# Special Kähler normal coordinates around a point

Let $(M,\omega)$ be a compact Kähler manifold and suppose there are holomorpic vector fields vanishing at a point $p$. As a consequence we have a group $G_{p}$ of biholomorpisms fixing $p$. Let $T_{p}$ a maximal torus of $G_{p}$. Suppose $\omega$ is invariant for the action of $T_{p}$. Can we find a neighborhood $\mathfrak{U}$ of $p$ and a set of holomorphic coordinates $z$ centered at $p$ such that the action of $T_{p}$ is linear (unitary) and coordinates $z$ are Kähler normal coordinates i.e. $$\omega=i\partial\overline{\partial}\left(\frac{|z|^{2}}{2}+\mathcal{O}(|z|^{4})\right)$$

Separately these two requests can be satisfied, but I can't prove that can hold simultaneously. Is there a reference for this? Or can someone give me a hint?

I don't know a reference, but this desired normal form is, indeed, attainable. Here is the argument:

Assume given a Kähler form $\omega$ defined on a neighborhood of $0\in\mathbb{C}^n$ and that there is a torus (i.e., a connected compact abelian Lie group) $\mathbb{T}$ acting effectively and holomorphically on $\mathbb{C}^n$ as $\omega$-isometries and fixing $0\in\mathbb{C}^n$.

As the OP points out, it is known that there exist holomorphic coordinates $z = (z^r)$ centered on $0$ in which the action of $\mathbb{T}$ is linear, i.e., that there exist $n$ characters (i.e., homomorphisms) $\chi_r:\mathbb{T}\to S^1$ such that $$a^*(z^r) = \chi_r(a)\,z^r\qquad \text{(no sum)}$$ for all $a\in \mathbb{T}$.

Now, write $\omega = \mathrm{i}\partial\bar\partial\phi$, where $\phi$ is a Kähler potential for $\omega$ in a neighborhood of $0$. Since $\omega$ satisfies $a^*\omega = \omega$ for all $a\in\mathbb{T}$, by averaging $\phi$ with respect to $\mathbb{T}$, one can assume that $\phi$ is also $\mathbb{T}$-invariant. After discarding holomorphic and anti-holomorphic terms (which are $\mathbb{T}$-invariant and don't show up in $\omega$), it can be assumed that $\phi$ has a Taylor expansion up to terms vanishing to order $4$ of the form $$\phi \equiv_4 h_{p\bar q}\,z^p\overline{z^q} + a_{pq}^r\,z^pz^q\overline{z^r} + \overline{a_{pq}^r}\,\overline{z^p}\overline{z^q}z^r$$ where $\overline{h_{p\bar q}} = h_{q\bar p}$ is positive definite, and $a_{pq}^r = a_{qp}^r$.

Now, because $\phi$ is $\mathbb{T}$-invariant, it follows that $h_{p\bar q} = 0$ unless $\chi_p = \chi_q$. Thus, by collecting the $z^j$ that share the same character and making a 'block-form' linear transformation that preserves these subsets, one can arrange that the $z^p$ have been chosen so that $h_{p\bar q} = \delta_{pq}$, i.e., that the quadratic term in $\phi$ is simply $\tfrac12\bigl(|z^1|^2+\cdots+|z^n|^2\bigr)$.

Now, again because of the $\mathbb{T}$-invariance of $\phi$, one sees that $a_{pq}^r = 0$ unless $\chi_r = \chi_p\chi_q$. Now, consider the change of variables of the form $$w^r = z^r + 2a_{pq}^r\,z^pz^q.$$ Because $a_{pq}^r$ vanishes unless $\chi_r = \chi_p\chi_q$, it follows that $a^*(w^r) = \chi_r(a)\,w^r$ for all $r$, so the $w$-coordinates also linearize the action of $\mathbb{T}$. Moreover, one clearly has $$\phi \equiv_4 \tfrac12\bigl(|w^1|^2+\cdots+|w^n|^2\bigr).$$

• I have tor try now but do you think it is possible to get the fourth order term of the form $$\sum_{i,j,k,l}a_{ik}^{jl}z^{i}\overline{z^{j}}z^{k}\overline{z^{l}}$$ or there is some obstruction? Oct 11, 2014 at 11:31
• Yes, the $\mathbb{T}$-invariant terms of type $(3,1)$ and $(1,3)$ in $\phi$ can be removed by a change of variables of the form $$w^s = z^s + a_{pqr}^s\,z^pz^qz^r,$$ for the same reasons that one can remove the terms of type $(2,1)$ and $(1,2)$. (Of course, the $(4,0)$ and $(0,4)$ terms can be discarded.) Oct 11, 2014 at 12:14