# Contraction by the Fundamental Form of A Hermitian Metric

Let $M$ be a complex manifold, with a Hermitian metric $g$ which we will consider as a $C^\infty(M)$-bi-module map $$g:\Omega^1(M) \otimes_{C^{\infty}(M)} \Omega^1(M) \to C^\infty(M),$$ where $\Omega^1(M)$ is the module of one forms of $M$. Moreover, let $\omega$ be the fundamental form of $g$. Now contraction by $\omega$ is a map $$\omega \llcorner :\Omega^1(M) \wedge \Omega^1(M) \to C^\infty(M).$$ Am I naive in thinking that there might be some simple relationship between these two maps? For example, $$(\omega \llcorner) \circ \wedge = g,$$ where $$\wedge:\Omega^1(M) \otimes_{C^{\infty}(M)} \Omega^1(M) \to \Omega^1(M) \wedge \Omega^1(M)$$ is the obvious map. Does it help if I assume that $g$ is Kaehler, ie d$\omega = 0$?

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Sorry but this isn't really appropriate for this site. You should simply write out the definition of the two maps in, say, local co-ordinates and compare them. And, if you still have any questions, I would recommend posting them on math.stackexchange.com. –  Deane Yang Apr 30 '12 at 14:24
Ok. But can I take that as a yes? –  Ago Szekeres Apr 30 '12 at 14:26
No, you can't take my answer as a yes. –  Deane Yang Apr 30 '12 at 15:02
By the way, what do you get if you apply your map to $(v,v)$, where $v$ is a non-vanishing vector field? –  Deane Yang Apr 30 '12 at 15:11
Yes I see now, a zero on the left side, and a non-zero value on the right side. Thanks. –  Ago Szekeres Apr 30 '12 at 15:58