SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Lemma 2. Assume that there are two latent variable models  $((B^{(i)},\eta^{(i)},t_i)_{i\in I},f)$  and  $((\widetilde{B}^{(i)},\widetilde{\eta}^{(i)},\widetilde{t}_i)_{i\in I},\widetilde{f})$  as defined above such that  $f(Z^{(i)})\stackrel {\mathcal{D}}{=}\widetilde{f} (\widetilde{Z}^{(i)})$ . Then their latent dimension  $d$  agree and  $h = \widetilde{f}^{-1}\circ f$  exists and is a diffeomorphism, i.e., bijective and differentiable with differentiable inverse. Moreover, it satisfies the relation
$$
\begin{array}{l} \frac {1}{2} z ^ {\top} (\Theta^ {(i)} - \Theta^ {(0)}) z - \eta^ {(i)} (r ^ {(i)}) ^ {\top} z \\ = \frac {1}{2} h (z) ^ {\top} \left(\widetilde {\Theta} ^ {(i)} - \widetilde {\Theta} ^ {(0)}\right) h (z) - \widetilde {\eta} ^ {(i)} \left(\widetilde {r} ^ {(i)}\right) ^ {\top} h (z) + c ^ {(i)} \tag {19} \\ \end{array}
$$
for all  $z$  and  $i$  and some constant  $c^{(i)}$ . If  $\eta^{(i)} = \widetilde{\eta}^{(i)} = 0$  this simplifies to
$$
\frac {1}{2} z ^ {\top} \left(\Theta^ {(i)} - \Theta^ {(0)}\right) z = \frac {1}{2} h (z) ^ {\top} \left(\widetilde {\Theta} ^ {(i)} - \widetilde {\Theta} ^ {(0)}\right) h (z) + c ^ {(i)}. \tag {20}
$$
Proof. Note first that  $X^{(0)} \stackrel{\mathcal{D}}{=} f(Z^{(0)}}$  implies since  $f$  is an embedding by assumption that the dimension of the support of  $X^{(0)}$  is the same as the latent dimension and this dimension is identifiable from the observations, e.g., by considering the tangent space. As  $Z^{(0)}$  and  $\widetilde{Z}^{(0)}$  have full support and  $f$  and  $\tilde{f}$  are by assumption embeddings, the equality  $f(Z^{(0)}) \stackrel{\mathcal{D}}{=} \widetilde{f}(\widetilde{Z}^{(0)})$  implies that the two image manifolds agree and  $h = \widetilde{f}^{-1} \circ f$  is a differentiable map. Moreover, we find that  $h_{*}Z^{(i)} = \widetilde{Z}^{(i)}$  where  $h_{*}$  denotes the pushforward map. The variables  $Z^{(i)}$  are Gaussian with distribution  $Z^{(i)} \sim N(\mu^{(i)}, \Sigma^{(i)})$ . This implies that its density  $p^{(i)}$  can be written as
$$
\ln \left(p ^ {(i)} (z)\right) = - \frac {n}{2} \ln (2 \pi) - \frac {1}{2} \ln | \Sigma^ {(i)} | - \frac {1}{2} \left(z - \mu^ {(i)}\right) ^ {\top} \Theta^ {(i)} \left(z - \mu^ {(i)}\right). \tag {21}
$$
A similar relation holds for  $\widetilde{p}^{(i)}(z)$ , the density of  $\widetilde{Z}^{(i)}$ . Finally, we have the relations
$$
p ^ {(i)} (z) = \widetilde {p} ^ {(i)} (h (z)) | \det J _ {h} (z) | \tag {22}
$$
which is a consequence of  $h_* Z^{(i)} = \widetilde{Z}^{(i)}$ . Here  $J_h$  denotes the Jacobian matrix of partial derivatives. Thus we conclude that
$$
\begin{array}{l} - \frac {n}{2} \ln (2 \pi) - \frac {1}{2} \ln | \Sigma^ {(i)} | - \frac {1}{2} (z - \mu^ {(i)}) ^ {\top} \Theta^ {(i)} (z - \mu^ {(i)}) \\ = - \frac {n}{2} \ln (2 \pi) - \frac {1}{2} \ln | \widetilde {\Sigma} ^ {(i)} | - \frac {1}{2} \left(h (z) - \widetilde {\mu} ^ {(i)}\right) ^ {\top} \widetilde {\Theta} ^ {(i)} \left(h (z) - \widetilde {\mu} ^ {(i)}\right) + \ln | \det J _ {h} (z) |. \tag {23} \\ \end{array}
$$
