Conformal invariance of Cotton tensor in dimension 3

Question

If $(M, g)$ is a Riemannian $3$-manifold with Ricci curvature $Rc$ and scalar curvature $S$, the Schouten tensor is defined by $$ P = Rc - \frac{S}{4}g $$ and the Cotton tensor is $C = -DP$, where $DP $ is the exterior covariant derivative of $P$: $$ DP(X,Y,Z) = -(\nabla P)(X, Y, Z) + (\nabla P)(X, Z, Y) $$ This is an $\mathbb R$-linear operator on the bundle of covariant $2$-tensors.

I'm trying to solve the following problem:

Suppose $\tilde g = e^{2f}g$ for some $f \in C^\infty(M)$. If $C$ and $\tilde C$ denote the Cotton tensors of $g$ and $\tilde g$ respectively, then $C = \tilde C$.

I know the Schouten tensor of $\tilde g$ satisfies the following conformal transformation law: $$ \tilde P = P - \nabla^2 f + (df \otimes df) - \frac 1 2 |df|^2_g g $$ and I was able to show $D\left(|df|^2_g g\right) \equiv 0$ using Riemannian normal coodinates, so if I can show $D\left(\nabla^2 f\right)$ and $D(df\otimes df)$ both vanish identically, then I'm done. But I'm having trouble doing this, and I'm not sure where the dimension of $M$ is coming into the picture. Any suggestions?

Answer 1

You can find a proof of conformal invariant of the Cotton tensor in several places. For instance:

Sergiu Moroianu, The Cotton tensor and Chern-Simons invariants in dimension 3: an introduction, Bul. Acad. Ştiinţe Repub. Mold. Mat. 2015, no. 2(78), 3–20. Proposition 14,

or, for a more hands-on computation, in

Introduction to Conformal Geometry, I think, written by Sean Curry.

Answer 2

I have encountered this problem when reading John Lee's book. Inspiring by the write-up of Haim Grebnev, I also show $\tilde{C} = C + \text{grad}f \lrcorner W$, where $W$ is Weyl tensor. In coordinate, $\tilde{C}_{ijk} = C_{ijk} + W_{ijk}^lf_l$, where $df = f_ldx^l$. My computation might be easier.

I use (7.21) and (7.43) from Lee's book:

$$ \begin{align} R_{p q j}\,^m \beta_m & = \beta_{j ; p q}-\beta_{j ; q p} \quad \quad &(1)\\ \tilde{\Gamma}_{i j}^k&=\Gamma_{i j}^k+f_{; i} \delta_j^k+f_{; j} \delta_i^k-g^{k l} f_{; l} g_{i j}\quad \quad &(2) \end{align} $$

Some notations: Let $\tilde{P} = P + \Lambda$, where $\Lambda = -\nabla^2 f+(d f \otimes d f)-\frac{1}{2}\langle d f, d f\rangle_g^2 g$. Let $\tilde{\Gamma}_{i j}^k = \Gamma_{i j}^k + S_{ij}^k$, where $S_{ij}^k = f_{; i} \delta_j^k+f_{; j} \delta_i^k-g^{k l} f_{; l} g_{i j}$. In coordinate, $\Lambda$ can be written as $\Lambda_{i j}=-f_{;i j}+f_{;i} f_{;j}-\frac{1}{2} |df|_g g_{i j}$.

$$ \begin{aligned}(\tilde{\nabla} \tilde{P})_{i j ; k} & =\partial_k \tilde{P}_{i j}-\Gamma_{k i}^\lambda P_{\lambda j}-\Gamma_{k j}^\lambda P_{i \lambda} \\ & =\partial_k \tilde{P}_{i j}-\left(\Gamma_{k i}^\lambda+S_{k j}^\lambda\right) \tilde{P}_{\lambda j}-\left(\Gamma_{k j}^\lambda+S_{k j}^\lambda\right) \tilde{P}_{\lambda i} \\ & =\tilde{P}_{i j; k}-S_{k i}^\lambda \tilde{P}_{\lambda j}-S_{k j}^\lambda \tilde{P}_{\lambda_i} \\ & =P_{i j ; k}+\Lambda_{i j, k}-S_{k i}^\lambda \tilde{P}_{\lambda j}-S_{k j}^\lambda \tilde{P}_{\lambda i}\\ \end{aligned} $$ where LHS indices are respect to $\tilde{g}$ and RHS indices are repect to $g$. Thus, $$ \begin{align} \tilde{C}_{ijk}&=C_{i j k}+(\Lambda)_{i j ; k}-(\Lambda)_{i k, j} -S_{k_i}^\lambda \tilde{P}_{\lambda j}+S_{i j}^\lambda \tilde{P}_{\lambda k} \\ & = C_{i j k}+(\Lambda)_{i j ; k}-(\Lambda)_{i k, j} -S_{k_i}^\lambda P_{\lambda j}+S_{i j}^\lambda P_{\lambda k} -S_{k_i}^\lambda \Lambda_{\lambda j}+S_{i j}^\lambda \Lambda_{\lambda k} \end{align} $$

