SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$H = SO(2)\times \mathbb{R}$
Figure 11: We can view  $SE(3)$  as an  $SO(2) \times \mathbb{R}$ -principal bundle, where the projection map  $p: SE(3) \to \mathcal{R}$  is  $p(R, t) = (R[0,0,1]^T, t \times (R[0,0,1]^T)$ , and the inverse of  $p$  is  $p^{-1}(x) = \{(R, t) | (R, t) \eta = x\}$ . We use the coordinate frames (red axis denotes  $Z$ -axis, green axis denotes  $X$ -axis, and purple axis denotes  $Y$ -axis) to denote the element in  $SE(3)$  because we can use the position of the coordinate origin to denote the translation  $t$  and use  $X$ -axis,  $Y$ -axis, and  $Z$ -axis to represent the first, second and third columns in rotation  $R$ . When we say next "the coordinate frame on the line/ray" we will mean that its origin is on the line/ray. By this convention, the coordinate frames representing the element in  $H = SO(2) \times \mathbb{R}$  are the frames whose  $Z$ -axis aligns with  $[0,0,1]^T$  and whose origin is  $[0,0,t]^T$  for any  $t \in \mathbb{R}$ , i.e., frames on the yellow line in the left of the figure. For one ray  $x = (\pmb{d}, \pmb{m})$  (illustrated as the chosen blue ray), the coordinate frames on the ray  $x$  whose  $Z$ -axis aligns with the ray  $\pmb{d}_x$  are in  $p^{-1}(x)$ . As shown in the figure, there exists a bijection (gray double arrow line) between  $p^{-1}(x)$  and  $H = SO(2) \times \mathbb{R}$ .  $p^{-1}(x)$  is isomorphic to  $H = SO(2) \times \mathbb{R}$ .
Example 4. Projection, section map and twist function for  $\mathbb{R}^3$  and  $SE(3)$ : According to Ex. 2, we can consider a bundle with total space as  $SE(3)$ , base space as  $\mathbb{R}^3$ , and the fiber as  $SO(3)$ . For any  $g = (R,t) \in SE(3)$ , the projection map  $p: SE(3) \to \mathbb{R}^3$  projects  $g$  as  $p(R,t) = t$ . For any  $\boldsymbol{x} \in \mathbb{R}^3$ , we can define the section map  $s: \mathbb{R}^3 \to SE(3)$  as  $s(\boldsymbol{x}) = (I,\boldsymbol{x})$ . The twist function  $\mathrm{h}: SE(3) \times \mathbb{R}^3 \to SO(3)$  is that  $\mathrm{h}(g,\boldsymbol{x}) = s(g\boldsymbol{x})^{-1}gs(\boldsymbol{x}) = R$  for any  $\boldsymbol{x} \in \mathbb{R}^3$  and any  $g = (R,t) \in SE(3)$ . This twist function is independent of  $\boldsymbol{x}$  due to the fact that  $SE(3) = \mathbb{R}^3 \rtimes SO(3)$  is a semidirect product group as stated in [18].
Example 5. Projection, section map, and twist function for  $\mathbb{S}^2$  and  $SO(3)$ : As shown in Ex. 3,  $SO(3)$  can be viewed as a principal bundle with the base space as  $\mathbb{S}^2$  and the fiber as  $SO(2)$ . With the rotation  $R \in SO(3)$  parameterized as  $R = R_Z(\alpha)R_Y(\beta)R_Z(\gamma)$ , the projection  $p: G \to G / H$  maps  $R$  as follows:
$$
\begin{array}{l} p (R) = R _ {Z} (\alpha) R _ {Y} (\beta) R _ {Z} (\gamma) [ 0, 0, 1 ] ^ {T} \\ = R _ {Z} (\alpha) R _ {Y} (\beta) [ 0, 0, 1 ] ^ {T} \\ = \left[ \sin (\beta) \cos (\alpha), \sin (\beta) \sin (\alpha), \cos (\beta) \right] ^ {T}. \\ \end{array}
$$
For any  $\pmb{d} \in \mathbb{S}^2$ , the section map  $s: \mathbb{S}^2 \to SO(3)$  of  $p$  should satisfy that  $p \circ s = id_{\mathbb{S}^2}$  as mentioned above, i.e.,  $s(\pmb{d})[0, 0, 1]^T = \pmb{d}$ . For instance, we could define the section map  $s$  as:
$$
s (\boldsymbol {d}) = R _ {Z} \left(\alpha_ {\boldsymbol {d}}\right) R _ {Y} \left(\beta_ {\boldsymbol {d}}\right),
$$
where  $\alpha_{d}$  and  $\beta_{d}$  satisfies that
$$
\boldsymbol {d} = \left[ \sin \left(\beta_ {\boldsymbol {d}}\right) \cos \left(\alpha_ {\boldsymbol {d}}\right), \sin \left(\beta_ {\boldsymbol {d}}\right) \sin \left(\alpha_ {\boldsymbol {d}}\right), \cos \left(\beta_ {\boldsymbol {d}}\right) \right] ^ {T}.