Calling this relation  $E^i$ , we consider the difference  $E^0 - E^i$ . Then the determinant term cancels, and we obtain for some constant  $c^{(i)}$
$$
\begin{array}{l} \frac {1}{2} (z - \mu^ {(i)}) ^ {\top} \Theta^ {(i)} (z - \mu^ {(i)}) - \frac {1}{2} z ^ {\top} \Theta^ {(0)} z \\ = \frac {1}{2} (h (z) - \widetilde {\mu} ^ {(i)}) ^ {\top} \widetilde {\Theta} ^ {(i)} (h (z) - \widetilde {\mu} ^ {(i)}) - \frac {1}{2} h (z) ^ {\top} \widetilde {\Theta} ^ {(0)} h (z) + c ^ {(i)}. \\ \end{array}
$$
Expanding the quadratic expression on the left-hand side we obtain
$$
\frac {1}{2} (z - \mu^ {(i)}) ^ {\top} \Theta^ {(i)} (z - \mu^ {(i)}) - \frac {1}{2} z ^ {\top} \Theta^ {(0)} z = \frac {1}{2} z ^ {\top} (\Theta^ {(i)} - \Theta) z - z ^ {\top} \Theta^ {(i)} \mu^ {(i)} + \frac {1}{2} (\mu^ {(i)}) ^ {\top} \Theta^ {(i)} \mu^ {(i)}. \tag {25}
$$
Finally we simplify the last term using (18)
$$
\Theta^ {(i)} \mu^ {(i)} = \eta^ {(i)} (B ^ {(i)}) ^ {\top} B ^ {(i)} (B ^ {(i)}) ^ {- 1} e _ {t _ {i}} = \eta^ {(i)} (B ^ {(i)}) ^ {\top} e _ {t _ {i}} = \eta^ {(i)} r ^ {(i)} \tag {26}
$$
Apply the same simplification on the right hand side. Finally, we collect the constant terms independent of  $z$  in  $c^{(i)}$ , to obtain
$$
\begin{array}{l} \frac {1}{2} z ^ {\top} (\Theta^ {(i)} - \Theta^ {(0)}) z - \eta^ {(i)} (r ^ {(i)}) ^ {\top} z \\ = \frac {1}{2} h (z) ^ {\top} \left(\widetilde {\Theta} ^ {(i)} - \widetilde {\Theta} ^ {(0)}\right) h (z) - \widetilde {\eta} ^ {(i)} \left(\widetilde {r} ^ {(i)}\right) ^ {\top} h (z) + c ^ {(i)} \tag {27} \\ \end{array}
$$
which is (19) as desired.
A second key ingredient for our identifiability results is that the specific structure of the interventions that each target a single node which implies that the difference of precision matrices  $\Theta^{(i)} - \Theta^{(0)}$  has rank at most 2. This is also the key ingredient used (in a very different manner) in the proof of the main result in [84]. Here we highlight this simple result because we make use of it frequently.
Lemma 3. The precision matrices satisfy the relation
$$
\Theta^ {(i)} - \Theta^ {(0)} = r ^ {(i)} \otimes r ^ {(i)} - s ^ {(i)} \otimes s ^ {(i)}. \tag {28}
$$
Proof. Using (17) we find
$$
\begin{array}{l} \Theta^ {(i)} - \Theta^ {(0)} = (B ^ {(i)}) ^ {\top} B ^ {(i)} - B ^ {\top} B \\ = \left(B ^ {(i)}\right) ^ {\top} \left(\sum_ {k = 1} ^ {d} e _ {k} e _ {k} ^ {\top}\right) B ^ {(i)} - B ^ {\top} \left(\sum_ {k = 1} ^ {d} e _ {k} e _ {k} ^ {\top}\right) B \\ = \sum_ {k = 1} ^ {d} \left(B ^ {(i)}\right) ^ {\top} e _ {k} \left(\left(B ^ {(i)}\right) ^ {\top} e _ {k}\right) ^ {\top} - \sum_ {k = 1} ^ {d} B ^ {\top} e _ {k} \left(B ^ {\top} e _ {k}\right) ^ {\top} \tag {29} \\ = \left(\left(B ^ {(i)}\right) ^ {\top} e _ {t _ {i}}\right) \otimes \left(\left(B ^ {(i)}\right) ^ {\top} e _ {t _ {i}}\right) - \left(B ^ {\top} e _ {t _ {i}}\right) \otimes \left(B ^ {\top} e _ {t _ {i}}\right) \\ = r ^ {(i)} \otimes r ^ {(i)} - s ^ {(i)} \otimes s ^ {(i)}. \\ \end{array}
$$
Here we used the fact that all rows except row  $t_i$  of  $B^{(i)}$  and  $B^{(0)}$  agree as the intervention targets a single node, i.e.,  $(B^{(i)})^\top e_k = (B^{(0)})^\top e_k$  for  $k \neq t_i$ .
A.3 Warm-up
To provide some intuition how identifiability arises, we first provide the proof for a simpler result with a much stronger assumption on the set of intervention targets and the type of interventions. But note importantly that we don't relax the assumption of non-linearity of  $f$ .
Theorem 4. Let  $f(Z^{(i)}) \stackrel{\mathcal{D}}{=} \widetilde{f}(\widetilde{Z}^{(i)})$  for two sets of parameters  $((B^{(i)}, \eta^{(i)}, t_i)_{i \in I}, f)$  and  $((\widetilde{B}^{(i)}, \widetilde{\eta}^{(i)}, \widetilde{t}_i)_{i \in I}, \widetilde{f})$  as in the above model. We assume that there are  $2n$  perfect pairwise different interventions (i.e.,  $Z^{(i)} \neq Z^{(i')}$  for  $i \neq i'$ ), without shifts (i.e.,  $\eta^{(i)} = \widetilde{\eta}^{(i)} = 0$ ). Finally, we assume that there are 2 interventions for each node, and we know the intervention targets of the interventions, i.e., we assume that  $t_i = \widetilde{t}_i$  for all  $i$ . Then,  $Z^{(i)}$  and  $\widetilde{Z}^{(i)}$  agree up to scaling, i.e.,  $Z^{(i)} = D\tilde{Z}^{(i)}$  for a diagonal matrix  $D \in \mathrm{Diag}(n)$ .
Remark 4. Actually this result can be generalized to a setting where the latent SCM is non-parametric [92].
Proof. We consider a node  $k$  and two interventions  $i_1$  and  $i_2$  which target node  $k = t_{i_1} = t_{i_2}$  (by assumption those interventions exist and  $i_1$  and  $i_2$  agree for both representations). Since we assumed that the interventions are perfect, we have in our notation that
$$
r ^ {(i _ {1})} = \lambda_ {i _ {1}} e _ {k}, \quad r ^ {(i _ {2})} = \lambda_ {i _ {2}} e _ {k} \tag {30}
$$
for two constants  $\lambda_{i_1} \neq \lambda_{i_2} > 0$  and similarly for  $\widetilde{r}^{(i_j)}$ . Indeed, assuming positivity is not a restriction since  $\epsilon$  follows a symmetric distribution (this is also contained in the parametrisation  $(D^{(i)})^{\frac{1}{2}}$ ) and  $\lambda_{i_1} \neq \lambda_{i_2}$  is necessary to ensure that the interventional distributions differ. Combining (20) from Lemma 2 with Lemma 3 we obtain the relation
$$
\begin{array}{l} z ^ {\top} \left(r ^ {(i _ {j})} \otimes r ^ {(i _ {j})} - s ^ {(i _ {j})} \otimes s ^ {(i _ {j})}\right) z = z ^ {\top} \left(\Theta^ {(i _ {j})} - \Theta^ {(0)}\right) z \\ = h (z) ^ {\top} \left(\widetilde {\Theta} ^ {(i _ {j})} - \widetilde {\Theta} ^ {(0)}\right) h (z) + 2 c ^ {(i _ {j})} \tag {31} \\ = h (z) ^ {\top} \left(\widetilde {r} ^ {(i _ {j})} \otimes \widetilde {r} ^ {(i _ {j})} - \widetilde {s} ^ {(i _ {j})} \otimes \widetilde {s} ^ {(i _ {j})}\right) h (z) + 2 c ^ {(i _ {j})}. \\ \end{array}
$$
for  $j = 1,2$