Further computation, $$ \begin{align} S_{k i}^\lambda P_{\lambda j}&=f_{; k} P_{i j}+f_{;i} P_{k j}-g^{\lambda l} P_{\lambda j} f_{; l} g_{k_i} \\ \delta_{i j}^\lambda P_{\lambda k}&=f_{; i} P_{j k}+f_{; j} P_{i k}-g^{\lambda l} P_{\lambda k} f_{; l} g_{i j} \\ S_{k i}^\lambda f_{;\lambda j} &=f_{; k} f_{; i j}+f_{;j} f_{; k j}+g^{\lambda l} f_{; l} f_{;\lambda j} g_{k i}\\ S_{i j}^\lambda f_{;\lambda k}&=f_{;j} f_{;j k}+f_{;j} f_{;i k}+g^{\lambda l} f_{;l} f_{;\lambda k} g_{i j} \\ f_{;\lambda} f_{;j} S_{k i}^\lambda &=f_{;k}, f_{;i} f_{; j}+f_{; i} f_{; k} f_{; j}-f_{; \lambda} f_{; j} g^{\lambda l} f_{; l} g_{k i}\\ f_{;\lambda} f_{; k} S_{i j}^\lambda &=f_{;i}f_{; j} f_{;k}+f_{;j} f_{;i} f_{;k}-f_{;\lambda} f_{;k} g^{\lambda l} f_{;l} g_{i j} \\ S_{k i}^\lambda g_{\lambda j} &=f_{;k} g_{i j}+f_{;i} g_{k j}-f_{; j} g_{k i} \\ S_{i j}^\lambda g_{\lambda k} &=f_{;i} g_{j k}+f_{;j} g_{i k}-f_{;k} g_{i j} \end{align} $$ $$\begin{align} S_{i j}^\lambda \Lambda_{\lambda k}-S_{k i}^\lambda \Lambda_{\lambda j} = f_{;k} f_{;ij}-f_{;j} f_{;i k}+g^{\lambda l} f_{;l} f_{;\lambda j} g_{k i}-g^{\lambda l} f_{;l} f_{; \lambda k} g_{i j} \end{align}$$ For $\nabla \Lambda$, $$ \begin{align} \Lambda_{ij;k} &= -f_{;i;jk}+ f_{;ik} f_{;j}+f{;i} f_{; j k}-g^{\lambda l} f_{;l} f_{; \lambda k} f_{i j} \\ \Lambda_{ik;j} &= -f_{;i;kj}+ f_{;ij} f_{;k}+f_{;i} f_{;k j}-g^{\lambda l} f_{;l} f_{;\lambda j} g_{i k} \end{align} $$ Therefore, $$ \begin{align} \Lambda_{i j, k}-\Lambda_{i k, j}+S_{i j}^\lambda \Lambda_{i k}-S_{k i}^\lambda \Lambda_{\lambda j}&=f_{;i; k j}-f_{;i;j k} \\ & = R_{kji}{}^l f_{;l} \end{align} $$ where we use (1). Note that $(P \oslash g)_{\lambda kij}=P_{\lambda k} g_{i j}+P_{i j} g_{\lambda k}-P_{\lambda k} g_{i \lambda}-P_{j \lambda} g_{i k}$, where $\oslash$ denotes Kulkarni-Nomizu product. Then we have $$ \begin{aligned} & S_{i j}^\lambda P_{\lambda k}-S_{k i}^\lambda P_{\lambda j}+(P \oslash g)_{\lambda k i j} g^{\lambda l} f_{; l} \\ = & f_{; j} P_{i k}-f_{; k} P_{i j}+P_{i j} g_{\lambda k} \lambda^{\lambda l} f_{; l}-P_{ik} g_{i \lambda} g^{\lambda l} f_{;l} \\ = & 0 \end{aligned} $$

Thus $$ \begin{align} \tilde{C}_{ijk}&=C_{ijk}+ W_{kji}{}^l f_{;l} \\ & = C_{ijk}+ W_{ijk}{}^l f_{;l} \end{align} $$ where the last equality is due to the symmetric property of W

Answer 3

I also wrote up notes regarding how the Cotton tensor changes under conformal transformations at:

https://sites.math.washington.edu/~hgrebnev/D&Writings/Z_PDF_Documents_I/Jack%20Lee%20Riemannian%20Geometry%20Notes.pdf

In the proof I use some standard results from chapter 7 of John M. Lee’s book Introduction to Riemannian Manifold (2nd Ed). The proof is indeed quite long. In order to minimize the amount of calculation that needs to be written down, I make use of symmetry in certain quantities a couple of times to argue that they cancel out in the end.