## Conformal transformations of compact Riemannian manifolds

[Updated on 4/10/2011 for some more results about non-positive scalar curvature case. ]

In this note, we study the conformal transformations of compact Riemannian manifolds. The main results are again due to Obata.

 Definition 1 A diffeomorphism ${f: (M,g)\rightarrow (M',g')}$ between two Riemannian manifolds is said to be conformal if $\displaystyle f^*g'= e^{2\phi} g$ for some function ${\phi}$ on ${M}$. If ${\phi}$ is constant, then ${f}$ is homothetic. Especially, if ${\phi=0}$, then it is an isometry.

We now want to compare their curvatures. Suppose ${e_i}$ is an orthnormal frame for ${g}$ and ${\omega^i}$ be its dual coframe. The connection 1-form ${\omega^i_j\in \Omega^1(M)}$ is then given by (using Einstein notation)

$\displaystyle \nabla _X e_i:=\omega_i^j(X)e_j.$

It is easy to see that

$\displaystyle \nabla _X \omega^i =- \omega^i _j (X)\omega^j.$

We can regard the Riemannian curvature tensor as an endomorphism-valued 2-form:

$\displaystyle Rm(\cdot, \cdot)\in \Omega^2( M;End(TM)).$

i.e.

$\displaystyle Rm(X, Y)Z=-Rm(Y, X)Z\in TM$

is linear in all slots and is antisymmetric in ${X, Y}$. In local frame, the curvature 2-forms ${\Omega_i^j}$ are given by

$\displaystyle \Omega_i^j (X, Y)e_j:=Rm(X, Y)e_i. \ \ \ \ \ (1)$

We have the

 Theorem 2 (Cartan’s structure equations) $\displaystyle d\omega^i = \omega ^j\wedge \omega^i_j. \ \ \ \ \ (2)$ $\displaystyle \Omega_i^j= d\omega _i^j -\omega _i^k \wedge \omega_k^j. \ \ \ \ \ (3)$

Proof:

$\displaystyle \begin{array}{rcl} d\omega ^i (X, Y) = (\nabla _X \omega^i) (Y)- (\nabla _Y \omega^i )(X) &=& - \omega^i _j (X)\omega^j(Y)+\omega^i _j (Y)\omega^j(X)\\ &=& \omega^j\wedge \omega_j^i (X, Y). \end{array}$

For the second equation, first note that

$\displaystyle \nabla^2 e_i(X, \cdot)\stackrel{\triangle}= \nabla_X \nabla e_i= \nabla _X (\omega_i^j \otimes e_j)=\nabla _X \omega_i^j\otimes e_j+\omega_i^j\otimes \nabla _X e_j.$

So we have

$\displaystyle \begin{array}{rcl} \Omega_i^j (X, Y) &=& \langle\nabla ^2_{X, Y}e_i-\nabla^2_{Y, X}e_i,e_j\rangle\\ &=& \langle (\nabla_X\omega _i^k)(Y)e_k+ \omega _i^k(Y)\nabla _Xe_k -(\nabla_Y\omega _i^k)(X)e_k+ \omega _i^k(X)\nabla _Ye_k ,e_j\rangle\\ &=& \langle [(\nabla_X\omega _i^k)(Y)-(\nabla_Y\omega _i^k)(X)]e_k,e_j\rangle\\ &&+\langle \omega _i^k(Y)\nabla _Xe_k- \omega _i^k(X)\nabla _Ye_k ,e_j\rangle\\ &=& d\omega_i^k(X, Y)\langle e_k,e_j\rangle -( \omega _i^k (X)\omega_k^l(Y)-\omega _i^k (Y)\omega_k^l(X))\langle e_l, e_j\rangle \\ &=& d\omega _i ^j(X, Y)- \omega _i^k\wedge \omega_k^j(X, Y). \end{array}$

$\Box$

Now we have enough tools to compare the two curvatures. For two conformally equivalent metrics ${g}$ and ${\widetilde g=e^{2\phi}g}$. Suppose ${e_i}$ is an orthnormal frame for ${g}$ and ${\omega^i}$ be its dual coframe, then ${\widetilde e_i=e^{-\phi}e_i}$ is an orthnormal frame and ${\widetilde \omega ^i=e^\phi \omega^i}$ is an orthnormal coframe for ${\widetilde g}$.

