Next, using the fact that $K$ is constant (in the given coordinates), and inserting the explicit form of Christoffel symbols we compute

$$ 
\begin{aligned}
\nabla_a (\mathring{K}^2)_b^a &= K_c^a \Gamma^{c}_{ad} K_b^d - K_c^a \Gamma^{d}_{ab} K_b^c \\
&= \frac{1}{2} K^{ac} (g_{ca,d} + g_{cd,a} - g_{ad,c}) K_b^d - \frac{1}{2} (K^2)^{ad} (g_{da,b} + g_{db,a} - g_{ab,d}).
\end{aligned}
$$

In both terms the second and third derivatives of $g$ cancel by symmetry of $K^{ac}$ and of $(K^2)^{ad}$. We now write sums explicitly after using that $g$ and $K$ are diagonal:

$$ \nabla_a (\mathring{K}^2)_b^a = \frac{1}{2} K_b^d K^{ac} g_{ac,d} - \frac{1}{2} (K^2)^{ac} g_{ac,b} = \frac{1}{2} \sum_a (k_b - k_a) k_a \partial_b h_a, \quad b = 1, 2, 3. $$

Using that $\partial_b h_a = 2x_b \lambda'(k_{a+1} - k_{a-1})$ with indices understood modulo 3, we compute by explicitly writing down the terms in the sum and factorizing to get

$$ \nabla_a (\mathring{K}^2)_b^a = x_b \lambda'(k_1 - k_2)(k_2 - k_3)(k_3 - k_1). $$

Since $x_b \lambda'$ is non-zero for $0 < |x|^2 < 1/2$ and since we assumed that the prescribed data has distinct eigenvalues, we find as announced that $\nabla_a (\mathring{K}^2)_b^a \neq 0$ for $0 < |x|^2 < 1/2$. $\square$

**Data with non-trivial derivatives.** In the course of proving the classification the momentum constraint reduces to an equation of the form

$$ \partial_a \gamma \mathring{K}_b^a = \sum_I \chi_I \partial_b \zeta_I, \qquad (6.10) $$

where $\gamma, \chi_I, \zeta_I$ are some scalar functions of the singularity data. The following lemma states that this equation implies that both sides vanish separately. Its proof is analogous to that of lemma 6.1 and we give it in section 6.3.

**Lemma 6.6 (Extrinsic curvature and derivatives of scalars).** Let $\mathcal{H}$ be a 3-manifold. Let $\gamma, \chi_I, \zeta_I$ be a finite collection of continuous ultralocal scalar fields such that for any singularity data $(g, K, \phi_0, \phi_1)$ on $\mathcal{H}$ the relation (6.10) holds at all points $x \in \mathcal{H}$ such that $K(x)$ has three distinct eigenvalues. Then $\gamma$ is a constant: it only depends on the data through the signature of $g$.

**Derivatives of scalars are independent.** Thanks to lemma 6.6 the momentum constraint reduces from (6.10) down to the vanishing of a sum of terms $\chi_I \partial_b \zeta_I$, $b = 1, 2, 3$, with $\chi_I, \zeta_I$ being ultralocal scalars. Expanding the derivatives gives a linear combination of $\partial_b \theta, \partial_b \phi_0, \partial_b \phi_1$, and the following lemma states that these derivatives are linearly independent in a suitable sense. Its proof is analogous to that of lemma 6.1 and we give it in section 6.3.

**Lemma 6.7 (Linear independence of derivatives of scalars).** Let $\mathcal{H}$ be a 3-manifold and let $\mu, \nu, \varkappa$ be continuous ultralocal scalar fields such that for any singularity data $(g, K, \phi_0, \phi_1)$ on $\mathcal{H}$ one has

$$ \mu \partial_b \theta + \nu \partial_b \phi_0 + \varkappa \partial_b \phi_1 = 0, \quad b = 1, 2, 3, \qquad (6.11) $$

at all points $x \in \mathcal{H}$ such that $K(x)$ has three distinct eigenvalues. Then one has $\mu = \nu = \varkappa = 0$.
## 6.3 Technical steps for the classification of singularity scattering maps

**Proof of lemma 6.6.** Fix a signature for the metric once and for all. We use the singularity data set on $\mathcal{H}$ that we constructed in the proof of lemma 6.1 for the case of a single prescribed value $(\underline{K}, \underline{\phi}_0, \underline{\phi}_1)$. We recall now solely the aspects that we need in this proof. Outside a ball $B \subset \mathcal{H}$ which we identify (by a diffeomorphism) with the unit ball in $\mathbb{R}^3$, the data are trivial. Inside the ball, we have four layers, in which all scalar functions are radial, in the sense that they only depend on $|x|^2$. The data at 0 matches the prescribed value $(\underline{K}, \underline{\phi}_0, \underline{\phi}_1)$. In the first three layers,

$$g = \Omega^{-2/3} \operatorname{diag}(\pm 1, 1, 1), \qquad \check{K} = \Omega \check{K}, \qquad (6.12)$$

with a suitably chosen radial function $\Omega > 0$. In particular, $\Omega(0) = 1$ to reproduce the prescribed data at 0, while at the outer edge of the third layer $\Omega$ is equal to a value (6.4) for which $\Omega = \phi_0$. In the last layer,

$$g = \phi_0^{-2/3} \operatorname{diag}(\pm 1, 1, 1), \qquad \check{K} = \phi_0 \alpha \check{K}, \qquad (6.13)$$

with a suitably chosen radial function $\alpha$ interpolating smoothly from $\alpha = 1$ at the inner edge of the layer to $\alpha = 0$ in a neighborhood of the boundary of the ball, say for $|x|^2 \ge 5/6$. We can select $\alpha$ so that it is positive for $|x|^2 < 5/6$.

Crucially, $\check{K}$ is proportional to $\check{K}$ throughout, and tends to 0 at $|x|^2 = 5/6$. Choose now $K$ to have pairwise distinct eigenvalues, so that $K(x)$ also does in the region $|x|^2 < 5/6$. By assumption, we thus have

$$\sum_I \chi_I \partial_b \zeta_I = \partial_a \gamma \check{K}_b^a = \left(k_b - \frac{1}{3}\right) \partial_b \gamma, \quad b = 1, 2, 3, \qquad (6.14)$$

in this region. Since all scalars are radial we find

$$2x_b \sum_I \chi_I \zeta_I' = 2x_b \left(k_b - \frac{1}{3}\right) \gamma', \quad b = 1, 2, 3, \qquad (6.15)$$

where primes denote $|x|^2$ derivatives. Away from the coordinate planes we can divide by $2x_b$ and take the difference of two of these equations to get $(k_b - k_a)\gamma' = 0$ for all $a, b = 1, 2, 3$. Since $K(x)$ is non-degenerate for $|x|^2 < 5/6$, we learn that $\gamma' = 0$ in this region minus the coordinate planes. Since $\gamma$ is a radial function we finally get that $\gamma$ is a constant on $0 < |x|^2 < 5/6$. By continuity, $\gamma(0)$ and $\gamma(5/6)$ also take the same value. The scalar field thus takes the same value for the prescribed data as for the trivial data $\check{K} = 0$, $\phi_0 = 1/\sqrt{12\pi}$, $\phi_1 = 0$. Let us call this constant value $\gamma_0$. Now $\gamma - \gamma_0$ obeys the conditions of lemma 6.3 hence $\gamma = \gamma_0$ throughout $\mathcal{H}$ for arbitrary data sets, as we wanted to prove.

