For a compact 2D manifold, does there exist a traceless symmetric $\sigma_{ab}: \nabla_{[a}\sigma_{b]c} = 0?$

Question

Let $S$ be a smooth, compact, 2-dimensional manifold with a positive-definite Riemannian metric $g_{ab}$ with a compatible covariant derivative $\nabla_a$.

I want to show that there exists a unique traceless, symmetric tensor $\sigma_{ab}$ satisfying

$$\nabla_{[a}\sigma_{b]c}=0.$$

This question is actually related to Theorem 5 in Geroch's work Asymptotic Structure of Space-Time. Unfortunately, I cannot find a free version of it. But I have already translated the essential part of Theorem 5. In Geroch's proof, he claimed that $\sigma_{ab}=0$, so the uniqueness is proved. His argument can be given below:

Let $\xi_a$ be any conformal Killing vector field on $S$ such that

$$\nabla_{(a}\xi_{b)}=kg_{ab}$$

for some function $k$ on $S$. Form a new tensor $\sigma_{ab}\xi_c$ by tensor product, and evaluate,

$$\nabla_{[a}(\sigma_{b]c}\xi_d)=\sigma_{c[b}\nabla_{a]}\xi_d.$$

Contract both sides with $g^{cd}$ to obtain,

$$\nabla_{[a}(\sigma_{b]c}\xi^c)=\sigma_{c[b}(\nabla_{a]}\xi_d)g^{cd}.$$

Here, he suggested that one write $\nabla_{a}\xi_d$ as a sum of its symmetric and antisymmetric parts. Using the property of being a conformal Killing vector, one can show that the symmetric part gives $\sigma_{c[b}kg_{a]d}g^{cd}=k\sigma_{[ab]}=0$, as $\sigma$ is symmetric. For the antisymmetric part, he said that you would get a multiple of $g^{cd}\sigma_{cd}$, which is zero, too.

But I come across some problem with the antisymmetric part. The antisymmetric part does not possess any particular property, so I cannot really obtain his result.

Answer 1

The proof to Geroch's claim uses the fact that the manifold is 2-dimensional. Thanks to @JamalS. In this case, any antisymmetric tensor, such as $\nabla_{[a}\xi_{b]}$, is proportional to the volume element $\epsilon_{ab}$. Let $\nabla_{[a}\xi_{b]}=\alpha\epsilon_{ab}$ for some function $\alpha$ on $S$.

Let us consider the contribution of $\nabla_{[a}\xi_{b]}$ to $\nabla_{[a}(\sigma_{b]c}\xi^c)$:

$\sigma_{c[b}(\nabla_{[a]}\xi_{d]})g^{cd}=\alpha\sigma_{c[b}\epsilon_{a]}{}^c$.

Now, choose an orthonormal basis $\{(e_1)^a,(e_2)^a\}$ such that $\epsilon_{12}=1$, and $\sigma_{11}+\sigma_{22}=0$, as $\sigma_{ab}$ is traceless. So consider

$\sigma_{c[1}\epsilon_{2]}{}^c=\frac{1}{2}(\sigma_{11}\epsilon_2{}^1+\sigma_{21}\epsilon_2{}^2-\sigma_{12}\epsilon_1{}^1-\sigma_{22}\epsilon_1{}^2)=\frac{1}{2}(\sigma_{11}\epsilon_2{}^1-\sigma_{22}\epsilon_1{}^2)=\frac{1}{2}(\sigma_{11}+\sigma_{22})\epsilon_2{}^1=0$.

This completes the proof.

Another method is to set $\sigma_{c[b}\epsilon_{a]}{}^c=\beta\epsilon_{ab}$. Contracting both sides with $\epsilon^{ab}$ to obtain $\beta=\alpha\sigma_{cb}g^{bc}/2=0$. Thanks @Valter Moretti!

It turns out the dimensionality plays an important role in proving Geroch's claim, which I ignored completely before.