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I am trying to understand the proof of lemma 3.1, in this paper

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?

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  • $\begingroup$ originally, a metric $g$ is a pairing between vectors, but it also induces a metric on convectors or any other tensor field; this is the first one. For the second one, you can apply the one form $dz_i$ to the vector field valued 1-form $\nabla \tau_k$ and the result is a 1-form which you want to be zero. $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Commented Nov 27, 2014 at 14:30
  • $\begingroup$ In my question, I mean, why we have first and second equality. Why left hand side gives right hand side $g(dz_i,d\tau_k)=dz_i(\nabla\tau_k)$? $\endgroup$
    – Alon
    Commented Nov 27, 2014 at 14:35

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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 $\nabla \tau_k \cdot z_i = 0$.

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  • $\begingroup$ Thanks Eric, can you tell me , please why $\nabla \tau_k = JV_k$ and about second $g(dz_i,\theta_k)=0$, because we can not get it from your method of solution? $\endgroup$
    – Alon
    Commented Nov 27, 2014 at 16:02
  • $\begingroup$ why we have $g(dz_i,\theta_k)=0$. Do you have a refference? $\endgroup$
    – Alon
    Commented Nov 28, 2014 at 15:46

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