**Proof of lemma 6.7.** As usual, lemma 5.1 implies that $\mu, \nu, \varkappa$ are functions of $(\theta, \phi_0, \phi_1)$. We prove $\varkappa = 0$, $\nu = 0$, and $\mu = 0$, in this order, by applying the continuity lemma 6.3 after proving these identities for data with any prescribed value $(\underline{K}, \underline{\phi}_0, \underline{\phi}_1)$ such that $\phi_0 \ne 0$ and $\underline{K}$ has pairwise distinct eigenvalues. We take $\underline{K}$ diagonal without loss of generality. Let us show $\varkappa = 0$. As in previous proofs, we construct the data in layers in the unit ball $B \subset \mathbb{R}^3$
and work in the standard basis. We only describe the first layer, as the construction can easily be completed, using the same layers as in lemma 6.1, to a data set on B that is trivial near the boundary. In the first layer, say $|x|^2 \le 1/2$, we keep $K = \mathcal{K}$ and $\phi_0 = \phi_0$ constant but vary $\phi_1$ and the metric, which we choose to be diagonal and of determinant $\pm 1$, that is, $g = \text{diag}(\pm e^{h_1}, e^{h_2}, e^{h_3})$ with $h_1 + h_2 + h_3 = 0$. For this data, the momentum constraint simplifies further than (6.9), namely we have $-\frac{1}{2} \sum_a k_a \partial_b h_a = 8\pi \phi_0 \partial_b \phi_1$ ($b = 1, 2, 3$), whose solution is

$$ \phi_1 = \frac{-1}{16\pi\phi_0} \sum_a k_a h_a + \text{constant.} \qquad (6.16) $$

The functions $h_a(|x|^2)$ are arbitrary except for $h_1 + h_2 + h_3 = 0$ and for the fact that they must vanish together with all their derivatives at 0 and $1/2$, so we can arrange that $\phi'_1 \ne 0$ in the interval $0 < |x|^2 < 1/2$. On the other hand, (6.11) reads $0 = \kappa \partial_b \phi_1 = 2x_b \kappa \phi'_1$, so we conclude that $\kappa$ vanishes near 0, except along the coordinate planes. By continuity, $\kappa$ vanishes at 0, namely it vanishes for the prescribed data. By lemma 6.3, $\kappa = 0$ identically.

Next, to prove $\nu = 0$, we consider exactly the data set on $B \subset \mathbb{R}^3$ used for lemma 6.1, but specify further the conformal factor $\Omega$ used there. Recall that in the first layer $\tilde{K} = \Omega \tilde{K}$ where $\tilde{K}$ is the prescribed value and $\Omega = \Omega(|x|^2)$ interpolates between $\Omega(0) = 1$ and some value at the outer boundary $|x|^2 = 1/4$ of the layer, with all derivatives vanishing at these end-points. We choose $\Omega$ such that it takes the value 1 again for some $|x|^2 \in (0, 1/4)$, but with a non-zero derivative. Then the data at this point is equal to the prescribed value, but $\partial_b \phi_0 \ne 0$. On the other hand, by construction of the data set, $\tilde{K}$ is everywhere a non-negative multiple of the given data, in other words $\theta$ is constant. Thus, (6.11) reads $\nu \partial_b \phi_0 = 0$ hence $\nu = 0$. Since this holds for arbitrary prescribed data, we conclude that $\nu = 0$ identically.

Finally, proving $\mu = 0$ requires building singularity data sets with variable $\theta$, and we can ignore how $\phi_0, \phi_1$ vary since we already showed $\nu = \kappa = 0$. We use the conformally flat data constructed in (6.5). In particular, this data set has $\tilde{K} = |\phi_0| \tilde{H}$ with

$$ \tilde{H}_b^a = \alpha(|x|^2) \frac{\tilde{K}_b^a}{\tilde{K}_b^a + \beta(|x|^2)} \left( x^a x_b - \frac{1}{3} \delta_b^a |x|^2 \right). \qquad (6.17) $$

Contrarily to what we do below (6.5), we now take $\alpha = \alpha(|x|^2)$ and $\beta = \beta(|x|^2)$ to be different radial functions. Specifically, we fix some generic point $y \in B$ (we determine later the genericity condition) and impose some values for $\alpha, \beta, \alpha', \beta'$ at that particular point:

$$ \alpha(|y|^2) = 1, \quad \beta(|y|^2) = 0, \quad \alpha'(|y|^2) = \frac{\tilde{K}_b^a}{\tilde{K}_b^a} y^b y_a, \quad \beta'(|y|^2) = - \text{Tr}(\tilde{K}^2). \qquad (6.18) $$

Then, $\tilde{H}(y) = \tilde{K}$ while $\partial_b \tilde{H}(y)_c^a = 2y_b(\alpha' \tilde{K}_c^a + \beta'(y^a y_c - \frac{1}{3} \delta_c^a |y|^2))$, so

$$ \partial_b \text{Tr}(\tilde{H}^2) = 2 \text{Tr}(\tilde{H} \partial_b \tilde{H}) = 4x_b (\alpha' \text{Tr}(\tilde{K}^2) + \beta' y^a \tilde{K}_a^c y_c) = 0. \qquad (6.19) $$

On the other hand, we have

$$ \begin{aligned} \partial_b \text{Tr}(\tilde{H}^3) &= 3 \text{Tr}(\tilde{H}^2 \partial_b \tilde{H}) \\ &= 6y_b \left( \alpha' \text{Tr}(\tilde{K}^3) + \beta' \left( y^a (\tilde{K}^2)_a^c y_c - \frac{1}{3} |y|^2 \text{Tr}(\tilde{K}^2) \right) \right) \\ &= 2y_b (\underline{k}_1 - \underline{k}_2)(\underline{k}_2 - \underline{k}_3)(\underline{k}_3 - \underline{k}_1)((\underline{k}_2 - \underline{k}_3)y^1 y_1 + (\underline{k}_3 - \underline{k}_1)y^2 y_2 + (\underline{k}_1 - \underline{k}_2)y^3 y_3) \end{aligned} \qquad (6.20) $$
where we obtained the second line by explicitly writing down all terms, using $\mathrm{Tr} \hat{\mathcal{K}} = 0$, and factorizing. For generic $y$ the result is nonzero provided eigenvalues of $\hat{\mathcal{K}}$ are pairwise distinct.

We are interested in the Kasner angle, which one can get from (5.5), and using that $\hat{\mathcal{K}}$ is a positive multiple of $\tilde{H}$: $\cos(3\theta) = \frac{\mathrm{Tr} \hat{\mathcal{K}}^3}{\sqrt{6}(\mathrm{Tr} \hat{\mathcal{K}}^2)^{3/2}} = \frac{\mathrm{Tr} \tilde{H}^3}{\sqrt{6}(\mathrm{Tr} \tilde{H}^2)^{3/2}}$. Taking a derivative and evaluating at $y$ we get

$$ -3 \sin(3\theta) \partial_b \theta = \frac{\partial_b \mathrm{Tr}(\tilde{H}^3)}{\sqrt{6}(\mathrm{Tr} \tilde{H}^2)^{3/2}} \neq 0. \qquad (6.21) $$

As a consistency check we note that $\sin(3\theta)$ is indeed non-zero when $\theta \neq 0 \mod \pi/3$, namely when eigenvalues of $K$ are pairwise distinct. An important consequence of (6.21), however, is that $\partial_b \theta \neq 0$. Then (6.11) $\mu \partial_b \theta = 0$ implies that $\mu = 0$ at the point $y$, where data can take any prescribed value with $\phi_0 \neq 0$ (so that the conformal scaling makes sense) and pairwise distinct eigenvalues for $K$. We conclude by lemma 6.3 that $\mu = 0$ identically. This establishes lemma 6.7.

