May 24, 2026
We study the algebraic curve over \(\mathbb{F}_{q^2}\) defined by \[y^{q+1} = x^n(x^n+1),\] where \(n\) is a positive integer coprime to the characteristic. We first prove (when \(q\) is odd) that the nonsingular model of this curve is \(\mathbb{F}_{q^2}\)-maximal if and only if \(n \mid (q+1)\). Writing \(n = \frac{q+1}{m}\), we obtain a family of maximal curves parameterized by the divisors \(m\) of \(q+1\), which extends the previously studied case \(m=3\) corresponding to maximal curves with the third largest possible genus.
For this family, we determine the Weierstrass semigroups at several classes of rational points, including those lying above the branch points of the natural projection. These semigroups are described explicitly in terms of \(q\) and \(m\), and exhibit different behaviors depending on the arithmetic properties of \(m\).
Moreover, we determine the full automorphism group of the curve under a mild condition on the characteristic. Our results extend an earlier work on the case \(m=3\) and provide new insight into the structure of this family of maximal curves.
Let \(\mathbb{F}_{q^2}\) be the finite field with \(q^2\) elements, where \(q\) is a power of a prime \(p\). An algebraic curve \(\mathcal{X}\) defined over \(\mathbb{F}_{q^2}\) is called \(\mathbb{F}_{q^2}\)-maximal if the number of its \(\mathbb{F}_{q^2}\)-rational points achieves the Hasse–Weil upper bound \[\# \mathcal{X}(\mathbb{F}_{q^2}) = q^2 + 1 + 2q\, g(\mathcal{X}),\] where \(g(\mathcal{X})\) denotes the genus of \(\mathcal{X}\). Maximal curves over finite fields are a central topic in arithmetic geometry, owing to their rich geometric structure and applications in coding theory, cryptography, and finite geometry.
The Hermitian curve \[\mathcal{H}_{q+1} : \quad y^{q+1} = x^{q+1} + 1\] is the classical example of a maximal curve over \(\mathbb{F}_{q^2}\). It has genus \(g(\mathcal{H}_{q+1}) = q(q-1)/2\) and achieves Ihara’s bound; that is, any maximal curve over \(\mathbb{F}_{q^2}\) satisfies \(g(\mathcal{X}) \le q(q-1)/2\), with equality if and only if \(\mathcal{X}\cong \mathcal{H}_{q+1}\). Further families of maximal curves can be obtained as quotients or subcovers of the Hermitian curve, though not every maximal curve arises in this way.
A broad framework for the construction of maximal curves is given by Kummer extensions of the projective line of the form \[y^m = f(x),\] where \(f(x) \in \mathbb{F}_{q^2}[x]\) is a separable polynomial and \(\gcd(m, {\rm deg}f) = 1\). This class includes classical examples such as the Fermat and Hermitian curves, and their subfamilies often exhibit deep arithmetic properties.
In the present paper, we study the family of curves defined by the affine equation \[\label{maincurve} \overline{\mathcal{X}}_{n,q} : \quad y^{q+1} = x^n(x^n + 1),\tag{1}\] where \(n \ge 2\) is an integer coprime to \(p\).
The curve 1 enjoys a high degree of symmetry, owing to the exponent \(q+1\) on \(y\), which divides the order of the multiplicative group \(\mathbb{F}_{q^2}^\ast\). This property makes \(\overline{\mathcal{X}}_{n,q}\) a natural object of study from both algebraic and arithmetic perspectives.
The main goal of this paper is to give a complete classification of the maximal curves of type 1 . We determine explicit conditions on \(n\) and \(q\) under which the nonsingular model of \(\overline{\mathcal{X}}_{n,q}\) is \(\mathbb{F}_{q^2}\)-maximal, and we compute its main geometric invariants, including the Weierstrass semigroups at certain \(\mathbb{F}_{q^2}\)-rational points. Furthermore, we investigate the structure of the Weierstrass semigroups at the remaining rational points and, under a mild assumption on the characteristic, we determine the full automorphism group of the curve.
In particular, we compute the genus and the ramification structure of the natural projection \(x \colon \overline{\mathcal{X}}_{n,q} \to \mathbb{P}^1\), and we characterize the cases in which \(\overline{\mathcal{X}}_{n,q}\) is \(\mathbb{F}_{q^2}\)-covered by the Hermitian curve or by a quotient curve thereof. Consequently, we obtain new explicit examples of maximal curves.
From an applications perspective, the explicit determination of Weierstrass semigroups and the exact number of rational points on algebraic curves has significant implications in geometric coding theory and cryptography. In particular, in the construction of algebraic geometry codes (such as Goppa codes), this precise geometric data is essential for designing codes with optimal parameters, including maximized minimum distance and high dimension (see, e.g., [1]). By addressing these problems, our results provide the necessary computational and theoretical tools for these engineering applications, aligning with the core interests of coding theory over finite fields.
The paper is organized as follows. In Section 2, we determine the precise arithmetic conditions on \(n\) and \(q\) under which \(\overline{\mathcal{X}}_{n,q}\) is \(\mathbb{F}_{q^2}\)-maximal. In Section 3, we introduce the necessary notation and collect preliminary results on divisors and automorphisms of \(\overline{\mathcal{X}}_{n,q}\) needed in the subsequent sections. In Section 4, we determine the Weierstrass semigroups at several classes of rational points of \(\overline{\mathcal{X}}_{n,q}\). In Section 5, we study the case \(n = \frac{q+1}{4}\) and discuss how these results extend to the general case. Finally, in Section 6, we determine the full automorphism group of \(\overline{\mathcal{X}}_{n,q}\) under a mild assumption on the characteristic.
In this section, we determine precisely when the nonsingular model of the curve \[\overline{\mathcal{X}}_{n,q} :\quad y^{q+1} = x^{n}(x^{n}+1)\] is \(\mathbb{F}_{q^2}\)-maximal. Our result generalizes the results of [2], [3], where special cases of this family were studied.
Theorem 1. Let \(q\) be a prime power and let \(n \ge 1\) be an integer with \(\gcd(2n,q)=1\). Then the nonsingular model of the curve \[\overline{\mathcal{X}}_{n,q}:\; y^{q+1}=x^{n}(x^{n}+1)\] is \(\mathbb{F}_{q^2}\)-maximal if and only if \(n \mid (q+1)\).
Proof. Assume first that \(n \mid (q+1)\). Write \(q+1 = nm\).
Let \(\mathcal{H}\) be the Hermitian curve \(y^{q+1}=x^{q+1}+1\) over \(\mathbb{F}_{q^{2}}\). Consider the morphism \[\varphi:\mathcal{H}\longrightarrow \overline{\mathcal{X}}_{n,q}, \qquad (a,b)\mapsto (a^{m},ab).\] A straightforward computation shows that \(\varphi(a,b)\) satisfies \(y^{q+1}=x^{n}(x^{n}+1)\). Hence \(\overline{\mathcal{X}}_{n,q}\) is a quotient of the Hermitian curve. Since \(\mathcal{H}\) is \(\mathbb{F}_{q^{2}}\)-maximal and maximality is preserved under nonconstant \(\mathbb{F}_{q^2}\)-rational morphisms [4], the nonsingular model of \(\overline{\mathcal{X}}_{n,q}\) is also \(\mathbb{F}_{q^{2}}\)-maximal.
Conversely, suppose that \(\overline{\mathcal{X}}_{n,q}\) is \(\mathbb{F}_{q^{2}}\)-maximal. Consider the hyperelliptic curve \[\mathcal{X}_{n} : \quad y^{2} = x^{n}(x^{n} + 1).\] It is covered by \(\overline{\mathcal{X}}_{n,q}\) via the map \[(x,y) \longmapsto \bigl(x,\, y^{(q+1)/2}\bigr).\] A change of variables shows that \(\mathcal{X}_{n}\) is birational to \[\mathcal{C}_{n} : \quad y_{1}^{2} = x_1^{n} + 1,\] regardless of the parity of \(n\). If \(n\) is even, take \(y_{1} = y / x^{n/2}\). If \(n\) is odd, take \(y_{2} = y / x^{(n-1)/2}\); then the curve is birational to \(y_2^2 = x^{n+1} + x\), which is in turn birational to \(y_1^2 = x_1^n + 1\) via the substitution \[y_1 = y_2/x^{(n+1)/2}, \qquad x_1 = 1/x.\] Hence \(\mathcal{X}_{n} \cong \mathcal{C}_{n}\) over \(\mathbb{F}_{q^{2}}\).
Since maximality passes to quotients, \(\mathcal{C}_{n}\) is also \(\mathbb{F}_{q^{2}}\)-maximal. By [5], a hyperelliptic curve of the form \(y^{2} = x^{n} + 1\) is \(\mathbb{F}_{q^{2}}\)-maximal if and only if \(n \mid (q+1)\). Therefore \(n \mid (q+1)\), completing the proof. ◻
Remark 2. Suppose that \(n \mid (q+1)\) and write \(q+1 = mn\) for some positive integer \(m\). Then the curve is isomorphic to \[\mathcal{X}_{m,q} : \quad y^{q+1} + x^{2(q+1)/m} + x^{(q+1)/m} = 0.\] This form is particularly convenient, as the special case \(m=3\) has already been investigated in the literature (see [6] and [7], [8]).
Remark 3. We will not consider certain cases, since they are already well understood: \(m=1\) (birationally equivalent to the Hermitian curve), \(m=2\) (the maximal curve with second highest genus, namely \((q-1)^2/4\)), and \(m=3\) (the maximal curve studied in [7]).
In this section, we introduce the notation and collect the preliminary results on divisors and automorphisms of the curve \(\mathcal{X}_{m,q}\) needed in the subsequent sections. Let \(\mathcal{X}_{m,q}\) be the curve over \(\mathbb{F}_{q^2}\) defined by the equation \[y^{q+1} + x^{2(q+1)/m} + x^{(q+1)/m} = 0,\] where \(m\) is a divisor of \(q+1\) with \(2 < m < q+1\); write \(p=\operatorname{char}(\mathbb{F}_{q^2})\). Note that \(\mathcal{X}_{m,q}\) is a Kummer extension of \(\mathbb{P}^1\) of degree \(q+1\), and that \(\mathcal{X}_{m,q}\) is Galois covered by the Hermitian curve \(\mathcal{H}:u^{q+1}+v^{q+1}+1=0\). Indeed, the map \((u,v) \mapsto (u^m,uv)\) defines a covering from \(\mathcal{H}\) to \(\mathcal{X}_{m,q}\) with generic fiber of cardinality \(m\), so that \([\mathbb{F}_{q^2}(\mathcal{H}):\mathbb{F}_{q^2}(\mathcal{X}_{m,q})]=m\).
Let \(\xi\in\mathbb{F}_{q^2}\) be a primitive \((q+1)\)-th root of unity. One can verify that \(G \leq {\rm Aut}(\mathcal{H})\), where \[G \cong \left\{\begin{pmatrix} \xi^{k(q+1)/m} & 0 & 0 \\ 0 & \xi^{q+1 - k(q+1)/m} & 0 \\ 0 & 0 & 1 \end{pmatrix} \right\}_{k = 0, \ldots, m-1}.\] Since \(|G| = m\) and \(G\) fixes the covering map from \(\mathcal{H}\) to \(\mathcal{X}_{m,q}\), we conclude that \(\mathbb{F}_{q^2}(\mathcal{H})/\mathbb{F}_{q^2}(\mathcal{X}_{m,q})\) is a Galois extension, and that the nonsingular model of \(\mathcal{X}_{m,q}\) is precisely \(\mathcal{H}/G\).
