I'm trying to read the paper "Transversality in Elliptic Morse Theory" (https://people.math.ethz.ch/~salamon/PREPRINTS/trans.pdf). I'm stuck in the very beginning of the proof of the Carleman similarity principle on page 5 of the paper. My question below uses the notation of Theorem 2.2 on page 5.
$J$ is assumed to be of class $W^{1,p}$ with $p>2$. In the very first step of the proof, we choose $\Psi\in W^{1,p}(B_\delta,GL_{\mathbb R}{\mathbb C^n})$ such that $J(z)\Psi(z) = \Psi(z)i$ for all $z\in B_\delta$. I'm not able to see why we can choose such a $\Psi$. Is it completely obvious, or is there a reference where I can find it?
Let us see $\mathbb{C}^n$ as $\mathbb{R}^{2n}$ equipped with its canonical basis $\{e_1, \dots, e_n, f_1, \dots, f_n \}$. It is more or less a matter of convention, but the standard complex structure $i$ is completely determined by the relations $i(e_k) = f_k$ and $i(f_k) = - e_k$ (for all $k = 1, \dots, n$) and by linearity.
A necessary condition for the existence of linear maps $J, \Psi : \mathbb{R}^{2n} \to \mathbb{R}^{2n}$ such that $\Psi$ is invertible and such that $J\Psi = \Psi i$ is that $J$ be a complex structure i.e. $J^2 = -Id$. Indeed, from the hypotheses, we have $J = \Psi i \Psi^{-1}$ and thus $J^2 = \Psi i^2 \Psi^{-1} = \Psi(-Id) \Psi^{-1} = - Id$.
Conversely, given a complex structure $J$, there exists an invertible linear map $\Psi$ such that $J\Psi = \Psi i$. In fact, there are a lot of them. Indeed, pick any (ordered) set $S = \{ v_1, \dots, v_n \} \subset \mathbb{R}^{2n}$ having the following properties:
(i) $v_1 \neq 0$, and
(ii) for all $k \in \{2, \dots, n\}$, the vector $v_k$ is not in the linear span of the set $\{v_1, \dots, v_{k-1}, Jv_1, \dots, Jv_{k-1} \}$.
Define $\Psi$ as the unique linear map which satisfies $\Psi(e_k) = v_k$ and $\Psi(f_k) = Jv_k$. By construction, $\Psi$ is injective on $\{e_1, \dots, f_n\}$ and is thus an invertible linear map. Moreover, it satisfies $J \Psi = \Psi i$.
Let us come back to the situation of the paper. We have a continuous map which associates to $z \in B_{\epsilon}$ a complex structure $J(z)$. From what we did above, there exists $\Psi(0)$ such that $J(0)\Psi(0) = \Psi(0)i$. For $k=1, \dots, n$, set $v_k = \Psi(0)e_k$. By continuity of $J$, there exists $0 < \delta < \epsilon$ such that $S := \{v_1, \dots, v_n\}$ satisfies conditions (i) and (ii) above for all $J(z)$ with $z \in B_{\delta}$. Hence, we deduce the existence of a (unique) map $\Psi : B_{\delta} \to Gl(\mathbb{R}^{2n})$ such that $\Psi(z)e_k = v_k$ and such that $J(z)\Psi(z) = \Psi(z) i$ for all $k$ and all $z$.
We still don't know how regular this map $\Psi$ is. However, since it is unique (with respect to some choices we made), we expect to be able to obtain a rather explicit expression for it in terms of the given data (which are the 'constant map' $i$, the 'constant association' $\Psi(0) : e_k \to v_k$ and the rather regular map $J(z)$). From such an expression, we should be able to deduce the regularity of $\Psi$.
In fact, observe that $\Psi(z)$ is represented in the canonical basis by the matrix $(\Psi(z)e_1 \, \dots \, \Psi(z)f_n)$, that is
$$ \left( \Psi(0)e_1 \, \dots \, \Psi(0)e_n \; J(z)\Psi(0)e_1 \, \dots \, J(z)\Psi(0)e_n \right) = \left( v_1 \, \dots \, v_n \; \, J(z)v_1 \, \dots \, J(z)v_n \right) $$
where each entry is to be understood as a column vector. From there, it is rather straightforward to see that the $W^{1,p}$-norm of $\Psi(z)$ is controlled by the $W^{1,p}$-norm of $J(z)$, i.e. the former norm is bounded from above by the latter up to a multiplicative constant which depends on the constant vectors $v_k$. Therefore, $\Psi \in W^{1,p}(B_{\delta}, Gl(\mathbb{R}^{2n}))$ whenever $J \in W^{1,p}(B_{\delta}, Gl(\mathbb{R}^{2n}))$ .