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I'm working with dynamical system defined on $SO(3)$ and have to deal with the issue of "gimbal lock" (or similar) in all coordinate charts that I've found so far. It seems that this is a necessary issue, however I'm at a loss finding a convincing argument why it always has to occur.

Ideally I would be able to find a surjective (and likely not bijective) differentiable map from $\mathbb{R}^3$ to $SO(3)$.

What is the reason which prohibits such map to exist?

Edits as result of comments:

Suppose I find $f : \mathbb{R}^3 \to SO(3)$ sufficiently nice. Then I can pullback vector field describing dynamics from $SO(3)$ to $\mathbb{R}^3$ and study it (numerically) in an easier topology without loss of generality.

Demanding that $f$ is differentiable covering is certainly sufficient but likely not necessary.

Radost
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    $SO(3)$ is homeomorphic to $\Bbb RP^3$ (see here), and $\Bbb RP^3$ has $\Bbb S^3$ as the universal cover (consider antipode identification of $\Bbb S^3$). If $\Bbb R^3$ were cover $SO(3)$, then it would be universal cover ($\Bbb R^3$ is contractible, hence simply connected). Then by the uniqueness of the universal cover, $\Bbb R^3$ would be homeomorphic to $\Bbb S^3$, a contradiction as $\Bbb R^3$ is non-compact, though $\Bbb S^3$ is compact. – Sumanta Feb 18 '22 at 13:48
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    @User Why not an official answer? – Paul Frost Feb 18 '22 at 14:13
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    Do you want to find a arbitrary smooth surjection or a covering map? – Paul Frost Feb 18 '22 at 14:15
  • @PaulFrost coverings are great but the condition seems too strong as User's eloquent answer shows. It seems that condition that preimage of any neighborhood is a disjoint collection of homeomorphic things is more than I need. – Radost Feb 18 '22 at 14:18
  • @PaulFrost I'd like to find a chart that doesn't have any 'nasty' points. It seems to me that 'nasty' points arise in locations where 'chart' fails to be local diffeomorhpism (is that so?) – Radost Feb 18 '22 at 14:20
  • What do you mean by a chart? A chart on the manifold $SO(3)$? And what are nasty points? – Paul Frost Feb 18 '22 at 14:27
  • @PaulFrost Formalizing sufficiently well behaved seems to be the crux of the issue here. I'd like to pull dynamics from SO(3) to R^3 wlog, anything that can do that would be enough (cf. edits to question) – Radost Feb 18 '22 at 14:31
  • What about the exponential map $\exp : A\to S0(3)$ which send an $(3,3) $ antisymmetric matrix to a rotation. This is not a covering map, but it is nice surjective and the critical points are easy to describe (vectors of length $\ln (k2\pi)$) – Thomas Feb 18 '22 at 14:32
  • @Thomas This is what I arrived at, and thought is 'best possible'. A ball around 0 in axis-angle parametrisation has no singulariries but every time dynamics take you outside you have to jump back in to an equivalent point. It seems such jumps are unavoidable but I don't know why. – Radost Feb 18 '22 at 14:45
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    It depends on what you mean by "pull back dynamics," but it seems this implies that the map is a covering map: Suppose you have a vector field on $SO(3)$, and an integral curve $\gamma$ of this field. You ought to be able to lift this vector field, and with it an arbitrarily long part of the curve to $\mathbb{R}^3$ by a map that satisfies the condition you want. But if you can do this then the map has the path lifting property, and it is therefore a covering. – Geva Yashfe Feb 18 '22 at 15:15
  • @GevaYashfe Oh, of course! Thinking about integral curves is what I was missing. Together with Users answer this gives a general characterization of spaces which will experience similar issues (for example any parametrisation of S^2 will have to deal with similar issue). I'll write up an answer tomorrow if none of You is keen to do so :) – Radost Feb 18 '22 at 15:39
  • @Radost I think you can feel free to write an answer. :) – Geva Yashfe Feb 18 '22 at 15:43

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Thanks to @Geva Yashfe and @User comments I found more general answer to my problem.

When studying dynamical system on a manifold $X$ it is sometimes possible to parametrize $X$ and focus, without loss of generality, on a dynamical system in $\mathbb{R}^n$ instead. If that's the case then integral curves of the dynamical system on $X$ lift to curves in $\mathbb{R}^n$. It turns out such condition is strong enough to ensure that parametrisation is a covering (see this answer for details)

Since $\mathbb{R}^n$ is simply connected then, if it is a cover of $X$, it is also a universal cover. Universal covers are unique and thus it is the universal cover. To conclude: $X$ admits 'good' parametrisation without any 'problematic' points if and only if its universal cover is $\mathbb{R}^n$.

Since universal cover of $SO(3)$ is $S^3$ there is no such parametrisation.

(Another common example is $S^2$ which is its own universal cover and thus there is no 'good' parametrisation there either).

Radost
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