## 6.4 Structure of scattering maps

**Reduction to pointwise scattering maps.** Let us consider a spacelike or timelike ultralocal singularity scattering map $\mathbf{S}$ on some (unimportant) 3-manifold $\mathcal{H}$. Its restriction $\mathbf{S}_x$ to any one point $x \in \mathcal{H}$ can be described as follows. Any choice of local coordinates near $x$ identifies the space of possible values of $(g_-, K_-, \phi_{0-}, \phi_{1-})(x)$ to the finite-dimensional space $\mathbf{I}_{\text{point}}$ of tuples $(g, K, \phi_0, \phi_1)$ such that $\phi_0, \phi_1 \in \mathbb{R}$, $g$ is a quadratic form on $\mathbb{R}^3$ with signature $\pm$++ in the spacelike or timelike case, and $K$ is a matrix that is symmetric with respect to $g$ and that obeys $\mathrm{Tr} K = 1$ and $1 - \mathrm{Tr}(K^2) = 8\pi\phi_0^2$. Under this identification, $\mathbf{S}$ yields a map $\mathbf{S}_{\text{point}}: \mathbf{I}_{\text{point}} \to \mathbf{I}_{\text{point}}$ that is independent of the choice of $x$ and of local coordinates thanks to diffeomorphism invariance. Changing local coordinates acts with a matrix $A \in \mathrm{GL}(3, \mathbb{R})$ on both sides of the singularity, namely

$$ \mathbf{S}_{\text{point}}(A \cdot (g, K)) = A \cdot \mathbf{S}_{\text{point}}(g, K), $$

where $A$ acts in the obvious manner $g_{ab} \mapsto A_a^c A_b^d g_{cd}$ and $K_b^a \mapsto (A^{-1})_c^a A_b^d K_d^c$. We arrive at a first useful description of ultralocal singularity scattering maps.

**Lemma 6.8 (Reduction to pointwise scattering maps).** *Specifying an ultralocal singularity scattering map $\mathbf{S}$ is equivalent to specifying a $\mathrm{GL}(3, \mathbb{R})$-covariant map $\mathbf{S}_{\text{point}}: \mathbf{I}_{\text{point}} \to \mathbf{I}_{\text{point}}$ that preserves the momentum constraint in the following sense. For any data $(g_-, K_-, \phi_{0-}, \phi_{1-})$ on some three-manifold $\mathcal{H}$, and $(g_+, K_+, \phi_{0+}, \phi_{1+})$ its image under pointwise application of $\mathbf{S}_{\text{point}}$, one has:*

$$ \text{if } \nabla_{-a} K_{-b}^{a} = 8\pi \phi_{0-} \partial_{b} \phi_{1-} \text{ then } \nabla_{+a} K_{+b}^{a} = 8\pi \phi_{0+} \partial_{b} \phi_{1+}. $$

**Polynomial structure of extrinsic curvature.** We have seen in lemma 5.1 that scalars such as $\phi_{0+}$ and $\phi_{1+}$ are simply functions of $\theta, \phi_0, \phi_1$ that are even and $2\pi/3$-periodic (spacelike case) or $2\pi$-periodic (timelike case) in $\theta$, and that are $\theta$-independent for $\phi_0 = \pm 1/\sqrt{12\pi}$. The tensors $g_+$ and $K_+$ are likewise constrained by covariance under $\mathrm{GL}(3, \mathbb{R})$
(change of basis). We focus first on $K_+$ for definiteness, then we apply the same arguments to $(g_-)^{-1}g_+$, and finally to its logarithm after showing it exists and is real.

Let us work in an orthonormal basis of eigenvectors of $K_-$, namely a basis $v_1, v_2, v_3$ in which $g_- = \text{diag}(\pm 1, 1, 1)$ and $K_-$ is diagonal. The change of basis mapping one of the eigenvectors $v_a$ to its opposite does not affect $g_-$ and $K_-$ hence the image $(g_+, K_+, \phi_{0+}, \phi_{1+})$ of $(g_-, K_-, \phi_{0-}, \phi_{1-})$ under the scattering map is also unaffected. However, off-diagonal components of $K_+$ in the basis $v_1, v_2, v_3$ change sign under such a change of basis, so they must vanish. We learn that $K_+$ is diagonal in the same basis as $g_-$ and $K_-$. If $k_{1-}, k_{2-}, k_{3-}$ are all distinct the three matrices $\delta = \text{diag}(1, 1, 1)$, $K_- = \text{diag}(k_{1-}, k_{2-}, k_{3-})$, and $K_-^2 = \text{diag}(k_{1-}^2, k_{2-}^2, k_{3-}^2)$ span the space of all diagonal matrices, so $K_+$ is a linear combination of them. It is most convenient later on to work with the traceless $\hat{K}_+ = K_+ - \frac{1}{3}\delta$ and powers of $\hat{K}_- = K_- - \frac{1}{3}\delta$, and write

$$ \hat{K}_+ = \beta_0\delta + \beta_1\hat{K}_- + \beta_2\hat{K}_-^2 \quad (6.22) $$

for some functions $\beta_0, \beta_1, \beta_2$ of $(\theta_-, \phi_{0-}, \phi_{1-})$. Tracelessness of $\hat{K}_+$ imposes $\beta_0 = -\frac{1}{3}\beta_2 \text{Tr} \hat{K}_-^2$, of course, but it is more convenient for us to keep all three functions. At this stage of the argument, these functions are only defined when eigenvalues are all distinct, namely when $\theta_- \neq 0 \mod \pi/3$. Let us comment on periodicity. Exchanging the eigenvectors $v_2$ and $v_3$ maps $\theta_- \to -\theta_-$ and swaps $k_{2\pm} \leftrightarrow k_{3\pm}$, which must leave (6.22) invariant, so the functions $\beta_0, \beta_1, \beta_2$ are even in $\theta_-$. Likewise, they are $2\pi/3$ periodic in the spacelike case because the cyclic permutation $v_1 \to v_2 \to v_3 \to v_1$ permutes eigenvalues $k_{a-}$ and $k_{a+}$ in the same way and maps $\theta_- \to \theta_- + 2\pi/3$. In the timelike case this cyclic permutation is not available because $v_1$ is singled out as being timelike with respect to the metric $g_-$.

