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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?

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  • $\begingroup$ Maybe there's something I'm missing, but can you not find such connections by picking a Riemannian metric on the underlying smooth manifold and taking its Levi-Civita connection? $\endgroup$
    – Gunnar Þór Magnússon
    Commented Jan 27, 2022 at 12:29
  • $\begingroup$ Hi Gunnar! If I'm not misreading the cited result and the surrounding paragraphis in Huybrechts, then in order for the induced connection $\nabla$ to be hermitian, the Riemannian metric on the underlying smooth manifold would have to be compatible with the hermitian structure, and in that case, the connection on the holomorphic tangent bundle corresponding to the Levi-Civita connection on the real tangent bundle is a $(1,0)$-connection if and only if the metric is Kähler. $\endgroup$
    – Richard Lärkäng
    Commented Jan 27, 2022 at 14:38
  • $\begingroup$ Hi! The whole point is to find torsion-free non-hermitian connections, no? The Levi-Civita connection $\nabla$ of a random Riemannian metric won't be compatible with the complex structure, so it won't be hermitian, but it will be torsion-free. You then get a connection on the holomorphic tangent bundle by extending $\nabla$ to the complexified tangent bundle, injecting $T_M^{1,0}$ into that, and projecting down again according to the orthogonal splitting into holo- and antiholomorphic vector fields. I think that works, at least. $\endgroup$
    – Gunnar Þór Magnússon
    Commented Jan 31, 2022 at 9:07
  • $\begingroup$ Sorry for the confusion, I wrote "hermitian", but what I wanted to say was "compatible with the complex structure". I would like to find is a torsion-free (1,0)-connection on the holomorphic tangent bundle. I suppose it is implicit that the connection should be complex linear. If I interpret Proposition 4.A.8 in Huybrechts correctly, it seems like the connection on the holomorphic tangent bundle induced by the Levi-Civita connection (via the smooth isomorphism between the smooth and the holomorphic tangent bundle) is complex linear if and only if the metric induces a Kähler metric. $\endgroup$
    – Richard Lärkäng
    Commented Jan 31, 2022 at 10:09
  • $\begingroup$ It is not really clear to me that your suggested procedure, projecting onto $T^{1,0}_M$, really produces a complex linear $(1,0)$-connection, but maybe it does. In any case, the answer of Ben MacKay seems to provide a simple solution for finding the connections I want. $\endgroup$
    – Richard Lärkäng
    Commented Jan 31, 2022 at 10:10

1 Answer 1

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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.

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  • $\begingroup$ Ah, it was as simple as patching together local connections, thank you! I believed I had thought through in my head that this procedure would fail, but it seems like you are right that it actually works. $\endgroup$
    – Richard Lärkäng
    Commented Jan 28, 2022 at 12:58

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