smooth vs. analytic in the definition of almost-complex manifolds Let $A_{\infty}\hspace{-0.03 in}$ be a maximal $C^{\infty}\hspace{-0.02 in}$ atlas on $M\hspace{-0.03 in}$, and with that smooth structure on $M$,

suppose $\: j : TM\to TM\:$ is a smooth function such that
$\;\;\;$ for all vectors $v$ in TM, $\: \hspace{.03 in}j\hspace{.02 in}(\hspace{.05 in}j\hspace{.02 in}(v)) = -v \:$ and $\: \pi \hspace{.03 in}(\hspace{.05 in}j\hspace{.02 in}(v)) = \pi \hspace{.03 in}(v) \:$
$\;\;\;\;\;\;\;\;$ and
$\;\;\;$ for all points $p$ in $M\hspace{-0.03 in}$, the function $\: \hspace{.04 in}f : T_{\hspace{-0.03 in}p}\hspace{.03 in}M\to $ $T_{\hspace{-0.03 in}p}\hspace{.03 in}M$ $\:$ given by $\: \hspace{.04 in}f(v) = \hspace{.03 in}j\hspace{.02 in}(v) \:$ is a linear map
.
Does it follow that there is a unique subset $A_j$ of $A_{\infty}$ such that

$A_j$ is a maximal real analytic atlas on $M$ and $A_j$ makes $\hspace{.02 in}j$ real analytic?
(This question is due to wikipedia's definition of almost-complex manifold

only requiring smoothness, rather than real analyticity.)

 A: If $M$ has (real) dimension $2$, then the answer is yes. Every $2$-dimensional almost complex structure is integrable, and the Newlander-Nirenberg theorem says that it determines a unique complex structure, i.e., a maximal atlas of holomorphically compatible charts. This atlas is, in particular, real-analytic.
But in higher dimensions, the answer is no. For example, you could start with the standard integrable almost complex structure on $\mathbb C^2$, and then use a bump function to modify it in some open set to make it non-integrable there. The result is an almost complex structure $j$ whose Nijenhuis tensor is zero on an open set but not globally zero. If $j$ were real-analytic with respect to some analytic structure, then this would be impossible.
A: The Morrey-Grauert theorem does say that every smooth manifold has a unique analytic structure. I don't see any reason why $j$ would automatically be analytic, but the space of almost complex structures is a fiber bundle over $M$ with fiber the symmetric space $SO(2n)/U(n)$, which is in particular a manifold, so you should be able to perturb $j$ to be analytic.