Whenever two eigenvalues of $K_-$ coincide (say, $k_{1-} = k_{2-}$ for definiteness), the corresponding eigenvalues of $K_+$ also do, as we now prove. For this, change basis in $\text{Span}(v_1, v_2)$ from $v_1, v_2$ to another orthonormal pair of vectors $v'_1, v'_2$ with the same timelike/spacelike nature, namely with $g_-(v_i, v_j) = g_-(v'_i, v'_j)$ for $1 \le i, j \le 2$. This is an $O(2, \mathbb{R})$ or $O(1, 1, \mathbb{R})$ transformation depending on signature. The change of basis leaves $g_-$, $K_-$ invariant hence must leave $g_+$, $K_+$ invariant. In particular, $K_+$ remains diagonal, namely $v'_1, v'_2$ are also eigenvectors of $K_+$, which implies that $v_1$ and $v_2$ have the same eigenvalue under $K_+$.

From this fact, and assuming that singularity scattering maps map smooth data to twice differentiable data, it would be possible to prove that $\beta_0, \beta_1, \beta_2$ extend to continuous functions for all $(\theta_-, \phi_{0-}, \phi_{1-})$. The analysis is somewhat tedious but we will not need it: indeed, lemma 6.4 ensures that studying a singularity scattering map restricted to non-degenerate data is enough to fully characterize it. We will simply impose at the end that the scattering maps we find have well-defined limits when two Kasner exponents coincide.

**Polynomial structure of scattering maps.** The arguments above apply if we replace $K_+$ by the matrix $g_-^{-1}g_+$ with components $g_-^{ab}g_{+bc}$, and they lead to expressing this matrix as a linear combination of $\delta, K_-, K_-^2$ with coefficients that are possibly singular at $r_- = 0$. The real matrix $g_-^{-1}g_+$ is diagonal in the real basis $v_1, v_2, v_3$ hence it has real eigenvalues. They are non-zero since the matrix is invertible (with inverse $g_+^{-1}g_-$). Consider briefly the special case where the two spacelike eigenvalues $k_{2-} = k_{3-}$ of $K_-$ coincide. Then as
proven above for $K_+$, the entries (2,2) and (3,3) of $g_-^{-1}g_+$ are equal, from which we deduce $g_+(v_2, v_2) = g_+(v_3, v_3)$. Because $g_+$ is diagonal and has signature $-++$, exactly one of its diagonal entries must be negative, so by elimination $g_+(v_1, v_1) < 0$. We thus learn that eigenvalues of $g_-^{-1}g_+$ are all positive in this case $k_{2-} = k_{3-}$. To extend the result to any $K_-$, consider a smooth singularity data set $(g_-, K_-, \phi_{0-}, \phi_{1-})$ interpolating between a point where $k_{2-} = k_{3-}$ and a point with the desired value of $K_-$. Continuity of the metric $g_+$ implies that eigenvalues of $g_-^{-1}g_+$ vary continuously. Since they are positive at a point and cannot vanish, they are positive everywhere. We conclude that the matrix $g_-^{-1}g_+$ has positive eigenvalues only. The matrix thus admits a logarithm, to which the arguments above apply as well. We conclude that the matrix $\log(g_-^{-1}g_+)$ is a linear combination of $\delta, K_-, K_-^2$ too, as long as the $k_{a-}$ are pairwise distinct.

In practice, instead of $\delta, K_-, K_-^2$ we write matrices as linear combinations of two other sets of matrices. For $K_+$ we write $K_- = \frac{1}{3} + \mathring{K}_-$ and express $\delta, K_-, K_-^2$ as linear combinations of $\delta, \mathring{K}_-, \mathring{K}_-^2$ as stated above. For the metric we express these matrices further in terms of $\delta, \cos(\Theta_-), \cos(2\Theta_-)$ where $\Theta_- = \operatorname{diag}(\theta_-, \theta_- + 2\pi/3, \theta_- + 4\pi/3)$: we recall the relations (5.4)

$$ \mathring{K}_- = \frac{2r_-}{3} \cos \Theta_-, \quad \mathring{K}_-^2 = \frac{2r_-^2}{9} (\delta + \cos(2\Theta_-)). \qquad (6.23) $$

Importantly, the angle $\theta_-$, hence the matrices $\Theta_-$, $\cos\Theta_-$, and $\cos(2\Theta_-)$, are ill-defined at $r_- = 0$. Thus, a linear combination $\alpha_0\delta + \alpha_1\cos\Theta_- + \alpha_2\cos(2\Theta_-)$ only has a well-defined limit if the scalar fields $\partial_{\theta_-}\alpha_0, \alpha_1, \alpha_2$ vanish at $r_- = 0$. We deduce the following lemma.

**Lemma 6.9 (Polynomial structure of scattering maps).** *Any ultralocal singularity scattering map obeys*

$$ g_+ = \exp(\alpha_0 \delta + \alpha_1 \cos \Theta_- + \alpha_2 \cos(2\Theta_-))g_-, \quad \mathring{K}_+ = \beta_0 + \beta_1 \mathring{K}_- + \beta_2 \mathring{K}_-^2, \qquad (6.24) $$

in which $\alpha_0, \alpha_1, \alpha_2, \beta_0, \beta_1, \beta_2$ are scalar functions, like $\phi_{0+}, \phi_{1+}$, namely functions of $\theta_-, \phi_{0-}, \phi_{1-}$ that are even and periodic in $\theta_-$ with period $2\pi/3$ in the spacelike case and $2\pi$ in the timelike case. In addition, $\partial_{\theta_-}\alpha_0, \alpha_1, \alpha_2$ vanish at $r_- = 0$.

The tracelessness $\operatorname{Tr} \mathring{K}_+ = 0$ translates to $\beta_0 = -\beta_2 \operatorname{Tr}(\mathring{K}_-^2)/3$ but we do not need this for now. We also do not analyse yet how $\beta_0, \beta_1, \beta_2$ behave at $r_- = 0$. The exponential in (6.24) is defined by its power series; it yields a matrix, whose upper index we lower using $g_-$, so as to obtain the (0,2) tensor $g_+$. We easily compute the inverse metric and the ratio $\omega$ of volume factors, which simplifies because $\cos\Theta_-$ and $\cos(2\Theta_-)$ are traceless:

$$ g_+^{-1} = \exp(-\alpha_0 \delta - \alpha_1 \cos \Theta_- - \alpha_2 \cos(2\Theta_-)) g_-^{-1}, \quad \omega := \sqrt{|g_+|/|g_-|} = e^{3\alpha_0/2}. \qquad (6.25) $$