Note that since  $k = t_{i_1} = t_{i_2}$ , we have  $s^{(i_1)} = s^{(i_2)} = B^\top e_k$  by definition. Thus, we subtract the relation in the last display for  $i_1$  from the relation for  $i_2$  and find using (30),
$$
\begin{array}{l} \left(\lambda_ {i _ {1}} ^ {2} - \lambda_ {i _ {2}} ^ {2}\right) z _ {k} ^ {2} = \left(z ^ {\top} r ^ {(i _ {1})}\right) ^ {2} - \left(z ^ {\top} r ^ {(i _ {2})}\right) ^ {2} \\ = \left(h (z) ^ {\top} \widetilde {r} ^ {(i _ {1})}\right) ^ {2} - \left(h (z) ^ {\top} \widetilde {r} ^ {(i _ {2})}\right) ^ {2} + 2 \left(c ^ {(i _ {1})} - c ^ {(i _ {2})}\right) \tag {32} \\ = \left(\widetilde {\lambda} _ {i _ {1}} ^ {2} - \widetilde {\lambda} _ {i _ {2}} ^ {2}\right) h _ {k} (z) ^ {2} + 2 \left(c ^ {(i _ {1})} - c ^ {(i _ {2})}\right). \\ \end{array}
$$
for all  $z$ . Since  $\widetilde{\lambda}_{i_1} \neq \widetilde{\lambda}_{i_2} > 0$  we can divide and obtain
$$
\frac {\lambda_ {i _ {1}} ^ {2} - \lambda_ {i _ {2}} ^ {2}}{\widetilde {\lambda} _ {i _ {1}} ^ {2} - \widetilde {\lambda} _ {i _ {2}} ^ {2}} z _ {k} ^ {2} = h _ {k} (z) ^ {2} + 2 \frac {c ^ {(i _ {1})} - c ^ {(i _ {2})}}{\widetilde {\lambda} _ {i _ {1}} ^ {2} - \widetilde {\lambda} _ {i _ {2}} ^ {2}}. \tag {33}
$$
Since  $h$  is surjective, we have that  $h_k$  is also surjective which implies that the range of the right hand side is  $\left[2(c^{(i_1)} - c^{(i_2)}) / (\widetilde{\lambda}_{i_1}^2 -\widetilde{\lambda}_{i_2}^2),\infty\right)$ . The range of the left hand side is depending on the sign of the prefactor and is either  $(-\infty ,0]$  or  $[0,\infty)$ . Since the ranges have to agree we conclude that  $c^{(i_1)} = c^{(i_2)}$  and
$$
h _ {k} (z) = \pm \sqrt {\frac {\lambda_ {i _ {1}} ^ {2} - \lambda_ {i _ {2}} ^ {2}}{\widetilde {\lambda} _ {i _ {1}} ^ {2} - \widetilde {\lambda} _ {i _ {2}} ^ {2}}} z _ {k}. \tag {34}
$$
Since  $k$  was arbitrary, this ends the proof.
A.4 Proof of identifiability up to linear maps
The proof of our main result relies on two essentially geometric lemmas and a slight extension of Lemma 2 which we discuss now. The first lemma alone is sufficient to handle the case where interventions change the scaling matrix  $D$ , while the second is required to handle interventions that change  $B$  and the third allows us to consider pure shift interventions.
Lemma 4. Let  $g: \mathbb{R}^d \to \mathbb{R}$  be a continuous and surjective function,  $Q$  a quadratic form on  $\mathbb{R}^d$  and  $\alpha \in \mathbb{R}$  a constant such that
$$
g (z) ^ {2} = z \cdot Q z + \alpha . \tag {35}
$$
Then  $Q$  has rank 1 and  $g$  is linear, i.e.,  $g(z) = v \cdot z$  for some  $v \in \mathbb{R}^d$ .
