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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:

http://mathoverflow.net/questions/75841/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

http://ocw.mit.edu/courses/mathematics/18-966-geometry-of-manifolds-spring-2007/lecture-notes/

Thanks for helping to clear out my confusion.

flag	asked Apr 23 2012 at 0:10 Gigou 85●4

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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 2012 at 2:15

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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 – Paul Skerritt Apr 23 2012 at 7:47

Thank you both! – Gigou Apr 23 2012 at 12:51

Are the connection 1-forms of a hermitian connection on line bundle imaginary or not?

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