The pre-image of \((0:0:1)\) under the covering map is the set \(\{(0:\eta:1) \mid \eta^{q+1}+1=0\}\). The group \(G\) partitions these points into \(\frac{q+1}{m}\) orbits. By [9], there are \(\frac{q+1}{m}\) places centered at \((0:0:1)\). These places will be denoted by \(P_0^1, \ldots, P_0^{(q+1)/m}\), and we define \[\mathcal{O}_0 = \{P_0^1, \ldots, P_0^{(q+1)/m}\} \quad \text{and} \quad D_0=\sum_{i=1}^{(q+1)/m}P_0^i.\]
Analogously, the number of places centered at \((1:0:0)\) is \[\begin{cases} \dfrac{q+1}{m} & \text{if } m \text{ is odd,} \\[6pt] \dfrac{2(q+1)}{m} & \text{if } m \text{ is even.} \end{cases}\] These places are denoted by \(P_{\infty}^1, \ldots, P_{\infty}^{(q+1)/m}\) or \(P_{\infty}^1, \ldots, P_{\infty}^{2(q+1)/m}\) according to whether \(m\) is odd or even, and we define \[\mathcal{O}_{\infty} = \begin{cases} \{P_{\infty}^1, \ldots, P_{\infty}^{(q+1)/m}\} & \text{if } m \text{ is odd,} \\[4pt] \{P_{\infty}^1, \ldots, P_{\infty}^{2(q+1)/m}\} & \text{if } m \text{ is even,} \end{cases}\] \[D_{\infty} = \begin{cases} \displaystyle\sum_{i=1}^{(q+1)/m}P_{\infty}^i & \text{if } m \text{ is odd,} \\[6pt] \displaystyle\sum_{i=1}^{2(q+1)/m}P_{\infty}^i & \text{if } m \text{ is even.} \end{cases}\]
A point \(P=(a:b:1) \in \mathcal{X}_{m,q}(\overline{\mathbb{F}_{q^2}})\) with \(a \neq 0\) will be denoted by \(P_{(a,b)}\). Since this point is nonsingular, it is the center of a unique place of \(\overline{\mathbb{F}_{q^2}}(\mathcal{X}_{m,q})\), which will also be denoted by \(P_{(a,b)}\). By Kummer theory, the only ramified places of the extension \(\overline{\mathbb{F}_{q^2}}(\mathcal{X}_{m,q})/\overline{\mathbb{F}_{q^2}}(x)\) are those of \(\mathcal{O}_0\) (with ramification index \(m\)), those of \(\mathcal{O}_{\infty}\) (with ramification index \(m\) or \(\frac{m}{2}\), according to whether \(m\) is odd or even), and \(P_{(a,0)}\) with \(a^{(q+1)/m}+1 = 0\) (with ramification index \(m\)). By [1], we obtain \[g(\mathcal{X}_{m,q}) = \begin{cases} \dfrac{q^2-q+2m-2}{2m} & \text{if } m \text{ is odd,} \\[6pt] \dfrac{q^2-2q+2m-3}{2m} & \text{if } m \text{ is even.} \end{cases}\]
The extension \(\mathbb{F}_{q^2}(\mathcal{H})/\mathbb{F}_{q^2}(\mathcal{X}_{m,q})\) is also a Kummer extension. Indeed, \(\mathbb{F}_{q^2}(\mathcal{H}) = \mathbb{F}_{q^2}(\mathcal{X}_{m,q})(u,v)\), where \(u^m = x\) and \(v = y/u \in \mathbb{F}_{q^2}(\mathcal{X}_{m,q})(u)\). By [1], this extension is unramified when \(m\) is odd, and when \(m\) is even it is ramified only at \(P_{\infty}^i\), \(i = 1, \ldots, 2(q+1)/m\), with ramification index \(2\) at each place. In particular, for any place \(P_{(a,b)} \in \mathbb{P}_{\mathbb{F}_{q^2}(\mathcal{X}_{m,q})}\) and any place \(Q_{(A,B)} \in \mathbb{P}_{\mathbb{F}_{q^2}(\mathcal{H})}\) lying above \(P_{(a,b)}\), and any function \(f \in \mathbb{F}_{q^2}(\mathcal{X}_{m,q})\), we have \(v_{P_{(a,b)}}(f) = v_{Q_{(A,B)}}(f)\). This allows us to compute the Weierstrass semigroups at the places \(P_{(a,b)}\) using expansions in terms of a local parameter at \(Q_{(A,B)}\).
Throughout the paper, we denote \[\mathcal{O} = \mathcal{O}_0 \cup \mathcal{O}_{\infty} \cup \mathcal{O}_m,\] where \(\mathcal{O}_m := \{P_{(a,0)} : a^{(q+1)/m} + 1 = 0\}\), and \(\mathcal{O}' = \{P_{(a,b)} : 2a^{(q+1)/m} + 1 = 0\}\).
Note that, in this article, we consider automorphisms as certain maps acting on the (finite) points in the curve, not as maps acting on the function field.
Proposition 4. The automorphism group \({\rm Aut}(\mathcal{X}_{m,q})\) contains a subgroup \(G\) of order \(2(q+1)^2/m\), isomorphic to a semidirect product of \(\mathbb{Z}/2 \ltimes (\mathbb{Z}/\frac{q+1}{m} \times \mathbb{Z}/(q+1))\), generated by \(A \cup \{\theta_2\}\), where \[A := \{\theta_{\gamma,\delta}(x,y)=(\gamma x, \delta y) \mid \gamma^{(q+1)/m}=\delta^{q+1}=1\}, \quad \theta_2(x,y) = \left(\frac{y^m}{x}, y \right).\]
Proof. The maps \(\theta_{\gamma,\delta}\) clearly preserve the equation of the curve, as does \(\theta_2\), since \[y^{q+1} + \left(\frac{y^m}{x}\right)^{2(q+1)/m} + \left(\frac{y^m}{x}\right)^{(q+1)/m} = y^{q+1} + \frac{y^{2(q+1)}}{x^{2(q+1)/m}} + \frac{y^{q+1}}{x^{(q+1)/m}}\] \[= \frac{y^{q+1}}{x^{2(q+1)/m}} \left( x^{2(q+1)/m} + y^{q+1} + x^{(q+1)/m} \right) = 0.\] Moreover, \(\theta_2^2 = \mathrm{id}\), and \[\theta_2(\theta_{\gamma,\delta}(x,y)) = \theta_2(\gamma x, \delta y) = \left(\frac{\delta^m}{\gamma} \cdot \frac{y^m}{x}, \delta y\right) = \theta_{\delta^m/\gamma,\,\delta}(\theta_2(x,y)),\] with \((\delta^m/\gamma)^{(q+1)/m} = \delta^{q+1}/\gamma^{(q+1)/m} = 1\). Hence \(G\) is the semidirect product of \(\{\mathrm{id},\theta_2\}\) acting on \(A\). ◻
The elements of \(A\) act transitively on \(\mathcal{O}_0\), \(\mathcal{O}_{\infty}\), and \(\mathcal{O}_m\), while \(\theta_2\) maps \(\mathcal{O}_0\) to \(\mathcal{O}_m\). Therefore, \[H(P_0^i) = H(P_{(a,0)}), \quad i = 1, \ldots, \frac{q+1}{m},\quad a^{(q+1)/m}+1=0.\]
Proposition 5. Let \(m\) be odd and \(a, b \in \mathbb{F}_{q^2}^*\). We have \[(x-a) = \begin{cases} \displaystyle\sum_{\xi^{q+1}=1} P_{(a,\xi b)} - m D_{\infty} & \text{if } a^{(q+1)/m}+1\neq 0, \\[6pt] (q+1)P_{(a,0)} - m D_{\infty} & \text{if } a^{(q+1)/m}+1 = 0, \end{cases}\] \[(x) = m\sum_{i=1}^{(q+1)/m}P_0^i - m D_{\infty},\] \[(y) = \sum_{i=1}^{(q+1)/m}P_0^i + \sum_{a^{(q+1)/m}+1=0}P_{(a,0)} - 2D_{\infty},\] \[(y-b) = E_b - 2D_{\infty},\] where \(E_b\) is an effective divisor of degree \(\frac{2(q+1)}{m}\), with \[\mathrm{Supp}(E_b) = \{P_{(a,b)} \mid b^{q+1}+a^{2(q+1)/m}+a^{(q+1)/m}=0\}\] and \[v_{P_{(a,b)}}(E_b) = \begin{cases} 2 & \text{if } 2a^{(q+1)/m}+1=0, \\ 1 & \text{otherwise.} \end{cases}\]
Proof. The formulas for \((x)\) and \((x-a)\) follow from the ramification indices of the extension \(\mathbb{F}_{q^2}(\mathcal{X}_{m,q})/\mathbb{F}_{q^2}(x)\). The formula for \((y)\) follows from \(y^{q+1}=-x^{(q+1)/m}(x^{(q+1)/m}+1)\). For \((y-b)\), we note that \(a\) is a multiple root of \(x^{2(q+1)/m}+x^{(q+1)/m}+b^{q+1}\) if and only if \(2a^{(q+1)/m}+1=0\). ◻
Proposition 6. Let \(m\) be even and \(a, b \in \mathbb{F}_{q^2}^*\). We have \[(x-a) = \begin{cases} \displaystyle\sum_{\xi^{q+1}=1} P_{(a,\xi b)} - \frac{m}{2} D_{\infty} & \text{if } a^{(q+1)/m}+1\neq 0, \\[6pt] (q+1)P_{(a,0)} - \frac{m}{2} D_{\infty} & \text{if } a^{(q+1)/m}+1 = 0, \end{cases}\] \[(x) = m\sum_{i=1}^{(q+1)/m}P_0^i - \frac{m}{2} D_{\infty},\] \[(y) = \sum_{i=1}^{(q+1)/m}P_0^i + \sum_{a^{(q+1)/m}+1=0}P_{(a,0)} - D_{\infty},\] \[(y-b) = E_b - D_{\infty},\] where \(E_b\) is an effective divisor of degree \(\frac{2(q+1)}{m}\) with \[\mathrm{Supp}(E_b) = \{P_{(a,b)} \mid b^{q+1}+a^{2(q+1)/m}+a^{(q+1)/m}=0\}\] and \[v_{P_{(a,b)}}(E_b) = \begin{cases} 2 & \text{if } 2a^{(q+1)/m}+1=0, \\ 1 & \text{otherwise.} \end{cases}\]
Proof. The proof follows analogously to that of Proposition 5. ◻
In this section, we present the Weierstrass semigroups at the places of \(\mathcal{O}\). In the first subsection, we study the genera of some numerical semigroups, and in the second one we determine that these are indeed the Weierstrass semigroups at the places of \(\mathcal{O}\), by constructing explicit functions with the desired pole divisor.
Denote \[S(q,m) \mathrel{:=}\left\langle q+1,\, q,\, \ldots,\, q+1-\left\lfloor\frac{m}{2}\right\rfloor \right\rangle + \left\langle q+1-m \right\rangle,\] \[T(q,m) \mathrel{:=}\begin{cases} \left\langle q+1 \right\rangle + \left\langle q, q-2, \ldots, q+1-m \right\rangle & \text{if } m \text{ is odd,} \\[4pt] \left\langle q, q-1, \ldots, q+1-\tfrac{m}{2} \right\rangle + \left\langle \tfrac{q+1}{2} \right\rangle & \text{if } m \text{ is even.} \end{cases}\]
Proposition 7. The genus of \(S(q,m)\) is at most \(g = g(\mathcal{X}_{m,q})\).
Proof. Let \(a \mathrel{:=}(q+1)-m = \min\left(S(q,m) \smallsetminus \{0\}\right)\). Following the ideas in [10], we will present a function \(f \colon \{0,\ldots,a-1\} \to S(q,m)\) such that, for each \(v \in \{0,\ldots,a-1\}\), \(f(v) \equiv v \pmod{a}\). In particular, \(S(q,m) \supseteq f(v) + a\mathbb{N}\). We will then obtain \[\begin{align} g\left(S(q,m)\right) &= \left|\bigcup_{v=1}^{a-1} \left\{ w \in \mathbb{N} \mid w \notin S(q,m),\; w \equiv v \pmod{a} \right\}\right| \\ &= \sum_{v=1}^{a-1} \left|\left\{ k \in \mathbb{N} \mid v+ka \notin S(q,m) \right\}\right| \\ &\le \sum_{v=1}^{a-1} \left|\left\{ k \in \mathbb{N} \mid v+ka < f(v) \right\}\right| \\ &= \sum_{v=1}^{a-1} \left\lfloor \frac{f(v)}{a} \right\rfloor, \end{align}\] and \(f\) will be chosen so that \(\sum_{v=1}^{a-1}\lfloor\frac{f(v)}{a}\rfloor = g\).