Proof. Assume that  $Q$  has the eigendecomposition  $Q = \sum_{i=1}^{r} \lambda_i v_i v_i^\top$  where  $r$  is the rank of  $Q$  and  $\lambda_i \neq 0$  and  $v_i$  has unit norm. Clearly  $g$  surjective implies  $r > 0$ . Since  $g$  is surjective we find that the range of the left hand side is  $[0, \infty)$  and therefore  $z \cdot Qz \geq -\alpha$ . This implies that  $\lambda_i \geq 0$  for all  $i$  and therefore, the range of the right hand side is  $[\alpha, \infty)$  which implies  $\alpha = 0$ . Now we fix any  $c > 0$  and consider  $E_c = \{z \in \mathbb{R}^d | z \cdot Qz = c\}$ . It is easy to see that  $E_c$  is the product of an ellipsoid and  $\mathbb{R}^{d-r}$  and thus is connected if  $r \geq 2$ . We find by (35) that  $g(z) = \pm \sqrt{c}$  for  $z \in E_c$  and if  $g(z) \in \{-\sqrt{c}, \sqrt{c}\}$  then  $z \in E_c$ . By continuity of  $g$  and since  $E_c$  is connected this implies that  $g(z) = \sqrt{c}$  for all  $z \in E_c$  or  $g(z) = -\sqrt{c}$  for all  $z \in E_c$  contradicting the surjectivity of  $g$ . We conclude that  $r = 1$ , i.e.,  $Q = \lambda_1 v_1 v_1^\top = \tilde{v}_1 \tilde{v}_1^\top$  with  $\tilde{v}_1 = \sqrt{\lambda_1} v_1$ . Thus we find that
$$
g (z) ^ {2} = | \tilde {v} _ {1} \cdot z | ^ {2}. \tag {36}
$$
Since  $g$  is continuous, this implies that  $g(z) = \pm \tilde{v}_1 \cdot z$  or  $g(z) = \pm |\tilde{v}_1 \cdot z|$ . As only the first functions are surjective we conclude that  $g(z) = \pm \tilde{v}_1 \cdot z$  and the claim follows with  $w = \pm \tilde{v}_1$ .
To handle shift interventions we need a slightly stronger version of the lemma above which contains an additional linear term.
Corollary 1. Let  $g: \mathbb{R}^d \to \mathbb{R}$  be a continuous and surjective function,  $Q$  a quadratic form on  $\mathbb{R}^d$ ,  $w \in \mathbb{R}^d$  a vector, and  $\alpha \in \mathbb{R}$  a constant such that
$$
g (z) ^ {2} = z \cdot Q z + w \cdot z + \alpha . \tag {37}
$$
Then  $Q$  has rank 1 and  $g$  is affine, i.e.,  $g(z) = v \cdot z + c$  for some  $v \in \mathbb{R}^d$  and  $c \in \mathbb{R}$ .
Proof. Assume as before that  $Q$  has the eigendecomposition  $Q = \sum_{i=1}^{r} \lambda_i v_i v_i^\top$  where  $r$  is the rank of  $Q$  and  $\lambda_i \neq 0$  and  $|v_i| = 1$ . First, we show that  $w \in \langle v_1, \ldots, v_r \rangle$  where  $\langle . \rangle$  denotes the span. Indeed, suppose that this is not the case. Then we could find  $z_0$  such that  $z_0 \cdot w < 0$  and  $z_0 v_i = 0$  for  $1 \leq i \leq r$ , and therefore  $z_0 \cdot Qz_0 = 0$ . Plugging  $z = \lambda z_0$  for  $\lambda \to \infty$  in (37) the right-hand side tends to  $-\infty$  while the left-hand side is non-negative. This is a contradiction and therefore  $w \in \langle v_1, \ldots, v_r \rangle$  holds. This implies that there is  $w'$  such that  $Qw' = w/2$  and thus
$$
z \cdot Q z + w \cdot z + \alpha = \left(z + w ^ {\prime}\right) \cdot Q \left(z + w ^ {\prime}\right) + \alpha - w ^ {\prime} \cdot Q w ^ {\prime}. \tag {38}
$$
Then we consider  $\widetilde{g} (\widetilde{z}) = g(\widetilde{z} -w^{\prime}),\widetilde{\alpha} = \alpha -w^{\prime}\cdot Qw^{\prime}$  which satisfy
$$
\begin{array}{l} \widetilde {g} ^ {2} (\widetilde {z}) = g ^ {2} \left(\widetilde {z} - w ^ {\prime}\right) (39) \\ = \left(\widetilde {z} - w ^ {\prime} + w ^ {\prime}\right) \cdot Q \left(\widetilde {z} - w ^ {\prime} + w ^ {\prime}\right) + \alpha - w ^ {\prime} \cdot Q w ^ {\prime} (40) \\ = \widetilde {z} \cdot Q \widetilde {z} + \widetilde {\alpha}. (41) \\ \end{array}
$$
Using Lemma 4 we conclude that  $Q$  has rank 1 and  $\widetilde{g}(\widetilde{z}) = v \cdot z$  for some  $v$ . But then  $g(z) = \widetilde{g}(z + w') = z \cdot v + w' \cdot v$  which concludes the proof.
