Proof. This follows from the Hurwitz–Chevalley–Weil formula (see [Reference Moonen and OortMO13, Proposition 5.9]). Alternatively, see [Reference LooijengaLoo07, Lemma 4.2].

Proof. Write $\Lambda _{\mathbf{C}} = \Lambda \otimes _{\mathbf{Z}} {\mathbf{C}} = {\rm H}^1(C({\mathbf{C}}),{\mathbf{C}})$ . For each embedding $\phi \colon K \to {\mathbf{C}}$ , the restriction of the hermitian form

\begin{align*}\phi \left (5/\eta\, \right ) \cdot E_{\mathbf{C}}(x, \bar{y}) \colon \Lambda _{\mathbf{C}} \times \Lambda _{\mathbf{C}} \to {\mathbf{C}}\end{align*}

to $(\Lambda _{\mathbf{C}})_{\phi } \, \subset \Lambda _{\mathbf{C}}$ coincides with $ {\mathfrak{h}}^{\phi }$ by [Reference de Gaay FortmanGF24, Lemma 11.3]. Moreover, the hermitian form

\begin{align*} i \cdot E_{\mathbf{C}}(x, \bar{y}) \colon {\rm H}^1(C({\mathbf{C}}),{\mathbf{C}}) \times {\rm H}^1(C({\mathbf{C}}),{\mathbf{C}}) \to {\mathbf{C}} \end{align*}

is positive definite on $ {\rm H}^0(C, \Omega^1) = {\rm H}^{1,0}(C) = {\rm H}^{0,-1}(JC)$ and negative definite on ${\rm H}^{-1,0}(JC) = {\rm H}^{0,1}(C)$ (see [Reference VoisinVoi02, Théorème 6.32]). As $\Im (\sigma _1(\eta )) \gt 0$ and $\Im (\sigma _2(\eta )) \lt 0$ , we have $\Im (\sigma _1(5/\eta )) \lt 0$ and $\Im (\sigma _2(5/\eta )) \gt 0$ . Consequently, the hermitian form

\begin{align*} \sigma _1(5/\eta )\cdot E_{\mathbf{C}}(x, \bar{y}) = {\mathfrak{h}}^{\sigma _1}(x,y) \colon {\rm H}^1(C({\mathbf{C}}),{\mathbf{C}})_\zeta \times {\rm H}^1(C({\mathbf{C}}),{\mathbf{C}})_\zeta \to {\mathbf{C}} \end{align*}

is negative definite on ${\rm H}^{0,-1}(JC)_\zeta$ and positive definite on ${\rm H}^{-1,0}(JC)_{\zeta }$ , so that $ {{\rm sign}}({\mathfrak{h}}^{\sigma _1}) = (2,1)$ by Lemma 3.1. Similarly, the hermitian form

\begin{align*} \sigma _2(5/\eta )\cdot E_{\mathbf{C}}(x, \bar{y}) = {\mathfrak{h}}^{\sigma _2}(x,y) \colon {\rm H}^1(C({\mathbf{C}}),{\mathbf{C}})_{\zeta^2} \times {\rm H}^1(C({\mathbf{C}}),{\mathbf{C}})_{\zeta^2} \to {\mathbf{C}} \end{align*}

is positive definite on ${\rm H}^{0,-1}(JC)_{\zeta^2}$ and negative definite on ${\rm H}^{-1,0}(JC)_{\zeta^2}$ . Hence, using Lemma 3.1 again, we conclude that ${{\rm sign}}({\mathfrak{h}}^{\sigma _2}) = (3,0)$ .

Proof. Note that $\mathcal {N}_0$ is connected because $(\mathbf{C}^2 - \{0\})^2$ is connected. For \begin{align}v = ((a_1, b_2&), \dotsc, &\\(a_5, b_5)) \in \mathcal N_0\end{align} , define $F_v = \prod_{j=1}^5 (b_j \cdot x - a_j \cdot y)$ . The polynomial $F_v$ is a smooth binary quintic, equipped with the marking $m_v$ defined by the ordering $\{[a_1 \colon b_1], \dotsc, [a_5 \colon b_5])\}$ of the roots $[a_j \colon b_j] \in \mathbf{P}^1(\mathbf{C})$ of $F_v$ . The association $v \mapsto (F_v, m_v)$ defines the bijection alluded to in the first statement of the lemma. The rest is clear.