$$
Specifically, when  $\pmb{d} = [0,0,1]^T$ ,  $\alpha_{\pmb{d}} = 0$  and  $\beta_{\pmb{d}} = 0$ ; when  $\pmb{d} = -[0,0,1]^T$ ,  $\alpha_{\pmb{d}} = 0$  and  $\beta_{\pmb{d}} = \pi$ . As defined, the twist function  $\mathrm{h}:SO(3)\times \mathbb{S}^2\to SO(2)$  is that  $\mathrm{h}(R,d) = s(Rd)^{-1}Rs(d)$ .
Example 6. Projection, section map, and twist function for  $\mathcal{R}$  and  $SE(3)$ : The final example is  $SE(3)$  with  $\mathcal{R}$  as the base space and  $SO(2) \times \mathbb{R}$  as the fiber, which is the focus of this work, as shown
Figure 12: For a ray  $x = (d, m)$ , we need to choose an element  $(R, t) \in SE(3)$  as the representative element  $s(x)$  such that  $s(x)[0, 0, 1]^T, [0, 0, 0]^T) = x$ . This figure shows one example of the section map  $s$  from ray space to  $SE(3)$ . This map also serves as the section provided in this paper. The axes of the coordinate frame in the figure represent  $R = s_a(d) = R_Z(\alpha_d)R_Y(\beta_d)$ , where the green axis, purple axis, and red axis represent 1st, 2nd and 3rd column in the rotation matrix  $R$ , respectively. The origin of the frame,  $t = s_b(d, m) = d \times m$ , denotes the translation.
$$
d = [ \sin (\beta) \cos (\alpha), \sin (\beta) \sin (\alpha), \cos (\beta) ] ^ {T}
$$
in figure 11. According to the group action defined in Eq. 5, the projection map  $p: SE(3) \to \mathcal{R}$  is:
$$
p \left(\left(R, \boldsymbol {t}\right)\right) = \left(R, \boldsymbol {t}\right) \eta = \left(R [ 0, 0, 1 ] ^ {T}, \boldsymbol {t} \times \left(R [ 0, 0, 1 ] ^ {T}\right)\right).
$$
This represents a ray direction  $\pmb{d}$  with the 3rd column of a rotation matrix and the moment  $\pmb{m}$  with the cross product of the translation and the ray direction. We can construct a section  $s: G / H \to G$  using the Plücker coordinate:
$$
s \left(\left(\boldsymbol {d}, \boldsymbol {m}\right)\right) = \left(s _ {a} (\boldsymbol {d}), s _ {b} (\boldsymbol {d}, \boldsymbol {m})\right),
$$
where  $s_a(\pmb{d}) \in SO(3)$  is a rotation that  $s_a(\pmb{d})[0,0,1]^T = \pmb{d}$ , i.e.,  $s_a$  is a section map from  $\mathbb{S}^2$  to  $SO(3)$  as shown in Ex. 5; and  $s_b(\pmb{d},\pmb{m}) \in \mathbb{R}^3$  is a point on the ray  $(\pmb{d},\pmb{m})$ . In this paper, we define the section map as  $s((\pmb{d},\pmb{m})) = (R_Z(\alpha_d)R_Y(\beta_d),\pmb{d} \times \pmb{m})$ , where  $\alpha_d$  and  $\beta_d$  satisfy that  $\pmb{d} = R_Z(\alpha_d)R_Y(\beta_d)[0,0,1]^T$ , which is the same as Ex. 5. Figure 12 displays the visualization of the section map.
Given the section map, for any  $g = (R_g, t_g) \in SE(3)$  and  $x = (\pmb{d}_x, \pmb{m}_x) \in \mathcal{R}$ , we have the twist function  $h: SE(3) \times \mathcal{R} \to SO(2) \times \mathbb{R}$  is  $h(g, x) = s^{-1}(gx)gs(x) = (\mathrm{h}_a(R_g, d_x), \mathrm{h}_b(g, x))$ , where  $h_a: SO(3) \times \mathbb{S}^2 \to SO(2)$  is the twist function corresponding to  $s_a$ , as shown in Ex. 5, and  $h_b(g, x) = \langle R_g s_b(x) + t_g - s_b(gx), R_g d_x \rangle$ . With the above section  $s$  defined in this paper, the twist function  $h: SE(3) \times \mathcal{R} \to SO(2) \times \mathbb{R}$  is
$$
\mathrm {h} (g, x) = s ^ {- 1} (g x) g s (x) = \left(R _ {Z} \left(R _ {g}, \boldsymbol {d} _ {x}\right), \langle \boldsymbol {t} _ {g}, \left(R _ {g} \boldsymbol {d} _ {x}\right) \rangle\right),
$$
where  $R_Z(R_g, \pmb{d}_x) = R_Y^{-1}(\beta_{R_g \pmb{d}_x}) R_Z^{-1}(\alpha_{R_g \pmb{d}_x}) R_g R_Z(\alpha_{\pmb{d}_x}) R_Y(\beta_{\pmb{d}_x})$ .
To understand the twist function clearly, we visualize a twist induced by a translation in  $SE(3)$  in figure 13,
A.3 Associated Vector Bundle
Given the principal bundle  $G$ , we can construct the associated vector bundle by replacing the fiber  $H$  with the vector space  $V$ , where  $V \simeq \mathbb{R}^n$  and  $H$  acts on  $V$  through a group representation
Figure 13: When we translate a ray  $x = (d, m)$  with  $g = (I, t) \in SE(3)$ , we will find that  $gs(x)$  does not agree with  $s(gx)$ . As defined in figure 12, we have  $s_b(x) \perp d$  and  $s_b(gx) \perp d$ . Following the geometry of the figure, we obtain that  $h_b(g, x) = \langle t, d \rangle [0, 0, 1]^T$ , i.e.,  $h(g, x) = s(gx)^{-1} gs(x) = (I, \langle t, d \rangle [0, 0, 1]^T) = (0, \langle t, d \rangle)$ .
$\rho : H \to GL(V)$ . The group representation corresponds to the type of geometric quantity in the vector space  $V$ , for example, the scalar, the vector, or the higher-order tensor.
The quotient space  $E = G \times_{\rho} V / H$  is defined through the right action of  $H$  on  $G \times V$ :  $(g, v)h = (gh, \rho(h)^{-1}v)$  for any  $h \in H$ ,  $g \in G$  and  $v \in V$ . With the defined projection map  $p: G \times_{\rho} V \to G / H$ :  $p([g, v]) = gH$ , where  $[g, v] = \{(gh, \rho(h)^{-1}v) | h \in H\}$ , the element in  $G \times_{\rho} V$ , we obtain the fiber bundle  $E = G \times_{\rho} V$  associated to the principal bundle  $G$ . For more background and details of the associated vector bundle, we recommend referring to the following sources: [52] and [18].
The feature function  $f: U \subseteq G / H \to V$  can encode the local section of the associated vector bundle  $s_v: U \subseteq G / H \to G \times_{\rho} V$ :  $s_v(x) = [s(x), f(x)]$ , where  $s$  is the section map of the principal bundle as defined in Sec. A.2. The group  $G$  acting on the field  $f$  as shown in [18]:
$$
\left(\mathcal {L} _ {g} f\right) (x) = \rho \left(\mathrm {h} \left(g ^ {- 1}, x\right)\right) ^ {- 1} f \left(g ^ {- 1} x\right), \tag {6}
$$
where  $\mathrm{h}: G \times G / H \to H$  is the twist function as defined in Sec. A.2.
A.4 Equivariant Convolution Over Homogeneous Space
The generalized equivariant convolution over homogeneous space, as stated in [18], that maps a feature field  $f^{l_{in}}$  over homogeneous space  $G / H_1$  to a feature  $f_{out}^{l'}$  over homogeneous space  $G / H_2$  by convolving with a kernel  $\kappa$  is defined as:
$$
f ^ {l _ {\text {o u t}} ^ {\prime}} (x) = \int_ {G / H _ {1}} \kappa \left(s _ {2} (x) ^ {- 1} y\right) \rho_ {i n} \left(\mathrm {h} _ {1} \left(s _ {2} (x) ^ {- 1} s _ {1} (y)\right)\right) f ^ {l _ {i n}} (y) d y, \tag {7}
$$
where  $l_{in}$  and  $l_{out}^{\prime 2}$  denote the input and output feature types, respectively.  $\rho_{in}$  is the group representation of  $H_{1}$  corresponding to the feature type  $l_{in}$ ,  $s_1$  is the section map from  $G / H_{1}$  to  $G$  (see Sec. A.2),  $s_2$  is the section map from  $G / H_{2}$  to  $G$  (see Sec. A.2),  $\mathrm{h}_1$  is the twist function corresponding to  $s_1$  (see Sec. A.2).
The convolution is equivariant with respect to  $G$ , that is
$$
\mathcal {L} _ {g} ^ {\text {o u t}} f ^ {l _ {\text {o u t}} ^ {\prime}} = \kappa * \mathcal {L} _ {g} ^ {\text {i n}} f ^ {l _ {\text {i n}}},
$$
In this context, the feature type indicates the specific geometric quantity in vector spaces  $V_{in}$  and  $V_{out}$ .  $V_{in}$  corresponds to the stabilizer  $H_{1}$  and  $V_{out}$  corresponds to the stabilizer  $H_{2}$ .  $H_{1}$  and  $H_{2}$  can be distinct; therefore, to differentiate the types of features corresponding to different stabilizers, we utilize  $l$  and  $l'$  as notations for the feature types.
if and only if  $\kappa(h_2x) = \rho_{out}(h_2)\kappa(x)\rho_{in}(\mathrm{h}_1^{-1}(h_2,x))$  for any  $h_2 \in H_2$ , where  $\rho_{out}$  is the group representation of  $H_2$  corresponding to the feature type  $l_{out}'$ .
In the following examples, we will illustrate three instances where the input and output homogeneous spaces, denoted as  $G / H_{1}$  and  $G / H_{2}$ , respectively, are identical, meaning that  $H_{1} = H_{2}$ . These examples involve convolutions from  $\mathbb{R}^3$  to  $\mathbb{R}^3$ , from  $\mathbb{S}^2$  to  $\mathbb{S}^2$ , and from  $\mathcal{R}$  to  $\mathcal{R}$ . Furthermore, we will show an example where  $H_{1}$  and  $H_{2}$  differ, explicitly focusing on the convolution from  $\mathcal{R}$  to  $\mathbb{R}^3$ . Example 7.  $SE(3)$  equivariant convolution from  $\mathbb{R}^3$  to  $\mathbb{R}^3$ : If we use the section map as stated in Ex. 4, we will find that  $\mathrm{h}(s(x)^{-1}s(y)) = I$ , therefore convolution 7 becomes:
$$
\begin{array}{l} f ^ {l _ {o u t}} (x) = \int_ {\mathbb {R} ^ {3}} \kappa (s (x) ^ {- 1} y) f ^ {l _ {i n}} (y) d y \\ = \int_ {\mathbb {R} ^ {3}} \kappa (y - x) f ^ {l _ {i n}} (y) d y \\ \end{array}
$$
and  $\kappa$  should satisfy
$$
\begin{array}{l} \kappa (R x) = \rho_ {o u t} (R) \kappa (x) \rho_ {i n} (\mathrm {h} ^ {- 1} (R, x)) \\ = \rho_ {o u t} (R) \kappa (x) \rho_ {i n} \left(\mathrm {h} ^ {- 1} (R)\right) \\ = \rho_ {o u t} (R) \kappa (x) \rho_ {i n} ^ {- 1} (R) \\ \end{array}
$$
for any  $R \in SO(3)$ . When the feature type  $l_{in}$  and  $l_{out}$  corresponds to the irreducible representation, we have
$$
\kappa (R x) = D _ {l _ {o u t}} (R) \kappa (x) D _ {l _ {i n}} (R) ^ {- 1}
$$
where  $D^{l_{in}}$  and  $D^{l_{out}}$  are the Wigner-D matrices, i.e. irreducible representations corresponding to the feature types  $l_{in}$  and  $l_{out}$ , which is the same as the analytical result in [63].
