How draw axes & figure with TikZ using "mathematician's axes"?

Question

Typically when drawing a 3D figure on paper (or a black/white board), a mathematician draws the z-axis to point due north on the page, the y-axis to point due east, and the x-axis to point due southwest (as if coming out of the page). And then the 3D objects are drawn on that; e.g., when a sphere is drawn on those axes, it looks head-on like a circle, without distortion, so that its cross-section in the yz-plane is a perfect circle.

In the answer https://tex.stackexchange.com/a/447401/13492 by @Max to my question Draw lower (southern) hemisphere and great semicircle with "mathematician's" axes orientation, he shows how to create a 3d set of axes that look like that, using a Cabinet projection.

However, when the Cabinet projection is applied to a sphere, it distorts the sphere's shape, as he shows.

Is there a way to create with TikZ a 3D drawing that does draw the axes as I've described, but does not distort solid object such as a sphere?

Answer 1

No, TikZ cannot do that. However, that's not the fault of TikZ.

\documentclass[fleqn]{article}
\usepackage{amsmath,marvosym}
\begin{document}
You wish to have a coordinate system that 
\begin{enumerate}
 \item preserves shapes and\label{preserve}
 \item has $\vec e_y$ point east, $\vec e_z$ point north and $\vec e_x$ point
 south west.\label{directions}
\end{enumerate}
The first requirement means that the coordinate axes are orthogonal,
\begin{equation}\label{eq:orthogonality}
 \vec e_x\cdot \vec e_y~=~\vec e_x\cdot \vec e_z~=~\vec e_y\cdot \vec e_z
~=~0\;.
\end{equation}
So we wish to find a two--dimensional projection of these vectors that fulfill
the requirement \ref{directions}. Decompose the vectors in two--dimensional
projections on the paper plane $\vec e_i^{(\|)}$ and the orthogonal complements
$\vec e_i^{(\perp)}$. Clearly, the $\vec e_i^{(\perp)}$ are just
one--dimensional objects, which we will just call $e_i^{(\perp)}$. Requirement \ref{directions} implies that
\begin{equation}
 \vec e_x^{(\|)} \cdot \vec e_y^{(\|)}~=~
\vec e_x^{(\|)} \cdot \vec e_z^{(\|)}~=:~\xi~\ne~0\;.
\end{equation}
Due to the orthogonality relations \eqref{eq:orthogonality}, this means that
\begin{equation}
 e_x^{(\perp)} \cdot e_y^{(\perp)}~=~
 e_x^{(\perp)} \cdot e_z^{(\perp)}~=~-\xi~\ne~0\;.
\end{equation}
None of the $e_i^{(\perp)}$ may vanish as otherwise there won't be an $x$--axis,
and
\begin{equation}
e_y^{(\perp)}~=~e_z^{(\perp)}~=~-\frac{\xi}{e_x^{(\perp)}}\;. 
\end{equation}
However, requirement \ref{directions} 
implies that $\vec e_y^{(\|)} \cdot \vec e_z^{(\|)}=0$, so
\begin{equation}
 \vec e_y\cdot \vec e_z~=~\vec e_y^{(\|)} \cdot \vec e_z^{(\|)}
 +e_y^{(\perp)} \cdot e_z^{(\perp)}~=~0+\left(\frac{\xi}{e_x^{(\perp)}}\right)^2
 ~\ne~0\;.\qquad \text{\Huge\Lightning}
\end{equation}
\end{document}