Ngô Quốc Anh

October 4, 2011

Locally conformally flat manifolds and Weyl and Cotton tensors

Filed under: Riemannian geometry — Ngô Quốc Anh @ 20:45

The purpose of this note is to prove the following

Theorem. A Riemannian manifold $(M^n, g)$ is locally conformally flat if and only if

• for $n \geqslant 4$, the Weyl tensor vanishes;
• for $n=3$, the Cotton tensor vanishes.

To this purpose, let us briefly recall some definitions

The Weyl tensor. The Weyl tensor can be defined using the following formula

$\displaystyle W = \text{Rm} - \frac{\text{Scal}}{{2(n - 1)n}}g \odot g - \frac{1}{{n - 2}}\left( {\text{Ric} - \frac{\text{Scal}}{n}g} \right) \odot g$

where $n\geqslant 3$ and $\odot$ denotes the Kulkarni–Nomizu product of two symmetric (0,2) tensors. Writing the Weyl tensor in this way means that the Weyl tensor is actually a (0,4) tensor. It can be seen that the Weyl tensor can be rewritten in this form

$\displaystyle W = \text{Rm} - \frac{1}{{n - 2}}\left( {\text{Ric} - \frac{g}{{2(n - 2)}}\text{Scal}} \right) \odot g$

where the part

$\displaystyle S = \frac{1}{{n - 2}}\left( {{\text{Ric}} - \frac{g}{{2(n - 2)}}{\text{Scal}}} \right) \odot g$

is called the Schouten tensor. We have the following result

Proposition. If $n \geqslant 3$, then

$\displaystyle {\nabla ^l}{W_{lijk}} = \frac{{n - 3}}{{n - 2}}{C_{ijk}}$

where

$\displaystyle {C_{ijk}} = {\nabla _k}{S_{ij}} - {\nabla _i}{S_{kj}}.$

The Cotton tensor. The (0,3) tensor $C$ above is called the Cotton tensor. Apparently, if either the Weyl tensor or the Ricci tensor vanishes, so does the Cotton tensor.

The Weyl and Cotton tensors under the conformal changes of metric. It is well-known that these tensors are invariant under the conformal changes of metric, that is,

$\displaystyle W = \widetilde W, \quad C = \widetilde C$

under the change $\widetilde g = e^{2f}g$ for some smooth function $f$ (see this note).

The Riemmanian curvature tensor under the conformal changes of metric. We list here the following rule

$\displaystyle {e^{ - 2f}}\widetilde {\text{Rm}} = \text{Rm} - \left( {{\nabla _i}{\nabla _j}f - {\nabla _i}f{\nabla _j}f + \frac{1}{2}|\nabla f{|^2}{g_{ij}}} \right) \odot g.$

See this note for further details.

Locally conformally flat manifolds. Roughly speaking, this is to say at each point $p \in M$, there exists a neighborhood $U$ of $p$ such that the conformal class of $g$ contains the flat metric in $U$, that is to say

$\displaystyle \widetilde{\text{Rm}} =0.$

We are now in a position to prove the theorem.

Proof of Theorem. We first assume that $M$ is locally conformally flat, that is, $\widetilde{\text{Rm}} =0$. If $n\geqslant 4$, using the formula for $W$ we have

$\displaystyle W = \widetilde W = \widetilde {\text{Rm}} - \frac{{\text{Scal}_{\widetilde g}}}{{2(n - 1)n}}\widetilde g \odot \widetilde g - \frac{1}{{n - 2}}\left( {\widetilde {\text{Ric}} - \frac{{\text{Scal}_{\widetilde g}}}{n}\widetilde g} \right) \odot \widetilde g = 0$

since the Riemmanian curvature tensor vanishes. If $n=3$, we use the formula for $C$, we obtain

$\displaystyle {C_{ijk}} = {\widetilde C_{ijk}} = {\widetilde {{\text{Ric}}}_{ij,k}} - {\widetilde {{\text{Ric}}}_{ik,j}} = 0$

since the Ricci tensor vanishes.

Conversely, if the Weyl tensor vanishes, we have

$\displaystyle 0 = \text{Rm} - \frac{1}{{n - 2}}\left( {\text{Ric} - \frac{g}{{2(n - 2)}}\text{Scal}} \right) \odot g.$

Under the conformal change, for some $f$, we have

$\displaystyle {e^{ - 2f}}\widetilde {Rm} = \left[ {\frac{1}{{n - 2}}\left( {\text{Ric} - \frac{g}{{2(n - 2)}}\text{Scal}} \right) - \left( {{\nabla _i}{\nabla _j}f - {\nabla _i}f{\nabla _j}f + \frac{1}{2}|\nabla f{|^2}{g_{ij}}} \right)} \right] \odot g.$

Since the mapping $\odot: S^2 M \to CM$ given by $\odot (h) = h \odot g$ is injective, it suffices to show that the following equation

$\displaystyle \frac{1}{{n - 2}}\left( {\text{Ric} - \frac{g}{{2(n - 2)}}\text{Scal}} \right) = \left( {{\nabla _i}{\nabla _j}f - {\nabla _i}f{\nabla _j}f + \frac{1}{2}|\nabla f{|^2}{g_{ij}}} \right).$

is locally solvable. This can be done using the following whose proof is postponed.

Lemma. Provided the Weyl tensor vanishes, equation

$\displaystyle {\nabla _i}{\nabla _j}f - {\nabla _i}f{\nabla _j}f + \frac{1}{2}|\nabla f{|^2}{g_{ij}} = {S_{ij}}$

is locally solvable if and only if the following integrability condition is satis ed

$\displaystyle {\nabla _k}{S_{ij}} = {\nabla _i}{S_{kj}}.$

That is, if and only if the Cotton tensor vanishes.

Recall that when $n\geqslant 4$; the condition follows from the Weyl tensor vanishes. This concludes the proof.

Source

1. what is the relationship between cotton tensor and Bak tensor?

Comment by mathsnail — October 6, 2011 @ 22:21

• Sorry, I don’t know. I haven’t heard any Bak tensor.

Comment by Ngô Quốc Anh — October 7, 2011 @ 1:16

• I means the Bach tensor, in some reference, both of them are same, but in wiki, Bach tensor is defined by

$B_{ij}:=S_{kl}W^{kl}_{ij}+\nabla^k\nabla_iS_{jk}-\nabla^k\nabla_kS_{ij}$

do you know the difference?

Comment by mathsnail — October 8, 2011 @ 16:08

2. $B_{ij}:=S_{kl}W^{kl}_{ij}+\nabla^k\nabla_iS_{jk}-\nabla^k\nabla_kS_{ij}$

Comment by mathsnail — October 8, 2011 @ 16:11

• The Bach tensor is a tensor of rank 2 while the Cotton tensor has rank 3, so they are different. While the Cotton tensor plays an important role in differential geometry, I just guess the Bach tensor is just a conformally invariant tensor.

Comment by Ngô Quốc Anh — October 9, 2011 @ 9:27