Proof. The equality $P(\mu )\left (\pi _1(\mathcal{N}_0,m_0)\right ) = P\Gamma _\theta$ follows from $\mu \left (\pi _1(\mathcal{N}_0,m_0)\right ) = \Gamma _\theta$ . For the latter, see [Reference Yamazaki and YoshidaYY84, Propositions 4.2 and 4.3]. The group $P{{\rm Aut}}(W) \cong {\rm PO}_3(\mathbf{F}_5)$ is isomorphic to ${\mathfrak{S}}_5$ , and $\overline {\rho }_\Gamma \colon {\mathfrak{S}}_5 \to P\Gamma /\Gamma _\theta$ is an isomorphism (cf. [Reference Yamazaki and YoshidaYY84, Propositions 4.2 and 4.3] and [Reference Apéry and YoshidaAY98, p. 10]).

Proof. Indeed, $\mathcal{F}_0$ is connected by Corollary 3.6. The isomorphism $P\Gamma _\theta \setminus \mathcal{F}_0 \cong \mathcal{N}_0$ of covering spaces of $X_0({\mathbf{C}})$ follows from the isomorphism $P\Gamma /P\Gamma _\theta \cong {\mathfrak{S}}_5$ as quotients of $\pi _1(X_0({\mathbf{C}}), F_0)$ , which was shown in Proposition 3.5. The lemma follows.

Proof. The irreducibility of $\Delta$ is well known; see e.g. [Reference VoisinVoi02, Section 14.1.1]. Let $(p,F)$ be a point on the incidence variety $\mathcal{I} = \left \{ (p, F) \in {\mathbf{P}}^1 \times X_s \mid p \in {{\rm Sing}}(F) \right \}$ , choose a hyperplane away from $p$ , and view $p \in {\mathbf{C}}$ and $F$ as a polynomial $f(x)$ . Since $f^{\prime\prime}(p)$ is non-zero, the tangent map

\begin{align*}Tq \colon T_{(p,f)} (\mathcal{I}) \to T_f(X_s) = X({\mathbf{C}})\end{align*}

of the projection $q \colon \mathcal{I} \to X_s$ is injective, and its image consists of the linear subspace $X_p \, \subset X$ of binary quintics that contain $p$ as a root. If $f$ has $k$ double points $p_j$ , where $k \in \left \{ 1,2 \right \}$ , then $\Delta$ is locally isomorphic to the union $\cup _{j=1}^k X_{p_j}$ . These $X_{p_j}$ intersect transversally, and we are done.

Proof. Let $F_1 \in \Delta _1$ be a binary quintic with one node. Consider the universal family ${\mathscr{C}} \to X_s$ of quintic covers of ${\mathbf{P}}^1$ ramified along a stable binary quintic, whose restriction to $X_0$ is (11). Let $D \, \subset X_s({\mathbf{C}})$ be an open disc transverse to $\Delta _1$ at $F_1$ . For $F \in D^\ast$ , one obtains a monodromy transformation $T \colon {\rm H}^1(C_F({\mathbf{C}}), {\mathbf{Z}}) \to {\rm H}^1(C_F({\mathbf{C}}), {\mathbf{Z}})$ induced by a vanishing cycle, and $T$ has order ten by [Reference Deligne and MostowDM86, Proposition 9.2]. Similarly, if $F_2 \in \Delta _2$ has two nodes, we may choose an embedding $D^2 \, \subset X_s({\mathbf{C}})$ of the polydisc $D^2 = D \times D$ transversal to $\Delta _2$ at $F_2$ . Since distinct nodes have orthogonal vanishing cycles, the local monodromy transformations commute.

Proof. In view of [Reference BeauvilleBea09, Lemma 7.2] (see also [Reference FoxFox57] and [Reference Deligne and MostowDM86, Section 8.1]), this follows from Lemma 3.10.

Proof. Consider the natural action of $G({\mathbf{C}})$ on $\mathcal{N}_s$ , and the action of $G({\mathbf{C}})$ on $P\Gamma _\theta \setminus \mathcal{F}_s$ induced by the action of $G({\mathbf{C}})$ on $\mathcal{F}_s$ . With respect to these actions, the isomorphism of ramified covering spaces $P\Gamma _\theta \setminus \mathcal{F}_s \cong \mathcal{N}_s$ of $X_s({\mathbf{C}})$ that results from Lemma 3.7(ii) is $G({\mathbf{C}})$ -equivariant. In particular, the natural ramified covering map $\mathcal{F}_s \to \mathcal{N}_s$ is $G({\mathbf{C}})$ -equivariant, and so it suffices to show that $G({\mathbf{C}})$ acts freely on $\mathcal{N}_s$ .

