[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 1A diffeomorphism between two Riemannian manifolds is said to be conformal if

for some function on . If is constant, then is homothetic. Especially, if , then it is an isometry.

We now want to compare their curvatures. Suppose is an orthnormal frame for and be its dual coframe. The connection 1-form is then given by (using Einstein notation)

It is easy to see that

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

i.e.

is linear in all slots and is antisymmetric in . In local frame, the curvature 2-forms are given by

We have the

Theorem 2 (Cartan’s structure equations)

*Proof:*

For the second equation, first note that

So we have

Now we have enough tools to compare the two curvatures. For two conformally equivalent metrics and . Suppose is an orthnormal frame for and be its dual coframe, then is an orthnormal frame and is an orthnormal coframe for .

Proposition 3

*Proof:* Exercise. Use the first Cartan’s formula to show that , and then apply the second Cartan’s formula.

Applying the above to (1), we have (see also here)

and the scalar curvatures are related by

In particular, for a surface

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

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

Proposition 4The scalar curvatures of two conformally equivalent metrics are related by

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 and are compact Riemannian manifolds, has nonpositive scalar curvature and has nonnegative scalar curvature, but not both identically zero. Then there is no conformal transformation between them.

*Proof:* If not, is such a map, then . We have

Thus is a superharmonic function on , which must then be constant by maximum principle (it can’t have nontrivial minimum). So we have

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

The same proof also shows that

Theorem 6 (Obata)Suppose 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 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. homothetic)For a conformal map between two compact Riemannian manifolds with non-positive but not identically zero scalar curvatures, suppose , then is homothetic if and only if

for some constant , i.e. their scalar curvatures are the same up to scaling. In this case, we actually have , i.e. .

*Proof:* If is homothetic, i.e. is actaully , then by Proposition 4, we clearly have .

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

and we will prove the result on their double covers. (In fact, if is the orientable cover, since is a diffeomorphism, is the orientable double cover of , in this case is just the identity, \underline{however} let us distinguish the two double cover by and respectively, we give the metric and the metric , then is NOT necessarily an isometry, but is conformal. )

So now, by Proposition 4, as ,

A natural step to do will be multiplying both side by and integrating them (and do integration by parts), then . \underline{If} we can show that , then it would force , which is what we want. Unfortunately, this is not successful, because the will become (readers may try)

which may not be positive is . Instead, by a simple trial, it would become clear that the next sensible thing to do is to multiply (4) by , noting that . Doing this, the LHS becomes

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

Thus we conclude that . From RHS, we deduce . But then if , since is not identically zero, then , so must be .

Especially if , 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 and are compact manifolds with negative constant scalar curvatures, then all conformal map between them are homothetic, with

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