I am trying to understand the proof of lemma 3.1, [in this paper](http://arxiv.org/pdf/0903.2413v1)


In proof, they say that $g(dz\_i,d\tau\_k)=dz\_i(\nabla\tau\_k)=0$ I don't understand first and second equality.In second they say, $g(dz\_i,\theta\_k)=0$ by using $J(dz\_i)={\sqrt {-1}}dz\_i$, but how?


Also, why $\nabla \tau\_k = JV\_k$. Can someone explain for me?



Moment map coordinates in tours action
In the last paragraph of the previous page, they say that $\nabla \tau\_k$ is the gradient of $\tau\_k$. This means that if $\tilde g : TM \to T^\*M$ denotes the isomorphism induced by the metric (i.e. $\tilde g(v)(w) = g(v,w)$) then
$$
\tilde g (\nabla \tau\_k) = d\tau\_k.
$$
Then
$$
g(dz\_i, d\tau\_k) = g(\tilde g^{-1} dz\_i, \tilde g^{-1} d \tau\_k) = g(\tilde g^{-1} dz\_i, \nabla\tau\_k) = dz\_i(\nabla \tau\_k) = \nabla \tau\_k \cdot z\_i.
$$
They define the coordinates $z\_i$ to be on the quotient manifold and they say that $\nabla \tau\_k = JV\_k$ is tangent to the orbits. This means that the $z\_i$ is constant in the $\nabla \tau\_k$ direction, which gives $\nabla \tau\_k \cdot z\_i = 0$.


