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Let $G$ be a bounded homogeneous domain in $\mathbb{C}^{n}$ and let $z\in G$.

Assume that $f$ is a holomorphic function on $G$, which is isotropy symmetric, i.e. $f\circ \varphi=f$ for any biholomorphism $\varphi:G\to G$, such that $\varphi(z)=z$.

Q1. Is it true that $f$ is a constant function?

Seems to be true for a ball. Note, that the isotropy group of a bounded homogeneous domain is not trivial, as proven here. Edit: due to Holonomia's comment this is true for bounded symmetric domains and false in general. Hence, let me restate the second question:

Q2. Can we characterize bounded symmetric domains as such that any isotropy-invariant holomorphic function on them must be a constant?

Note, that any bounded homogeneous domain can be viewed as a Siegel domain of genus 2, but for a point of a form $(w,0)$ isotropy includes maps $(z,w)\to (z,e^{i\theta}w)$, and so any isotropy-symmetric holomorphic function must be independent of second variable. Thus it is enough to answer Q1 and Q2 only for tube domains.

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Q3. Is there any symmetry condition such that a holomorphic function which satisfy this condition must be a constant?

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    $\begingroup$ In case $G$ is symmetric you can prove that $f$ is constant by using the so called circle realization i.e. $0 \in G$ and the maps $f_{\theta}: Z \to e^{i \theta} Z$ belongs to the isotropy at $0$. Instead, it seems to me that the function $f(z_1,z_2,z_3,u) = z_3^2$ of the non-symmetric homogeneous Siegel domain $S_1$ in $C^4$ constructed by Piatetski-Shapiro is invariant by the isotropy group at $(1,1,0,0)$. See page 26 of "Automorphic functions and the Geometry of Classical domains" for the definition of $S_1$ and the computation of the relevant automorphisms. $\endgroup$
    – Holonomia
    Commented Mar 22, 2017 at 9:58
  • $\begingroup$ @Holonomia I am not sure I understand your notations for your first statement: what is $Z$? Is the action of $S^1$ sufficient to prove that a function is a constant? About your second statement, I am not sure I understand what exactly is the group that they have calculated? Is this exactly the group of all automorphisms, or are these only transformations of some particular form (they call them linear)? $\endgroup$
    – erz
    Commented Mar 22, 2017 at 19:50
  • $\begingroup$ $Z$ is an arbitrary point of $G$. Yes, this action of $S^1$ is enough: take the Taylor development at $0$ of $f$ and observe that any homogeneous term must be invariant by $S^1$. $\endgroup$
    – Holonomia
    Commented Mar 22, 2017 at 22:27

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