 Proposition 3 $\displaystyle \begin{array}{rcl} \widetilde {\Omega_i^j}&=& \Omega_i^j+ \nabla_{e_k}\nabla_{e_i} \phi\omega^k \wedge \omega ^j - \nabla_{e_k}\nabla_{e_j} \phi\omega^k \wedge \omega ^i\\ &&+|\nabla \phi|^2 \omega ^i \wedge \omega^j+ d\phi \wedge (e_j(\phi) \omega^i -e_i(\phi) \omega^j). \end{array}$

Proof: Exercise. Use the first Cartan’s formula to show that ${\widetilde \omega_i^j= \omega_i^j +e_i(\phi) \omega^j}$, and then apply the second Cartan’s formula. $\Box$

$\displaystyle \begin{array}{rcl} \widetilde {Rc}(\widetilde e_l, \widetilde e_i)&=& \sum_m \langle \widetilde {Rm} (\widetilde e_m , \widetilde e_l)\widetilde e_i, \widetilde e_m\rangle\\ &=&\sum_m \widetilde {\Omega}_i^m (\widetilde e_m , \widetilde e_l)\\ &=&e^{-2\phi} (Rc(e_l,e_i) +(2-n) \nabla _{e_l}\nabla_{e_i} \phi- \delta _{li}\Delta \phi +|\nabla \phi|^2 (2-n) e_l( \phi)e_i(\phi)) \end{array}$

and the scalar curvatures are related by

$\displaystyle \begin{array}{rcl} \widetilde R&=& \sum_i \widetilde{Rc}(\widetilde e_i, \widetilde e_i)\\ &=& e^{-2\phi} (R- 2(n-1)\Delta \phi-(n-1)(n-2) |\nabla \phi|^2). \end{array}$

In particular, for a surface

$\displaystyle R(e^\phi g)= e^{-\phi}(R(g) -\Delta_g \phi).$

This formula also shows that all surfaces are (locally) conformally flat, in the sense that locally the metric is conformally equivalent to the flat metic, as we can always locally solve the equation

$\displaystyle \Delta_g \phi-R(g)=0,$

so that the new metric has constant curvature ${0}$, thus it must be flat locally. Let me summarize the result in

 Proposition 4 The scalar curvatures of two conformally equivalent metrics are related by $\displaystyle R(e^{2\phi}g) = e^{-2\phi} (R(g)- 2(n-1)\Delta_g \phi-(n-1)(n-2) |\nabla_g \phi|^2).$

We now give an application of the above formula. We will show that the signs of scalar curvatures are preserved by a conformal transformation:

 Theorem 5 (Obata) Suppose ${(M,g)}$ and ${(M',g')}$ are compact Riemannian manifolds, ${(M,g)}$ has nonpositive scalar curvature and ${(M',g')}$ has nonnegative scalar curvature, but not both identically zero. Then there is no conformal transformation between them.

Proof: If not, ${f}$ is such a map, then ${f^*g'= e^{2\phi}g}$. We have

$\displaystyle -2(n-1)\Delta_g \phi= e^{2\phi }R(g')- R(g) +(n-1)(n-2)|\nabla \phi|^2\geq 0.$

Thus ${\phi }$ is a superharmonic function on ${M}$, which must then be constant ${c}$ by maximum principle (it can’t have nontrivial minimum). So we have

$\displaystyle R(g') =e^{-2c}R(g)$

which is impossible except when both of them are identically zero. $\Box$

The same proof also shows that

 Theorem 6 (Obata) Suppose ${f:(M,g) \rightarrow (M',g')}$ between two compact manifolds is conformal. If one of them has zero scalar curvature and the other has non-negative (or non-positive) scalar curvature, then ${f}$ is homothetic.

For the non-positive scalar curvature case, by the above results, we only have to consider the case where both of the scalar curvatures are not identically zero:

 Theorem 7 (Obata, preserves scal. curv.${\Rightarrow}$ homothetic) For a conformal map ${f: (M,g)\rightarrow (M',g')}$ between two compact Riemannian manifolds with non-positive but not identically zero scalar curvatures, suppose ${f^*g'=e^{2\phi}g}$, then ${f}$ is homothetic if and only if $\displaystyle f^*R(g')= e^{-2k} R(g)$ for some constant ${k}$, i.e. their scalar curvatures are the same up to scaling. In this case, we actually have ${\phi=k}$, i.e. ${f^*g'=e^{2k}g}$.

