper). After passing to the limit, $u$ solves

$\displaystyle \frac{{4(n - 1)}}{{n - 2}}\int_M {\nabla u\nabla \varphi d{v_g}} + \int_M {Ru\varphi d{v_g}} = {\mu _{2n/(n - 2)}}\int_M {{u^{\frac{{n + 2}}{{n - 2}}}}\varphi d{v_g}} .$

Using standard regularity theorems, we can conclude that $u$ solves

$\displaystyle - \frac{{4(n - 1)}}{{n - 2}}\Delta u + Ru = {\mu _{2n/(n - 2)}}{u^{\frac{{2n}}{{n - 2}}}}.$

By normalizing, we can change $\mu_{2n/(n-2)}$ to any constant we want. However, it is necessary to check whether latex u>0\$ or not. The maximum principle simply tells us that either $u>0$ everywhere in $M$ or $u \equiv 0$. To rule out the latter, we make use of the estimate $\displaystyle\mu_{2n/(n-2)} \leqslant n(n-1)\omega_n^{2/n}$ and here is the only place we need this. To achieve that goal, thanks to the strongly convergence $\|u_i\|_2 \to \|u\|_2$, it is necessary to bound $\|u_i\|_2$ away from zero. For small $\varepsilon >0$, Aubin first wrote as follows

$\displaystyle 1 = \left\| {{u_i}} \right\|_q^2 \leqslant \left\| {{u_i}} \right\|_N^2 \leqslant (K{(n,2)^2} + \varepsilon )\left\| {\nabla {u_i}} \right\|_2^2 + A(\varepsilon )\left\| {{u_i}} \right\|_2^2.$

Using the equation, we arrive further at

$\displaystyle 1 \leqslant \left\| {{u_i}} \right\|_N^2 \leqslant \frac{{n - 2}}{{4(n - 1)}}(K{(n,2)^2} + \varepsilon )\left( {{\mu _q} - \int_M {Ru_i^2d{v_g}} } \right) + A(\varepsilon )\left\| {{u_i}} \right\|_2^2.$

Consequently,

$\displaystyle 1 \leqslant \frac{{n - 2}}{{4(n - 1)}}(K{(n,2)^2} + \varepsilon ){\mu _q} + \left( {\frac{{n - 2}}{{4(n - 1)}}(K{{(n,2)}^2} + \varepsilon )\mathop {\sup }\limits_M |R| + A(\varepsilon )} \right)\left\| {{u_i}} \right\|_2^2.$

Now thanks to

$\displaystyle {\mu _q} \to {\mu _{2n/(n - 2)}} < \frac{{4(n - 1)}}{{n - 2}}K{(n,2)^{-2}}$

we can choose $\varepsilon>0$ small enough in such a way that

$\displaystyle \frac{{n - 2}}{{4(n - 1)}}(K{(n,2)^2} + \varepsilon ){\mu _q} < 1.$

This shows that $\|u_i\|_2$ is bounded from below by a positive constant.

As far as we know, the condition $\mu_{2n/(n-2)} is crucial when solving the Yamabe problem. We note that $\mu_{2n/(n-2)}$ achieves its maximum value $n(n-1)\omega_n^{2/n}$ only when $M$ is $(\mathbb S^n, g_{\text{car}})$ the standard unit sphere with standard metric. Since the stereographic projection is a conformal mapping, $(\mathbb S^n, g_{\text{car}})$ is conformal to the standard Euclidean space $(\mathbb R^n, ds^2)$ with standard Euclidean metric. Consequently, any conformally flat manifold is globally conformal to $(\mathbb S^n, g_{\text{car}})$. Regardless to this globally set, either the manifold is locally conformal flat or nonlocally conformal flat.  The latter case under the condition $n\geqslant 6$ was proved by Aubin.

To conclude the paper, Aubin proved the following

Theorem 3. If $(M, g)$ ($n \geqslant 6$) is a compact nonlocally conformally flat Riemannian manifold, then

$\displaystyle\mu_{2n/(n-2)}

Therefore, the cases that $n=3, 4, 5$ and that $M$ is locally conformally flat are still open. To prove this result, Aubin constructed test functions, i.e. he used the following

$\displaystyle \phi_\varepsilon (x)=\left( \frac{\varepsilon}{\varepsilon^2 + |x|^2}\right)^{(n-2)/2}.$

Eventually, the Weyl tensor comes up in his asymptotic expansion for the test functions, i.e. the sign for the second term is

$-W_{ijkl}W^{ijkl}$

provided $n \geqslant 6$. When the manifold is not compactly locally conformaly flat, the Weyl tensor never vanishes, see this. This helped him to make calculation possible. However, as Aubin had already pointed out that the Yamabe has not only the critical exponent $\frac{n+2}{n-2}$ but also has the scalar function in the linear term.

In the latter notes, we show how Schoen overcame this difficulty using the positive mass theorem.