Let $M$ be a complex manifold. Consider a connection $\nabla$ on the holomorphic tangent bundle $T^{1,0}M$. The torsion of $\nabla$ is defined as the torsion of the induced connection $D$ on the real tangent bundle, $$T_\nabla(\alpha,\beta) = T_D(\alpha,\beta) := D_\alpha \beta – D_\beta \alpha – [\alpha,\beta]$$ for smooth vector fields $\alpha$ and $\beta$. The connection $\nabla$ is torsion-free if $T_\nabla = 0$.
Assume now that $\nabla$ is torsion-free. It is well-known that if $\nabla$ is also hermitian, i.e., compatible with a hermitian metric $h$ on $T^{1,0}M$, then it is the Chern connection of the metric, i.e., $\nabla$ is a $(1,0)$-connection in the sense that $\nabla^{0,1}=\bar\partial$, and $h$ provides a Kähler metric on $M$, see e.g., Huybrechts, Complex Geometry, Proposition 4.A.7.
I am interested in whether one can find torsion-free $(1,0)$-connections on $T^{1,0} M$ also on non Kähler manifolds? Such connections would thus necessarily not be hermitian by the above result. Or is there some obstruction?
Best Answer
I will write in terms of the holomorphic frame bundle, i.e. the bundle of choices of complex linear bases of tangent spaces, a holomorphic principal $\operatorname{GL}_n$-bundle. Any sum $\gamma=\sum h_a \gamma_a$ with $\sum h_a=1$ of $(1,0)$-connection forms on the holomorphic frame bundle is a $(1,0)$-connection. If all $\gamma_a$ are torsion-free, so is $\gamma$. To see this, take the soldering forms $\omega=(\omega^{\mu})$ on the frame bundle, and then the structure equations of a connection are $d\omega=-\gamma\wedge\omega+\frac{1}{2}a\omega\wedge\omega+b\omega\wedge\bar\omega+c\bar\omega\wedge\bar\omega$, with $a,b,c$ the components of the torsion. (Note that $c$ is the Nijenhuis tensor, so $c=0$ on a complex manifold.) If these vanish for all of the $\gamma_a$, they still vanish for $\gamma$. Therefore if the $h_a$ form a partition of unity, for local choices of torsion-free $(1,0)$-connections, then $\gamma$ is a torsion-free $(1,0)$-connection.