## 6.5 Scaling of trace-free extrinsic curvature

**Simplifying the momentum constraint.** Our next step is to plug the polynomial form (6.24) into the momentum constraint. Since $K_+$ has a constant trace, its trace-free part $\mathring{K}_+ = K_+ - \frac{1}{3}\delta$ has the same divergence as $K_+$ and the constraint reads $\nabla_{+a} \mathring{K}_{+b}^a =$
$8\pi \phi_{0+} \partial_b \phi_{1+}$. The Levi-Civita connections of two metrics $g_+$ and $g_-$ differ by a tensor, whose components are

$$ (\Gamma_+ - \Gamma_-)^c_{ab} = \Gamma^c_{+ab} - \Gamma^c_{-ab} = \frac{1}{2}(g_+^{-1})^{cd}(\nabla_{-a}g_{+bd} + \nabla_{-b}g_{+da} - \nabla_{-d}g_{+ab}), $$

where we wrote the inverse metric of $g_+$ explicitly as $g_+^{-1}$ for emphasis. Using the well-known identity $\Gamma^a_{ac} = \partial_c(\log|g|^{1/2})$ we find $(\Gamma_+ - \Gamma_-)^a_{ac} = \Gamma^a_{+ac} - \Gamma^a_{-ac} = \partial_c(\log \omega)$. Combining the above, we compute

$$ \begin{aligned} \nabla_{+a} \mathring{K}_{+b}^{a} &= \nabla_{-a} \mathring{K}_{+b}^{a} + \mathring{K}_{+b}^{c} (\Gamma_{+} - \Gamma_{-})^{a}_{ac} - \mathring{K}_{+c}^{a} (\Gamma_{+} - \Gamma_{-})^{c}_{ab} \\ &= \nabla_{-a} \mathring{K}_{+b}^{a} + \mathring{K}_{+b}^{a} \partial_a (\log \omega) - \frac{1}{2} \mathring{K}_{+c}^{a} (g_{+}^{-1})^{cd} \nabla_{-b} g_{+da} = \omega^{-1} \nabla_{-a} (\omega \mathring{K}_{+b}^{a}) - \frac{1}{2} X_b. \end{aligned} \tag{6.26a} $$

Here, we used the symmetry of $\mathring{K}_{+c}^{a}(g_{+}^{-1})^{cd}$ and cancelled two terms in $(\Gamma_+ - \Gamma_-)^c_{ab}$, while we introduced a notation for the last term:

$$ X_b := \mathring{K}_{+c}^{a}(g_{+}^{-1})^{cd} \nabla_{-b} g_{+da}. \tag{6.26b} $$

**Most terms involve derivatives of scalars.** The term $X_b$ defined in (6.26b) can be recast (using $\nabla_{-}g_{-} = 0$) as the trace of a product of matrices:

$$ X_b = \mathring{K}_{+c}^{a}(g_{+}^{-1}g_{-})_d^c \nabla_{-b}(g_{-}^{-1}g_{+})_a^d = \mathrm{Tr}(\mathring{K}_{+}(g_{+}^{-1}g_{-})\nabla_{-b}(g_{-}^{-1}g_{+})). $$

Given their explicit polynomial forms in lemma 6.9, $\mathring{K}_+$ and $g_+^{-1}g_-$ commute, so $(g_+^{-1}g_-)\nabla_{-b}(g_-^{-1}g_+)$ can be replaced within the trace by $\nabla_{-b}\log(g_-^{-1}g_+)$. Explicitly,

$$ X_b = \mathrm{Tr}(\mathring{K}_+ \nabla_{-b} \log(g_-^{-1} g_+)) = \mathrm{Tr}((\beta_0 + \beta_1 \mathring{K}_- + \beta_2 \mathring{K}_-^2) \nabla_{-b} (\alpha_0 \delta + \alpha_1 \cos \Theta_- + \alpha_2 \cos(2\Theta_-))). \tag{6.27} $$

By writing $\cos \Theta_-$ and $\cos(2\Theta_-)$ in terms of $\delta$, $\mathring{K}_-$, $\mathring{K}_-^2$ using (6.23) we obtain polynomial expressions in $\mathring{K}_-$. Expanding further, the derivative $\nabla_{-b}$ can either act on scalars $\alpha_n$, giving terms of the form $\mathrm{Tr}(\dots)\partial_b \alpha_n$, or act on powers of $\mathring{K}_-$, giving terms of the form $\mathrm{Tr}(\mathring{K}_-^n \nabla_{-b} \mathring{K}_-)$ times a scalar. Since $\mathrm{Tr}(\mathring{K}_-^n \nabla_{-b} \mathring{K}_-) = \partial_b \mathrm{Tr}(\mathring{K}_-^{n+1})/(n+1)$ is the derivative of a scalar, all terms in $X_b$ take the form (scalar) $\partial_b(\mathrm{scalar})$.

The momentum constraint on the “+” side of the singularity states that $\nabla_{+a} \mathring{K}_{+b}^{a}$ is also of the form (scalar) $\partial_b(\mathrm{scalar})$, so (6.26a) can be written as

$$ \nabla_{-a} (\omega \mathring{K}_{+b}^{a}) = \sum_{I} \chi_{I} \partial_{b} \zeta_{I} \tag{6.28} $$

for some collection of scalar fields $\chi_I$ and $\zeta_I$ whose precise expression is not useful yet.

**Scaling of trace-free extrinsic curvature.** To get rid of derivatives of scalar fields in (6.28), we consider particular configurations $(g_-, K_-, \phi_{0-}, \phi_{1-})$ constructed in lemma 6.5. These data sets are such that $(\theta_-, \phi_{0-}, \phi_{1-})$ is constant in some domain $\Omega \subset \mathcal{H}$ and is equal to any prescribed value in $\Delta/=$. This set, defined in (6.6), consists of values such that the corresponding eigenvalues $k_1, k_2, k_3$ are pairwise distinct and $\phi_{0-} \neq 0$. Since all
scalsars are functions of $\theta_-$, $\phi_0$, $\phi_1$, the first derivative $\partial_a$ of any scalar then vanishes at $x$. In addition, the data sets are such that $\nabla_{-a}(\hat{K}^2)_b^a \neq 0$ on $\Omega$.

For these data sets, the right-hand side of (6.28) vanishes in the domain $\Omega$. We compute its left-hand side in $\Omega$ by plugging the polynomial form (6.24), then dropping all derivatives of scalar fields since they vanish for this configuration:

$$ \nabla_{-a}(\omega \hat{K}_{+b}^a) = \nabla_{-a}(\omega \beta_0 \delta_b^a + \omega \beta_1 \hat{K}_{-b}^a + \omega \beta_2 (\hat{K}_{-b}^2)_b^a) = \omega \beta_1 \nabla_{-a} \hat{K}_{-b}^a + \omega \beta_2 \nabla_{-a} (\hat{K}_{-b}^2)_b^a. $$

The momentum constraint is $\nabla_{-a} \hat{K}_{-b}^a = 8\pi\phi_0 - \partial_b \phi_1$, which vanishes at $x$ in the given configuration. This eliminates the first term above and we learn that

$$ \omega \beta_2 \nabla_{-a} (\hat{K}_{-b}^2)_b^a = 0. $$

For the data sets given by lemma 6.5, $\nabla_{-a}(\hat{K}^2)_b^a \neq 0$, so we learn that $\beta_2 = 0$. Because $\hat{K}_+$ is traceless we deduce $\beta_0 = 0$. Altogether,

$$ \beta_0 = \beta_2 = 0, \quad \hat{K}_+ = \beta_1 \hat{K}_- \quad \text{when } K_- \text{ has three different eigenvalues.} $$

By a continuity argument identical to the proof of lemmas 6.3 and 6.4 we could prove that this conclusion holds even when eigenvalues are degenerate, but we do not need this.