Example 8.  $SO(3)$  equivariant spherical convolution from  $\mathbb{S}^2$  to  $\mathbb{S}^2$ : For spherical convolution, when we substitute the section in Eq. 7 with the section we defined in Ex. 5, the convolution integral takes the following form:
$$
\begin{array}{l} f ^ {l _ {\text {o u t}}} (\alpha , \beta) \\ = \int_ {\alpha^ {\prime} \in [ 0, 2 \pi), \beta^ {\prime} \in [ 0, \pi)} \kappa (R _ {Y} ^ {- 1} (\beta) R _ {Z} ^ {- 1} (\alpha) R _ {Z} (\alpha^ {\prime}) R _ {Y} (\beta^ {\prime}) [ 0, 0, 1 ] ^ {T}) \\ \end{array}
$$
$$
\rho_ {i n} \left(\mathrm {h} \left(R _ {Y} ^ {- 1} (\beta) R _ {Z} ^ {- 1} (\alpha) R _ {Z} \left(\alpha^ {\prime}\right) R _ {Y} \left(\beta^ {\prime}\right)\right)\right) f ^ {l _ {i n}} \left(\alpha^ {\prime}, \beta^ {\prime}\right) d \alpha^ {\prime} \sin \left(\beta^ {\prime}\right) d \beta^ {\prime}
$$
where  $[0,0,1]^T$  is the fixed original point as stated in Ex. 3,  $\rho_{in}$  is the group representation of  $SO(2)$  corresponding to the feature type  $l_{in}$ . When  $\rho_{in}$  and  $\rho_{out}$  are the irreducible representations of  $SO(2)$ ,  $\rho_{in}$  and  $\rho_{out}$  can be denoted as  $\rho_{in}(\theta) = e^{-il_{in}\theta}$  and  $\rho_{out}(\theta) = e^{-il_{out}\theta}$ .
To simplify the notation, we utilize  $R(\theta)$  to represent  $R_Z(\theta) \in SO(2)$ , where  $\theta \in [0, 2\pi)$ . When considering the cases where  $x = [0, 0, 1]^T$ ,  $h(R(\theta)x) = R(\theta)$ ; when  $x = -[0, 0, 1]^T$ ,  $h(R(\theta)x) = R(-\theta)$ ; and when  $x \in \mathbb{S}^2 - \{[0, 0, 1]^T, -[0, 0, 1]^T\}$ ,  $h(R(\theta)x) = R(-\theta) = I$ . Therefore, the kernel  $\kappa$  should satisfy the following conditions:  $\kappa(R(\theta)x) = e^{-il_{out}\theta}\kappa(x)$  for any  $R(\theta) \in SO(2)$  and any  $x \in \mathbb{S}^2 - \{[0, 0, 1]^T, -[0, 0, 1]^T\}$ ;  $\kappa(x) = e^{-i(l_{out} - l_{in})\theta}\kappa(x)$  for  $x = [0, 0, 1]^T$ ; and  $\kappa(x) = e^{-i(l_{out} + l_{in})\theta}\kappa(x)$  for  $x = -[0, 0, 1]^T$ .
Specifically, when the input and output are scalar feature fields over the sphere, convolution reads
$$
\begin{array}{l} f ^ {o u t} (\alpha , \beta) \\ = \int_ {\alpha^ {\prime} \in [ 0, 2 \pi), \beta^ {\prime} \in [ 0, \pi)} \kappa \left(R _ {Y} ^ {- 1} (\beta) R _ {Z} ^ {- 1} (\alpha) R _ {Z} \left(\alpha^ {\prime}\right) R _ {Y} \left(\beta^ {\prime}\right) \eta\right) \\ f ^ {i n} \left(\alpha^ {\prime}, \beta^ {\prime}\right) d \alpha^ {\prime} s i n \left(\beta^ {\prime}\right) d \beta^ {\prime} \\ \end{array}
$$
$\kappa$  has such constraint:
$$
\kappa (R (\theta) x) = \kappa (x)
$$
for any  $R(\theta) \in SO(2)$ , which is consistent with the isotropic kernel of the convolution in [22].