Define \(f(0) = 0\). For \(v \in \{1,\ldots,a-1\}\), we define \(f(v)\) by cases.
Suppose that \(v \ge m - \lfloor\frac{m}{2}\rfloor\). Denote \(k \mathrel{:=}\lceil\frac{v}{m}\rceil \ge 1\). We claim that \(v + ka \in S(q,m)\). It follows that \(k\!\left(m-\lfloor\tfrac{m}{2}\rfloor\right) \le v \le km\). The second inequality comes from the definition of \(k\). For the first inequality, if \(k=1\), we already have \(m - \lfloor\frac{m}{2}\rfloor \le v\) by hypothesis. Now suppose \(k \ge 2\); then \[\begin{align} v - k\!\left(m-\left\lfloor\tfrac{m}{2}\right\rfloor\right) \ge v - k\tfrac{m+1}{2} = (v-km) + k\tfrac{m-1}{2} \ge -(m-1) + k\tfrac{m-1}{2} \ge 0. \end{align}\] Hence there exist \(e_0, \ldots, e_{\lfloor m/2 \rfloor} \in \mathbb{N}\) such that \(\sum_{i=0}^{\lfloor m/2 \rfloor} e_i = k\) and \(\sum_{i=0}^{\lfloor m/2 \rfloor} e_i \!\left(m - \lfloor\tfrac{m}{2}\rfloor + i\right) = v\). Thus, \[\begin{align} v + ka &= \sum_{i=0}^{\lfloor m/2 \rfloor} e_i \!\left(m - \left\lfloor\tfrac{m}{2}\right\rfloor + i + a\right) = \sum_{i=0}^{\lfloor m/2 \rfloor} e_i \!\left((q+1) - \left\lfloor\tfrac{m}{2}\right\rfloor + i\right) \in S(q,m). \end{align}\] So we may define \(f(v) \mathrel{:=}v + ka\).
Suppose that \(v < m - \lfloor\frac{m}{2}\rfloor\). We claim that \(v + \left(\frac{q+1}{m}+1\right)a \in S(q,m)\). It follows that \(\frac{q+1}{m}\!\left(m-\lfloor\tfrac{m}{2}\rfloor\right) \le v+a \le \frac{q+1}{m} \cdot m\). The second inequality is \(v + a < m + a = q+1\). For the first inequality, we proceed by cases.
If \(m\) is even, then \[v + a - \tfrac{q+1}{m}\!\left(m-\left\lfloor\tfrac{m}{2} \right\rfloor\right) = v + (q+1) - m - \tfrac{q+1}{2} \ge \tfrac{q+1}{2} - m \ge 0,\] since \(m\) is a proper divisor of \(q+1\).
If \(m\) is odd and \(m \neq \frac{q+1}{2}\), then, using \(m > 1\), \[\begin{align} v + a - \tfrac{q+1}{m}\!\left(m-\left\lfloor\tfrac{m}{2} \right\rfloor\right) &= v + (q+1) - m - (q+1)\tfrac{1+m^{-1}}{2} \\ &\ge v + (q+1) - m - (q+1)\tfrac{1+3^{-1}}{2} \\ &= v + \tfrac{q+1}{3} - m \ge 0. \end{align}\]
If \(m\) is odd and \(m = \frac{q+1}{2}\), then \[v + a - \tfrac{q+1}{m}\!\left(m-\left\lfloor\tfrac{m}{2} \right\rfloor\right) = v + m - 2\cdot\tfrac{m+1}{2} = v - 1 \ge 0.\]
Hence there exist \(e_0,\ldots,e_{\lfloor m/2\rfloor} \in \mathbb{N}\) such that \(\sum_{i=0}^{\lfloor m/2\rfloor} e_i = \frac{q+1}{m}\) and \[\sum_{i=0}^{\lfloor m/2\rfloor} e_i \!\left(m-\lfloor\tfrac{m}{2}\rfloor+i\right) = v+a.\] Thus, as in the previous case, we may define \(f(v) \mathrel{:=}v + \left(\frac{q+1}{m}+1\right)a \in S(q,m)\).
Now we have \[\sum_{v=1}^{a-1} \left\lfloor\frac{f(v)}{a}\right\rfloor = \sum_{v=1}^{m-\lfloor m/2\rfloor-1} \left(\frac{q+1}{m}+1\right) + \sum_{v=m-\lfloor m/2\rfloor}^{a-1} \left\lceil\frac{v}{m}\right\rceil.\] Note that \[\begin{align} \sum_{v=m-\lfloor m/2\rfloor}^{a-1} \left\lceil\frac{v}{m}\right\rceil &= \sum_{k=1}^{\frac{q+1}{m}-1} k \cdot \sum_{v=m-\lfloor m/2\rfloor}^{a-1} \left[k = \left\lceil\tfrac{v}{m}\right\rceil\right] \\ &= 1 \cdot \sum_{v=m-\lfloor m/2\rfloor}^{m} 1 + \sum_{k=2}^{\frac{q+1}{m}-1} k \cdot \sum_{v=(k-1)m+1}^{km - [k=\frac{q+1}{m}-1]} 1 \\ &= \left(\left\lfloor\tfrac{m}{2}\right\rfloor+1\right) + \sum_{k=2}^{\frac{q+1}{m}-1} k\!\left(m - \left[k=\tfrac{q+1}{m}-1\right]\right) \\ &= \left(\left\lfloor\tfrac{m}{2}\right\rfloor+1\right) + m\!\left(\binom{(q+1)/m}{2}-1\right) - \left(\tfrac{q+1}{m}-1\right). \end{align}\] Therefore, \[\begin{align} \sum_{v=1}^{a-1}\left\lfloor\frac{f(v)}{a}\right\rfloor &= \left(\frac{q+1}{m}+1\right) \left(m-\left\lfloor\tfrac{m}{2}\right\rfloor-1\right) + \left(\left\lfloor\tfrac{m}{2}\right\rfloor+1\right) + m\!\left(\binom{(q+1)/m}{2}-1\right) - \left(\tfrac{q+1}{m}-1\right), \end{align}\] which equals \(g\) after straightforward simplification. ◻
Proposition 8. The genus of \(T(q,m)\) is at most \(g = g(\mathcal{X}_{m,q})\).
Proof. The proof is similar to the previous one, though the details are more involved. Let \(a\) be the smallest element of \(T(q,m) \smallsetminus \{0\}\). When \(m\) is even, for \(v \ge 1\) we define \[f(v) \mathrel{:=}v + \left(2\left\lceil\frac{a-v}{m/2}\right\rceil - 1\right)a.\] When \(m\) is odd, for \(v \ge 1\) we define \[f(v) \mathrel{:=}\begin{cases} v + \left\lceil\dfrac{v}{m}\right\rceil a & \left(v \ge m\right), \\[6pt] v + a & \left(v < m,\;v \in 2\mathbb{N}\right), \\[6pt] v + \left(\dfrac{q+1}{m}+1\right)a & \left(v < m,\;v \notin 2\mathbb{N}\right). \end{cases}\]
Let us show that \(f(v) \in T(q,m)\) for each \(1 \le v \le a-1\).
Case: \(m\) is even. We have \(a = \frac{q+1}{2}\). Write \(a - v = k\frac{m}{2} - s\), where \(k = \lceil\frac{a-v}{m/2}\rceil \ge 1\) and \(0 \le s \le \frac{m}{2}-1\). We have that \(k \le a-v \le k\frac{m}{2}\). Indeed, \[a - v - k = k\left(\tfrac{m}{2}-1\right) - s \ge (k-1)\left(\tfrac{m}{2}-1\right) \ge 0,\] and \(k\frac{m}{2} \ge \frac{a-v}{m/2}\cdot\frac{m}{2}\) by definition of \(k\). As before, there exist \(e_0, \ldots, e_{m/2-1} \in \mathbb{N}\) such that \(\sum_{i=0}^{m/2-1} e_i = k\) and \(\sum_{i=0}^{m/2-1} e_i\!\left(\frac{m}{2}-i\right) = a-v\). Therefore \(T(q,m)\) contains \[\sum_{i=0}^{m/2-1} e_i\!\left(q+1-\tfrac{m}{2}+i\right) = k(q+1) - a + v = (2k-1)a + v = f(v).\]
Case: \(m\) is odd. We have \(a = (q+1)-m\). We show by induction on \(v\) that \(v + \lceil\frac{v}{m}\rceil a \in T(q,m)\) for each \(v \in \{m+1, \ldots, a-1\}\). Write \(v = km - s\), with \(k = \lceil\frac{v}{m}\rceil\) and \(0 \le s \le m-1\). We have two cases:
Case \(k = 2\). So \(m+1 \le v \le 2m\). When \(v = 2m\), \[v + 2a = 2(a+m) = 2(q+1) \in T(q,m).\] Otherwise \(v \le 2m-1\). If \(v\) is odd, write \(v = m + 2i\) for some \(i \in \{1,\ldots,\frac{m-1}{2}\}\); then \[v + 2a = (a+m) + (a+2i) = (q+1) + (q+1-(m-2i)) \in T(q,m).\] If \(v\) is even, write \(v = m-1+2i\) for some \(i \in \{1,\ldots,\frac{m-1}{2}\}\); then \[v + a = (a+m-1) + (a+2i) = (q+1-1) + (q+1-(m-2i)) \in T(q,m).\]
Case \(k \ge 3\). So \(v \ge (k-1)m+1 \ge 2m+1\), hence \(v - m \ge m+1\). By the inductive hypothesis, \((v-m) + (k-1)a \in T(q,m)\), and therefore \[v + ka = (v-m) + (k-1)a + (a+m) \in T(q,m).\]
Thus \(f(v) \in T(q,m)\) for each \(m+1 \le v \le a-1\). When \(v = m\), we have \[f(v) = m + a = q+1 \in T(q,m).\]
If \(1 \le v < m\) and \(v\) is even, then \(f(v) = v + a = q+1-(m-v) \in T(q,m)\).
If \(1 \le v < m\) and \(v\) is odd, we consider two cases:
Case \(\frac{q+1}{2} \neq m\). Then \(v - m + a \le a-1\) and \[v - m + a \ge 1 - m + a = 1 + (q+1) - 2m \ge 1 + 3m - 2m = 1 + m.\] So, as shown above, \(f(v) = v - m + a + ka \in T(q,m)\), where \[k = \left\lceil\frac{v-m+a}{m}\right\rceil = \left\lceil\frac{v}{m}\right\rceil - 1 + \frac{a}{m} = \frac{a}{m}.\] Thus, \[v + \left(\frac{q+1}{m}+1\right)a = v + \left(\frac{a+m}{m}+1\right)a = (v-m+a+ka) + (a+m) \in T(q,m).\]
Case \(\frac{q+1}{2} = m\). Then, \[v + \left(\frac{q+1}{m}+1\right)a = v + 3m = (2m-1) + (v+m+1) = (q+1-1) + (q+1-(m-1-v)) \in T(q,m).\]
It remains to show that \(\sum_{v=1}^{a-1}\lfloor\frac{f(v)}{a}\rfloor = g\).