The second lemma is similar, but slightly simpler.
Lemma 5. Assume that there is a quadratic form  $Q$ , a vector  $0 \neq w \in \mathbb{R}^d$ ,  $w' \in \mathbb{R}^d$ ,  $\alpha \in \mathbb{R}$  and a continuous function  $g: \mathbb{R}^d \to \mathbb{R}$  such that
$$
g (z) (w \cdot z) = z \cdot Q z + w ^ {\prime} \cdot z + \alpha . \tag {42}
$$
Then  $g(z) = v \cdot z + c$  for some  $v \in \mathbb{R}^d$ ,  $c \in \mathbb{R}$ , i.e.,  $g$  is an affine function.
Proof. By rescaling we can assume that  $w$  has unit norm. Plugging in  $z = 0$  we find that  $\alpha = 0$ . Denote by  $\Pi_w$  the orthogonal projection on  $w$ , i.e.,  $\Pi_w v = (w \cdot v)w$  and by  $\Pi_w^\perp$  the projection on the orthogonal complement of  $w$ , i.e.,  $z = \Pi_w z + \Pi_w^\perp z$  for all  $z \in \mathbb{R}^d$ . Then we find for all  $z$
$$
0 = g \left(\Pi_ {w} ^ {\perp} z\right) w \cdot \Pi_ {w} ^ {\perp} z = \left(\Pi_ {w} ^ {\perp} z\right) \cdot Q \Pi_ {w} ^ {\perp} z + \left(\Pi_ {w} ^ {\perp} z\right) \cdot w ^ {\prime}. \tag {43}
$$
By scaling  $z$  we find that both terms on the right hand side vanish and thus for all  $z$
$$
\left(\Pi_ {w} ^ {\perp} z\right) \cdot Q \Pi_ {w} ^ {\perp} z = 0, \tag {44}
$$
$$
\left(\Pi_ {w} ^ {\perp} z\right) \cdot w ^ {\prime} = 0. \tag {45}
$$
This implies (using also the symmetry of  $Q$ )
$$
\begin{array}{l} g (z) w \cdot \Pi_ {w} z = g (z) w \cdot z \\ = z \cdot Q z + w ^ {\prime} \cdot z \\ = \left(\Pi_ {w} z + \Pi_ {w} ^ {\perp} z\right) \cdot Q \left(\Pi_ {w} z + \Pi_ {w} ^ {\perp} z\right) + w ^ {\prime} \cdot \left(\Pi_ {w} z + \Pi_ {w} ^ {\perp} z\right) \tag {46} \\ = (\Pi_ {w} z) \cdot Q \Pi_ {w} z + 2 (\Pi_ {w} z) Q \Pi_ {w} ^ {\perp} z + (\Pi_ {w} ^ {\perp} z) \cdot Q \Pi_ {w} ^ {\perp} z + \Pi_ {w} z \cdot w ^ {\prime} \\ = (\Pi_ {w} z) \cdot (Q (\Pi_ {w} z + 2 \Pi_ {w} ^ {\perp} z) + w ^ {\prime}). \\ \end{array}
$$
Now we use  $\Pi_w z = (w \cdot \Pi_w z) w$  and find
$$
g (z) w \cdot \Pi_ {w} z = \left(w \cdot \Pi_ {w} z\right) w \cdot \left(Q \left(\Pi_ {w} z + 2 \Pi_ {w} ^ {\perp} z\right) + w ^ {\prime}\right). \tag {47}
$$
$$
g (z) = w \cdot Q \left(\Pi_ {w} z + 2 \Pi_ {w} ^ {\perp} z\right) + w \cdot w ^ {\prime} \tag {48}
$$
for all  $z$  such that  $\Pi_w z \neq 0$ . By continuity we conclude that this holds for all  $z$ , i.e.,  $g$  is affine.
Again we need a slight generalization of this lemma.
