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I'm studying the SE(3) group for an engineering application and I'm curious about the difference between $SE(3)$ and $SO(3) \times \mathbb{R}^3$.

Practically, I'd like to optimize in the tangent space of $SE(3)$ and differentiate through its $\mathrm{Exp}$ operator. I'm wondering if I can get away with just differentiating through the $\mathrm{Exp}$ of $SO(3)$ and leaving the translation component untouched.

The paper I linked has the following phrase on page 44

  • SE(3) is diffeomorphic to SO(3) × R3 as a manifold, where each element is described by 3 · 3 + 3 = 12 coordinates (see §7.2).

  • SE(3) is not isomorphic to SO(3) × R3 as a group, since the group multiplications of both groups are different. It is said that SE(3) is a semidirect product of the groups SO(3) and R3.

I'm not sure what the implication is of "diffeomorphic as a manifold" and "not isomorphic as a group".

Is $\mathrm{Exp}(SO(3)) \times \mathbb{R}^3$ still a valid retraction of $SE(3)$ ? Why?

Shaun
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aaa
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    The tangent space of $SO(3)$ is ${\frak so}(3)$, so you mean $\exp{\frak so}(3)\times\Bbb R^3$, not $\exp SO(3)\times\Bbb R^3$. Of course $\exp{\frak so}(3)=SO(3)$, so this is just $SO(3)\times\Bbb R^3$. There's no point to saying it's a "retract" because it's diffeomorphic - i.e. you can parametrize $\mathbb{SE}(3)$ using $SO(3)\times\Bbb R^3$ and the parametrization is differentiable (although does not respect group operations). [More generally, if $G$ is any Lie group, it has a maximal compact group $M$ unique up to conjugacy, and $G\simeq M\times\Bbb R^r$ are diffeomorphic for some $r$.] – anon Aug 31 '22 at 01:32
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    "Practically, I'd like to optimize in the tangent space of SE(3) and differentiate through its Exp operator." I do not understand what this sentence is saying. If the next sentence is asking if every rigid motion is uniquely expressible as the exponential of an "infinitessimal rotation" (i.e. element of the lie algebra ${\frak so}(3)$, i.e. traceless antisymmetric $3\times 3$ matrix) followed by a translation, that is correct. (With the caveat the lie algebra element itself is not unique even if its exponential is, basically just like $e^{2\pi i}=e^0$.) – anon Aug 31 '22 at 01:34
  • Thank you for correcting the typo, I indeed meant $\mathfrak{so}(3)$. Can you please elaborate on "caveat the lie algebra element itself is not unique even if its exponential is"? – aaa Aug 31 '22 at 02:16
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    For example, let $J$ be a $90^\circ$ rotation around an axis (but the axis is its nullspace), so $J\in{\frak so}(3)$. Then the identity element is uniquely expressible as a product of a rotation and a translation: the rotation is the identity, and the translation is by the zero vector. But the identity rotation is not itself uniquely expressible as an exponential, since $I_3=\exp(0J)=\exp(2\pi J)$, for example. Or the element $\exp(\pi J)$ for instance is also expressible as $\exp(-\pi J)$, though $\pi J$ and $-\pi J$ are different elements of the lie algebra. – anon Aug 31 '22 at 02:30
  • Thank you! But this isn't unique to $SO(3) \times \mathbb{R}^3$ right? This is due to the fact that $S(3)$ is a double cover of $SO(3)$? – aaa Aug 31 '22 at 04:20
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    @aaa no I don't think that has anything to do with it. Your problem here is that $\exp$ is not injective which it still won't be on the double cover. In general $\exp$ only gives a "local" diffeomorphism and only close to the identity – Callum Aug 31 '22 at 09:51
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    It is true that ${\rm Spin}(3)$ (which also goes by $S^3$, ${\rm Sp}(1)$, or ${\rm SU}(2)$; I've not seen someone use the notation $S(3)$ before) is a double-cover of ${\rm SO}(3)$. I'm guessing you're thinking of how different elements of ${\rm Spin}(3)$ can represent the same element of ${\rm SO}(3)$. But that's not what I'm talking about here. No Lie group with a nontrivial compact subgroup has an injective exponential map (also see). – anon Aug 31 '22 at 15:53
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    This applies even for simply connected groups (i.e. those with no nontrivial covers), including $S^3$ itself. After all, $S^3$ includes the circle group $S^1\subset\Bbb C$ and I just mentioned $e^{2\pi i}=e^0$. (While $S^1$ is covered by $\Bbb R$, the group $S^3$ has no nontrivial cover because it is simply connected.) – anon Aug 31 '22 at 15:57
  • @runway44 thanks for the detailed response! – aaa Aug 31 '22 at 20:06

1 Answers1

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Summary of comments.