When \(m\) is even, \[\begin{align} \sum_{v=1}^{a-1}\left\lfloor\frac{f(v)}{a}\right\rfloor &= \sum_{v=1}^{a-1}\left(2\left\lceil\frac{a-v}{m/2}\right\rceil-1\right) \\ &= \sum_{k=1}^{\frac{q+1}{m}}\left(2k-1\right) \cdot\sum_{v=1}^{a-1} \left[k=\left\lceil\tfrac{a-v}{m/2}\right\rceil\right] \\ &= \sum_{k=1}^{\frac{q+1}{m}}\left(2k-1\right) \cdot\sum_{w=1}^{a-1} \left[k=\left\lceil\tfrac{w}{m/2}\right\rceil\right] \\ &= \sum_{k=1}^{\frac{q+1}{m}}\left(2k-1\right) \cdot\sum_{w=(k-1)(m/2)+1}^{k(m/2)-[k=\frac{q+1}{m}]}1 \\ &= \sum_{k=1}^{\frac{q+1}{m}}\left(2k-1\right) \cdot\left(\frac{m}{2}-\left[k=\tfrac{q+1}{m}\right]\right) \\ &= \left(2\binom{(q+1)/m+1}{2}-\frac{q+1}{m}\right)\frac{m}{2} - \left(2\,\frac{q+1}{m}-1\right), \end{align}\] which equals \(g\) after straightforward simplification.
When \(m\) is odd, \[\begin{align} \sum_{v=1}^{a-1}\left\lfloor\frac{f(v)}{a}\right\rfloor &= \sum_{v=m}^{a-1}\left\lceil\frac{v}{m}\right\rceil + \sum_{v=1}^{m-1} \left(\frac{q+1}{m}+1\right)^{[v\notin 2\mathbb{N}]} \\ &= 1 + \sum_{k=2}^{\frac{q+1}{m}-1} k \cdot\sum_{v=m+1}^{a-1} \left[k=\left\lceil\tfrac{v}{m}\right\rceil\right] + \left(\frac{q+1}{m}+1\right)\cdot\frac{m-1}{2} + 1\cdot\frac{m-1}{2} \\ &= 1 + m\!\left(\binom{(q+1)/m}{2}-1\right) - \left(\frac{q+1}{m}-1\right) + \left(\frac{q+1}{m}+1\right)\cdot\frac{m-1}{2} + \frac{m-1}{2}, \end{align}\] which equals \(g\) after straightforward simplification. ◻
We can now determine the Weierstrass semigroups of the places of \(\mathcal{O}\). This will be done by providing explicit functions with the desired pole divisor.
Proposition 9. Let \(P_0^i\), \(i=1,\ldots,\frac{q+1}{m}\), be a place centered at \((0:0:1)\), and let \(a\in\mathbb{F}_{q^2}\) be such that \(a^{(q+1)/m}+1=0\). We have \[H(P_0^i) = H(P_{(a,0)}) = \left\langle q+1-m,\, q+1-\left\lfloor\tfrac{m}{2}\right\rfloor,\, \ldots,\, q+1 \right\rangle.\]
Proof. We already know that \(H(P_0^i) = H(P_{(a,0)})\), since these places lie in the same orbit under the group \(G\) defined in Proposition 4. We will then show that the result holds for a place \(P_{(a,0)}\).
With the information on the divisors of \(x\) and \(y\) given in Propositions 5 and 6, one can verify that \[\left(\frac{y^i}{x-a}\right)_{\infty} = (q+1-i)P_{(a,0)}, \quad i=0, \ldots, \left\lfloor \frac{m}{2} \right\rfloor.\] Additionally, we have \[\left(\frac{y^m}{x(x-a)}\right) = m\sum_{\substack{\overline{a}^{(q+1)/m}+1 = 0 \\ \overline{a} \neq a}} P_{(\overline{a},0)} - (q+1-m)P_{(a,0)}.\] Therefore, \[\left\langle q+1-m,\, q+1-\left\lfloor\tfrac{m}{2}\right\rfloor,\, \ldots,\, q+1 \right\rangle \subseteq H(P_{(a,0)}).\] From Proposition 7, we conclude that \(S(q,m)\) and \(H(P_{(a,0)})\) have the same genus, and since one contains the other, the result follows. ◻
The Weierstrass semigroups at the places of \(\mathcal{O}_{\infty}\) depend on the parity of \(m\).
Lemma 1. Let \(m\) be even and let \(P_{\infty}^i\) be a place centered at \((1:0:0)\). There exist a local parameter \(\pi\) at \(P_{\infty}^i\) and elements \(\alpha_i, \beta_i \in \mathbb{F}_{q^2}^*\) such that \(\alpha_i^{2(q+1)/m} = -1\) and \[x = \alpha_i \pi^{-m/2} + \beta_i \pi^{(q+1-m)/2} + \mathrm{h.o.t.}\]
Proof. From Proposition 6, we can choose \(\pi = \frac{1}{y}\) for any \(i = 1, \ldots, \frac{2(q+1)}{m}\). Setting \(y = \pi^{-1}\), we expand \(x = \alpha_i \pi^{-m/2} + a_{-m/2+1}\pi^{-m/2+1} + \cdots\). Note that \(m\) even implies \(p \neq 2\). From the equation of the curve, we obtain \[\begin{align} \pi^{-(q+1)} &+ \left(\alpha_i \pi^{-m/2} + a_{-m/2+1}\pi^{-m/2+1} + \cdots \right)^{\frac{2(q+1)}{m}} \\ &+ \left(\alpha_i \pi^{-m/2} + a_{-m/2+1}\pi^{-m/2+1} + \cdots \right)^{\frac{q+1}{m}} = 0. \end{align}\] By expanding the binomial terms, we get \[\begin{align} \pi^{-(q+1)} &+ \Bigl(\alpha_i^{2(q+1)/m}\pi^{-(q+1)} + \tfrac{2(q+1)}{m}\,\alpha_i^{2(q+1)/m-1}a_{-m/2+1} \pi^{-(q+1)+1} + \textrm{h.o.t.}\Bigr) \\ &+ \bigl(\alpha_i^{(q+1)/m}\pi^{-(q+1)/2} + \textrm{h.o.t.}\bigr) = 0. \end{align}\] We thus conclude that \(\alpha_i^{2(q+1)/m} = -1\). Also, the smallest exponent of \(\pi\) in the second binomial expansion is \(-\frac{q+1}{2}\), so we must have \[\frac{2(q+1)}{m}\,\alpha_i^{2(q+1)/m-1}a_{-m/2+1} = 0 \Longrightarrow a_{-m/2+1} = 0.\] Consequently, the term with the second smallest exponent of \(\pi\) in the first binomial expansion is \(\frac{2(q+1)}{m}\,\alpha_i^{2(q+1)/m-1}a_{-m/2+2}\pi^{-(q+1)+2}\), and we also obtain \(a_{-m/2+2} = 0\). Proceeding inductively, we get \[a_{-m/2+1} = \cdots = a_{(q+1-m)/2-1} = 0.\] Finally, we define \(\beta_i \mathrel{:=}a_{(q+1-m)/2}\), which satisfies \[\frac{2(q+1)}{m}\,\alpha_i^{2(q+1)/m-1}\beta_i + \alpha_i^{(q+1)/m} = 0.\] ◻
We will now obtain the Weierstrass semigroups at the places centered at \((1:0:0)\). We first note that \(\alpha_i \neq \alpha_j\) whenever \(i \neq j\). Indeed, as observed in the determination of \(\beta_i\) in Lemma 1, the expansion of \(x\) in terms of \(\pi\) is uniquely determined by \(\alpha_i\). Since any element of \(\mathbb{F}_{q^2}(\mathcal{X}_{m,q})\) can be written as a rational function in \(x\) and \(y\), having \(\alpha_i = \alpha_j\) would imply that the rings of regular functions at \(P_{\infty}^i\) and \(P_{\infty}^j\) coincide, hence \(P_{\infty}^i = P_{\infty}^j\).
Proposition 10. Let \(m\) be even and let \(P_{\infty}^i\), \(i = 1, \ldots, \frac{2(q+1)}{m}\), be a place centered at \((1:0:0)\). We have \[H(P_{\infty}^i) = \left\langle \frac{q+1}{2},\, q+1 - \frac{m}{2},\, \ldots,\, q+1 \right\rangle.\]
Proof. We already noted in Proposition 4 that all these places have the same Weierstrass semigroup. Therefore, we show the result for \(i = 1\). Let \[F(T) = b_0 + b_1 T + \cdots + b_{2(q+1)/m-1} T^{2(q+1)/m-1}\] be the polynomial of degree \(\frac{2(q+1)}{m}-1\) whose roots are \(\alpha_2, \ldots, \alpha_{2(q+1)/m}\). Define \[f = \sum_{i=0}^{2(q+1)/m-1} b_i\, x^i\, y^{q+1-(i+1)m/2}.\] By expanding in terms of \(\pi\), we conclude that \[f = F(\alpha_i)\pi^{-(q+1-m/2)} + G(\alpha_i,\beta_i)\pi^{-(q+1-m)/2} + \textrm{h.o.t.}\] Therefore, \(v_{P_{\infty}^1}(f) = -(q+1-\frac{m}{2})\), and \(v_{P_{\infty}^i}(f) \geq -\frac{q+1-m}{2}\) for \(i = 2, \ldots, \frac{2(q+1)}{m}\).
By the strict triangle inequality for valuations, \(v_{P_0^i}(f) = q+1-\frac{m}{2}\) and \(f\) has no poles outside \(\mathcal{O}_{\infty}\). Comparing degrees, we conclude that \[(f) = \left(q+1-\tfrac{m}{2}\right)\sum_{i=1}^{(q+1)/m} P_0^i - \left(q+1-\tfrac{m}{2}\right) P_{\infty}^1 - \tfrac{q+1-m}{2} \sum_{i=2}^{2(q+1)/m} P_{\infty}^i.\] Setting \(g_1 \mathrel{:=}\frac{f}{x^{(q+1)/m-1}}\), we have \[(g_1) = \frac{m}{2}\sum_{i=1}^{(q+1)/m} P_0^i - \frac{q+1}{2} P_{\infty}^1.\] Analogously, we define a function \(g_i\) whose divisor is \((g_i) = \frac{m}{2}\sum_{j=1}^{(q+1)/m} P_0^j - \frac{q+1}{2} P_{\infty}^i\). We then obtain \[\left(\frac{y^j \cdot f}{g_2 \cdots g_{2(q+1)/m}}\right)_{\infty} = \left(q+1-\tfrac{m}{2}+j\right) P_{\infty}^1.\] Finally, since \(P_{\infty}^1\) is a rational place, we have \(q+1 \in H(P_{\infty}^1)\), and the Riemann–Roch theorem gives \((q+1)P \sim (q+1)P_{\infty}^1\) for any rational place \(P\). Therefore, \[\left\langle \frac{q+1}{2},\, q+1 - \frac{m}{2},\, \ldots,\, q+1 \right\rangle \subseteq H(P_{\infty}^1).\] The equality of both numerical semigroups follows from Proposition 8. ◻
Proposition 11. Let \(m\) be odd. We have \[H(P_{\infty}^i) = \left\langle q+1-m,\, q+1-m+2,\, \ldots,\, q,\, q+1 \right\rangle.\]
Proof. The first part of this proof is very similar to that of Lemma 1. However, we need to adjust the local parameter at \(P_{\infty}^1\): since \(v_{P_{\infty}^1}(x) = -m\) and \(p \nmid m\), we may apply Hensel’s lemma in the completed local ring at \(P_{\infty}^1\) to write \(x = \pi^{-m}\) for some local parameter \(\pi\). We then obtain \[y = \beta_1\pi^{-2} + \gamma_1\pi^{q+1} + \textrm{h.o.t.},\] with \(\beta_1^{q+1} = -1\).
The function \(\pi\) is a local parameter at each \(P_{\infty}^i\), and the coefficients \(\beta_i\) are related by multiplication by a primitive \((q+1)\)-th root of unity (as can also be seen from the covering map \(\mathcal{H} \rightarrow \mathcal{X}\)). In particular, the elements \(\beta_i^m\) are pairwise distinct. We define the polynomial \(F\) whose roots are \(\bigl(\beta_i^{-m}\bigr)_i\).