Corollary 2. Assume that there is a quadratic form  $Q$ , a vector  $0 \neq w \in \mathbb{R}^d$ ,  $w' \in \mathbb{R}^d$ ,  $\alpha, \beta \in \mathbb{R}$  and a continuous function  $g: \mathbb{R}^d \to \mathbb{R}$  such that
$$
g (z) \left(w \cdot z + \beta\right) = z \cdot Q z + w ^ {\prime} \cdot z + \alpha . \tag {49}
$$
Proof. Define  $\widetilde{g} (\widetilde{z}) = g(\widetilde{z} -\beta w|w|^{-2})$  . Then
$$
\begin{array}{l} \widetilde {g} (\widetilde {z}) (w \cdot \widetilde {z}) = g (\widetilde {z} - \beta w | w | ^ {- 2}) (w \cdot (\widetilde {z} - \beta w | w | ^ {- 2}) + \beta) \\ = (\widetilde {z} - \beta w | w | ^ {- 2}) \cdot Q (\widetilde {z} - \beta w | w | ^ {- 2}) + w ^ {\prime} \cdot (\widetilde {z} - \beta w | w | ^ {- 2}) + \alpha \tag {50} \\ = \widetilde {z} \cdot Q \widetilde {z} + \widetilde {w} \cdot \widetilde {z} + \widetilde {\alpha} \\ \end{array}
$$
for some  $\widetilde{w}$  and  $\widetilde{\alpha}$ . So, we conclude from Lemma 5 that  $\widetilde{g}$  is affine and thus the same is true for  $g$ .
We now discuss a small extension of Lemma 2 by showing that we can complete the squares in (19) which then allows us to obtain a relation that is very similar to (20).
Lemma 6. Assume the general setup as introduced above and also  $\Theta^{(i)}\neq \Theta$  . Then there is a vector  $\mu^{\prime (i)}$  and a constant  $c$  such that
$$
\left(z - \mu^ {(i)}\right) ^ {\top} \Theta^ {(i)} \left(z - \mu^ {(i)}\right) - z ^ {\top} \Theta^ {(0)} z = \left(z - \mu^ {\prime (i)}\right) ^ {\top} \left(\Theta^ {(i)} - \Theta^ {(0)}\right) \left(z - \mu^ {\prime (i)}\right) + c. \tag {51}
$$
Proof. Using the symmetry of the precision matrices and expanding all the terms, we can express the difference of the left-hand side and the right-hand side of (51) as
$$
\begin{array}{l} (z - \mu^ {(i)}) ^ {\top} \Theta^ {(i)} (z - \mu^ {(i)}) - z ^ {\top} \Theta^ {(0)} z - (z - \mu^ {\prime (i)}) ^ {\top} (\Theta^ {(i)} - \Theta^ {(0)}) (z - \mu^ {\prime (i)}) - c \\ = - 2 z ^ {\top} \Theta^ {(i)} \mu^ {(i)} + 2 z ^ {\top} \left(\Theta^ {(i)} - \Theta^ {(0)}\right) \mu^ {\prime (i)} + \left(\mu^ {(i)}\right) ^ {\top} \Theta^ {(i)} \mu^ {(i)} - \left(\mu^ {\prime (i)}\right) ^ {\top} \left(\Theta^ {(i)} - \Theta^ {(0)}\right) \mu^ {\prime (i)} - c. \tag {52} \\ \end{array}
$$
Hence, we can set  $c$  at the end such that the constant terms cancel and all we have to show is that there exists  $\mu^{(i)}$  such that
$$
(\Theta^ {(i)} - \Theta) \mu^ {\prime (i)} = \Theta^ {(i)} \mu^ {(i)}. \tag {53}
$$
This is clearly not true for arbitrary  $\mu^{(i)}$  because  $\Theta^{(i)} - \Theta$  has only rank 2. However, such a  $\mu^{\prime (i)}$  does exist for  $\mu^{(i)} = \eta^{(i)}(B^{(i)})^{-1}e_{t_i}$  (see equation (18)). Indeed, we can simplify using (17) and Lemma 3
$$
(\Theta^ {(i)} - \Theta) \mu^ {\prime (i)} = r ^ {(i)} \left(\left(r ^ {(i)}\right) ^ {\top} \mu^ {\prime (i)}\right) - s ^ {(i)} \left(\left(s ^ {(i)}\right) ^ {\top} \mu^ {\prime (i)}\right) \tag {54}
$$
$$
\Theta^ {(i)} \mu^ {(i)} = \eta^ {(i)} \left(B ^ {(i)}\right) ^ {\top} B ^ {(i)} \left(B ^ {(i)}\right) ^ {- 1} e _ {t _ {i}} = \eta^ {(i)} \left(B ^ {(i)}\right) ^ {\top} e _ {t _ {i}} = \eta^ {(i)} r ^ {(i)}. \tag {55}
$$
Now there are two cases. When  $r^{(i)}$  and  $s^{(i)}$  are collinear (but not identical, by assumption) we can choose  $\mu'^{(i)} = \lambda r^{(i)}$  for a suitable  $\lambda$ . Otherwise, we can pick any vector  $\mu'^{(i)}$  such that  $\mu'^{(i)} s^{(i)} = 0$ ,  $\mu'^{(i)} r^{(i)} = \eta^{(i)}$ . This is always possible for non-collinear vectors (project  $r^{(i)}$  on the orthogonal complement of  $s^{(i)}$ ).