Example 9.  $SE(3)$  equivariant convolution from  $\mathcal{R}$  to  $\mathcal{R}$ : In our case, the equivariant convolution from ray space to ray space is also based on the generalized equivariant convolution over a homogeneous space. See Sec. 3.4.1 for the details. We solve the constraint of the kernel here:
$$
\kappa (h x) = \rho_ {\text {o u t}} (h) \kappa (x) \rho_ {\text {i n}} \left(\mathrm {h} ^ {- 1} (h, x)\right), \tag {8}
$$
for any  $h\in SO(2)\times \mathbb{R}$
The irreducible group representation  $\rho_{in}$  for the corresponding feature type  $l_{in} = (\omega_{in}^{1},\omega_{in}^{2})$  , where  $\omega_{in}^{1}\in \mathbb{N}$  and  $\omega_{in}^{2}\in \mathbb{R}$  , can be written as  $\rho_{in}(\gamma ,t) = e^{-i(\omega_{in}^{1}\gamma +\omega_{in}^{2}t)}$  for any  $h = (\gamma ,t)\in SO(2)\times \mathbb{R}$  and the irreducible group representation  $\rho_{out}(\gamma ,t) = e^{-i(\omega_{out}^{1}\gamma +\omega_{out}^{2}t)}$  for the feature type  $l_{out} = (\omega_{out}^{1},\omega_{out}^{2})$  , where  $\omega_{out}^{1}\in \mathbb{N}$  and  $\omega_{out}^{2}\in \mathbb{R}$  , for any  $h = (\gamma ,t)\in SO(2)\times \mathbb{R}$
To simplify the notation, we utilize  $R(\gamma)$  to represent  $R_Z(\gamma) \in SO(2)$ , where  $\gamma \in [0, 2\pi)$ . For any  $h = (\gamma, t) \in SO(2) \times \mathbb{R}$  and any  $x = (d_x, m_x) \in \mathcal{R}$ , we have  $\mathrm{h}(h, x) = s(hx)^{-1}hs(x) = (R_Z(R(\gamma), d_x), \langle t[0, 0, 1]^T, d_x \rangle)$  according to Ex. 6. Since  $SO(2) \times \mathbb{R}$  is a product group, we can have  $\kappa(x) = \kappa_1(x)\kappa_2(x)$ , where
$$
\kappa_ {1} ((\gamma , t) x) = \rho_ {\text {o u t}} ((\gamma , 0)) \kappa_ {1} (x) \rho_ {i n} ^ {- 1} \left(\left(R _ {Z} (R (\gamma), \boldsymbol {d} _ {x}), 0\right)\right) \tag {9}
$$
$$
\kappa_ {2} ((\gamma , t) x) = \rho_ {o u t} ((0, t)) \kappa_ {2} (x) \rho_ {i n} ^ {- 1} ((0, \langle t [ 0, 0, 1 ] ^ {T}, \boldsymbol {d} _ {x} \rangle)) \tag {10}
$$
Now we solve the constraint for the kernel  $\kappa_{1}$ :
One can check that for any  $\pmb{d}_x \in \mathbb{S}^2 - \left\{ [0, 0, 1]^T, -[0, 0, 1]^T \right\}$ ,  $R_Z(R(\gamma), \pmb{d}_x) = I$ ; when  $\pmb{d}_x = [0, 0, 1]^T$ ,  $R_Z(R(\gamma), \pmb{d}_x) = R(\gamma)$ ; and when  $\pmb{d}_x = -[0, 0, 1]^T$ ,  $R_Z(R(\gamma), \pmb{d}_x) = R(-\gamma)$ .
Therefore, we obtain the constraint that
$$
\kappa_ {1} ((\gamma , t) x) = e ^ {- i \omega_ {o u t} ^ {1} \gamma} \kappa_ {1} (x) \tag {11}
$$
when  $\pmb{d}_x\in \mathbb{S}^2 -\left\{[0,0,1]^T, - [0,0,1]^T\right\}$
$$
\kappa_ {1} ((\gamma , t) x) = e ^ {- i \left(\omega_ {o u t} ^ {1} - \omega_ {i n} ^ {1}\right) \gamma} \kappa_ {1} (x) \tag {12}
$$
when  $\pmb{d}_x = [0, 0, 1]^T$ ;
$$
\kappa_ {1} ((\gamma , t) x) = e ^ {- i \left(\omega_ {o u t} ^ {1} + \omega_ {i n} ^ {1}\right) \gamma} \kappa_ {1} (x) \tag {13}
$$
when  $\pmb{d}_x = -[0,0,1]^T$
The solution for Eq. 11 is that  $\kappa_{1}(x) = f(d(\eta, x), \angle([0,0,1]^{T}, \boldsymbol{d}_{x})) e^{-i\omega_{out}^{1} \text{atan2}([0,1,0]\boldsymbol{d}_{x}, [1,0,0]\boldsymbol{d}_{x})}$ , where  $\text{atan2}$  is the 2-argument arctangent function, and  $f$  is an arbitrary function that maps  $(d(\eta, x), \angle([0,0,1]^{T}, \boldsymbol{d}_{x}))$  to the complex domain.