To this end, note that $\mathcal{N}_s$ admits a natural ${\mathbf{C}}^\ast$ -quotient map

(21) \begin{align} \mathcal{N}_s \to P_s, \end{align}

where $P_s \, \subset {\mathbf{P}}^1({\mathbf{C}})^5$ is the space of stable ordered five-tuples in ${\mathbf{P}}^1({\mathbf{C}})$ introduced in Section 1, and where ${\mathbf{C}}^\ast$ acts on $\mathcal{N}_s$ by $\lambda \cdot (v_1, v_2, \dotsc , v_5) = (\lambda v_1, v_2, \dotsc , v_5) = (v_1, \lambda v_2, \dotsc , v_5) = \cdots = (v_1, v_2, \dotsc , \lambda v_5) \in \mathcal{N}_s$ for $\lambda \in {\mathbf{C}}^\ast$ and $(v_1, v_2, \dotsc , v_5) \in \mathcal{N}_s$ (see the description of $\mathcal{N}_s$ in Section 3.5), and (21) is equivariant for the natural homomorphism $G({\mathbf{C}}) \to {\rm PGL}_2({\mathbf{C}})$ . Let $g \in {\rm GL}_2({\mathbf{C}})$ and $x \in \mathcal{N}_s$ such that $gx = x$ . Since any element of $ {\rm PGL}_2({\mathbf{C}})$ that fixes three distinct points on ${\mathbf{P}}^1({\mathbf{C}})$ is the identity, we have that ${\rm PGL}_2({\mathbf{C}})$ acts freely on $P_s$ . Therefore, $g \in {\mathbf{C}}^\ast \, \subset {\rm GL}_2({\mathbf{C}})$ . Let $F \in X_s({\mathbf{C}})$ be the image of $x \in \mathcal{N}_s$ ; then

\begin{align*}F(x,y) = gF(x,y) = F(g^{-1}(x,y)) = F(g^{-1}x, g^{-1}y) = g^{-5}F(x,y).\end{align*}

Thus, we have $g^{5} = 1 \in {\mathbf{C}}^\ast$ , and we conclude that $g \in D \, \subset {\rm GL}_2({\mathbf{C}})$ .

Proof. Recall that $P_0 \, \subset {\mathbf{P}}^1({\mathbf{C}})^5$ is the set of $(x_1, \dotsc , x_5) \in {\mathbf{P}}^1({\mathbf{C}})^5$ such that all $x_i$ are distinct. In accordance with [Reference Deligne and MostowDM86], define $Q = G({\mathbf{C}}) \setminus \mathcal{N}_0 = {\rm PGL}_2({\mathbf{C}}) \setminus P_0$ and $Q_{{\rm st}} := G({\mathbf{C}}) \setminus \mathcal{N}_s = {\rm PGL}_2({\mathbf{C}}) \setminus P_s$ . Fix a base point $0 \in Q$ whose image in ${\mathfrak{S}}_5 \setminus Q$ coincides with the image of $F_0 \in X_0({\mathbf{C}})$ under the canonical map $X_0({\mathbf{C}}) \to \mathcal{M}_0({\mathbf{C}}) = {\mathfrak{S}}_5 \setminus Q$ .

By Lemma 3.7, we have that $\mathcal{F}_0$ is a covering space of $\mathcal{N}_0$ , with Galois group $P\Gamma _\theta$ . In [Reference Deligne and MostowDM86], Deligne and Mostow define $\widetilde Q \to Q$ to be the covering space corresponding to the monodromy representation $\pi _1(Q,0) \to P\Gamma$ ; since the image of this homomorphism is $P\Gamma _\theta$ (see Proposition 3.5), it follows that $G({\mathbf{C}}) \setminus \mathcal{F}_0 \cong \widetilde Q$ as covering spaces of $Q$ . Consequently, if $\widetilde Q_{{\rm st}} \to Q_{{\rm st}}$ denotes the Fox completion (cf. [Reference FoxFox57], [Reference Deligne and MostowDM86, Section 8.1]) of the spread

\begin{align*}\widetilde Q \to Q \hookrightarrow Q_{{\rm st}},\end{align*}

then there is an isomorphism $G({\mathbf{C}}) \setminus \mathcal{F}_s \cong \widetilde Q_{{\rm st}}$ of branched covering spaces of $ Q_{{\rm st}}$ . We obtain the following commutative diagram, in which the horizontal arrows on the left are isomorphisms.

