I have a problem that someone with more experiece and knowledge of differential geometry can hopfully solve.
While studying a problem I was working on, I encountered a specific "manifold" in 4D space which held a certain significance, and I wanted to explore functions that have special boundary condition at the custom manifold, and for that purpose I tried to build a new coordinate system that naturally describe the shape I have in mind.
The manifold is defined in 4D space $\vec{x}= (w,x,y,z)$ using the equation: \begin{equation} \mathcal{M}=\{(w,x,y,z)\in \mathbb{R}^4,F(\vec{x}) = w z-xy = 0\}, \end{equation}
and I was looking for a coordinate system
\begin{equation} \left(F(\vec{x}),B(\vec{x}),C(\vec{x}),D(\vec{x})\right) \end{equation}
for which the first coordinate is the function $F$, and like cartesian and polar coordinates, will produce 4 perpendicular unit vectors as the gardiants of the different coordinates. For example, in polar coordiantes we have two unit vectors $\hat{r} \propto \vec{\nabla}{r} , \hat{\theta} \propto \vec{\nabla}{\theta}$, which are perpendicular to each other at every point in space (possibly excluding the origin point).
For my new system I took $F(\vec{x})$ as the first coordinate, which created a vector perpendicular to the manifold $\mathcal{M}$
\begin{equation} A = w z-xy, \quad \vec{\nabla}A = (z, -y, -x, w). \end{equation}
Next I tried to seek the rest of the coordiantes by finding other vectors in the tangent space that are perpendicular to each other. I found the following vectors:
\begin{align} &B = \frac{w^2 + x^2}{2} - \frac{y^2 + z^2}{2}, &\quad& \vec{\nabla}B = (+w , +x , -y , -z)& \\ &C = wy+xz, &\quad& \vec{\nabla}C = (+y , +z , +w , +x)& \\ \end{align}
which are both perpendicular to each other and to the vector $\vec{\nabla}A$. However, the forth vector presented a problem. The remaining vector in the tangent space is automatically determined by the rest of them as \begin{equation} \vec{\nabla}D \propto (x , -w , z , -y), \end{equation} for which I couldn't find a scalar function $D $ as a coordinate. This last vector is quite interesting, as it is similar to the $\vec{\theta}$ vector in polar coordinates $(-y,x)$. I tried to find a normalization that will work for finding the missing coordinate, but I couldn't make any progress.
I tried to search for other vectors systems, but each try always ended with one or more vectors that I cannot shape into a gradient of a scalar function.
TLDR
I tried to "invent" a new proper coordinate system for 4D when the first coordinate is $wz-xy$, and couldn't find one.
Now I have three questions:
How can I be sure whether a vector field $\vec{V}(\vec{x})$ can be renormalized into a gradient of a scalar function? Is there a normalization function $f(\vec{x})$ that will "fix" the vector field and make it the result of a gradient operator $f \cdot \vec{V} = \vec{\nabla}D$ ?
Is there a better method to search for the other coordinates, given the first one?
Is there a deeper reason to why I couldn't come up with a complete solution? Are there some expressions that cannot be a part of a coordinate system? Am I going against some fundamental theorem of differential geometry?