# Torsion-free and compatible connection for non-symmetric non-degenerate $(0,2)$ tensor field?

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:

1. Torsion-free: $$\Gamma^a_{bc} = \Gamma^a_{(bc)}$$ (symmetric in lower indices).
2. 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?

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.
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. $$\Box$$