The map $\widetilde {Q}_{{\rm st}} \to {\mathbf{C}} H^2$ is an isomorphism by [Reference Deligne and MostowDM86, (3.11)]. It follows that $Q_{{\rm st}} \to P\Gamma _\theta \setminus {\mathbf{C}} H^2$ and ${\mathfrak{S}}_5 \setminus Q_{{\rm st}} \to P\Gamma \setminus {\mathbf{C}} H^2$ are isomorphisms as well. Therefore, we are done if we can show that the composition $\mathcal{F}_0 \to \widetilde Q \to {\mathbf{C}} H^2$ agrees with the period map $\mathcal{P}$ of equation (22). This follows from [Reference Deligne and MostowDM86, (2.23) and (12.9)].

Proof. By Lemma 3.2, we have that $\mathfrak{h}$ has signature $(2,1)$ with respect to the embedding $\sigma _1 \colon K \hookrightarrow {\mathbf{C}}$ , and signature $(3,0)$ with respect to the embedding $\sigma _2 \colon K \hookrightarrow {\mathbf{C}}$ , where $\sigma _1$ and $\sigma _2$ are defined in (9). Therefore, the result follows from [Reference de Gaay FortmanGF24, Theorem 2.5] and [Reference de Gaay FortmanGF24, Example 2.12].

Proof. We have $\mathcal{P}_s(\mathcal{F}_0) \, \subset {\mathbf{C}} H^2 - {\mathscr{H}}$ by [Reference de Gaay FortmanGF24, Proposition 11.12], because the Jacobian of a smooth curve cannot contain a non-trivial abelian subvariety whose induced polarization is principal. Therefore, we have $\mathcal{P}_s^{-1}({\mathscr{H}}\ ) \, \subset \mathcal{F}_s - \mathcal{F}_0$ . Since $\mathcal{F}_s$ is irreducible (it is smooth by Corollary 3.11 and connected by Lemma 3.7(i)), the analytic space $\mathcal{P}_s^{-1}({\mathscr{H}}\ )$ is a divisor. Since $\mathcal{F}_s - \mathcal{F}_0$ is also a divisor by Corollary 3.11, we have $\mathcal{P}_s^{-1}({\mathscr{H}}\ ) = \mathcal{F}_s - \mathcal{F}_0$ .

Proof.

1. (i) We know that $\mathcal{P}_s$ induces an isomorphism $G({\mathbf{C}}) \setminus \widetilde \Delta \xrightarrow {\sim } {\mathscr{H}}\ $ by Theorem 4.1 and Proposition 4.3. This map must identify the smooth (respectively singular) locus of one analytic variety with the smooth (respectively singular) locus of the other, from which the result follows.

2. (ii) This follows from Lemma 3.10 and Corollary 3.11 (compare [Reference Allcock, Carlson and ToledoACT02b, (3.10)] and [Reference Allcock, Carlson and ToledoACT10, Lemma 10.3]).

Proof. By [Reference BeauvilleBea10, Theorem 4.2], any finite subgroup of ${\rm PGL}_2({\mathbf{C}})$ is isomorphic to ${\mathbf{Z}}/n$ , $D_n$ (the dihedral group of order $2n$ ), ${\mathfrak{A}}_4$ , ${\mathfrak{S}}_4$ or ${\mathfrak{A}}_5$ , and there is only one conjugacy class for each of these groups. Let $H$ be any of these groups, considered as a subgroup of ${\rm PGL}_2({\mathbf{C}})$ . Assume that, with respect to the action of $H$ on the finite subsets of ${\mathbf{P}}^1({\mathbf{C}})$ , one has

\begin{align*} H \cdot \left \{ z_1, \dotsc , z_5 \right \} = \left \{ z_1, \dotsc , z_5 \right \} \, \subset {\mathbf{P}}^1({\mathbf{C}}) \quad {\rm for} \quad z_1, \dotsc , z_5 \in {\mathbf{P}}^1({\mathbf{C}}) \quad {\rm distinct.} \end{align*}

This gives a homomorphism $\rho \colon H \to {\mathfrak{S}}_5$ as follows: for an element $j \in \left \{ 1,2,3,4,5 \right \}$ , we let $\rho (h)(j) \in \left \{ 1,2,3,4,5 \right \}$ be the element with $z_{\rho (h)(j)} = h \cdot z_j$ .