Let us now first discuss a rough sketch of the proof of our main result. This allows us to present the main ideas without discussing all the technical details required for the full proof of the result. We restate the theorem for convenience.
Theorem 3. Assume that  $X^{(i)}$  is generated according to a model  $((B^{(i)},\eta^{(i)},t_i)_{i\in \bar{I}},f)$  such that the Assumptions 1-4 hold. Then we have identifiability up to linear transformations, i.e., if  $((\widetilde{B}^{(i)},\widetilde{\eta}^{(i)},\widetilde{t}_i),\widetilde{f})$  generates the same observed distributions  $\widetilde{f} (\widetilde{Z}^{(i)})\stackrel {\mathcal{D}}{=}X^{(i)}$ , then their latent dimensions  $d$  agree and there is an invertible linear map  $T$  such that
Proof Sketch. For simplicity of the sketch we ignore shift interventions here. We assume that we have two sets of parameters  $((B^{(i)},t_i)_{i\in \bar{I}},f),(\widetilde{B}^{(i)},\widetilde{t}_i)_{i\in \bar{I}},\widetilde{f})$  generating the same observations. We can relabel the interventions  $i$  and the variables  $\widetilde{Z}_k$  such that  $\widetilde{t}_i = i$  and such that  $1,2,\ldots$  is a valid causal order of the graph. The general idea is to show by induction that  $(h_1(z),\dots,h_k(z))$  is a linear function of  $z$ . Let us sketch how this is achieved. The key ingredients here are Lemma 2 and Lemma 3. Applying (20) from Lemma 2 in combination with Lemma 3 we obtain
$$
\frac {1}{2} z ^ {\top} \left(\Theta^ {(k + 1)} - \Theta^ {(0)}\right) z = \frac {1}{2} h (z) ^ {\top} \left(\widetilde {r} ^ {(k + 1)} \otimes \widetilde {r} ^ {(k + 1)} - \widetilde {s} ^ {(k + 1)} \otimes \widetilde {s} ^ {(k + 1)}\right) h (z) + c ^ {(k + 1)}. \tag {56}
$$
By our assumption that the variables  $\widetilde{Z}^{(i)}$  follow the causal order of the underlying graph, only the entries  $\widetilde{r}_r^{(k + 1)}$  for  $r\leq k + 1$  are non-zero and similarly for  $\widetilde{s}^{(k + 1)}$ . Therefore, the right-hand side is a quadratic form of  $(h_1(z),\ldots ,h_{k + 1}(z))$ . Now we apply our induction hypothesis which states that  $(h_1(z),\dots,h_k(z))$  is a linear function of  $z$ . Then we find expanding the quadratic form on the right-hand side that there is a matrix  $Q$ , a vector  $v$ , a constant  $\delta$  (which we here assume to be non-zero) and another quadratic form  $Q^{\prime}$  such that
$$
\begin{array}{l} \frac {1}{2} z ^ {\top} Q ^ {\prime} z = \frac {1}{2} z ^ {\top} Q z + (v \cdot z) h _ {k + 1} (z) + \delta h _ {k + 1} ^ {2} (z) + c ^ {(k + 1)} \\ = \frac {1}{2} z ^ {\top} Q z + \delta \left(h _ {k + 1} (z) + \frac {(v \cdot z)}{2 \delta}\right) ^ {2} - \frac {(v \cdot z) ^ {2}}{4 \delta} + c ^ {(k + 1)}. \tag {57} \\ \end{array}
$$
Now we can apply Corollary 1 to conclude that  $h_{k + 1}(z)$  is an affine function of  $z$ . Adding the shifts and considering also the case  $\delta = 0$  adds several technical difficulties, which we handle in the full proof.