Why does gimbal lock occur “in two circles”?

Question

Let $S^1=ℝ/2\pi ℤ$. Consider the Euler angle parametrization $$ T^3 \to SO(3),\\\ (ξ, υ, ζ)\mapsto R^Z_{ζ}R^Y_{υ}R^X_{ξ}, $$ where $T^3\cong S^1\times S^1\times S^1$ is the 3-torus and $$ R^X_ξ=\begin{pmatrix}1&0&0\\\ 0& \cos ξ&-\sin ξ\\\ 0& \sinξ &\cos ξ\end{pmatrix} $$ represents the rotation around the $x$-axis etc. This map has singular values whenever $υ=\pm \pi/2$, a phenomenon which is known as gimbal lock.

It is well known that we cannot circumvent the existence of singular values with such a three-angle parametrization: If a continuous map $T^3\to SO(3) $ would have only regular values, it would have to be a covering map, because it's proper. But the only cover of $SO(3)\cong \mathbb RP^3$ is $S^3$, contradiction.

However, couldn't it be possible that there exists a different map $T^3\to SO(3)$ which posesses „less“ singular values?

Calculating the image of the singular values $S^1\times \{\pm \pi/2\}\times S^1$, we get (unless I miscalculated) $$ \begin{pmatrix}0 & \sin(ξ-ζ) & \cos(ξ-ζ) \\\ 0 & \cos(ξ-ζ) & -\sin(ξ-ζ) \\\ -1 & 0 & 0\end{pmatrix},\qquad \begin{pmatrix}0 & -\sin(ξ+ζ) & -\cos(ξ+ζ) \\\ 0 & \cos(ξ+ζ) & -\sin(ξ+ζ) \\\ 1 & 0 & 0\end{pmatrix}, $$ where $ξ, ζ$ range over $S^1$. This is homeomorphic to $S^1\sqcup S^1$.

Question

• Is there a continuous surjection $T^3 \to SO(3)$ whose (image of the) set of singular values is „topologically simpler“ than $S^1\sqcup S^1$, e.g. by being of the form $S^1\times D$ or $D$ where $D$ is discrete?
• If not, is there a theorem putting qualitative restrictions on what this subspace of $SO(3)$ can look like?

Answer 1

As explained in this MathOverflow answer, there is a way to realise any closed orientable connected 3-manifold (in particular, $T^3$) as a branched covering of $S^3$. By projecting further to $\mathbb{RP}^3\cong\mathrm{SO}(3)$, one can therefore reduce the singular set to finitely many points.

However, this completely destroys one useful property of the Euler angle parametrisation, namely the fact that it is a group homomorphism when restricted to the standard ”coordinate“ 1-parameter subgroups of $T^3$. If one adds this restriction, I believe, one wouldn't be able to do better than with some 1-dimensional singular set; your computation essentially shows this for the case where the 1-parameter subgroups in $\mathrm{SO}(3)$ obtained this way are the standard generators.