Note that $\rho$ is injective, as $h \cdot z_i = z_i$ for each $i$ implies $h = {{\rm id}}$ . Therefore,

\begin{align*} H \in \left \{ {\mathbf{Z}}/2, {\mathbf{Z}}/3, {\mathbf{Z}}/4, {\mathbf{Z}}/5, {\mathbf{Z}}/6, D_3, D_4, D_5, {\mathfrak{A}}_4, {\mathfrak{S}}_4, {\mathfrak{A}}_5 \right\}\!. \end{align*}

Next, assume that $H = {\rm Stab}_{{\rm PGL}_2({\mathbf{C}})}(x)$ for the five-element subset $x = \left \{ z_1, \dotsc , z_5 \right \} \, \subset {\mathbf{P}}^1({\mathbf{C}})$ . Suppose that $\phi \in H$ is an element of order three. Note that there must be three distinct elements $z_i \in x$ with $\phi (z_i) \neq z_i$ . We may assume that these are $z_1, z_2$ and $z_3$ . Moreover, we may assume that $\phi (z_1) = z_2$ , $\phi (z_2) = z_3$ and $\phi (z_3) = z_1$ . By replacing $\phi$ by $g \phi g^{-1}$ for some $g \in {\rm PGL}_2({\mathbf{C}})$ , we may assume that $z_1=1, z_2 = \zeta _3$ and $z_3 = \zeta _3^2$ , and that $\phi (z) = \zeta _3 \cdot z$ for $z \in {\mathbf{P}}^1({\mathbf{C}})$ . This gives $x = \left \{ 1, \zeta _3, \zeta _3^2, z_4, z_5 \right \} \, \subset {\mathbf{P}}^1({\mathbf{C}})$ . As $\phi (z_4) \neq z_5$ , we have $\phi (z_4) = z_4$ and $\phi (z_5) = z_5$ , so that

(35) \begin{align} x = \left \{ 1, \zeta _3, \zeta _3^2, 0, \infty \right \} \, \subset {\mathbf{P}}^1({\mathbf{C}}). \end{align}

Let $\nu \in {\rm PGL}_2({\mathbf{C}})$ be the element with $\nu (z)=1/z$ for $z \in {\mathbf{P}}^1({\mathbf{C}})$ . Then (35) implies that $\nu$ is contained in $H$ , and one readily observes that $H = D_3$ . We conclude that

\begin{align*} H = {\rm Stab}_{{\rm PGL}_2({\mathbf{C}})}(x) \in \left \{ {\mathbf{Z}}/2, {\mathbf{Z}}/4, {\mathbf{Z}}/5, D_3, D_4, D_5 \right\}\!. \end{align*}

It remains to exclude ${\mathbf{Z}}/5$ and $D_4$ . Suppose that $H$ contains an element $\phi$ of order five. As above, one can readily show that one may assume that

\begin{align*} x = \left \{ 1, \zeta _5, \zeta _5^2, \zeta _5^3, \zeta _5^4 \right \} \, \subset {\mathbf{P}}^1({\mathbf{C}}), \quad \phi (z) = \zeta _5 \cdot z \ {\rm for} \ z \in {\mathbf{P}}^1({\mathbf{C}}). \end{align*}

This implies that the element $\nu \in {\rm PGL}_2({\mathbf{C}})$ as defined above is contained in $H$ , and $H = D_5$ . Finally, assume $H$ contains an element $\phi$ of order four. We may assume that

\begin{align*} x = \left \{ 1, i, -1, -i, 0 \right \}, \quad \phi (z) = i \cdot z \ {\rm for} \ z \in {\mathbf{P}}^1({\mathbf{C}}). \end{align*}

As a consequence, we have $H = {\mathbf{Z}}/4$ , and the proof is finished.

Proof. We have an injection $ (P_s/{\mathfrak{S}}_5)({\mathbf{R}}) \hookrightarrow P_s/{\mathfrak{S}}_5$ which is equivariant for the embedding ${\rm PGL}_2({\mathbf{R}}) \hookrightarrow {\rm PGL}_2({\mathbf{C}})$ . In particular, ${\rm PGL}_2({\mathbf{R}})_x \, \subset {\rm PGL}_2({\mathbf{C}})_x$ for every $x \in (P_s/{\mathfrak{S}}_5)({\mathbf{R}})$ . Note that none of the groups appearing in Proposition 6.3 have subgroups isomorphic to $D_2 = {\mathbf{Z}}/2 \rtimes {\mathbf{Z}}/2$ or $D_4 = {\mathbf{Z}}/2 \rtimes {\mathbf{Z}}/4$ . Consider the involution $ \nu = (z \mapsto 1/z) \in {\rm PGL}_2({\mathbf{R}})$ . We will prove the proposition by using the following steps.

