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It is known that any complex manifold, $M$ admits a Hermitian metric, i.e., a Riemannian metric, $g$, which satisfies \begin{equation} g_p(J_pX,J_pY)=g_p(X,Y) \end{equation} at each point $p\in M$, where $X,Y\in T_p M$ and $J$ is the complex structure.

The proof of this is that given any Riemannian metric $\hat{g}$ on $M$, one can obtain a Hermitian metric $g$ using \begin{equation} g_p(X,Y)\equiv \frac{1}{2}(\hat{g}_p(X,Y)+\hat{g}_p(J_pX,J_pY)). \end{equation}

My question is, does there always exist a diffeomorphism (i.e., a local coordinate transformation) that takes us from $\hat{g}$ to $g$?

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$\newcommand{\Reals}{\mathbf{R}}\newcommand{\dd}{\partial}$No: For example, let $M$ be $\Reals^{2}$ equipped with the standard complex structure $J(\dd_{x}) = \dd_{y}$ and $J(\dd_{y}) = -\dd_{x}$. If $E$ and $G$ are positive functions and $$ \widehat{g} = E(x, y)\, dx^{2} + G(x, y)\, dy^{2}, $$ then $$ \widehat{g}(J \cdot, J\cdot) = G(x, y)\, dx^{2} + E(x, y)\, dy^{2}, $$ so $$ g = \tfrac{1}{2}\bigl(G(x, y) + E(x, y)\bigr)\, (dx^{2} + dy^{2}). $$ For suitable non-constant functions $E$ and $G$, the metric $g$ has constant curvature (as does any metric diffeomorphism-equivalent to $g$), while $\widehat{g}$ does not have constant curvature.

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