Proof: If ${f}$ is homothetic, i.e. ${f^*g'}$ is actaully ${e^{2k} g}$, then by Proposition 4, we clearly have ${R(g')= e^{-2k}R(g)}$.

For the converse, first of all we can assume both ${M}$ and ${M'}$ are orientable, otherwise we can lift ${f}$ is their orientable double cover: (notations abused)

$\displaystyle \overline f: (\widetilde M, g)\rightarrow (\widetilde M', g')$

and we will prove the result on their double covers. (In fact, if ${\pi: \widetilde M\rightarrow M}$ is the orientable cover, since ${f}$ is a diffeomorphism, ${\pi'=f\circ \pi: \widetilde M \rightarrow M'}$ is the orientable double cover of ${M'}$, in this case ${\overline f}$ is just the identity, \underline{however} let us distinguish the two double cover by ${\widetilde M}$ and ${\widetilde M'}$ respectively, we give ${\widetilde M}$ the metric ${\pi^* g}$ and ${\widetilde M'}$ the metric ${\pi'^* g'}$, then ${\overline f= \textrm{id}}$ is NOT necessarily an isometry, but is conformal. )

So now, by Proposition 4, as ${R(g')= e^{-2k}R(g)}$,

$\displaystyle R(g)( e^{2(\phi-k)}-1) =-2(n-1)\Delta \phi-(n-1)(n-2)|d\phi|^2. \ \ \ \ \ (4)$

A natural step to do will be multiplying both side by ${e^{2(\phi-k)}-1}$ and integrating them (and do integration by parts), then ${LHS\leq 0}$. \underline{If} we can show that ${RHS\geq 0}$, then it would force ${e^{\phi-k}=1}$, which is what we want. Unfortunately, this is not successful, because the ${RHS}$ will become (readers may try)

$\displaystyle (6-n)\int {\text{\{something positive\}}}+(n-1)(n-2)\int |d\phi|^2$

which may not be positive is ${n>6}$. Instead, by a simple trial, it would become clear that the next sensible thing to do is to multiply (4) by ${e^{2n(\phi-k)}-1}$, noting that ${(e^{2(\phi-k)}-1)(e^{2n(\phi-k)}-1)\geq 0}$. Doing this, the LHS becomes

$\displaystyle L HS= \int_M R(g) (e^{2(\phi-k)}-1)(e^{2n(\phi-k)}-1)dV_g\leq 0.$

On the other hand, by integration by parts, the RHS is

$\displaystyle \begin{array}{rcl} RHS &=& -2(n-1)\int _M(e^{2n(\phi-k)}-1)\Delta \phi dV_g\\ &&-(n-1)(n-2)\int _M(e^{2n(\phi-k)}-1)|d\phi|^2dV_g\\ &=& 4n(n-1)\int _M e^{2n(\phi-k)}|d\phi |^2dV_g\\&&-(n-1)(n-2)\int _M(e^{2n(\phi-k)}-1)|d\phi|^2dV_g\\ &=& (3n+2)(n-1)\int _M e^{2n(\phi-k)}|d\phi |^2dV_g+(n-1)(n-2)\int _M|d\phi|^2dV_g\\ &\geq &0. \end{array}$

Thus we conclude that ${LHS=RHS =0}$. From RHS, we deduce ${\phi=const}$. But then if ${\phi\neq k}$, since ${R(g) }$ is not identically zero, then ${LHS<0}$, so ${\phi}$ must be ${k}$. $\Box$

Especially if ${k=0}$, ${f}$ is an isometry. So we have

 Corollary 8 (Obata) A conformal map between two compact Riemannian manifolds with nonpositive but not vanishing scalar curvatures is an isometry if and only if it preserves the scalar curvature.

In case of constant scalar curvatures,

 Corollary 9 (scaling factor = ratio of s.c.) If ${(M,g)}$ and ${(M',g')}$ are compact manifolds with negative constant scalar curvatures, then all conformal map ${f}$ between them are homothetic, with $\displaystyle f^* g'= \frac{R(g)}{R(g')}g.$

 Corollary 10 (Rigidity) If ${f}$ is a conformal map of a compact Riemannian manifold with negative constant scalar curvature, then it is an isometry.