**Constant scaling of densitized trace-free extrinsic curvature.** Now that we know $\hat{K}_+ = \beta_1 \hat{K}_-$ (for non-degenerate data) we can recalculate the left-hand side of (6.28) without assuming that scalar fields have vanishing derivative. We get

$$ \nabla_{-a}(\omega \hat{K}_{+b}^a) = \nabla_{-a}(\omega \beta_1 \hat{K}_{-b}^a) = \partial_a(\omega \beta_1) \hat{K}_{-b}^a + 8\pi\omega\beta_1\phi_0 - \partial_b\phi_1, $$

so (6.28) takes the form

$$ \partial_a(\omega\beta_1) \hat{K}_{-b}^a = -8\pi\omega\beta_1\phi_0 - \partial_b\phi_1 + \sum_I \chi_I \partial_b\zeta_I. $$

This identity takes the form (6.10) analyzed in lemma 6.6, so we learn that the scalar coefficient $\omega\beta_1$ in front of $\hat{K}_{-b}^a$ is an overall constant that only depends on the signature (and the scattering map), so

$$ \hat{K}_+ = \gamma\omega^{-1}\hat{K}_- \qquad (6.29) $$

for some constant $\gamma \in \mathbb{R}$. Note that the conclusion of lemma 6.6 does not involve any non-degeneracy assumption: the identity holds for all data. This gives an alternate proof of our corollary 5.5 that does not rely on the full classification.

A useful consequence of (6.29) is

$$ r_+ = \sqrt{\frac{2}{3} \operatorname{Tr} \hat{K}_+^2} = |\gamma| \omega^{-1} \sqrt{\frac{2}{3} \operatorname{Tr} \hat{K}_-^2} = |\gamma| \omega^{-1} r_- \qquad (6.30) $$

For $r_+ \neq 0$ (hence $\gamma \neq 0$ and $r_- \neq 0$ due to the above equation), one can write

$$ \cos \Theta_+ = \frac{3 \hat{K}_+}{2 r_+} = (\operatorname{sgn} \gamma) \frac{3 \hat{K}_-}{2 r_-} = (\operatorname{sgn} \gamma) \cos \Theta_-, \quad \text{hence} \quad \begin{cases} \theta_+ = \theta_- & \text{if } \gamma > 0, \\ \theta_+ = \theta_- + \pi & \text{if } \gamma < 0. \end{cases} \qquad (6.31) $$

We emphasize that while angles are only defined modulo $2\pi/3$ in the spacelike case because the three eigenvectors are indistinguishable, their difference $\theta_+ - \theta_-$ is actually well-defined modulo $2\pi$ in both the spacelike and timelike case because one can compare eigenvalues of $K_+$ and $K_-$ on the same eigenvectors.
## 6.6 Completion of the classification

**Rigidly conformal case.** As a warmup, we derive the classification of rigidly conformal and ultralocal maps announced in proposition 5.2. Specifically, we temporarily restrict ourselves to ultralocal scattering maps for which $g_+$ and $g_-$ have the same conformal class, namely $\alpha_1 = \alpha_2 = 0$. Then the expression (6.27) vanishes thanks to $\mathrm{Tr} \, \hat{K}_- = 0$,

$$X_b = \mathrm{Tr}(\beta_1 \hat{K}_- \nabla_{-b}(\alpha_0 \delta)) = \mathrm{Tr}(\hat{K}_- \delta) \beta_1 \partial_b \alpha_0 = 0,$$

so (6.26a), together with the momentum constraints, gives

$$8\pi\phi_{0+}\partial_b\phi_{1+} = \nabla_{+a}\hat{K}_{+b}^{a} = \gamma\omega^{-1}\nabla_{-a}\hat{K}_{-b}^{a} = 8\pi\gamma\omega^{-1}\phi_{0-}\partial_b\phi_{1-}.$$

The scalars $\omega, \phi_{0+}, \phi_{1+}$ are some functions of the scalars $(\theta_{-}, \phi_{0-}, \phi_{1-})$ given by the data. The chain rule for $\phi_{1+} = \phi_{1+}(\theta_{-}, \phi_{0-}, \phi_{1-})$ yields

$$8\pi(\phi_{0+} \partial_{\theta_{-}} \phi_{1+}) \partial_{b} \theta_{-} + 8\pi(\phi_{0+} \partial_{\phi_{0-}} \phi_{1+}) \partial_{b} \phi_{0-} + 8\pi(\phi_{0+} \partial_{\phi_{1-}} \phi_{1+} - \gamma \omega^{-1} \phi_{0-}) \partial_{b} \phi_{1-} = 0.$$

By lemma 6.7, the coefficients of $\partial_b \theta_{-}$, $\partial_b \phi_{0-}$, $\partial_b \phi_{1-}$ must vanish separately, namely

$$\phi_{0+} \partial_{\theta_{-}} \phi_{1+} = \phi_{0+} \partial_{\phi_{0-}} \phi_{1+} = \phi_{0+} \partial_{\phi_{1-}} \phi_{1+} - \gamma \omega^{-1} \phi_{0-} = 0. \qquad (6.32)$$

Then, there are two very different cases, $\gamma = 0$ and $\gamma \neq 0$.

* If $\gamma = 0$, we have $\hat{K}_+ = 0$ so $r(\phi_{0+}) = 0$ namely $\phi_{0+} = \epsilon/\sqrt{12\pi}$ with $\epsilon = \pm 1$. This sign is constant since we require scattering maps to map sufficiently regular data to (at least) continuous data. Since $\phi_{0+} \neq 0$, (6.32) simply states that $\phi_{1+}$ is a constant, while $\omega$ is completely unconstrained. This yields the isotropic scattering map given in (5.7), with $\lambda^3 = \omega$:

$$\mathbf{S}_{\lambda, \varphi, \epsilon}^{\mathrm{iso,rc}} : (g, K, \phi_0, \phi_1) \mapsto \left( \lambda^2 g, \frac{1}{3} \delta, \frac{\epsilon}{\sqrt{12\pi}}, \varphi \right).$$

* If $\gamma \neq 0$, then the last equation in (6.32) prevents $\phi_{0+}$ from vanishing unless $\phi_{0-} = 0$. Thus, we learn that $\partial_{\theta_{-}} \phi_{1+} = \partial_{\phi_{0-}} \phi_{1+} = 0$ for $\phi_{0-} \neq 0$, and, by continuity of $\phi_{1+}$, for $\phi_{0-} = 0$ as well. In other words, $\phi_{1+} = F(\phi_{1-})$ for some $F: \mathbb{R} \to \mathbb{R}$. The last equation in (6.32) reads

$$\phi_{0+} F'(\phi_{1-}) = \gamma \omega^{-1} \phi_{0-}, \qquad (6.33)$$

which implies that $F'$ is nowhere vanishing (since it is independent of $\phi_{0-}$).

We then have to solve (6.33) and the Hamiltonian constraint

$$1 - 12\pi\phi_{0+}^2 = \frac{3}{2} \mathrm{Tr}(\hat{K}_+^2) = \frac{3}{2}\gamma^2\omega^{-2} \mathrm{Tr}(\hat{K}_-^2) = \gamma^2\omega^{-2}(1 - 12\pi\phi_{0-}^2)$$

for $\phi_{0+}$ and $\omega$. Eliminating $\phi_{0+}$ using (6.33) gives

$$\gamma^{-2}\omega^2 = 1 + 12\pi\phi_{0-}^2(F'(\phi_{1-})^2 - 1). \qquad (6.34)$$
It is then immediate to solve (6.33) for $\phi_{0+}$. Denoting $\mu := |\omega/\gamma|^{1/3}$, given in terms of $\phi_{0-}$, $\phi_{1-}$ by (6.34), and denoting $\epsilon = \operatorname{sgn} \gamma = \pm 1$, we find

$$ 
\begin{aligned}
g_+ &= \omega^{2/3} g_- = |\gamma|^{2/3} \mu^2 g_-, & \qquad \hat{K}_+ &= \epsilon \mu^{-3} \hat{K}_-, \\
\phi_{0+} &= \epsilon \mu^{-3} \frac{\phi_{0-}}{F'(\phi_{1-})}, & \qquad \phi_{1+} &= F(\phi_{1-}),
\end{aligned} 
\qquad (6.35) 
$$

which is nothing by the anisotropic rigidly conformal scattering map defined in (5.8).

