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I read in a few places that for a hermitian line bundle endowed with a hermitian connection, the local connection 1-forms are purely imaginary-valued. I tried to prove this, and came up with something contradictory. In fact, what I did is exactly what Pr. O'Farrill did in this post:

Connections with compatible Hermitian products on complex line bundles

His last equation $\beta + \overline{\beta}=dh$ tells me that the real part of the connection 1-form $\beta$ generally does not vanish.

On the other hand, I found that contrary statement in Demailly's Complex analytic and differential geometry book p.299, and in D. Auroux's online lecture notes. See Lecture 10 at

Thanks for helping to clear out my confusion.

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The connection 1-form depends on the choice of local trivialisation for the line bundle. If you choose a hermitian trivialisation, in the sense that $h \equiv 1$, then the connection 1-form is imaginary. –  Johannes Nordström Apr 23 '12 at 2:15
Johannes is correct - remember (in Auroux's notation) the connection form $A$ on a Hermitian vector bundle is expressed with respect to a local frame of sections $\lbrace e_i \rbrace$, as explained in lecture 8. In the case that the sections are chosen orthonormal ($\langle e_i, e_j\rangle=\delta_{ij})$, $A$ becomes skew-Hermitian. This condition ensures that $\nabla$ is compatible with the Hermitian metric ($X\langle s,t\rangle=\langle \nabla_Xs,t\rangle+ \langle s,\nabla_Xt\rangle$). For a line bundle obviously the frame just consists of a local section, and the connection is pure imaginary –  user17945 Apr 23 '12 at 7:47
Thank you both! –  Gigou Apr 23 '12 at 12:51

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