The Lifiting Theorem by Béthuel [1,Lem.4.2] states
Theorem Assume $\mathbb{T}^3 \simeq \Omega=[-\pi n, \pi n]$ is the 3-dimensional torus obtained by identifying opposing edges.
Assume further $v \in H^1(\mathbb{T}^3,\mathbb{C}) \simeq H^1_{per}(\Omega, \mathbb{C})$ with \begin{equation} \vert v(x) \vert \geq \frac{1}{2} ~~~~~ \forall x \in \mathbb{T}^3. \end{equation}
Then one can rewrite $v=\vert v \vert \exp(i \varphi)$ with $\varphi \in H^1(\mathbb{T}^3,\mathbb{R})$.
Proof As $\vert v \vert \geq \frac{1}{2}$ we can write $v= \vert v \vert w$ with $\vert w \vert=1$.
Question 1 Why is $\vert v \vert \geq \frac{1}{2}$ necessary in order to obtain $v= \vert v \vert w$? Can't I just do that for every function that maps to $\mathbb{C}$?
Then we calculate $d(w \times dw) =0 $ and therefore the Hodge-de-Rham decomposition is written as \begin{equation} w \times dw = d \varphi + \sum_{j=1}^3 \alpha_j d x_j \end{equation} where $\alpha_j \in \mathbb{R}$ and $\varphi \in H^1(\mathbb{T}^3,\mathbb{C})$.
Question 2 How does he calculate $d(w \times dw)$? Why is that even possible? We have no regularity of $w$ so far and I thought the differential form \begin{equation} dw(x) = \sum_{i=1}^3 w_{x_i}(x) dx_i(x) \in (\mathbb{R}^n)^\ast\end{equation} is not defined for non-smooth functions $w$.
Everything else then follows from properties of the Hodge-de-Rham decomposition.
[1] Béthuel, F., P. Gravejat und J. C. Saut: Travelling waves for the Gross- Pitaevskii equation. II. Comm. Math. Phys., 285(2):567–651, 2009.
Nobody said it was necessary. The statement "Since $A$, it follows that $B$" does not contain an implication that $A$ is necessary for $B$. Rather, it says that $A$ is sufficient for $B$.
The restriction $|v|\ge \frac12$ is important later in the proof (although the particular value $\frac12$ could as well be replaced by any positive number).
What the authors really meant could be "As $v\ne 0$, we can write $v=|v|w$ with $|w|=1$, where $w$ is uniquely determined by $v$".
Actually we do, and this is where $|v|\ge 1/2$ is used. The map $z\mapsto z/|z|$ is Lipschitz on the set $\{z:|z|\ge 1/2\}$. It is well-known that composition with a Lipschitz function preserves first-order Sobolev spaces. Thus, $w\in H^1(\mathbb T^3)$.
As explained above, $dw$ makes sense: it is a form with $L^2$ coefficients. For the meaning of $d(w\times dw)$, see the text after Remark 4.2: it is the Jacobian of $w$, thought of as a map into the plane. It's formed by multiplying first-order derivatives of $w$, and therefore is a well-defined $L^1$ function.
Since $w$ takes values on the circle, the $2\times 3$ matrix of its partial derivatives has rank $1$, and this is why the Jacobian vanishes.