This concludes the classification in proposition 5.2 of ultralocal scattering maps that are rigidly conformal.

**Isotropic case.** We return to general ultralocal scattering maps, in which $\alpha_1, \alpha_2$ may be nonzero. The value of the constant $\gamma$ plays a key role again in the classification. We treat in this paragraph the case $\gamma = 0$, namely $\hat{K}_+ = 0$: the asymptotic profile on the “+” side of the singularity undergoes isotropic scaling.

In this case, the Hamiltonian constraint forces $\phi_{0+} = \epsilon/\sqrt{12\pi}$ for some $\epsilon = \pm 1$. Because we require scattering maps to map smooth data to (at least) continuous data, for such data $\phi_{0+}$ cannot jump between the values $\pm 1/\sqrt{12\pi}$, namely $\epsilon(x)$ is independent of $x \in \mathcal{H}$. By lemma 6.2 we learn that $\epsilon$ only depends on the scattering map and not on the data. Next, since $K_+ = \frac{1}{3}\delta$ is constant and $\phi_{0+} \neq 0$, the momentum constraint states that $\partial_b\phi_{1+} = 0$. Again we have a space-independent scalar $\phi_{1+}$, which by lemma 6.2 can only depend on the scattering map. Finally, the metric is not constrained beyond the polynomial structure given in lemma 6.9. This yields the isotropic scattering $S_{\alpha_0, \alpha_1, \alpha_2, \varphi, \epsilon}^{\text{iso}}$ of (5.9):

$$ (g_+, K_+, \phi_{0+}, \phi_{1+}) = \left( \exp\left(\alpha_0 \delta + \alpha_1 \cos\Theta_- + \alpha_2 \cos(2\Theta_-)\right) \right) g_-, \frac{1}{3}\delta, \frac{\epsilon}{\sqrt{12\pi}}, \varphi, \qquad (6.36) $$

where $\partial_{\theta_-}\alpha_0 = \alpha_1 = \alpha_2 = 0$ for $r_- = 0$ (namely $\phi_{0-} = \pm 1/\sqrt{12\pi}$).

**Anisotropic case.** We now turn to the case $\gamma \neq 0$, using the same method as for the rigidly conformal maps. A convenient form for the trace-free part $\hat{K}_+$ is

$$ \hat{K}_+ = \frac{2}{3} r_+ \cos \Theta_+ = \frac{2}{3} \epsilon r_+ \cos \Theta_-, \quad \text{with } \epsilon = \operatorname{sgn} \gamma = \pm 1, \qquad (6.37) $$

where the second equality is obvious for $r_+ = 0$ and is (6.31) otherwise. As we will see momentarily, inserting this expression of $\hat{K}_+$ in the momentum constraint reduces it down to a short sum of terms of the form (scalar)$\partial_b(\text{scalar})$. The chain rule rewrites the sum as a linear combination of $\partial_b\theta_-$, $\partial_b\phi_{0-}$, $\partial_b\phi_{1-}$, whose coefficients must all vanish by lemma 6.7. This vanishing gives three equations on derivatives of $\alpha_0, \alpha_1, \alpha_2, \phi_{0+}, \phi_{1+}$ with respect to $\theta_-, \phi_{0-}, \phi_{1-}$, and we eventually get the solutions $S_{\Phi, c, \epsilon}^{\text{ani}}$ defined by (5.12).

Let us begin by calculating using (6.37) the remainder term $X_b$ given in (6.27):

$$ X_b = \frac{2}{3} \epsilon r_+ \operatorname{Tr}(\cos \Theta_- \nabla_{-b} (\alpha_0 \delta + \alpha_1 \cos(\Theta_-) + \alpha_2 \cos(2\Theta_-))). $$
Upon expanding derivatives we encounter the traces: Tr(cosΘ$_{-}$) = 0 and

$$
\mathrm{Tr}((\cos \Theta_{-})^{2}) = \frac{1}{2} \mathrm{Tr}(\cos(2\Theta_{-}) + \delta) = \frac{3}{2},
$$

$$
\mathrm{Tr}(\cos \Theta_{-} \cos(2\Theta_{-})) = \frac{1}{2} \mathrm{Tr}(\cos(3\Theta_{-}) + \cos \Theta_{-}) = \frac{3}{2} \cos(3\theta_{-}),
$$

$$
\mathrm{Tr}(\cos \Theta_{-} \partial_{b} \cos \Theta_{-}) = \frac{1}{2} \partial_{b} \mathrm{Tr}((\cos \Theta_{-})^{2}) = \frac{1}{2} \partial_{b} \left(\frac{3}{2}\right) = 0,
$$

$$
\mathrm{Tr}(\cos \Theta_{-} \partial_{b} \cos(2\Theta_{-})) = \frac{4}{3} \partial_{b} \mathrm{Tr}((\cos \Theta_{-})^{3}) = \partial_{b} \cos(3\theta_{-}).
$$

Then $X_b$ simplifies to

$$
\begin{align*}
X_b &= \frac{2}{3}\epsilon r_+ \left( \frac{3}{2}\partial_b \alpha_1 + \frac{3}{2}\cos(3\theta_-)\partial_b \alpha_2 + \alpha_2 \partial_b \cos(3\theta_-) \right) \\
&= \epsilon r_+ \left( \partial_b(\alpha_1 + \cos(3\theta_-)\alpha_2) - \frac{1}{3}\alpha_2 \partial_b \cos(3\theta_-) \right) = 16\pi r_+ \partial_b \xi + \epsilon r_+ \alpha_2 \sin(3\theta_-) \partial_b \theta_-,
\end{align*}
$$

where we introduced (with a factor chosen to simplify later expressions)

$$
\xi = \frac{\epsilon}{16\pi} (\alpha_1 + \cos(3\theta_-)\alpha_2). \qquad (6.38)
$$

Using the divergence given in (6.26a) and our calculation of $X_b$, the momentum
constraint reads

$$
8\pi\phi_{0+}\partial_b\phi_{1+} - 8\pi\gamma\omega^{-1}\phi_{0-}\partial_b\phi_{1-} = \frac{-1}{2}X_b = \frac{-1}{2}(16\pi r_+\partial_b\xi + \epsilon r_+\alpha_2\sin(3\theta_-)\partial_b\theta_-),
$$

hence, dividing by $r_+ = |\gamma|\omega^{-1}r_-$ provided it is nonzero,

$$
\partial_b \xi + \frac{\phi_{0+}}{r_+} \partial_b \phi_{1+} = \epsilon \frac{\phi_{0-}}{r_-} \partial_b \phi_{1-} - \frac{1}{16\pi} \epsilon \alpha_2 \sin(3\theta_-) \partial_b \theta_-.
$$

