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

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.

• It may be useful to know in $d= 2$, your $k = \nabla_a \xi^a$, i.e. the conformal Killing equation is, $\nabla_{(a}\xi_{b)} = g_{ab}\nabla_c \xi^c$. – JamalS Nov 6 '17 at 12:33
• @JamalS Thanks for reminding of the dimensionality! I actually know how to prove it! I will write down the solution. – Drake Marquis Nov 6 '17 at 12:56

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.

• I was about answering!!! The point is that dim M = 2 – Valter Moretti Nov 6 '17 at 13:33
• @ValterMoretti Haha, I took your thunder! But thanks for trying to answer my question and your comment! Dimensionality sometimes plays an important role, which I always overlooked. So I do not need Geroch's suggestion on the contribution of the antisymmetric part. But still, his suggestion might be wrong. Right? – Drake Marquis Nov 6 '17 at 13:50
• Geroch's suggestion is correct. Dealing with a generic basis and using the fact that the components are only two you indeed find that the antisymmetric part is proportional to $\sigma_{ab}g^{ab}$... – Valter Moretti Nov 6 '17 at 14:00
• @ValterMoretti Yap! Just got it! Thank you so much! – Drake Marquis Nov 6 '17 at 14:13
• The nontrivial issue is why there is a confirmal Killing vector in a neighborhood of every point... – Valter Moretti Nov 6 '17 at 14:32