In order to find a manifestly unique expression for the connection coefficients in terms of a given (possibly non-symmetric) tensor $g_{ab}$ usually two additional properties are assumed:
- Torsion-free: $\Gamma^a_{bc} = \Gamma^a_{(bc)}$ (symmetric in lower indices).
- Compatibility: $\nabla_a g_{bc} =0$.
Now, by expanding out the equation of the metric compatibility for all three permutations of the indices, subtracting the equations and use of the symmetry of the connection the known Levi-Civita Christoffel symbols $\Gamma^a_{bc}$ in terms of the derivative of the metric are obtained.
In classical GR the metric is symmetric which is needed for the final (known) result of the Levi-Civita Christoffel symbols.
Is it straightforward (to show) to find a corresponding result for non-symmetric theories with a non-symmetric $(0,2)$ tensor field?
Related: Can a metric in General Relativity, Supergravity, String Theory, etc., be asymmetric?
Best Answer
TL;DR: Given a (possibly non-symmetric) non-degenerate covariant (0,2) tensor field $$\mathbb{g}~\in~\Gamma(T^{\ast}M\otimes T^{\ast}M),\tag{1}$$ then a compatible torsionfree tangent-bundle connection $\nabla$ might not exists or be unique. See e.g. the Theorem below.
Theorem. Given a non-degenerate 2-form $$\frac{1}{2}\omega_{ij}\mathrm{d}x^i\wedge\mathrm{d}x^j~=~\omega~\in~\Omega^2(M)~\equiv~\Gamma(\bigwedge\!{}^2T^{\ast}M)\tag{2}$$ on a paracompact manifold $M$. Define $$\frac{1}{3!}\Omega_{ijk}\mathrm{d}x^i\wedge\mathrm{d}x^j\wedge\mathrm{d}x^k~=~\Omega~:=~\mathrm{d}\omega~\in~\Omega^3(M).\tag{3}$$ Then a compatible torsionfree tangent-bundle connection $\nabla$ exists iff $\Omega=0$. In the affirmative case, there are infinitely many such connections.
Remark. Such connections are called Fedosov connections, cf. e.g. this Phys.SE post.
Sketched proof of Theorem. One may show (via partition of unity) that there exists a torsionfree connection $\nabla^{(0)}$ on a paracompact manifold $M$. We search for a compatible torsionfree connection $\nabla$. The difference in Christoffel symbols $$\Delta\Gamma^k_{ij}~:=~\Gamma^k_{ij}-\Gamma^{(0)k}_{ij}~=~(i\leftrightarrow j)\tag{4}$$ is a $(1,2)$ tensor. If we define a $(0,3)$-tensor $$\Delta\Gamma_{k,ij}~:=~\omega_{k\ell}\Delta\Gamma^{\ell}_{ij}~=~(i\leftrightarrow j),\tag{5}$$ the compatibility condition $$\nabla_k\omega_{ij}~\stackrel{?}{=}~0\tag{6}$$ then turn into the tensor equation $$\begin{align} -\nabla^{(0)}_k\omega_{ij}~\stackrel{?}{=}~&(\nabla_k-\nabla^{(0)}_k)\omega_{ij}\cr ~=~&-\Delta\Gamma^{\ell}_{ki}\omega_{\ell j}-\omega_{i \ell}\Delta\Gamma^{\ell}_{kj}\cr ~=~&\Delta\Gamma_{j,ki}-\Delta\Gamma_{i,kj}.\end{align}\tag{7}$$ If we define a $(0,3)$-tensor $$ S_{k,ij}~:=~\Delta\Gamma_{k,ij}-\frac{1}{3}\left(\nabla^{(0)}_i\omega_{kj}+\nabla^{(0)}_j\omega_{ki} \right)~=~(i\leftrightarrow j),\tag{8}$$ then eq. (7) becomes $$\begin{align} \frac{1}{3}\Omega_{ijk} ~=~&\frac{1}{3}\sum_{ijk~{\rm cycl.}}\partial_i\omega_{jk}\cr ~=~&\frac{1}{3}\sum_{ijk~{\rm cycl.}}\nabla^{(0)}_i\omega_{jk}\cr \stackrel{?}{=}~&S_{i,kj}-S_{j,ki}.\end{align}\tag{9}$$ If $g_{ij}$ is a positive definite diagonal metric (in one coordinate system), then eq. (9) implies $$ 0~\leq~\Omega_{ijk}(g^{-1})^{i\ell}(g^{-1})^{jm}(g^{-1})^{kn}\Omega_{\ell mn}~\stackrel{?}{=}~0,\tag{10}$$ which in turn implies that $$\Omega~=~0.\tag{11}$$ Conversely if $\Omega=0$, then eq. (9) becomes $$S_{i,kj}~\stackrel{?}{=}~S_{j,ki},\tag{12} $$ i.e., that $$S~\stackrel{?}{\in}~{\rm Sym}^3T^{\ast}M\tag{13}$$ is a totally symmetric tensor. But there are infinitely many such totally symmetric tensors $S$, cf. my Phys.SE answer here. So the sought-for connection $\nabla$ is far from unique. $\Box$
References: