SlowGuess/NeurIPS_2023_OCR huggingface dataset:
In particular, the equality  $P_{X^e} = B^e \# P_{S_e}$  shows that the representation in Line 4 of Algorithm 1 exists. Now, we show that it is unique up to signed permutation by applying results on identifiability of linear ICA. Since  $G^e$  has full column rank by Condition (C2) and  $[(I - A)^{-1}]_{S_e, S_e}$  is invertible, the matrix  $B^e$  also has full column rank. Let  $P_{X^e} = \widehat{B}^e \# P^e$  be any representation, where  $\widehat{B}^e \in \mathbb{R}^{d_e \times \widehat{s}_e}$  and  $P^e$  is an  $\widehat{s}_e$ -dimensional probability measure with independent, non-degenerate marginals  $P_i^e$ . Due to Condition (C1), all probability measures  $P_i$  are non-Gaussian and non-degenerate and therefore we have by Eriksson and Koivunen (2004, Theorem 3 and 4) the identities
$$
\widehat {B} ^ {e} = B ^ {e} R ^ {e} \Lambda^ {e} \quad \text {a n d} \quad P ^ {e} = \Lambda^ {e} \left(R ^ {e}\right) ^ {\top} \# P _ {S _ {e}}, \tag {3}
$$
where  $\Lambda^e$  is an  $s_e \times s_e$  diagonal matrix with nonzero entries and  $R^e$  is an  $s_e \times s_e$  permutation matrix. In particular, we have  $\widehat{s}_e = s_e$ , which means that  $\widehat{B}^e \in \mathbb{R}^{d_e \times s_e}$  and that  $P^e$  is an  $s_e$ -dimensional probability measure. Line 4 also requires that each marginal  $P_i^e$  has unit variance. This removes the scaling indeterminacy in (3) and we have
$$
\hat {B} ^ {e} = B ^ {e} R ^ {e} D ^ {e} \quad \text {a n d} \quad P ^ {e} = D ^ {e} (R ^ {e}) ^ {\top} \# P _ {S _ {e}},
$$
where  $D^{e}$  is a diagonal matrix with entries  $D_{ii}^{e} \in \{\pm 1\}$ . In particular, this means that the distributions  $P^{e}$  and  $P_{S_e}$  coincide up to permutation and sign of the marginals.
The matching in Line 6 identifies which components of  $P^e$  are shared. By Condition (C1), two components of different domains  $P_i^e$  and  $P_j^f$  are shared if and only if they coincide up to sign, that is, if and only if  $d(P_i^e, P_j^f) = 0$  or  $d(P_i^e, -P_j^f) = 0$ . If their distribution coincides up to sign, then either  $d(P_i^e, P_j^f) = 0$  or  $d(P_i^e, -P_j^f) = 0$  but not both since Condition (C1) requires the distribution of the error variables to be non-symmetric. We conclude that in each domain  $e \in [m]$  there exists an  $s_e \times s_e$  signed permutation matrix  $Q^e$  such that
$$
d \left(\left[ \left(Q ^ {e}\right) ^ {\top} \# P ^ {e} \right] _ {i}, \left[ \left(Q ^ {f}\right) ^ {\top} \# P ^ {f} \right] _ {i}\right) = 0 \tag {4}
$$
for all  $i = 1,\dots ,\widehat{\ell}$  and for all  $f\neq e$  . In particular,  $\widehat{\ell} = \ell$  and  $\widehat{\mathcal{L}} = \mathcal{L}$
It remains to show that  $\widehat{B} = B\Psi$  and  $\widehat{P} = \Psi^{\top}\# P$  for a signed permutation block matrix  $\Psi \in \Pi$ . By Equation (4), the distributions  $[(Q^{e})^{\top}\# P^{e}]_{\mathcal{L}}$  and  $[(Q^{e})^{\top}\# P^{e}]_{\mathcal{L}}$  coincide, which means that
$$
\left(Q ^ {e}\right) ^ {\top} \# P ^ {e} = \left(Q ^ {e}\right) ^ {\top} D ^ {e} \left(R ^ {e}\right) ^ {\top} \# P _ {S _ {e}} = \left( \begin{array}{c c} \Psi_ {\mathcal {L}} ^ {\top} & 0 \\ 0 & \Psi_ {I _ {e}} ^ {\top} \end{array} \right) \# \left( \begin{array}{c} P _ {\mathcal {L}} \\ P _ {I _ {e}} \end{array} \right), \tag {5}
$$
where  $\Psi_{\mathcal{L}}$  is an  $\ell \times \ell$  signed permutation matrix and  $\Psi_{I_e}$  is an  $|I_e|\times |I_e|$  signed permutation matrix. Importantly, the matrix  $\Psi_{\mathcal{L}}^{\top}$  does not depend on the domain  $e\in [m]$ . Hence, the matrix  $\Phi^{e}:= R^{e}D^{e}Q^{e}$  is a signed permutation matrix with block structure as in Equation (5). Moreover, we have
$$
\widehat {B} ^ {e} Q ^ {e} = B ^ {e} R ^ {e} D ^ {e} Q ^ {e} = B ^ {e} \Phi^ {e} = \left(B _ {\mathcal {L}} ^ {e} \Psi_ {\mathcal {L}} \mid B _ {[ s _ {e} ] \backslash \mathcal {L}} ^ {e} \Psi_ {I _ {e}}\right),
$$
which means that the matrix  $\widehat{B}$  can be factorized as
$$
\begin{array}{l} \widehat {B} = \left( \begin{array}{c c c c} {[ \widehat {B} ^ {1} Q ^ {1} ] _ {\widehat {\mathcal {L}}}} & {[ \widehat {B} ^ {1} Q ^ {1} ] _ {[ \widehat {s} _ {1} ] \setminus \widehat {\mathcal {L}}}} & & \\ \vdots & & \ddots & \\ {[ \widehat {B} ^ {m} Q ^ {m} ] _ {\widehat {\mathcal {L}}}} & & & {[ \widehat {B} ^ {m} Q ^ {m} ] _ {[ \widehat {s} _ {m} ] \setminus \widehat {\mathcal {L}}}} \end{array} \right) \\ = \left( \begin{array}{c c c c} B _ {\mathcal {L}} ^ {1} \Psi_ {\mathcal {L}} & B _ {[ s _ {1} ] \backslash \mathcal {L}} ^ {1} \Psi_ {I _ {1}} & & \\ \vdots & & \ddots & \\ B _ {\mathcal {L}} ^ {m} \Psi_ {\mathcal {L}} & & & B _ {[ s _ {m} ] \backslash \mathcal {L}} ^ {m} \Psi_ {I _ {m}} \end{array} \right) \\ = \left( \begin{array}{c c c c} B _ {\mathcal {L}} ^ {1} & B _ {[ s _ {1} ] \backslash \mathcal {L}} ^ {1} & & \\ \vdots & \ddots & & \\ B _ {\mathcal {L}} ^ {m} & & B _ {[ s _ {m} ] \backslash \mathcal {L}} ^ {m} \end{array} \right) \cdot \left( \begin{array}{c c c c} \Psi_ {\mathcal {L}} & & & \\ & \Psi_ {I _ {1}} & & \\ & & \ddots & \\ & & & \Psi_ {I _ {m}} \end{array} \right) = B \cdot \Psi , \\ \end{array}
$$
where  $\Psi \in \Pi$ . Similarly, we have for all  $e \in [m]$ ,
$$
\left(Q ^ {e}\right) ^ {\top} \# P ^ {e} = \left(\Phi^ {e}\right) ^ {\top} \# P _ {S _ {e}} = \left( \begin{array}{c} \left(\Psi_ {\mathcal {L}}\right) ^ {\top} \# P _ {\mathcal {L}} \\ \left(\Psi_ {I _ {e}}\right) ^ {\top} \# P _ {I _ {e}} \end{array} \right).
$$
$$
\begin{array}{l} \widehat {P} = \left( \begin{array}{c} [ (Q ^ {1}) ^ {\top} \# P ^ {1} ] _ {\widehat {\mathcal {L}}} \\ [ (Q ^ {1}) ^ {\top} \# P ^ {1} ] _ {[ \widehat {s} _ {1} ] \backslash \widehat {\mathcal {L}}} \\ \vdots \\ [ (Q ^ {m}) ^ {\top} \# P ^ {m} ] _ {[ \widehat {s} _ {m} ] \backslash \widehat {\mathcal {L}}} \end{array} \right) = \left( \begin{array}{c} (\Psi_ {\mathcal {L}}) ^ {\top} \# P _ {\mathcal {L}} \\ (\Psi_ {I _ {1}}) ^ {\top} \# P _ {I _ {1}} \\ \vdots \\ (\Psi_ {I _ {m}}) ^ {\top} \# P _ {I _ {m}} \end{array} \right) \\ = \left( \begin{array}{c c c c} \Psi_ {\mathcal {L}} & & & \\ & \Psi_ {I _ {1}} & & \\ & & \ddots & \\ & & & \Psi_ {I _ {m}} \end{array} \right) ^ {\top} \# \left( \begin{array}{c} P _ {\mathcal {L}} \\ P _ {I _ {1}} \\ \vdots \\ P _ {I _ {m}} \end{array} \right) = \Psi^ {\top} \# P. \\ \end{array}
$$
Before proving Lemma 4.3 and Theorem 4.4 we fix some notation. Let  $\mathcal{G}_m = (V \cup \mathcal{H}, D)$  be an  $m$ -domain graph. We denote by  $\operatorname{anc}(v) = \{k \in \mathcal{H} : \text{there is a directed path } k \to \dots \to v \text{ in } \mathcal{G}_m\}$  the ancestors of a node  $v \in V$ . For subsets  $W \subseteq V$ , we denote  $\operatorname{anc}(W) = \bigcup_{v \in W} \operatorname{anc}(w)$ . Moreover, for  $\mathcal{L} \subseteq \mathcal{H}$  and  $v \in V$ , we write shortly  $\mathrm{pa}_{\mathcal{L}}(v) = \mathrm{pa}(v) \cap \mathcal{L}$ .
Proof of Lemma 4.3. Let  $B \in \mathrm{Im}(\phi_{\mathcal{G}_m})$ . Then we can write  $B = G \cdot (I - A)^{-1}$  with
$$
G = \left( \begin{array}{c c c c} G _ {V _ {1}, \mathcal {L}} & G _ {V _ {1}, I _ {1}} & & \\ \vdots & & \ddots & \\ G _ {V _ {m}, \mathcal {L}} & & & G _ {V _ {m}, I _ {m}} \end{array} \right).
$$
Moreover, observe that, by the definition of an  $m$ -domain-graph, the matrix  $B_{V,\mathcal{L}}$  factorizes as
$$
B _ {V, \mathcal {L}} = G _ {V, \mathcal {L}} [ (I - A) ^ {- 1} ] _ {\mathcal {L}, \mathcal {L}}.
$$
Now, suppose that  $i$  and  $j$  are partial pure children of a fixed node  $k \in \mathcal{L}$ . Then  $\mathrm{pa}_{\mathcal{L}}(i) = \{k\} = \mathrm{pa}_{\mathcal{L}}(j)$ . In particular, the only entry that may be nonzero in the row  $G_{i,\mathcal{L}}$  is given by  $G_{ik}$  and the only entry that may be nonzero in the row  $G_{j,\mathcal{L}}$  is given by  $G_{jk}$ . Thus, we have
$$
B _ {i, \mathcal {L}} = \sum_ {q \in \mathcal {L}} G _ {i q} [ (I - A) ^ {- 1} ] _ {q, \mathcal {L}} = G _ {i k} [ (I - A) ^ {- 1} ] _ {k, \mathcal {L}}.
$$
Similarly, it follows that  $B_{j,\mathcal{L}} = G_{jk}[(I - A)^{-1}]_{k,\mathcal{L}}$ . This means that the row  $B_{j,\mathcal{L}}$  is a multiple of the row  $B_{i,\mathcal{L}}$  and we conclude that  $\mathrm{rank}(B_{\{i,j\},\mathcal{L}}) \leq 1$ . Equality holds due to the faithfulness condition (C4) which implies that  $B_{ik} \neq 0$  and  $B_{jk} \neq 0$ , i.e.,  $B_{\{i,j\},\mathcal{L}}$  is not the null matrix.
For the other direction suppose that  $\mathrm{rank}(B_{\{i,j\},\mathcal{L}}) = 1$ . By applying the Lindström-Gessel-Viennot Lemma (Gessel and Viennot, 1985; Lindström, 1973) equivalently as in Dai et al. (2022, Theorem 1 and 2), it can be seen that
$$
\operatorname {r a n k} \left(B _ {\{i, j \}, \mathcal {L}}\right) \leq \min  \left\{| S |: S \text {i s a v e r t e x c u t f r o m a n c} (\mathcal {L}) \text {t o} \{i, j \} \right\}, \tag {6}
$$
where  $S$  is a vertex cut from  $\operatorname{anc}(\mathcal{L})$  to  $\{i,j\}$  if and only if there exists no directed path in  $\mathcal{G}_m$  from  $\operatorname{anc}(\mathcal{L})$  to  $\{i,j\}$  without passing through  $S$ . Moreover, equality holds in (6) for generic (almost all) choices of parameters. Since we assumed rank faithfulness in Condition (C4) we exclude cases where the inequality is strict and therefore have equality. By the definition of an  $m$ -domain graph we have that  $\operatorname{anc}(\mathcal{L}) = \mathcal{L}$ . Thus, if  $\operatorname{rank}(B_{\{i,j\},\mathcal{L}}) = 1$ , there must be a single node  $k \in \mathcal{L}$  such that  $\{k\}$  is a vertex cut from  $\mathcal{L}$  to  $\{i,j\}$ . But then it follows that  $i$  and  $j$  have to be partial pure children of  $k$  by the definition of an  $m$ -domain graph and by using the assumption that there are no zero-rows in  $B_{\mathcal{L}}$ .
To prove Theorem 4.4 we need the following auxiliary lemma.
Lemma A.1. Let  $G = (V, D)$  be a DAG with topologically ordered nodes  $V = [p]$  and let  $M$  be a lower triangular matrix with entries  $M_{ii} \neq 0$  for all  $i = 1, \ldots, p$  and  $M_{ij} \neq 0$  if and only if there is a directed path  $j \to \dots \to i$  in  $G$ . Let  $Q_{\sigma_1}$  and  $Q_{\sigma_2}$  be permutation matrices. Then the matrix  $Q_{\sigma_1}MQ_{\sigma_2}$  is lower triangular if and only if  $\sigma_2 = \sigma_1^{-1}$  and  $\sigma_2 \in S(G)$ .
Proof of Lemma A.1. By the definition of a permutation matrix, we have
$$
\left[ Q _ {\sigma_ {1}} M Q _ {\sigma_ {2}} \right] _ {i j} = M _ {\sigma_ {1} (i) \sigma_ {2} ^ {- 1} (j)} \text {o r , e q u i v a l e n t l y ,} \left[ Q _ {\sigma_ {1}} M Q _ {\sigma_ {2}} \right] _ {\sigma_ {1} ^ {- 1} (i) \sigma_ {2} (j)} = M _ {i j}. \tag {7}
$$
First, suppose that  $\sigma_{2} = \sigma_{1}^{-1}$  and  $\sigma_{2} \in S(G)$ , and let  $i, j \in [p]$  such that  $\sigma_{2}(i) < \sigma_{2}(j)$ . Then, by the definition of  $S(G)$ , there is no directed path  $j \to \dots \to i$  in the graph  $G$  and therefore we have  $M_{ij} = 0$ . But this means that  $[Q_{\sigma_1}MQ_{\sigma_2}]_{\sigma_2(i)\sigma_2(j)} = 0$  and we conclude that the matrix  $Q_{\sigma_1}MQ_{\sigma_2}$  is lower triangular.
Now, assume that  $Q_{\sigma_1}MQ_{\sigma_2}$  is lower triangular, where  $\sigma_1$  and  $\sigma_2$  are arbitrary permutations on the set  $[p]$ . Since  $M$  has no zeros on the diagonal, we have  $M_{ii} = [Q_{\sigma_1}MQ_{\sigma_2}]_{\sigma_1^{-1}(i)\sigma_2(i)} \neq 0$  for all  $i = 1,\ldots ,p$ . It follows that  $\sigma_1^{-1}(i) \geq \sigma_2(i)$  for all  $i = 1,\dots ,p$  because  $Q_{\sigma_1}MQ_{\sigma_2}$  is lower triangular. But this is only possible if the permutations coincide on all elements, i.e., we have  $\sigma_2 = \sigma_1^{-1}$ . It remains to show that  $\sigma_2 = \sigma_1^{-1} \in S(G)$ . For any edge  $j \to i \in D$  we have that  $M_{ij} \neq 0$ . Recalling Equation (7) this means that  $[Q_{\sigma_1}MQ_{\sigma_2}]_{\sigma_2(i)\sigma_2(j)} \neq 0$ . But since  $Q_{\sigma_1}MQ_{\sigma_2}$  is lower triangular this can only be the case if  $\sigma_2(j) < \sigma_2(i)$  which proves that  $\sigma_2 \in S(G)$ .
Proof of Theorem 4.4. Each latent node in  $\mathcal{L}$  has two partial pure children by Condition (C3). After removing zero-rows in Line 3 of Algorithm 2 it holds by Lemma 4.3 that  $\mathrm{rank}(B_{\{i,j\},\mathcal{L}}^*) = 1$  if and only if there is a latent node in  $\mathcal{L}$  such that  $i$  and  $j$  are both partial pure children of that latent node. Hence, each tuple  $(i_k,j_k)_{k\in \mathcal{L}}$  in Line 4 of Algorithm 2 consists of two partial pure children of a certain latent node. The requirement  $\mathrm{rank}(B_{\{i_k,i_q\},\mathcal{L}}^*) = 2$  ensures that each pair of partial pure children has a different parent.
By the definition of an  $m$ -domain-graph and the fact that  $B^{*} = \widehat{B}_{\mathcal{L}}$ , for  $I = \{i_{1},\ldots ,i_{\ell}\}$ , we have the factorization
$$
B _ {I, \mathcal {L}} ^ {*} = B _ {I, \mathcal {L}} \Psi_ {\mathcal {L}} = G _ {I, \mathcal {L}} (I - A) _ {\mathcal {L}, \mathcal {L}} ^ {- 1} \Psi_ {\mathcal {L}} = G _ {I, \mathcal {L}} \left(I - A _ {\mathcal {L}, \mathcal {L}}\right) ^ {- 1} \Psi_ {\mathcal {L}}, \tag {8}
$$
where  $G \in \mathbb{R}^{D_{\mathrm{VH}}}$ ,  $A \in \mathbb{R}^{D_{\mathcal{H}}}$  and  $\Psi_{\mathcal{L}}$  is a signed permutation matrix. Let  $Q_{1}$  and  $Q_{2}$  be permutation matrices and let  $\Lambda$  be a diagonal matrix with non-zero diagonal elements and let  $D$  be a diagonal matrix with entries in  $\{\pm 1\}$ . Then we can rewrite Equation (8) as
$$
B_{I,\mathcal{L}}^{*} = Q_{1}\underbrace{\Lambda(I - A_{\mathcal{L},\mathcal{L}})^{-1}D}_{=: M}Q_{2}.
$$
Now, we apply Lemma A.1. Since we assume throughout Section 4 that the nodes  $\mathcal{L}$  are topologically ordered, the matrix  $M$  is lower triangular with no zeros on the diagonal. Moreover, by Condition (C4) we have  $M_{ij} \neq 0$  if and only if there is a directed path  $j \to \dots \to i$  in  $\mathcal{G}_{\mathcal{L}}$ . In Line 6 we find other permutation matrices  $R_{1}$  and  $R_{2}$  such that
$$
W = R _ {1} B _ {I, \mathcal {L}} ^ {*} R _ {2} = \left(R _ {1} Q _ {1}\right) M \left(Q _ {2} R _ {2}\right)
$$
is lower triangular. Now, define the permutation matrices  $Q_{\sigma_1} = R_1Q_1$  and  $Q_{\sigma_2} = Q_2R_2$ . Then we have by Lemma A.1 that  $Q_{\sigma_1} = Q_{\sigma_2}^\top$  and that  $\sigma_2 \in S(\mathcal{G}_{\mathcal{L}})$ . Hence, the matrix  $W$  factorizes as
$$
W = Q _ {\sigma_ {2}} ^ {\top} M Q _ {\sigma_ {2}} = Q _ {\sigma_ {2}} ^ {\top} \Lambda (I - A _ {\mathcal {L}, \mathcal {L}}) ^ {- 1} D Q _ {\sigma_ {2}} = \widetilde {\Lambda} Q _ {\sigma_ {2}} ^ {\top} (I - A _ {\mathcal {L}, \mathcal {L}}) ^ {- 1} Q _ {\sigma_ {2}} \widetilde {D},
$$
where  $\widetilde{\Lambda}$  and  $\widetilde{D}$  are diagonal matrices with the entries given by permutations of the entries of  $\Lambda$  and  $D$ . Lines 7 and 8 address the scaling and sign matrices  $\widetilde{\Lambda}$  and  $\widetilde{D}$ . In particular, we have that  $\widetilde{W}' = D'Q_{\sigma_2}^\top (I - A_{\mathcal{L},\mathcal{L}})^{-1}Q_{\sigma_2}D'$  for another diagonal matrix  $D'$  with entries in  $\{\pm 1\}$ , since each entry on the diagonal of  $\widetilde{W}'$  is equal to 1. Thus, we have
$$
\begin{array}{l} \widehat {A} = I - \left(\widetilde {W} ^ {\prime}\right) ^ {- 1} \\ = I - \left(D ^ {\prime} Q _ {\sigma_ {2}} ^ {\top} \left(I - A _ {\mathcal {L}, \mathcal {L}}\right) ^ {- 1} Q _ {\sigma_ {2}} D ^ {\prime}\right) ^ {- 1} \\ = I - D ^ {\prime} Q _ {\sigma_ {2}} ^ {\top} (I - A _ {\mathcal {L}, \mathcal {L}}) Q _ {\sigma_ {2} D ^ {\prime}} \\ = D ^ {\prime} Q _ {\sigma_ {2}} ^ {\top} A _ {\mathcal {L}, \mathcal {L}} Q _ {\sigma_ {2}} D ^ {\prime}. \\ \end{array}
$$
Since  $Q_{\sigma_2}D'$  is a signed permutation matrix with  $\sigma_2 \in S(\mathcal{G}_{\mathcal{L}})$ , the first part of the theorem is proved. If  $G_{vk} > 0$  whenever  $v$  is a pure child of  $k$ , the matrix  $\widetilde{\Lambda}$  only has positive entries which means that  $D'$  is equal to the identity matrix. This proves the second part.
The graph in Figure 4 is an  $m$ -domain graph corresponding to the compact version in Figure 2 in the main paper.
Example B.1. Consider the  $m$ -domain graph in Figure 4. The linear structural equation model among the latent variables is determined by lower triangular matrices of the form
$$
A = \left( \begin{array}{c c c c c} 0 & 0 & 0 & 0 & 0 \\ a _ {2 1} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & a _ {4 3} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right).
$$
Moreover, the domain-specific mixing matrices  $G^{e}$  are of the form
$$
G ^ {1} = \left( \begin{array}{c c c c} g _ {1 1} ^ {1} & g _ {1 2} ^ {1} & g _ {1 3} ^ {1} & 0 \\ g _ {2 1} ^ {1} & 0 & g _ {2 3} ^ {1} & 0 \\ 0 & g _ {3 2} ^ {1} & g _ {3 3} ^ {1} & g _ {3 4} ^ {1} \\ g _ {4 1} ^ {1} & g _ {4 2} ^ {1} & g _ {4 3} ^ {1} & g _ {4 4} ^ {1} \end{array} \right) \quad \text {a n d} \quad G ^ {2} = \left( \begin{array}{c c c} g _ {1 1} ^ {2} & 0 & g _ {1 3} ^ {2} \\ g _ {2 1} ^ {2} & g _ {2 2} ^ {2} & g _ {2 3} ^ {2} \\ 0 & g _ {3 2} ^ {2} & 0 \\ g _ {4 1} ^ {2} & g _ {4 2} ^ {2} & g _ {4 3} ^ {2} \\ g _ {5 1} ^ {2} & g _ {5 2} ^ {2} & 0 \end{array} \right).
$$
Since the shared latent nodes are given by  $\mathcal{L} = \{1,2\}$ , we have
$$
G = \left( \begin{array}{c c c c c} g _ {1 1} ^ {1} & g _ {1 2} ^ {1} & g _ {1 3} ^ {1} & 0 & 0 \\ g _ {2 1} ^ {1} & 0 & g _ {2 3} ^ {1} & 0 & 0 \\ 0 & g _ {3 2} ^ {1} & g _ {3 3} ^ {1} & g _ {3 4} ^ {1} & 0 \\ g _ {4 1} ^ {1} & g _ {4 2} ^ {1} & g _ {4 3} ^ {1} & g _ {4 4} ^ {1} & 0 \\ g _ {1 1} ^ {2} & 0 & 0 & 0 & g _ {1 3} ^ {2} \\ g _ {2 1} ^ {2} & g _ {2 2} ^ {2} & 0 & 0 & g _ {2 3} ^ {2} \\ 0 & g _ {3 2} ^ {2} & 0 & 0 & 0 \\ g _ {4 1} ^ {2} & g _ {4 2} ^ {2} & 0 & 0 & g _ {4 3} ^ {2} \\ g _ {5 1} ^ {2} & g _ {5 2} ^ {2} & 0 & 0 & 0 \end{array} \right)
$$
$$
B = G \cdot (I - A) ^ {- 1} = \left( \begin{array}{c c c c c} a _ {2 1} g _ {1 2} ^ {1} + g _ {1 1} ^ {1} & g _ {1 2} ^ {1} & g _ {1 3} ^ {1} & 0 & 0 \\ g _ {2 1} ^ {1} & 0 & g _ {2 3} ^ {1} & 0 & 0 \\ a _ {2 1} g _ {3 2} ^ {1} & g _ {3 2} ^ {1} & a _ {4 3} g _ {3 4} ^ {1} + g _ {3 3} ^ {1} & g _ {3 4} ^ {1} & 0 \\ a _ {2 1} g _ {4 2} ^ {1} + g _ {4 1} ^ {1} & g _ {4 2} ^ {1} & a _ {4 3} g _ {4 4} ^ {1} + g _ {4 3} ^ {1} & g _ {4 4} ^ {1} & 0 \\ g _ {1 1} ^ {2} & 0 & 0 & 0 & g _ {1 3} ^ {2} \\ a _ {2 1} g _ {2 2} ^ {2} + g _ {2 1} ^ {2} & g _ {2 2} ^ {2} & 0 & 0 & g _ {2 3} ^ {2} \\ a _ {2 1} g _ {3 2} ^ {2} & g _ {3 2} ^ {2} & 0 & 0 & 0 \\ a _ {2 1} g _ {4 2} ^ {2} + g _ {4 1} ^ {2} & g _ {4 2} ^ {2} & 0 & 0 & g _ {4 3} ^ {2} \\ a _ {2 1} g _ {5 2} ^ {2} + g _ {5 1} ^ {2} & g _ {5 2} ^ {2} & 0 & 0 & 0 \end{array} \right).
$$
Figure 4: A 2-domain graph with 5 latent nodes and dimensions of the observed domains given by  $|V_{1}| = d_{1} = 4$  and  $|V_{2}| = d_{2} = 5$ . We denote  $V_{e} = \{v_{1}^{e},\dots,v_{d_{e}}^{e}\}$ , that is, the superscript indicates the domain a node belongs to.
Example B.2. Consider the  $m$ -domain graph in Figure 4. The partial pure children of node  $1 \in \mathcal{L}$  are given by  $\{v_2^1, v_1^2\}$  and the partial pure children of  $2 \in \mathcal{L}$  are given by  $\{v_3^1, v_3^2\}$ . Moreover, by continuing Example B.1, we have that the matrix  $B_{\mathcal{L}}$  is given by
$$
B _ {\mathcal {L}} = \left( \begin{array}{c c} a _ {2 1} g _ {1 2} ^ {1} + g _ {1 1} ^ {1} & g _ {1 2} ^ {1} \\ g _ {2 1} ^ {1} & 0 \\ a _ {2 1} g _ {3 2} ^ {1} & g _ {3 2} ^ {1} \\ a _ {2 1} g _ {4 2} ^ {1} + g _ {4 1} ^ {1} & g _ {4 2} ^ {1} \\ g _ {1 1} ^ {2} & 0 \\ a _ {2 1} g _ {2 2} ^ {2} + g _ {2 1} ^ {2} & g _ {2 2} ^ {2} \\ a _ {2 1} g _ {3 2} ^ {2} & g _ {3 2} ^ {2} \\ a _ {2 1} g _ {4 2} ^ {2} + g _ {4 1} ^ {2} & g _ {4 2} ^ {2} \\ a _ {2 1} g _ {5 2} ^ {2} + g _ {5 1} ^ {2} & g _ {5 2} ^ {2} \end{array} \right)
$$
It is easy to see that the two submatrices
$$
\left( \begin{array}{c c} g _ {2 1} ^ {1} & 0 \\ g _ {1 1} ^ {2} & 0 \end{array} \right) \quad \text {a n d} \quad \left( \begin{array}{c c} a _ {2 1} g _ {3 2} ^ {1} & g _ {3 2} ^ {1} \\ a _ {2 1} g _ {3 2} ^ {2} & g _ {3 2} ^ {2} \end{array} \right)
$$
have rank one. The first matrix corresponds to the partial pure children  $\{v_2^1, v_1^2\}$  in the graph in Figure 4 while the second matrix correspond to the partial pure children  $\{v_3^1, v_3^2\}$ . Note that the rank of any other  $2 \times 2$  submatrix is generically (i.e., almost surely) equal to 2.
C Discussion of the Assumptions
In this section, we discuss aspects of Conditions (C1)-(C3) that allow for identifiability. In particular, we discuss the necessity of pairwise different and non-Gaussian error distributions if one is not willing to make further assumptions. Moreover, we elaborate on the sparsity conditions on the mixing matrix and explain why some sparsity assumption is necessary.
Pairwise Different Error Distributions. Given any two potentially different  $m$ -domain graphs  $\mathcal{G}_m = (\mathcal{H} \cup V, D)$  and  $\widetilde{\mathcal{G}}_m = (\widetilde{\mathcal{H}} \cup V, \widetilde{D})$ , identifiability of the joint distribution in multi-domain causal representation models means that
$$
B _ {V _ {e}, S _ {e}} \# P _ {S _ {e}} = \widetilde {B} _ {V _ {e}, \widetilde {S} _ {e}} \# \widetilde {P} _ {\widetilde {S} _ {e}} \text {f o r a l l} e \in [ m ] \Rightarrow B \# P = \widetilde {B} \# \widetilde {P} \tag {9}
$$
for any representation  $(B, P)$  of a distribution in  $\mathcal{M}(\mathcal{G}_m)$  and any representation  $(\widetilde{B}, \widetilde{P})$  of a distribution in  $\mathcal{M}(\widetilde{\mathcal{G}}_m)$ , where the matrices  $B_{V_e, S_e}$  and  $\widetilde{B}_{V_e, \widetilde{S}_e}$  have full column rank for all  $e \in [m]$ . The left-hand side says that the marginal distributions in each domain are equal, while the right-hand side says that the joint distributions are equal. If there are  $m$ -domain graphs, such that the left-hand sides hold but the right-hand side is violated, then we say that the joint distribution is not identifiable.
We assume in this section that the marginal error distributions  $P_{i}$ ,  $i \in \mathcal{H}$  are non-Gaussian and have unit variance, but are not necessarily pairwise different or non-symmetric. Then the right-hand side holds if and only if the number of shared latent nodes in each graph is equal, i.e.,  $\ell = \widetilde{\ell}$ , and there is a signed permutation matrix  $\Psi$  such that  $B = \widetilde{B}\Psi$  and  $P = \Psi^\top \# \widetilde{P}$ . Here, the matrix  $\Psi$  does
not necessarily have a block structure. The equivalence is implied by the identifiability of the usual, one-domain linear ICA (see, e.g., Buchholz et al. (2022)) together with the fact that for  $\ell \neq \widetilde{\ell}$ , we have  $|\mathcal{H}| \neq |\widetilde{\mathcal{H}}|$  and, therefore, the distributions on the right-hand have support over different dimensional subspaces.
Theorem 3.1 shows that assumptions (C1) and (C2) are sufficient for identifiability of the joint distribution. In particular, we show that they imply identifiability in a stronger sense, namely, that it follows from the left-hand side that  $\ell = \widetilde{\ell}$  and  $B = \widetilde{B}\Psi$  and  $P = \Psi^\top \# \widetilde{P}$  for a signed permutation  $\Psi \in \Pi$  with block structure. The next proposition reveals necessary conditions for identifiability.
Proposition C.1. Let  $\mathcal{G}_m$  be an  $m$ -domain graph with shared latent nodes  $\mathcal{L} = [\ell]$ , and let  $P_X \in \mathcal{M}(\mathcal{G}_m)$  with representation  $(B, P)$ . Suppose that  $m \geq 2$  and that everything except the assumption about pairwise different error distributions in Conditions (C1) and (C2) is satisfied. Then, the joint distribution is not identifiable if one of the following holds:
(i) There is  $i,j\in \mathcal{L}$  such that  $P_{i} = P_{j}$  or  $P_{i} = -P_{j}$
(ii) There is  $i \in \mathcal{L}$  and  $j \in I_e$  for some  $e \in [m]$  such that  $P_i = P_j$  or  $P_i = -P_j$ .
(iii) For all  $e \in [m]$  there is  $i_e \in I_e$  such that  $P_{i_e} = P_{j_f}$  or  $P_{i_e} = -P_{i_f}$  for all  $e \neq f$ .