We then use the chain rule to write all ∂_b(scalar) in terms of ∂_bθ_-, ∂_bφ_0-, ∂_bφ_1- and
we write down the three equations stating that coefficients of these three derivatives must
match due to lemma 6.7:

$$
\partial_{\theta_{-}} \xi + \frac{\phi_{0+}}{r_{+}} \partial_{\theta_{-}} \phi_{1+} = - \frac{1}{16\pi} \epsilon \alpha_{2} \sin(3\theta_{-}), \quad \partial_{\phi_{0-}} \xi + \frac{\phi_{0+}}{r_{+}} \partial_{\phi_{0-}} \phi_{1+} = 0,
$$

$$
\partial_{\phi_{1-}} \xi + \frac{\phi_{0+}}{r_+} \partial_{\phi_{1-}} \phi_{1+} = \epsilon \frac{\phi_{0-}}{r_-}. \qquad (6.39)
$$

The first equation lets us rewrite in terms of ξ the terms that appear in the polynomial
form (6.24) of g+: first express cos(2Θ−) as cos(x − y) = cos x cos y + sin x sin y for x = 3Θ−
and y = Θ−, then use (6.38) and (6.39). This yields

$$
\begin{align*}
\alpha_0 \delta + \alpha_1 \cos \Theta_- + \alpha_2 \cos(2\Theta_-) &= \alpha_0 \delta + (\alpha_1 + \alpha_2 \cos(3\theta_-)) \cos \Theta_- + \alpha_2 \sin(3\theta_-) \sin \Theta_- \\
&= \alpha_0 \delta + 16\pi\epsilon\xi \cos \Theta_- - 16\pi\epsilon \left(\partial_{\theta_-}\xi + \frac{\phi_{0+}}{r_+} \partial_{\theta_-}\phi_{1+}\right) \sin \Theta_-.
\end{align*}
$$
On the other hand, (6.25) and (6.30) relate $\alpha_0$ to Kasner radii as $\exp(3\alpha_0/2) = \omega = |\gamma|r_- / r_+$. Overall, for nonzero $r_-, r_+$,

$$ 
\begin{aligned}
g_+ &= \exp(\alpha_0 \delta + \alpha_1 \cos \Theta_- + \alpha_2 \cos(2\Theta_-)) g_- \\
&= \left| \frac{\gamma r_-}{r_+} \right|^{2/3} \exp \left( 16\pi\epsilon\xi \cos \Theta_- - 16\pi\epsilon \left( \partial_{\theta_-} \xi + \frac{\phi_{0+}}{r_+} \partial_{\theta_-} \phi_{1+} \right) \sin \Theta_- \right) g_-.
\end{aligned} 
\quad (6.40) $$

The expressions we obtained for ($g_+, K_+, \phi_{0+}, \phi_{1+}$) coincide with those of the anisotropic scattering $S_{\Phi,c,\epsilon}^{\text{ani}}$ (5.12g) with $c = |\gamma|\omega^{-1/3}$ and $\Phi: (\theta_-, \phi_{0-}, \phi_{1-}) \mapsto (\phi_{0+}, \phi_{1+})$, but there remains to show that the map $\Phi$ is indeed an $\epsilon$-canonical transformation in the sense of definition 5.3. We prove the conditions in turn.

(i) Periodic. This condition simply states that $\phi_{0+}, \phi_{1+}$ are scalar fields.

(ii) Maximal-momentum preserving. We know $r_+ = |\gamma|\omega^{-1}r_-$ from (6.30), and $|\gamma|\omega^{-1}$ is nowhere vanishing since $\omega$ is the ratio of volume factors of two non-degenerate metrics. Thus, $r_+ = 0 \Leftrightarrow r_- = 0$ and $r_+/r_- = |\gamma|\omega^{-1}$ remains finite and non-zero and becomes $\theta$-independent at the boundary (the last point being because $\omega$ is a scalar field).

(iii) Volume preserving. Imposing that the last two equations in (6.39) are compatible in the sense that $\partial_{\phi_{0-}} \partial_{\phi_{1-}} \xi = \partial_{\phi_{1-}} \partial_{\phi_{0-}} \xi$, we obtain for $r_+, r_- \neq 0$ that

$$ \partial_{\phi_{0-}} \left( \frac{\phi_{0+}}{r_+} \right) \partial_{\phi_{1-}} \phi_{1+} - \partial_{\phi_{1-}} \left( \frac{\phi_{0+}}{r_+} \right) \partial_{\phi_{0-}} \phi_{1+} = \epsilon \partial_{\phi_{0-}} \left( \frac{\phi_{0-}}{r_-} \right) \partial_{\phi_{1-}} \phi_{1-}. \quad (6.41) $$

We included here the trivial factor $\partial_{\phi_{1-}} \phi_{1-}$ to illustrate that this equation states preservation of the two-form $d(\phi_0/r)d\phi_1$ up to an overall sign $\epsilon$.

(iv) Regular at boundaries. One conclusion in lemma 6.9 is that $\alpha_1 = \alpha_2 = 0$ at the boundaries $\phi_{0-} = \pm 1/\sqrt{12\pi}$ (because $\tilde{K}_- = 0$ has no preferred directions). We deduce $\xi = 0$, and its derivatives $\partial_{\theta_-} \xi, \partial_{\phi_{1-}} \xi$ along the boundaries thus vanish. Inserting this fact (and $\alpha_2 = 0$) into (6.39) yields

$$ \frac{\phi_{0+}}{r_+} \partial_{\theta_-} \phi_{1+} \to 0, \quad \frac{\phi_{0+}}{r_+} \partial_{\phi_{1-}} \phi_{1+} - \epsilon \frac{\phi_{0-}}{r_-} \to 0. \quad (6.42) $$

Since $\xi$ vanishes on both boundaries we can integrate on $I_0$ the second equation in (6.39) to get

$$ \int_{-1/\sqrt{12\pi}}^{1/\sqrt{12\pi}} \frac{\phi_{0+}}{r_+} \partial_{\phi_{0-}} \phi_{1+} d\phi_{0-} = 0. \quad (6.43) $$

This concludes the proof of the first part of theorem 5.4, that is, the only ultralocal scattering maps are $S_{\alpha_0, \alpha_1, \alpha_2, \varphi, \epsilon}^{\text{iso}}$ and $S_{\Phi, c, \epsilon}^{\text{ani}}$. The second part was proven already in section 5.3.

This also concludes our study of scattering maps per se. In the companion paper in [56], we study the class of plane-symmetric spacetimes [34, 35, 47, 61, 67, 68, 76, 80] and we apply our theory to the particular scenario of colliding plane-symmetric gravitational waves.
Acknowledgments

The authors wish to thank Paul Tod for pointing out to them the relevance of the reference [9]. The authors gratefully acknowledge support from the Simons Center for Geometry and Physics, Stony Brook University. This research was initiated when the first author (BLF) was a research fellow at Princeton University, while the second author (PLF) was a visiting fellow at the Courant Institute for Mathematical Sciences, New York University. The second author is also grateful to the Institut Mittag-Leffler where he attended the Program: "General relativity, geometry and analysis: beyond the first 100 years after Einstein".

**Open Access.** This article is distributed under the terms of the Creative Commons Attribution License (CC-BY 4.0), which permits any use, distribution and reproduction in any medium, provided the original author(s) and source are credited.

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