Step 1: Let $\tau \in {\rm PGL}_2({\mathbf{R}})$ . Consider a subset $S = \{x,y,z\} \, \subset {\mathbf{P}}^1({\mathbf{C}})$ stabilized by complex conjugation, such that $\tau (x) = x$ , $\tau (y) = z$ and $\tau (z) = y$ . There is a transformation $g \in {\rm PGL}_2({\mathbf{R}})$ that maps $S$ to either $\{-1, 0, \infty \}$ or $\{-1, i, -i\}$ , and that satisfies $g \tau g^{-1} = \nu = (z \mapsto 1/z) \in {\rm PGL}_2({\mathbf{R}})$ . In particular, $\tau^2 = {{\rm id}}$ .

Step 2: There is no $\phi \in {\rm PGL}_2({\mathbf{R}})$ of order four that fixes a point $x \in (P_s/{\mathfrak{S}}_5)({\mathbf{R}})$ .

Step 3: Define $\rho \in {\rm PGL}_2({\mathbf{R}})$ by $\rho (z) = \frac {-1}{z+1}$ . Let $x = (x_1, \dotsc , x_5) \in (P_s/{\mathfrak{S}}_5)({\mathbf{R}})$ . Let $\phi \in {\rm PGL}_2({\mathbf{R}})$ be of order three, with $\phi (x) = x$ . There is a transformation $g \in {\rm PGL}_2({\mathbf{R}})$ mapping $x$ to $(-1, \infty , 0, \omega , \omega^2)$ with $\omega$ a primitive third root of unity. The stabilizer of $x$ is the subgroup of ${\rm PGL}_2({\mathbf{R}})$ generated by $\rho$ and $\nu$ . In particular, we have ${\rm PGL}_2({\mathbf{R}})_x \cong D_3$ .

Recall that $ \lambda = \zeta _5 + \zeta _5^{-1} \in {\mathbf{R}}$ . Define $ \gamma \in {\rm PGL}_2({\mathbf{R}})$ by $\gamma (z) = \frac {(\lambda+1)z - 1}{z+1}$ for $ z \in {\mathbf{P}}^1({\mathbf{C}}).$

Step 4: Let $x = (x_1, \dotsc , x_5) \in (P_s/{\mathfrak{S}}_5)({\mathbf{R}})$ . Suppose $x$ is stabilized by a subgroup of ${\rm PGL}_2({\mathbf{R}})$ of order five. There is a transformation $g \in {\rm PGL}_2({\mathbf{R}})$ mapping $x$ to $z = (0, -1, \infty , \lambda+1, \lambda )$ and identifying the stabilizer of $x$ with the subgroup of ${\rm PGL}_2({\mathbf{R}})$ generated by $\gamma$ and $\nu$ . In particular, the stabilizer ${\rm PGL}_2({\mathbf{R}})_x$ of $x$ is isomorphic to $D_5$ .

By Steps 1–4 above, together with Proposition 6.3, we are done.

Proof. Using Step 1 in the proof of Proposition 6.4, one checks that the only possibilities for the element $x=(x_1, \dotsc , x_5) \in (P_s/{\mathfrak{S}}_5)({\mathbf{R}})$ are $(-1, 0, \infty , \beta , \beta ^{-1})$ , $(-1, i, -i, \beta , \beta ^{-1})$ , $ (-1, -1, \beta , 0, \infty )$ , $ (-1, -1, \beta , i, -i)$ , $(0,0, \infty , \infty , -1)$ and $(-1, i, i, - i, -i)$ for some $\beta \in {\mathbf{P}}^1({\mathbf{R}})$ .

Proof. The map (36) is well-defined because if $g, g' \in G$ are such that $\gamma \cdot x=g \cdot x=g' \cdot x$ then $(g^{\prime})^{-1}g \in G_x$ . Since the construction is symmetric in $\Gamma$ and $G$ , the analogous map $G_{\bar{x}}/G_x \to \Gamma _{[x]}/\Gamma _x$ is also well-defined. The latter is a left and right inverse of (36), and we are done.