Any Lie group $G$ has a maximal compact subgroup $M$ (unique up to conjugacy), and $G$ is diffeomorphic to $M\times\mathbb{R}^r$ where $r=\dim G-\dim M$. Diffeomorphic means there is a differentiable one-to-one correspondence, so we can parametrize $G$ using $M\times\mathbb{R}^r$. However, the product group structure on $M\times\mathbb{R}^r$ does not match the group structure of $G$ in general (they are not isomorphic as groups).

In particular, the special Euclidean group $\mathbb{SE}(3)$ of rigid motions is diffeomorphic to $\mathrm{SO}(3)\times\mathbb{R}^3$, but not isomorphic. This means every rigid motion is uniquely expressible as a rotation followed by a translation (or vice-versa, as long as you're consistent).

Indeed, we can identify $\mathbb{R}^3$ with a subgroup of $\mathbb{SE}(3)$ (namely, the group of translations), and then $\mathbb{SE}(3)$ is a semidirect product $\mathbb{R}^3\rtimes\mathrm{SO}(3)$. This means the subgroup $\mathbb{R}^3$ is normal (conjugation-invariant as a set).

This is better than what you can expect for a general Lie group $G$: in general, you can't expect the $\mathbb{R}^r$ to even correspond to any subgroup (although I expect it is the bijective image of the exponential map restricted to a vector subspace of the lie algebra) or correspond to a normal subset (with the Iwasawa decomposition for $\mathrm{SL}_2\mathbb{R}$, for example, the upper triangular matrices are not conjugation-invariant).

It is also true every rotation is expressible as an exponential of a traceless antisymmetric matrix, i.e. an element of the lie algebra $\mathfrak{so}(3)$, but not uniquely. Every element of $\mathbb{R}^3$ correspond to an element of $\mathfrak{so}(3)$ via the cross product: that is, if $v\in\mathbb{R}^3$, then $T_v(u)=v\times u$ is a linear transformation expressible as a $3\times 3$ matrix which is in $\mathfrak{so}(3)$, namely

$$ \begin{bmatrix} x \\ y \\ z \end{bmatrix} \quad\longleftrightarrow\quad \begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix} $$.

What helps us understand the exponential map $\exp:\mathfrak{so}(3)\to\mathrm{SO}(3)$ is the observation $\exp(T_v)$ will be a rotation around the (oriented) axis $v$ (according to the right-hand rule) by an angle of $\|v\|$. In particular, if we get $w$ from $v$ by increasing or decreasing $v$'s magnitude by $2\pi$, while maintaining its direction (or reversing so magnitude isn't negative), we find $\exp(T_u)=\exp(T_v)$; the exponential map is not injective, and the lie algebra element used to represent a rotation is not unique.

However, the lie algebra is unique if you restrict it to being $T_v$s for $v$ in the closed ball of radius $\pi$, except for antipodal points on the boundary define the same rotation under the exponential map.

Note the non-uniqueness is not related to the spin group $\mathrm{Spin}(3)$, which also goes by $S^3$ or $\mathrm{Sp}(1)$ or $\mathrm{SU}(2)$, though it is true for every element of $\mathrm{SO}(3)$ there are two distinct elements of $S^3$ corresponding to it. Even in $S^3$ itself, which is has no nontrivial cover i.e. is simply connected, elements ("versors" in the quaternion algebra) are expressible with Euler's formula and we see the vectors used to represent versors are not unique (unless, as mentioned, we restrict to the open ball of radius $\pi$, in which case only $-1$ is not represented).

anon
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