Proceeding as in the proof of Lemma 1, we obtain a function \(f\) such that \[\left(\frac{f}{x^{(q+1)/m-1}}\right) = m \sum_{i=2}^{(q+1)/m} P_{\infty}^i - (q+1-m)P_{\infty}^1.\] The remaining generators of the semigroup are obtained by multiplying by suitable powers of \(y\), and the equality of both numerical semigroups follows from Proposition 8. ◻
Remark 12. Based on Propositions 9, 10, and 11, the places of \(\mathcal{O}\) share the same Weierstrass semigroup if and only if \(m = (q+1)/2\) is even.
This is a key difference between our work and the case \(m = 3\) studied in [7]. In that paper, the authors showed that \(\mathcal{O}\) is always an orbit under the action of \(\textrm{Aut}(\mathcal{X}_{3,q})\), but in our case \(\mathcal{O}\) can only be an orbit if \(m = (q+1)/2\) and \(m\) is even (which is equivalent to \(q \equiv 3 \pmod{4}\)). In fact, in this case the curve \(\mathcal{X}_{(q+1)/2,q}\) admits more automorphisms than those found in Proposition 4.
Proposition 13. Let \(q \ge 7\) with \(q \equiv 3 \pmod{4}\), and let \(\delta_1, \delta_2, \delta_3 \in \mathbb{F}_{q^2}^*\) be
such that \[(\delta_3)^{q+1} = \tfrac{1}{16},\quad (\delta_2)^4 = -\tfrac{1}{16},\quad \delta_1 = -4^{-1} (\delta_2)^{-1}\] Then there is an automorphism \(\theta_4 \in {\rm Aut}(\mathcal{X}_{(q+1)/2,q})\) given by \[\label{theta4} \theta_4^{\delta_1,\delta_2,\delta_3}(x,y) = \theta_4(x, y) \mathrel{:=} \left(\frac{\delta_1 x}{y^{(q+1)/4}}+\frac{\delta_2 y^{(q+1)/4}}{x},\, \frac{\delta_3}{y}\right).\qquad{(1)}\] Moreover, if \((\delta_3)^{(q+1)/2} = \frac{1}{4}\) and if \(G \subseteq {\rm Aut}(\mathcal{X}_{(q+1)/2,q})\) is the subgroup generated by \(A \cup \{\theta_4\}\), where \[A \mathrel{:=}\{\theta_{\gamma,\delta}(x,y) = (\gamma x, \delta y) \mid \gamma^2 = \delta^{q+1} = 1\}\] then \(G\) has order \(8(q+1)\).
Proof. We first verify that \(\theta_4\) is an automorphism of the curve. Note that \[\delta_1^4 = \delta_2^4 = -\tfrac{1}{16}, \quad \delta_3^{q+1} = \tfrac{1}{16}, \quad 6\delta_1^2\delta_2^2 + 2\delta_1\delta_2 = -\tfrac{2}{16}, \quad 4\delta_1\delta_2 + 1 = 0.\] Therefore, \[\begin{align} &\left(\frac{\delta_3}{y}\right)^{q+1} + \left(\frac{\delta_1x}{y^{(q+1)/4}} + \frac{\delta_2y^{(q+1)/4}}{x}\right)^4 + \left(\frac{\delta_1x}{y^{(q+1)/4}} + \frac{\delta_2y^{(q+1)/4}}{x}\right)^2 \\ &= \frac{\delta_3^{q+1}x^4 + \delta_1^4x^8 + (4\delta_1^3\delta_2 + \delta_1^2)x^6y^{\frac{q+1}{2}} + (6\delta_1^2\delta_2^2 + 2\delta_1\delta_2)x^4y^{q+1} + (4\delta_1\delta_2^3 + \delta_2^2)x^2y^{\frac{3(q+1)}{2}} + \delta_2^4y^{2(q+1)}}{x^4y^{q+1}} \\ &= \frac{y^{2(q+1)}+2x^4y^{q+1}+x^8-x^4}{16x^4y^{q+1}} = \frac{y^{2(q+1)}+2x^4y^{q+1}+2x^2y^{q+1}+x^8+2x^6+x^4}{16x^4y^{q+1}} \\ &= \frac{(y^{q+1}+x^4+x^2)^2}{16x^4y^{q+1}}, \end{align}\] which shows that \(\theta_4 \in {\rm Aut}(\mathcal{X}_{(q+1)/2,q})\).
Since \(2^{-4} = \delta_3^{q+1} = (\delta_3^{(q+1)/4})^4\), the value \(2\delta_3^{(q+1)/4}\) is a fourth root of unity. We may verify that:
When \(2\delta_3^{(q+1)/4} \in \{1,-1\}\), then \(\theta_4\circ \theta_4 = \theta_{-2\delta_3^{(q+1)/4}, 1} \circ \theta_2 \notin A\) (in the notation of Proposition 4).
When \(2\delta_3^{(q+1)/4} \notin \{1,-1\}\), then \(8\delta_3^{(q+1)/4}\delta_2^2 \in \{1,-1\}\) and \(\theta_4\circ \theta_4 = \theta_{8\delta^{(q+1)/4}\delta_2^2, 1} \in A\).
In particular, \((\theta_4)^4 = \mathrm{id}\).
The same formulas for \(\theta_4 \circ \theta_4\) can be applied to \[\theta_{\gamma,\delta} \circ \theta_4^{\delta_1,\delta_2,\delta_3} = \theta_4^{\gamma\delta_1,\gamma\delta_2,\delta\delta_3}\] in place of \(\theta_4 = \theta_4^{\delta_1,\delta_2,\delta_3}\).
So, we have, in the case \(\delta_3^{(q+1)/2}=\frac{1}{4}\), \[\begin{align} \theta_{4}\circ\theta_{\gamma,\delta} & =\theta_{\gamma,\delta}^{-1}\circ(\theta_{\gamma,\delta}\circ\theta_{4})^{2}\circ(\theta_{4})^{-1}\\ & \in\theta_{\gamma,\delta}^{-1}\circ A\circ\{(\theta_4)^2,\mathrm{id}\}\circ(\theta_{4})^{-1}\\ & =A\circ\{\theta_{4},(\theta_{4})^{3}\}, \end{align}\] and so \(\left\langle \theta_{4}\right\rangle \circ A\subseteq A \circ \left\langle \theta_{4}\right\rangle\); so \(A\circ \left\langle \theta_{4}\right\rangle\) is a subgroup, and has exactly \(8(q+1)\) elements, because \(A\cap\left\langle \theta_{4}\right\rangle =\{\mathrm{id}\}\). ◻
One can verify that \(\theta_4(\mathcal{O}_m) \subseteq \mathcal{O}_{\infty}\), so in the case where \(m = (q+1)/2\) is even, \(\mathcal{O}\) can indeed be an orbit under the action of \(\textrm{Aut}(\mathcal{X}_{(q+1)/2,q})\). To confirm this, it remains to show that the Weierstrass semigroups at the rational places outside \(\mathcal{O}\) differ from those of \(\mathcal{O}\). This will be done in the following section.
Although \(\mathcal{O}\) is not always an orbit under the action of the automorphism group of \(\mathcal{X}_{m,q}\), the Weierstrass semigroups at the places outside of \(\mathcal{O}\) differ from those of \(\mathcal{O}\). This is analogous to the result obtained in [7] for the case \(m = 3\). Following their approach, we extend the argument to the curves \(\mathcal{X}_{m,q}\) in order to obtain the Weierstrass semigroups at the remaining rational places. As will be seen in the following subsections, this depends heavily on the values of \(m\) and on the characteristic of \(\mathbb{F}_q\). We first present the case \(m = 4\) in detail, and then discuss how to extend this to the general case.
Let \(P_{(a,b)}\), \(ab \neq 0\), be a rational place not in \(\mathcal{O}\). Let \(Q_{(A,B)}\), with \(A^{m} = a\) and \(AB = b\), be a place of the Hermitian function field lying over \(P_{(a,b)}\). Then, as observed in Section 3, \(e\!\left(Q_{(A,B)} \mid P_{(a,b)}\right) = 1\). Let \(T \mathrel{:=}(u-A)/A\) be a local parameter at \(Q_{(A,B)}\). Then \[\frac{x-a}{a} = \frac{u^{m}-A^{m}}{A^{m}} = (T+1)^{m} - 1 = \binom{m}{1}T + \binom{m}{2}T^{2} + \cdots + \binom{m}{m}T^{m},\] \[\begin{align} \frac{y-b}{b} &= \frac{uv - AB}{AB} = \frac{(u-A+A)(v-B+B) - AB}{AB} \\ &= \frac{(u-A)(v-B) + A(v-B) + B(u-A)}{AB} = T\frac{v-B}{B} + \frac{v-B}{B} + T. \end{align}\] From the equation of the curve, \[\begin{align} 0 &= u^{q+1} + v^{q+1} + 1 \\ &= (u-A+A)^{q+1} + (v-B+B)^{q+1} + 1 \\ &= (u-A)^{q+1} + A^{q}(u-A) + A(u-A)^{q} + A^{q+1} \\ &\quad + (v-B)^{q+1} + B^{q}(v-B) + B(v-B)^{q} + B^{q+1} + 1, \end{align}\] which gives \[\begin{align} 0 &= (u-A)^{q+1} + (v-B)^{q+1} + A^{q}(u-A) + B^{q}(v-B) + A(u-A)^{q} + B(v-B)^{q} \\ &= B^{q}(v-B) + A^{q+1}T + O\!\left(T^{q}\right), \end{align}\] so \[\begin{align} \frac{y-b}{b} &= (T+1)\frac{v-B}{B} + T \\ &= (T+1)\left(-\left(\frac{A}{B}\right)^{q+1}T + O\!\left(T^{q}\right)\right) + T \\ &= \left(1 - \left(\frac{A}{B}\right)^{q+1}\right)T - \left(\frac{A}{B}\right)^{q+1}T^{2} + O\!\left(T^{q}\right) \\ &= \left(\frac{2a^{(q+1)/m}+1}{1+a^{(q+1)/m}}\right)T + \frac{a^{(q+1)/m}}{1+a^{(q+1)/m}}\,T^{2} + O\!\left(T^{q}\right), \end{align}\] where \(O(T^{q})\) denotes an element with \(Q_{(A,B)}\)-valuation at least \(q\).
Defining \[\alpha \mathrel{:=}\frac{a^{(q+1)/m}}{1+a^{(q+1)/m}} = \frac{A^{q+1}}{1+A^{q+1}} = \frac{A^{q+1}}{-B^{q+1}},\] we have \[X \mathrel{:=}\frac{x-a}{a} = (T+1)^{m} - 1 = \binom{m}{1}T + \binom{m}{2}T^{2} + \cdots + \binom{m}{m}T^{m},\] \[Y \mathrel{:=}\frac{y-b}{b} = (1+\alpha)T + T^{2} + O\!\left(T^{q}\right).\] We also define \[\begin{align} t_0 &\mathrel{:=}(1+\alpha)\cdot\frac{x-a}{a} - m\cdot\frac{y-b}{b} \\ &= (1+\alpha)\sum_{i=2}^{m}\binom{m}{i}T^{i} - m\alpha T^{2} + O\!\left(T^{q}\right) \\ &= \left(\binom{m}{2} + \left(\binom{m}{2}-m\right)\alpha\right)T^{2} + (1+\alpha)\sum_{i=3}^{m}\binom{m}{i}T^{i} + O\!\left(T^{q}\right). \end{align}\]
Definition 14. Let \(\iota \in \overline{K}\) be a square root of \(-1\). Since \(4 \mid q+1\), we have \(\iota \in K\) and \(\iota \neq -\iota\). We define, for each \(k \in \mathbb{N}\): \[\mathcal{P}_{k}(s) \mathrel{:=} \frac{\iota(s+\iota)^{4k} - \iota(s-\iota)^{4k}}{2\cdot(s - s^{2})} \quad \text{and} \quad \mathcal{Q}_{k}(s) \mathrel{:=} \frac{(1-\iota)(s+\iota)^{4k-1} + (1+\iota)(s-\iota)^{4k-1}}{2\cdot(s-1)}.\]
One can verify that, for each \(k \in \mathbb{N}\), \(\mathcal{P}_{k}(s)\) and \(\mathcal{Q}_{k}(s)\) are polynomials in \(\mathbb{F}_{p}[s]\) having no common roots. Our goal is to find, for each \(\alpha\), the values \(k \in \mathbb{N}\) for which there exists \[f_{k} \in \left\langle X^{i} \cdot Y^{j} \right\rangle_{i,j \in \mathbb{N}}^{i/2+j \le k} \subseteq \mathcal{L}\!\left(k \cdot (X)_{\infty}\right) \label{eq:fk32L32k32Xinf}\tag{2}\] with \(T\)-expansion \[f_{k} = \mathcal{P}_{k}(\alpha) \cdot T^{4k-1} + \mathcal{Q}_{k}(\alpha) \cdot T^{4k} + O\!\left(T^{q}\right) \label{eq:fk32Pk32Qk}\tag{3}\] and \(4k < q\). We will use these elements to describe the Weierstrass semigroup at \(P_{(a,b)}\).
