# Conformal invariance of Cotton tensor in dimension 3

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?

• Any thoughts about the answer? Commented Mar 3, 2020 at 1:07
• Oh yes. I've worked it out since posting this question; it ended up being a very long computation involving conformal relations of the Schouten tensor, Christoffel symbols, and Riemannian curvature tensor, and applying all of them to the formula $$C_{ijk} = \partial_k P_{ij} - \partial_j P_{ik} - P_{\ell j} \Gamma_{ki}^\ell + P_{\ell k} \Gamma_{ji}^{\ell}$$ but after about 7 pages of computation and cancelation, I finally arrived at $\tilde C_{ijk} = C_{ijk} + W_{ijk}^\ell \partial_\ell f$. Since $W = 0$ in dimension 3, this gives us what we need. Commented Mar 4, 2020 at 2:18

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.

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.

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