Proof. The statements can be deduced from Proposition 6.4, Lemmas 6.5 and 6.6, and [Reference de Gaay FortmanGF24, Proposition 5.19]. To show how this works, let us introduce some notation. Let $\tilde F \in \mathcal{F}_s({\mathbf{R}})$ with image $F \in X_s({\mathbf{R}})$ . Let $f \in Y$ be the image of $[\tilde F] \in G({\mathbf{R}}) \setminus \mathcal{F}_s({\mathbf{R}})$ under the isomorphism $G({\mathbf{R}}) \setminus \mathcal{F}_s({\mathbf{R}}) \xrightarrow {\sim } Y$ of Theorem 5.8. Let $P\Gamma _f \, \subset P\Gamma$ be the stabilizer of $f \in Y$ in the group $P\Gamma$ . Let $k \in \left \{ 0,1,2 \right \}$ be the number of nodes of $F$ , and write $k=2a + b$ , where $a$ is the number of pairs of complex conjugate nodes and $b$ the number of real nodes. Then the image $x \in {\mathbf{C}} H^2$ of $\tilde F$ under (23) lies on exactly $k$ distinct mutually orthogonal hyperplanes $H \in \mathcal{H}$ , with $\mathcal{H}$ the set defined in (26). Since $F \in X_s({\mathbf{R}})$ , we have that $\tilde F \in \mathcal{F}_s^\alpha$ for some $\alpha \in P{\mathscr{A}}$ . We get $x \in \mathbf{R} H^n_\alpha$ .

Let $\mathcal{H}(x) \, \subset \mathcal{H}$ be the set of hyperplanes $H \in \mathcal{H}$ such that $x \in H$ . Then $a$ equals the number of pairs of hyperplanes $H_1, H_2 \in \mathcal{H}(x)$ with $\alpha (H_1) = H_2$ , and $b$ equals the number of hyperplanes $H \in \mathcal{H}(x)$ with $\alpha (H) = H$ . Define $B_f \, \subset P\Gamma _f$ as the group generated by reflections in all $H \in \mathcal{H}(x)$ such that $\alpha (H) = H$ . Consider the quotient map $p \colon \sqcup _{\alpha \in P{\mathscr{A}}} {\mathbf{R}} H^2_\alpha \to Y\!$ , let $\alpha _1, \dotsc , \alpha _\ell \in P{\mathscr{A}}\ $ be the elements such that $(x,\alpha _i) \sim (x,\alpha )$ , and define $Y_f = \cup _{i=1}^\ell p({\mathbf{R}} H^2_{\alpha _i}) \, \subset Y$ . The subgroup $P\Gamma _f \, \subset P\Gamma$ preserves the subset $Y_f \, \subset Y$ by [Reference de Gaay FortmanGF24, Lemma 5.9]. Moreover, by [Reference de Gaay FortmanGF24, Proposition 5.19.4], there is an isometry between $B_f \setminus Y_f$ and the union of $10^a$ copies of $\mathbf{B}^2({\mathbf{R}})$ . Let $\mathbf{B}$ be any one of these copies of ${\mathbf{B}}^2({\mathbf{R}})$ , and define

\begin{align*}S_f := {\rm Stab}_{P\Gamma _f/B_f}(\mathbf{B}),\end{align*}

the stabilizer of $\mathbf{B}$ in the group $P\Gamma _f / B_f$ . By construction of the orbifold $\overline {{\mathscr{M}}}_{\mathbf{R}}$ (see [Reference de Gaay FortmanGF24, Propositions 5.1 and 5.19]), the group $S_f$ is a representative of the isomorphism class of stabilizer groups of the orbifold $\overline {{\mathscr{M}}}_{\mathbf{R}}$ at the moduli point $[F] \in \mathcal{M}_s({\mathbf{R}})$ induced by $F$ . Clearly, the stabilizer $G({\mathbf{R}})_F \, \subset G({\mathbf{R}})$ of $F \in X_s({\mathbf{R}})$ yields the isomorphism class of stabilizer groups of the orbifold $\mathcal{M}_s({\mathbf{R}})$ at the moduli point $[F] \in \mathcal{M}_s({\mathbf{R}})$ . In particular, we need to compare the isomorphism classes of the groups $S_f$ and $G({\mathbf{R}})_F$ . To do so, we claim that there is a canonical isomorphism

(37) \begin{align} G({\mathbf{R}})_F \xrightarrow {\sim } P\Gamma _f / G(x). \end{align}

Indeed, the actions of the groups $P\Gamma$ and $G({\mathbf{R}})$ on $\mathcal{F}_s({\mathbf{R}})$ commute, so we can apply Lemma 6.6. Recall that $\tilde F \in \mathcal{F}_s({\mathbf{R}})$ has images $F \in X_s({\mathbf{R}})$ and $f \in Y$ , and that the map $\mathcal{F}_s({\mathbf{R}}) \to Y$ factors through an isomorphism $G({\mathbf{R}}) \setminus \mathcal{F}_s({\mathbf{R}}) \xrightarrow {\sim } Y$ . Moreover, if $P\Gamma _{\tilde F} \, \subset P\Gamma$ denotes the stabilizer of $\tilde F$ in $P\Gamma$ , then $P\Gamma _{\tilde F} = G(x)$ by Lemma 4.4. As the group $G({\mathbf{C}})$ acts freely on $\mathcal{F}_s$ by Lemma 3.12, the group $G({\mathbf{R}})$ acts freely on $\mathcal{F}_s({\mathbf{R}})$ . Thus, (37) follows from Lemma 6.6.