The solution for Eq. 12 is that when  $\omega_{out}^{1} = \omega_{in}^{1}$ ,  $\kappa_{1}(x) = C$ , where  $C$  is any constant value; when  $\omega_{out}^{1} \neq \omega_{in}^{1}$  and  $x = \eta$ ,  $\kappa_{1}(x) = 0$ ; when  $\omega_{out}^{1} \neq \omega_{in}^{1}$  and  $x \neq \eta$ ,  $\kappa_{1}(x) = f(d(\eta, x)) e^{-i(\omega_{out}^{1} - \omega_{in}^{1}) \tan 2([0, 1, 0] \mathbf{m}_{x}, [1, 0, 0] \mathbf{m}_{x})}$ , where  $f$  is an arbitrary function that maps  $d(x, \eta)$  to the complex domain.
The solution for Eq. 13 is that when  $\omega_{out}^{1} = -\omega_{in}^{1}$ ,  $\kappa_{1}(x) = C$ , where  $C$  is any constant value; when  $\omega_{out}^{1} \neq -\omega_{in}^{1}$  and  $x = -\eta$ ,  $\kappa_{1}(x) = 0$ ; when  $\omega_{out}^{1} \neq -\omega_{in}^{1}$  and  $x \neq -\eta$ ,  $\kappa_{1}(x) = f(d(\eta, x)) e^{-i(\omega_{out}^{1} + \omega_{in}^{1}) \tan 2([0, 1, 0]^{T} \boldsymbol{m}_{x}, [1, 0, 0]^{T} \boldsymbol{m}_{x})}$ , where  $f$  is an arbitrary function that maps  $d(x, \eta)$  to the complex domain.
Next, we will solve the constraint for the kernel  $\kappa_{2}$ , which is that  $\kappa_{2}((\gamma ,t)x) = e^{-i(\omega_{out}^{2} - \omega_{in}^{2}\langle [0,0,1]^{T},\pmb{d}_{x}\rangle)t}\kappa_{2}(x)$ .
When  $\pmb{d}_x = [0,0,1]^T$ , and  $\omega_{out}^2 \neq \omega_{in}^2$ ,  $\kappa_2(x) = 0$ ; When  $\pmb{d}_x = -[0,0,1]^T$  and  $\omega_{out}^2 \neq -\omega_{in}^2$ ,  $\kappa_2(x) = 0$ ; When  $\pmb{d}_x = [0,0,1]^T$ , and  $\omega_{out}^2 = \omega_{in}^2$ ,  $\kappa_2(x) = f(d(x,\eta))$ , where  $f$  is an arbitrary
function that maps  $d(x, \eta)$  to the complex domain; When  $\pmb{d}_x = -[0, 0, 1]^T$ , and  $\omega_{out}^2 = -\omega_{in}^2$ ,  $\kappa_2(x) = f(d(x, \eta))$ , where  $f$  is an arbitrary function that maps  $d(x, \eta)$  to the complex domain; when  $\pmb{d}_x \in \mathbb{S}^2 - \{[0, 0, 1]^T, -[0, 0, 1]^T\}$ ,
$$
\kappa_ {2} (x) = f \left(d (\eta , x), \angle \left([ 0, 0, 1 ] ^ {T}, \boldsymbol {d} _ {x}\right)\right) e ^ {- i \left(\omega_ {\text {o u t}} ^ {2} - \omega_ {\text {i n}} ^ {2} \langle [ 0, 0, 1 ] ^ {T}, \boldsymbol {d} _ {x} \rangle\right) g (x)}, \tag {14}
$$
where  $f$  is an arbitrary function that maps  $(d(\eta, x), \angle([0, 0, 1]^T, \boldsymbol{d}_x))$  to the complex domain;  $g(x) = [0, 0, 1](\boldsymbol{x}_Q - [0, 0, 0]^T)$ , where  $\boldsymbol{x}_Q$  represents the 3D coordinates of a point  $Q$ . This point  $Q$  can be defined as the intersection of  $x$  and  $\eta$  if  $x$  and  $\eta$  intersect. Alternatively, if  $x$  and  $\eta$  do not intersect,  $Q$  is determined as the intersection of  $\eta$  and the ray  $y$ , which is perpendicular to both  $x$  and  $\eta$ , and intersects with both  $x$  and  $\eta$ . Refer to Figure 16 for a visual representation. One can easily check that  $g((\gamma, t)x) = t + g(x)$ , as shown in figure 16, which makes the solution valid.