We obtain these elements by successively cancelling terms of equal \(T\)-valuation. We have \(\mathcal{P}_1(s) = 4(s+1)\) and \(\mathcal{Q}_1(s) = s^2 + 4s + 1\). Since \(\alpha \neq 1\), we may define \[f_{1} \mathrel{:=} \frac{(2\alpha+6) \cdot Y^{2} - (\alpha+1)^{2} t_{0}}{\alpha - 1},\] which satisfies 2 and 3 . If \(\mathcal{Q}_{1}(\alpha) = \alpha^{2} + 4\alpha + 1 \neq 0\), we may define \[\begin{align} \mathcal{Q}_{1}(\alpha)^{2} \cdot f_{2} \mathrel{:=}{} & -2^{8}(\alpha+1)^{4} \cdot f_{1} + 2^{10}(\alpha+1)^{2} \cdot Y^{3} + 2^{6}(\alpha+1)^{2}(\alpha^{2}-8\alpha+1) \cdot f_{1}Y \\ &- 2^{4}\mathcal{Q}_{1}(\alpha)\mathcal{Q}_{1}(-\alpha)\,f_{1}Y^{2} + \left(\mathcal{Q}_{2}(\alpha) + 16\alpha^{2}\mathcal{Q}_{1}(-\alpha)\right)f_{1}^{2}, \end{align}\] and \(f_2\) satisfies 2 and 3 . Still assuming \(\mathcal{Q}_{1}(\alpha) \neq 0\), we obtain \(f_{3}\) as a \(\mathbb{Z}[\alpha]\)-linear combination of \(f_{2},\, f_{1}^{2}Y,\, f_{1}^{2}Y^{2},\, f_{1}^{3},\, f_{1}f_{2}\), divided by \(\mathcal{Q}_{1}(\alpha)^{4}\).
Now, for \(k \geq 4\), given \(f_{1}, \ldots, f_{k-1}\) as in 2 and 3 , and assuming \(\mathcal{P}_{k-3}(\alpha) \neq 0\), \(\alpha + 1 \neq 0\), and \(\alpha^{2} + 1 \neq 0\), we may define \[f_{k} \mathrel{:=} \frac{\mathcal{P}_{k-2}(\alpha)\mathcal{P}_{2}(\alpha) \cdot f_{k-1}f_{1} + \mathcal{P}_{k-1}(\alpha)\mathcal{P}_{1}(\alpha) \cdot f_{k-2}f_{2}}{2^{2}(\alpha+1)(\alpha^{2}+1)^{3}\mathcal{P}_{k-3}(\alpha)},\] which also satisfies 2 and 3 .
For most places \(P\), there exists a minimal value of \(k\) for which \(\mathcal{P}_{k}(\alpha) = 0\). Each such value yields a different Weierstrass semigroup.
Lemma 2. Suppose that \(\alpha \notin \{\iota, -\iota\}\). Then the set \(\{k \ge 1 \mid \mathcal{P}_{k}(\alpha) = 0\}\) is nonempty, and we denote its minimum by \[I \mathrel{:=}\min\left\{ k \ge 1 \mid \mathcal{P}_{k}(\alpha) = 0 \right\}.\] Moreover, \(I\) is a divisor of \(\frac{q+1}{4}\) (in particular, \(4I \le q+1\)).
Proof. We have \(\mathcal{P}_{k}(\alpha) = 0\) if and only if \(\left(\frac{\alpha+\iota}{\alpha-\iota}\right)^{4k} = 1\), thus it suffices to show that \(\left(\frac{\alpha+\iota}{\alpha-\iota}\right)^{q+1} = 1\). In fact, \[\left(\frac{\alpha+\iota}{\alpha-\iota}\right)^{q} = \frac{\alpha^{q}+\iota^{-1+4\frac{q+1}{4}}}{\alpha^{q}-\iota^{-1+4\frac{q+1}{4}}} = \frac{\alpha^{q}-\iota}{\alpha^{q}+\iota},\] thus it suffices to show that \(\alpha^{q} = \alpha\). Since \(\alpha = \frac{a^{(q+1)/4}}{1+a^{(q+1)/4}}\), it suffices to show that \(a^{(q+1)/4} \in \mathbb{F}_{q}\).
Since \((a^{(q+1)/4})^{q-1} = a^{(q^{2}-1)/4} \neq 0\) is a fourth root of \(a^{q^{2}-1} = 1\), we have \(a^{(q^{2}-1)/4} \in \{1, -1, \iota, -\iota\}\). Given that \(b^{q+1} + a^{2(q+1)/4} + a^{(q+1)/4} = 0\) and \(b \in \mathbb{F}_{q^{2}} \smallsetminus \{0\}\), we have \[\left(a^{2\frac{q+1}{4}} + a^{\frac{q+1}{4}}\right)^{q-1} = \left(-b^{q+1}\right)^{q-1} = (-1)^{q-1} b^{q^{2}-1} = 1,\] so \(a^{2(q+1)/4} + a^{(q+1)/4} \in \mathbb{F}_{q}\). Thus, \[a^{2\frac{q+1}{4}} + a^{\frac{q+1}{4}} = \left(a^{2\frac{q+1}{4}} + a^{\frac{q+1}{4}}\right)^{q} = a^{2\frac{q^{2}+q}{4}} + a^{\frac{q^{2}+q}{4}},\] that is, \[\left(a^{\frac{q+1}{4}}\right)^{2} + a^{\frac{q+1}{4}} = \left(a^{\frac{q^{2}-1}{4}} a^{\frac{q+1}{4}}\right)^{2} + a^{\frac{q^{2}-1}{4}} a^{\frac{q+1}{4}},\] \[\begin{align} a^{\frac{q+1}{4}} - a^{\frac{q^{2}-1}{4}} a^{\frac{q+1}{4}} &= \left(a^{\frac{q^{2}-1}{4}} a^{\frac{q+1}{4}}\right)^{2} - \left(a^{\frac{q+1}{4}}\right)^{2} \\ &= \left(a^{\frac{q^{2}-1}{4}} a^{\frac{q+1}{4}} + a^{\frac{q+1}{4}}\right) \left(a^{\frac{q^{2}-1}{4}} a^{\frac{q+1}{4}} - a^{\frac{q+1}{4}}\right). \end{align}\]
Suppose, for a contradiction, that \(a^{(q^{2}-1)/4} \neq 1\).
Then \(a^{\frac{q^{2}-1}{4}} a^{\frac{q+1}{4}} + a^{\frac{q+1}{4}} = -1\) and \(a^{\frac{q^{2}-1}{4}} + 1 = -a^{-\frac{q+1}{4}}\). Also, \[\begin{align} 0 \neq \alpha^{2} + 1 &= \left(\frac{a^{\frac{q+1}{4}}}{1+a^{\frac{q+1}{4}}}\right)^{2} + 1 = \frac{a^{2\frac{q+1}{4}} + \left(1 + 2a^{\frac{q+1}{4}} + a^{2\frac{q+1}{4}}\right)}{1 + 2a^{\frac{q+1}{4}} + a^{2\frac{q+1}{4}}}, \end{align}\] so \(1 + 2(a^{\frac{q+1}{2}} + a^{\frac{q+1}{4}}) \neq 0\). Multiplying by \(a^{-(q+1)/2}\), we obtain \[\begin{align} 0 &\neq a^{-\frac{q+1}{2}} + 2 + 2a^{-\frac{q+1}{4}} \\ &= \left(a^{\frac{q^{2}-1}{4}} + 1\right)^{2} + 2 - 2\left(a^{\frac{q^{2}-1}{4}} + 1\right) \\ &= \left(a^{\frac{q^{2}-1}{2}} + 2a^{\frac{q^{2}-1}{4}} + 1\right) + 2 - \left(2a^{\frac{q^{2}-1}{4}} + 2\right) \\ &= a^{\frac{q^{2}-1}{2}} + 1, \end{align}\] so \(a^{\frac{q^{2}-1}{4}} \notin \{\iota, -\iota\}\), leaving only the possibility \(a^{\frac{q^{2}-1}{4}} = -1\). But then \(a^{\frac{q^{2}-1}{4}} + 1 = 0\), which gives \(-a^{-\frac{q+1}{4}} = 0\), contradicting \(a \neq 0\). ◻
We now determine the Weierstrass semigroups at every rational point of \(\mathcal{X}_{m,q}\). Let \(P_{(a,b)} \in \mathcal{X}_{m,q}(\mathbb{F}_{q^2})\) with \(ab \neq 0\) be a rational place outside \(\mathcal{O}\), and let \(P_{(\overline{a},0)}\) be a rational place of \(\mathcal{O}_m\), so that \(\overline{a}^{\frac{q+1}{m}} + 1 = 0\). For \(k \in \mathbb{N}\) and \(h_k \in \mathcal{L}(k \cdot (x)_\infty)\), define \[\label{div-Gk} G_k \mathrel{:=}\frac{h_k \cdot f_{P_{(\overline{a},0)}}^k}{f_{P_{(a,b)}}^k \cdot (x - \overline{a})^k},\tag{4}\] where \(f_{P_{(a,b)}}\) and \(f_{P_{(\overline{a},0)}}\) are chosen via the fundamental relations \[(f_{P_{(a,b)}}) = (q+1)P_{(a,b)} - (q+1)P_\infty^1,\] \[(f_{P_{(\overline{a},0)}}) = (q+1)P_{(\overline{a},0)} - (q+1)P_\infty^1.\] Since \((x - \overline{a}) = (q+1)P_{(\overline{a},0)} - (x)_\infty\), a straightforward divisor computation gives \[(G_k) = (h_k) - k(q+1)P_{(a,b)} + k(x)_\infty,\] and in particular, \[k(q+1) - v_{P_{(a,b)}}(h_k) \in H(P_{(a,b)}).\]
Proposition 15. Suppose \(\alpha \notin \{\iota, -\iota\}\) and \(\mathcal{Q}_1(\alpha) = \alpha^2 + 4\alpha + 1 \neq 0\). Then there exists a divisor \(I\) of \(\frac{q+1}{4}\) such that \(H(P_{(a,b)})\) is generated by \[\{q,\, q+1,\, q-1\} \cup \Bigl\{\, k(q+1) - (4k-1) - [k = I] \;\Big|\; k = 1, \ldots, \min\!\Bigl(I,\, \tfrac{q+1}{4} - 1\Bigr) \Bigr\},\] where \([k = I]\) denotes the Iverson bracket, equal to \(1\) if \(k = I\) and \(0\) otherwise.
