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in the paper: https://arxiv.org/pdf/2010.02449.pdf the author consider some classes of functions that are invariant to the action of the group $G= \mathbb{R}^3 \rtimes SO(3) \times S_n$ on $\mathbb{R}^{3 \times n}$.

In simple terms, I understand that they're asking some function $f$, taking a point cloud $X \in \mathbb{R}^{3 \times n}$ as input to be equivariant to translations, rotations and permutations.

In general, I know that a semi-direct product is established when we have $H,K \le G$, ($H$ normal), $HK=G$ and $H \cap K = \{e\}$. Then we can define the following map

\begin{align}\phi: H\times K &\rightarrow G \\ (h,k)&\mapsto hk \end{align}

to be a group isomorphism by introducing the group operation $(h_1,k_1)\cdot(h_2,k_2):= (h_1k_1h_2k_1^{-1},k_1k_2)$, then we write $G = H \rtimes K$.

So what I don't understand here is, why we need semi-direct product of $\mathbb{R}^3$ and $SO(3)$ and then direct product with $S_n$ to establish equivariance to the aforementioned transformations? Which is the point?

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You should read this small section: https://en.wikipedia.org/wiki/Direct_product_of_groups#Semidirect_products. It explains when a group decomposes as a direct product or as a semidirect product. It basically comes down to whether $hk=kh$ for all $h\in H$, $k\in K$ or not. If it helps you, I think of the $\rtimes$ as an inner (or internal) semidirect prouct, and of the $\times$ as an outer (or external) direct product in $\left(\Bbb R^3 \rtimes SO(3)\right) \times S_n$.

$\Bbb R^3 \rtimes SO(3) \cong E^+(3)$ is the group of rigid motions (i.e. orientation-preserving isometries) of $\Bbb R^3$. The semidirect product decomposition expresses that any an isometry can be expressed in a unique way as a rotation followed by a translation. The set of translations is a normal subgroup of $E^+(3)$, but $SO(3)$ isn't, which means we get a semidirect product but not a direct product. In particular, translations don't commute with rotations.

Then $\left(\Bbb R^3 \rtimes SO(3)\right) \times S_n = E^+(3)\times S_n$ means that we consider transformations given by permuting the points followed by moving the whole point cloud with a rigid motion. We use the direct product because:

  • Our transformations are uniquely given by a point permutation and a rigid motion.
  • Point permutations and rigid motions commute with each other.

Another way to think of the difference between direct and semidirect products might be to ask whether the elements in $H$ and $K$ "interact" with each other. Say we looked at $\Bbb R^3 \times SO(3)$ instead of $\Bbb R^3 \rtimes SO(3)$. This would be like having two separate Euclidean spaces and then translating one of them and rotating the other -> no interaction.

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  • $\begingroup$ many thanks for your answer. The condition $kh=hk$ isn't related to some abelianity then? So how does the abelianity come into play? $\endgroup$
    – James Arten
    Dec 18, 2021 at 14:23
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    $\begingroup$ @JamesArten You could say it's related, but it's not exactly the same. The condition means that $H$ and $K$ are somehow "mutually abelian" as subgroups of $H\times K$. That is, elements from $H$ commute with elements from $K$. But importantly, $H$ and $K$ need not themselves be abelian, i.e. $hh'\ne h'h$ and $kk'\ne k'k$ are allowed. If in addition $H$ and $K$ are both abelian, then $H\times K$ is abelian as well. In our example neither $S_n$ nor $E^+(3)$ are abelian, but they live their separate lives in $S_n\times E^+(3)$, i.e. their elements mutually commute. $\endgroup$
    – Milten
    Dec 18, 2021 at 14:44
  • $\begingroup$ I should add that when we think of $H$ as a subgroup of $H\times K$, we really mean $H\cong H\times \{e\}$ (and similar for $K$). So $hk=kh$ in this sense means $(h,e)(e,k) = (h,k) = (e,k)(h,e)$. $\endgroup$
    – Milten
    Dec 18, 2021 at 14:45

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