If  $x$  and  $\eta$  are intersected, i.e.,  $[0, 0, 1] \pmb{m}_x = 0$ ,
$$
g (x) = [ 0, 0, 1 ] \left(\boldsymbol {d} _ {x} \times \boldsymbol {m} _ {x} - \frac {[ 1 , 0 , 0 ] (\boldsymbol {d} _ {x} \times \boldsymbol {m} _ {x})}{[ 1 , 0 , 0 ] \boldsymbol {d} _ {x}} \boldsymbol {d} _ {x}\right)
$$
when  $[1,0,0]\pmb{d}_x\neq 0$
$$
g (x) = [ 0, 0, 1 ] \left(\boldsymbol {d} _ {x} \times \boldsymbol {m} _ {x} - \frac {[ 0 , 1 , 0 ] (\boldsymbol {d} _ {x} \times \boldsymbol {m} _ {x})}{[ 0 , 1 , 0 ] \boldsymbol {d} _ {x}} \boldsymbol {d} _ {x}\right)
$$
when  $[1,0,0]\pmb{d}_x = 0$
When  $x$  and  $\eta$  are not intersected,
$$
g (x) = [ 0, 0, 1 ] (\boldsymbol {d} _ {x} \times \boldsymbol {m} _ {x} - \frac {[ 1 , 0 , 0 ] (\boldsymbol {d} _ {x} \times \boldsymbol {m} _ {x}) [ 1 , 0 , 0 ] \boldsymbol {d} _ {x} + [ 0 , 1 , 0 ] (\boldsymbol {d} _ {x} \times \boldsymbol {m} _ {x}) [ 0 , 1 , 0 ] \boldsymbol {d} _ {x}}{([ 1 , 0 , 0 ] \boldsymbol {d} _ {x}) ^ {2} + ([ 0 , 1 , 0 ] \boldsymbol {d} _ {x}) ^ {2}} \boldsymbol {d} _ {x}).
$$
Regular Representation Here, we delve into the case where the output field type corresponds to the group representation of  $SO(2)\times \mathbb{R}$  that  $\rho (\gamma ,t) = \rho_1(\gamma)\otimes \rho_2(t)$  for any  $(\gamma ,t)\in SO(2)\times \mathbb{R}$ , where  $\rho_{2}$  is the regular representation. The regular representation of a group G is a linear representation that arises from the group action of G on itself by translation, that is when  $\rho_{2}:\mathbb{R}\to GL(V)$  is the regular representation, for any  $v\in V$ , for any  $t,t'\in \mathbb{R}$ , we have  $(\rho_{2}(t^{\prime})v)_{t} = v_{t - t^{\prime}}$ , in other words,  $v\in V$  can be viewed as a function defined on  $\mathbb{R}$  or an infinite dimensional vector. Then according to Ex. 6, the group  $SE(3)$  acting on the the field  $f$  would be:
$$
\begin{array}{l} \left(\mathcal {L} _ {g} f\right) (x) _ {t} = \left(\rho (\mathrm {h} \left(g ^ {- 1}, x\right)) ^ {- 1} f \left(g ^ {- 1} x\right)\right) _ {t} \\ = \rho_ {1} \left(\mathrm {h} _ {a} \left(R _ {g ^ {- 1}}, \boldsymbol {d} _ {x}\right)\right) ^ {- 1} f \left(g ^ {- 1} x\right) _ {t + \mathrm {h} _ {b} \left(g ^ {- 1}, x\right)} \\ = \rho_ {1} \left(R _ {Z} \left(R _ {g ^ {- 1}}, \boldsymbol {d} _ {x}\right)\right) ^ {- 1} f \left(g ^ {- 1} x\right) _ {t + \langle \boldsymbol {t} _ {g - 1}, \left(R _ {g - 1} \boldsymbol {d} _ {x}\right) \rangle} \\ \end{array}
$$
for any  $t\in \mathbb{R}$ $x\in \mathcal{R}$  and  $g\in SE(3)$