5
$\begingroup$

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. \begin{equation} \omega=i\partial\overline{\partial}\left(\frac{|z|^{2}}{2}+\mathcal{O}(|z|^{4})\right) \end{equation}

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?

$\endgroup$
4
$\begingroup$

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 $$ \phi \equiv_4 \tfrac12\bigl(|w^1|^2+\cdots+|w^n|^2\bigr). $$

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – student
    Oct 11 '14 at 11:31
  • $\begingroup$ 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.) $\endgroup$ Oct 11 '14 at 12:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.