If $F$ has no nodes ( $k=0$ ), then $G(x)$ is trivial by [Reference de Gaay FortmanGF24, Proposition 5.19.1], and $P\Gamma _f = S_f$ . Thus, $G({\mathbf{R}})_F \cong S_f$ in view of (37).

If $F$ has only real nodes, then $B_f = G(x)$ and $P\Gamma _f/G(x) =S_f$ . Thus, $G({\mathbf{R}})_F \cong S_f$ by (37).

Finally, suppose that $F$ has a pair of complex nodes ( $a=1$ and $b=0$ ). The zero set of $F$ defines a five-tuple $\underline {z}= (z_1, \dotsc , z_5) \in {\mathbf{P}}^1({\mathbf{C}})^5$ , well defined up to the ${\rm PGL}_2({\mathbf{R}}) \times {\mathfrak{S}}_5$ action on ${\mathbf{P}}^1({\mathbf{C}})$ , where $z_1 \in {\mathbf{P}}^1({\mathbf{R}})$ and $z_3 = \bar{z}_2 = z_5 = \bar{z}_4 \in {\mathbf{P}}^1({\mathbf{C}}) \setminus {\mathbf{P}}^1({\mathbf{R}})$ . Write $\underline {z} = (w, z, \bar{z}, z, \bar{z})$ with $w \in {\mathbf{P}}^1({\mathbf{R}})$ and $z \in {\mathbf{P}}^1({\mathbf{C}}) \setminus {\mathbf{P}}^1({\mathbf{R}})$ . There is a unique $T \in {\rm PGL}_2({\mathbf{R}})$ such that $T(w) = \infty$ and $T(z) = i$ . This gives $T(\underline z) = (\infty , i , -i, i, -i)$ . In particular, $F$ is unique up to isomorphism.

We have $G(x) \cong ({\mathbf{Z}}/10)^2$ , and as there are no real nodes, $B_f$ is trivial. With respect to the isometry ${\mathbf{C}} H^2({\mathbf{C}}) \xrightarrow {\sim } {\mathbf{B}}^2({\mathbf{C}})$ of [Reference de Gaay FortmanGF24, Lemma 5.16], the anti-holomorphic involutions $\alpha _j\colon {\mathbf{B}}^2({\mathbf{C}}) \to {\mathbf{B}}^2({\mathbf{C}})$ induced by the $\alpha _j \in P{\mathscr{A}}$ are $(t_1, t_2) \mapsto (\bar{t}_2 \zeta ^j , \bar{t}_1 \zeta ^j)$ , for $j \in {\mathbf{Z}}/10$ . The fixed point sets are given by ${\mathbf{B}}^2({\mathbf{R}})_{\alpha _j} = \{t_2 = \bar{t}_1 \zeta ^j\}\, \subset {\mathbf{B}}^2({\mathbf{C}})$ . The subgroup $E_f \, \subset G(x)$ that stabilizes ${\mathbf{B}}^2({\mathbf{R}})_{\alpha _j}$ is the group $E_f \cong {\mathbf{Z}}/10$ generated by the transformations $(t_1, t_2) \mapsto (\zeta t_1, \zeta ^{-1} t_2)$ . There is only one non-trivial $T \in {\rm PGL}_2({\mathbf{R}})$ fixing $\infty$ and preserving $\{i, -i\} \, \subset {\mathbf{P}}^1({\mathbf{C}})$ , and $T$ has order two, so $G({\mathbf{R}})_F = {\mathbf{Z}}/2$ . Therefore, by (37), we have an exact sequence of groups $ 0 \to G(x) \cong ({\mathbf{Z}}/10)^2 \to P\Gamma _f \to {\mathbf{Z}}/2 \to 0$ inducing an exact sequence $0 \to E_f \cong {\mathbf{Z}}/10 \to S_f \to {\mathbf{Z}}/2 \to 0$ . By Proposition 6.4 and Lemma 6.5, the proposition follows.