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Let $X$ be a $2n$-dimensional $C^{\infty}$-manifold, let $G$ be a finite group acting smoothly on $X$. Let $J: TX\to TX$ be a map such that $J^2=-id$ and $g_*Jg^{-1}_*=J$ for any $g\in G$.

We can extend $J$ to a $\mathbb{C}$-linear map $J: T_{\mathbb{C}}X\to T_{\mathbb{C}}X. $Let $T^{0,1}X$ be the $-1$ eigenspace of $J$. We call $J$ is integrable if $[T^{0,1}X,T^{0,1}X]\subset T^{0,1}X$.

My question is: do we have a $G$-equivariant Newlander-Nirenberg theorem? More precisely, if $J$ is integrable, then is it true that $X$ is a complex manifold with $G$-acting holomorphically?

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    $\begingroup$ Viewing $g$ as a smooth map $X \to X$, do you mean $g_*Jg_*^{-1} = J$? Also, an almost complex structure is a map $J : TX \to TX$, not $T_{\mathbb{C}}X \to T_{\mathbb{C}}X$ (of course, the former extends to the latter). $\endgroup$
    – Michael Albanese
    Commented Jan 23, 2023 at 23:16
  • $\begingroup$ @MichaelAlbanese You are right. I made the edit according to your comments. $\endgroup$
    – Zhaoting Wei
    Commented Jan 24, 2023 at 0:57
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    $\begingroup$ In that case, the answer to your question is yes. If $J$ is integrable, then $X$ is a complex manifold, and a map $g : X \to X$ is holomorphic if and only if $g_*J = Jg_*$. This condition on $g_*$ is equivalent to the Cauchy-Riemann equations for $g$ in smooth coordinates obtained as the real and imaginary parts of holomorphic coordinates, see here for the one-dimensional case. $\endgroup$
    – Michael Albanese
    Commented Jan 24, 2023 at 4:27

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