Proof. Let \(I\) be as in Lemma 2. By the discussion preceding this proposition, for each \(k \in \{1, \ldots, \min(I, \frac{q+1}{4} - 1)\}\) there exist functions \(f_k\) satisfying 2 and 3 whose \(T\)-valuation equals \((4k-1) + [k = I]\) (note that \(\mathcal{P}_k(\alpha) = 0\) implies \(\mathcal{Q}_k(\alpha) \neq 0\), and \(4k < q\)). Applying 4 and the remark following it yields \[k(q+1) - (4k-1) - [k=I] \;\in\; H(P_{(a,b)}).\] It is known that \(q, q+1 \in H(P)\) for every rational place \(P\) of a maximal curve (see, e.g., [8]). Setting \(h_k = Y\) when \(\alpha = -1\), or \(h_k = Y^2\) when \(\alpha \neq -1\), in 4 shows that \(q - 1 \in H(P_{(a,b)})\). It remains to verify that the stated generators produce a numerical semigroup of genus at most \(g(\mathcal{X}_{m,q})\); this is established in Lemma 3. ◻
Remark 16. Since \(\mathcal{P}_1(s) = 4(s+1)\), we have \(I = 1\) whenever \(\alpha = -1\). In particular, \[H(P_{(a,b)}) = H(P_{(\overline{a},0)})\] for \(2a^{(q+1)/4} + 1 = 0\) and \(\overline{a}^{(q+1)/4} + 1 = 0\).
Proposition 17. Suppose \(\alpha \in \{\iota, -\iota\}\). Then \(H(P_{(a,b)})\) is generated by \[\{q,\, q+1,\, q-1\} \cup \Bigl\{\, k(q+1) - (4k-1) \;\Big|\; k = 1, \ldots, \tfrac{q+1}{4} - 1 \Bigr\}.\]
Proof. A direct computation gives \[\mathcal{P}_k(\alpha) = \frac{\alpha(2\alpha)^{4k}}{2(\alpha+1)} = 2^{4k-2}(\alpha+1) \qquad \text{and} \qquad \mathcal{Q}_k(\alpha) = \frac{(1-\alpha)(2\alpha)^{4k-1}}{2(\alpha-1)} = 2^{4k-2}\alpha.\] As before, there exist functions \(f_1, f_2, \ldots\) satisfying 2 and 3 , though their construction differs in this case. Write \(f_k \mathrel{:=}2^{4k-2} g_k\), where \(f_1, f_2, f_3\) are as previously defined, and for \(k \geq 3\) the functions \(g_k\) satisfy the recurrence \[g_k \mathrel{:=}-4\, g_{k-1} - 2\alpha\, g_{k-2} g_1 Y + 2\alpha\, g_{k-2} g_1 Y^2 + 5\, g_{k-2} g_1^2 - 4\alpha\, g_{k-1} g_1.\] For each \(k \in \{1, \ldots, \frac{q+1}{4} - 1\}\), the function \(g_k\) lies in \(\mathcal{L}(k \cdot (X)_\infty)\) and has \(T\)-valuation equal to \(4k - 1\), so that \[k(q+1) - (4k-1) \;\in\; H(P_{(a,b)}).\] Applying 4 with \(h_k \in \{1, X, Y^2\}\) yields \(\{q+1, q, q-1\} \subseteq H(P_{(a,b)})\). To conclude that the stated generators produce the full Weierstrass semigroup, we apply Lemma 3 with \(I \mathrel{:=}\frac{q+1}{4}\). ◻
Proposition 18. Let \(P_{(a,b)}\) be a rational place with \(\mathcal{Q}_1(\alpha) = \alpha^2 + 4\alpha + 1 = 0\). Then \(H(P_{(a,b)})\) is generated by \[\{q+1,\, q,\, q-1,\, q-2\} \cup \{2(q+1) - 7\} \cup \{3(q+1) - 12\}.\]
Proof. Since \((\alpha+2)^2 = 3\) and \(\alpha \neq 1\), the characteristic of \(\mathbb{F}_q\) is not \(3\). Moreover, \(\alpha \neq -1\), so the elements \(1, Y, Y^2 \in \mathcal{L}(1 \cdot (X)_\infty)\) have \(T\)-valuations \(0\), \(1\), and \(2\), respectively, giving \(q+1, q, q-1 \in H(P_{(a,b)})\). Furthermore, \[f_1 = 4(\alpha+1)T^3 + O(T^q) \;\in\; \mathcal{L}(1 \cdot (X)_\infty),\] so \(q - 2 \in H(P_{(a,b)})\).
From the proof of Lemma 2 and the fact that \(\alpha^2 + 1 = -4\alpha \neq 0\), we have \(\alpha \in \mathbb{F}_q\). In particular, \(3\) is a square in \(\mathbb{F}_q\) and \(q \neq 7\). Define \[\begin{align} g_{2,7} &\mathrel{:=}\frac{1}{12}\Bigl[ (-56\alpha - 208)Y^2 + (-58\alpha - 218)Y^3 + (-33\alpha - 123)Y^4 \\ &\qquad\qquad + (10\alpha + 38)XY + (15\alpha + 57)XY^2 \Bigr] \;\in\; \mathcal{L}(2 \cdot (X)_\infty), \end{align}\] so that \(g_{2,7} = T^7 - \frac{\alpha+1}{2}T^8 + O(T^q)\). Since \(q > 7\), the function \(g_{2,7}\) has \(T\)-valuation \(7\), and hence \(2(q+1) - 7 \in H(P_{(a,b)})\).
Next, let \(g_{3,12}\) be an appropriate \(\mathbb{Z}[\alpha]\)-linear combination of \(Y, \ldots, Y^6, X, XY, XY^3, XY^4\), divided by \(3\), such that \(g_{3,12} \in \mathcal{L}(3 \cdot (X)_\infty)\) and \(g_{3,12} = T^{12} + O(T^q)\). If \(q \neq 11\), then \(q > 12\) and \(g_{3,12}\) has \(T\)-valuation \(12\), giving \(3(q+1) - 12 \in H(P_{(a,b)})\). If \(q = 11\), then \(3(q+1) - 12 = 24 = 2(q+1) \in H(P_{(a,b)})\), so the conclusion holds in either case.
It remains to apply Lemma 3 with \(I = 3\), for which we need \(3 \mid (q+1)/4\), i.e., \(q \equiv -1 \pmod{12}\). Since \(\alpha\) satisfies a quadratic equation over \(\mathbb{F}_p\) and \(q\) is not a perfect square (as \(q \equiv -1 \pmod 4\)), we have \(\alpha \in \mathbb{F}_q \cap \mathbb{F}_{p^2} = \mathbb{F}_p\). Thus \(3 = (\alpha+2)^2\) is a quadratic residue modulo \(p \neq 2, 3\), and the law of quadratic reciprocity gives \[(-1)^{\frac{3-1}{2} \cdot \frac{p-1}{2}} = \left(\frac{3}{p}\right)\left(\frac{p}{3}\right) = 1 \cdot (-1)^{[p \equiv -1\,(\mathrm{mod}\,3)]},\] so \([p \equiv -1 \pmod{4}] = [p \equiv -1 \pmod{3}]\), meaning \(p \equiv \pm 1 \pmod{12}\). Since \(q \equiv -1 \pmod{4}\), we conclude \(p \equiv q \equiv -1 \pmod{12}\). ◻
Lemma 3. Let \(I \in \{1, \ldots, \frac{q+1}{4}\}\) be a divisor of \(\frac{q+1}{4}\). The genus of the numerical semigroup \[Q(I) \mathrel{:=}\langle q-1,\, q,\, q+1 \rangle + \Bigl\langle k(q+1) - (4k-1) - [k=I] \;\Big|\; k = 1, \ldots, \min\!\Bigl(I,\, \tfrac{q+1}{4} - 1\Bigr) \Bigr\rangle\] is at most \(g = g(\mathcal{X}_{4,q})\).
Proof. If \((q, I) = (7, 2)\), then \(Q(I) = \langle 5, 6, 7, 8 \rangle\) has genus \(5 = g\). Henceforth assume \((q, I) \neq (7, 2)\). Since \[k(q+1) - (4k-1) - [k=I] = k(q-3) + 1 - [k=I],\] the smallest positive element of \(Q(I)\) is \(a \mathrel{:=}q - 2 - [I=1]\) (noting that \(\frac{q+1}{4} > 1\)).
We construct \(f \colon \{1, \ldots, a-1\} \to Q(I)\) such that \(f(v) \equiv v \pmod{a}\) for each \(v\) and \(\sum_{v=1}^{a-1} \lfloor f(v)/a \rfloor = g\).
Case \(I > 1\) (so \(a = q-2\) and \(\frac{q+1}{4} > 2\)). Let \(v \in \{1, \ldots, a-1\}\).
Subcase \(1 \le v \le 3\frac{q+1}{4} - 3\). Write \(v = \lceil v/3 \rceil \cdot 3 - w\) with \(w \in \{0,1,2\}\), and define \[\begin{align} f(v) &\mathrel{:=}\Bigl(\lceil \tfrac{v}{3} \rceil - 2\Bigr)(q+1) + \bigl(q+1 - [w \ge 1]\bigr) + \bigl(q+1 - [w \ge 2]\bigr) \\ &= \lceil \tfrac{v}{3} \rceil (q+1) - w = v + \lceil \tfrac{v}{3} \rceil \cdot a, \end{align}\] which lies in \(Q(I)\) when \(\lceil v/3 \rceil \ge 2\). For \(\lceil v/3 \rceil = 1\), we have \(f(v) \in \{q-1, q, q+1\} \subseteq Q(I)\).
Subcase \(a - \frac{q+1}{4} + 1 \le v \le a-1\) (equivalently, \(1 \le a - v \le \frac{q+1}{4} - 1\)). Let \(b_k \mathrel{:=}k(q-3) + 1 - [k=I]\) denote the generators above (so \(a = b_1\)). These satisfy \(b_k \equiv -k + 1 - [k=I] \pmod{a}\). Set \[s \mathrel{:=}\bigl(-b_{I - [I=(q+1)/4]}\bigr) \bmod a = I - 2\bigl[I = \tfrac{q+1}{4}\bigr] > 0,\] and write \(a - v = sQ + R\) with \(Q \in \mathbb{N}\) and \(R \in \{0, \ldots, s-1\}\). Define \[f(v) \mathrel{:=}Q\, b_{I-[I=(q+1)/4]} + \begin{cases} 0 & (R = 0), \\ b_{I-1} + b_2 & (I < \tfrac{q+1}{4},\;R = I-1), \\ b_{R+1} & (\text{otherwise}). \end{cases}\] When \(I < \frac{q+1}{4}\) and \(R = I-1\), we have \[b_{I-1} + b_2 = (I-1)(q-3) + 1 + 2(q-3) + 1 = (R+1)(q-3) + 1 + (q-2),\] so in this subcase \(f(v) = a\bigl(a - v + [R=I-1] - [R=0]\bigr) + v\). When \(I = \frac{q+1}{4}\) (and \(s = I-2\)), one checks that \(Q = [a-v \in \frac{q+1}{4} - \{1,2\}]\) and \([R=0] = [a-v = \frac{q+1}{4}-2]\), giving \[f(v) = a\bigl(a - v + Q - [R=0]\bigr) + v.\]
Summing \(\lfloor f(v)/a \rfloor\) over \(v \in \{1,\ldots,a-1\}\) and using the formulas above, one computes: \[\begin{align} \sum_{v=1}^{a-1} \Bigl\lfloor \frac{f(v)}{a} \Bigr\rfloor &= \sum_{v=1}^{3((q+1)/4-1)} \!\!\Bigl\lceil \tfrac{v}{3} \Bigr\rceil + \sum_{w=1}^{(q+1)/4-1} \Bigl(w + \bigl[w \equiv I{-}1 \pmod{I}\bigr] - \bigl[w \equiv 0 \pmod{I}\bigr]\Bigr) \\ &= 3\binom{(q+1)/4}{2} + \binom{(q+1)/4}{2} + \frac{(q+1)/4}{I} - \Bigl(\frac{(q+1)/4}{I} - 1\Bigr) = g \quad \bigl(I < \tfrac{q+1}{4}\bigr), \end{align}\] and similarly \[\sum_{v=1}^{a-1} \Bigl\lfloor \frac{f(v)}{a} \Bigr\rfloor = \sum_{v=1}^{3((q+1)/4-1)} \!\!\Bigl\lceil \tfrac{v}{3} \Bigr\rceil + \sum_{w=1}^{(q+1)/4-1} \Bigl(w + \bigl[w = \tfrac{q+1}{4}-1\bigr]\Bigr) = g \quad \bigl(I = \tfrac{q+1}{4}\bigr).\]
Case \(I = 1\) (so \(a = q-3\)). Let \(v \in \{1, \ldots, a-1\}\).
