Let $V$ be a finite-dimensional complex vector space with a linear action of a complex reductive group $G$. Suppose that $\omega\_0$ and $\omega\_1$ are two $G$-invariant complex symplectic bilinear forms on $V$. Is there always a $G$-equivariant isomorphism $\varphi:V\to V$ such that $\varphi^\*\omega\_1=\omega\_0$?


**Remark.** This is not true for real symplectic forms and compact Lie group actions [Dellnitz-Melbourne. *The equivariant Darboux theorem.* Lect. Appl. Math 29 (1993): 163-169].



Equivalence of $G$-invariant symplectic forms
It is true that any two $G$-invariant symplectic forms on a representation are equivalent. A proof of this fact is provided in F. Knop: "Classification of multiplicity free symplectic representations", J. Algebra 301 (2006) 531–553. See Thm 2.1 (b). Since the question is so natural, I suspect that thismust have been observed earlier, though.


The argument is roughly as follows: Let $(V,\omega)$ be a symplectic representation and let $U\subseteq V$ be irreducible. If $U$ carries an invariant symplectic form then one can find a copy of $U$ in $V$ such that $\omega|\_U\ne0$. Then $V=U\oplus U^\perp$ and one proceeds by induction. If $U$ is not symplectic then $V$ contains a copy of $U^\*$ such that $\omega$ restricted to $\overline U=U\oplus U^\*$ is non-degenerate. Then $V=\overline U\oplus \overline U^\perp$ and one proceeds by induction.


The reason why the proof breaks down over $\mathbb R$ is that on an an absolutely irreducible symplectic representation $V$ a scalar $a\in\mathbb R^\*$ transforms $\omega $ into $a^2\omega$. So $\omega$ and $-\omega$ are not equivalent.