Subcase \(v = 1\). Define \[f(1) \mathrel{:=}\Bigl(\tfrac{q+1}{4} - 3\Bigr)(q+1) + 3q = \Bigl(\tfrac{q+1}{4} + 1\Bigr)a + 1,\] which lies in \(Q(I)\) when \((q+1)/4 \ge 3\). For \((q+1)/4 = 2\), one verifies \(f(1) = 13 = (7-1) + 7 \in Q(I)\).
Subcase \(v \ge 2\). Write \(v - 1 = 4Q + R\) with \(Q \in \mathbb{N}\) and \(R \in \{0,1,2,3\}\), and define \[f(v) \mathrel{:=} \begin{cases} Q(q+1) + (q - 2 + R) & (R \neq 0), \\ (Q-1)(q+1) + (q-1) + q & (R = 0), \end{cases}\] where \(q - 2 + R \in \{q-1, q, q+1\}\) when \(R \neq 0\). In both cases \(f(v) \in Q(I)\) and \(f(v) = \lceil v/4 \rceil \cdot a + v\).
Therefore, \[\begin{align} \sum_{v=1}^{a-1} \Bigl\lfloor \frac{f(v)}{a} \Bigr\rfloor &= \Bigl(\tfrac{q+1}{4} + 1\Bigr) + \sum_{v=2}^{a-1} \Bigl\lceil \tfrac{v}{4} \Bigr\rceil = \Bigl(\tfrac{q+1}{4} + 1\Bigr) + 4\binom{(q+1)/4}{2} - 1 - \tfrac{a}{4} = g. \qedhere \end{align}\] ◻
The preceding subsection illustrates the difficulty of obtaining a general result for Weierstrass semigroups at the remaining rational points, as they depend heavily on both \(m\) and the characteristic of the field.
The general strategy is to consider linear combinations of monomials \(X^i Y^j\) lying in \(\mathcal{L}(k \cdot (X)_\infty)\) with a prescribed valuation at \(P_{(a,b)}\). For \(k = 1\), for instance, we consider the functions \(Y, \ldots, Y^{\lfloor m/2 \rfloor}, X\). Their \(T\)-coefficients up to \(T^{m-1}\) are given by the matrix \[\label{matrix} \begin{array}{cccccc} & & T^1 & T^2 & \cdots & T^{m-1} \\[4pt] Y & : & 1 + \alpha & \alpha & \cdots & 0 \\ Y^2 & : & 0 & (1 + \alpha)^2 & \cdots & 0 \\ & & & \vdots & & \\ Y^{\lfloor m/2 \rfloor} & : & 0 & 0 & \cdots & \star \\ X & : & m & \binom{m}{2} & \cdots & m \end{array}\tag{5}\] where \[\star = \begin{cases} \dfrac{m}{2}(1 + \alpha)\,\alpha^{m/2 - 1} & \text{if } m \text{ is even,} \\[6pt] \alpha^{(m-1)/2} & \text{if } m \text{ is odd.} \end{cases}\]
If the matrix \(M\) formed by these entries has full rank, then there exists a nonzero vector \([a_1, a_2, \ldots, b]\) such that the product \([a_1, a_2, \ldots, b]\, M\) has its first \(\lfloor m/2 \rfloor\) entries equal to zero (as produced by Gaussian elimination). This yields a function \(f \in \mathbb{F}_{q^2}(\mathcal{X}_{m,q})\) satisfying \[\lfloor m/2 \rfloor < v_{P_{(a,b)}}(f) < m.\] Consequently, \(H(P_{(a,b)})\) contains an element strictly between \((q+1) - m\) and \((q+1) - \lfloor m/2 \rfloor\), so \(H(P_{(a,b)}) \neq H(P_\infty^1)\). Moreover, if \(m\) is odd, then \(Y^2 \in \mathcal{L}((X)_\infty)\), giving \((q+1) - 2 \in H(P_{(a,b)}) \setminus H(P_\infty^1)\). If \(m\) is even, then \(H(P_{(a,b)})\) contains an element strictly between \((q+1) - m \ge (q+1)/2\) and \((q+1) - m/2\), again showing \(H(P_{(a,b)}) \neq H(P_\infty^1)\).
One therefore concludes that the Weierstrass semigroup at any place of \(\mathcal{O}\) differs from that at \(P_{(a,b)}\), provided \(M\) has full rank for every \(\alpha\). This will be the key argument in the following section for determining the automorphism group of \(\mathcal{X}_{m,q}\). Although we were unable to prove that the rank is always maximal, we establish it under a mild restriction on the characteristic of the field, which we assume henceforth.
Proposition 19. If \(m \ge 3\), \(\alpha \neq -1\), and \(p \nmid \binom{m}{2 + (m \bmod 2)}\), then the matrix \(M\) has full rank.
Proof. It suffices to exhibit a minor \(N\) of \(M\) of order \(\lfloor m/2 \rfloor + 1\) with nonzero determinant. If \(m\) is odd, let \(N\) be the submatrix formed by columns \(1, \ldots, \lfloor m/2 \rfloor - 1, m-2, m-1\) of \(M\); if \(m\) is even, take columns \(1, \ldots, m/2 - 2, m-3, m-2, m-1\). We claim that \[{\rm det}~N = \begin{cases} \dfrac{m}{4}\dbinom{m}{3} (1-\alpha)\,\alpha^{m-4}\,(\alpha+1)^{\binom{m/2-1}{2}+1} & (m \text{ even}), \\[8pt] \dbinom{m}{2}\alpha^{(m-3)/2}(\alpha+1)^{\binom{(m-1)/2}{2}} & (m \text{ odd}). \end{cases}\] Partition \(N\) into blocks \[N = \begin{bmatrix} A & B \\ C & D \end{bmatrix},\] where \(A\) and \(D\) are square, \(D\) has order \(3 - (m \bmod 2)\), and \(B = 0\) (which is possible since \(m \ge 3\)). Hence \({\rm det}~N = {\rm det}~A \cdot {\rm det}~D\). The matrix \(A\) is upper triangular with \[\begin{align} {\rm det}~A &= \prod_{i=1}^{\lfloor m/2 \rfloor - 2 + (m \bmod 2)} (\alpha+1)^i = \begin{cases} (\alpha+1)^{\binom{m/2-1}{2}} & (m \text{ even}), \\[4pt] (\alpha+1)^{\binom{(m-1)/2}{2}} & (m \text{ odd}). \end{cases} \end{align}\]
Case \(m\) odd. We compute \[\begin{align} {\rm det}~D &= {\rm det}\begin{bmatrix} \frac{m-1}{2}(\alpha+1)\alpha^{(m-3)/2} & \alpha^{(m-1)/2} \\ \binom{m}{2} & m \end{bmatrix} \\ &= \alpha^{(m-3)/2}\!\left(\binom{m}{2}(\alpha+1) - \binom{m}{2}\alpha\right) = \binom{m}{2}\alpha^{(m-3)/2}, \end{align}\] and therefore \[{\rm det}~N = \binom{m}{2}\,\alpha^{(m-3)/2}\,(\alpha+1)^{\binom{(m-1)/2}{2}}.\]
Case \(m\) even. We compute \[\begin{align} {\rm det}~D &= {\rm det}\begin{bmatrix} \binom{m/2-1}{1}(\alpha+1)\alpha^{\frac{m}{2}-2} & \alpha^{\frac{m}{2}-1} & 0 \\[4pt] \binom{m/2}{3}(\alpha+1)^3\alpha^{\frac{m}{2}-3} & \binom{m/2}{2}(\alpha+1)^2\alpha^{\frac{m}{2}-2} & \frac{m}{2}(\alpha+1)\alpha^{\frac{m}{2}-1} \\[4pt] \binom{m}{3} & \binom{m}{2} & m \end{bmatrix} \\ &= \frac{m^2(m-1)(m-2)}{24}(1-\alpha)\,\alpha^{m-4}(\alpha+1), \end{align}\] and therefore \[{\rm det}~N = \frac{m}{4}\binom{m}{3} (1-\alpha)\,\alpha^{m-4}\,(\alpha+1)^{\binom{m/2-1}{2}+1}.\]
Since \(\alpha \notin \{0, 1\}\) and \(\alpha \neq -1\), we have \({\rm det}N \neq 0\), so \(M\) has an invertible minor of order \(\lfloor m/2 \rfloor + 1\), and the result follows. ◻
In the case where \(p \mid \binom{m}{2 + (m \bmod 2)}\), one could examine the remaining minors of \(M\) to determine whether any has nonzero determinant. Since the characteristic condition in Proposition 19 is mild, we do not pursue this direction further.
From Proposition 19, the Weierstrass semigroup at any place of \(\mathcal{O}\) differs from that at \(P_{(a,b)}\) whenever \(2a^{(q+1)/m} + 1 \neq 0\) and \(p \nmid \binom{m}{2+(m \bmod 2)}\). If \(m > 4\), the same conclusion holds for places with \(2a^{(q+1)/m} + 1 = 0\), with no restriction on the characteristic.
Proposition 20. Let \(\mathcal{O}' = \{P_{(a,b)} : 2a^{(q+1)/m} + 1 = 0\}\). If \(m > 4\), then \(\mathcal{O}'\) is an orbit under the action of the full automorphism group of \(\mathcal{X}_{m,q}\).
Proof. One verifies directly that \(\mathcal{O}'\) is an orbit under the group \(G\) of Proposition 4 or 13. It therefore suffices to show that the Weierstrass semigroup at any place of \(\mathcal{O}'\) differs from that at every other rational place. Let \(P_{(a,b)} \in \mathcal{O}'\). With the notation of the beginning of this section, we have \(Y = -T^2 + O(T^q)\), so \[v_{P_{(a,b)}}(Y^i) = 2i \le m < q+1 \quad \text{for every } i \in \bigl\{0, \ldots, \lfloor \tfrac{m}{2} \rfloor\bigr\},\] and hence \[\Bigl\{q+1,\;(q+1)-2,\;\ldots,\;(q+1) - 2\bigl\lfloor\tfrac{m}{2}\bigr\rfloor\Bigr\} \subseteq H(P_{(a,b)}).\]
Step 1: \(H(P_{(a,b)}) \neq H(P_\infty^1)\).
\(m\) odd. Then \((q+1) - 2 \in H(P_{(a,b)})\), whereas \((q+1) - 2 \notin H(P_\infty^1)\), since \(2((q+1)-m) \ge q+1\).
\(m < (q+1)/2\) even. Then \((q+1) - m \in H(P_{(a,b)})\), but \((q+1) - m \notin H(P_\infty^1)\).
\(m = (q+1)/2\) even. We have \[(q+1) - 2\!\left(\frac{m}{2} - 1\right) = \frac{q+1}{2} + 2.\] Since \(m \neq 4\), this value lies strictly between \(\frac{q+1}{2}\) and \(q+1-\frac{m}{2}\), so it belongs to \(H(P_{(a,b)})\) but not to \(H(P_\infty^1)\).
Step 2: \(H(P_{(a,b)}) \neq H(P_0^1)\).
\(m\) odd. Then \((q+1) - (m-1) \in H(P_{(a,b)})\). Since \(m > 4\) implies \(m - 1 > \lfloor m/2 \rfloor\), we have \((q+1) - (m-1) \notin H(P_0^1)\).
\(m\) even. Then \((q+1) - (m-2) \in H(P_{(a,b)})\). Since \(m > 4\) implies \(m - 2 > \lfloor m/2 \rfloor\), we have \((q+1) - (m-2) \notin H(P_0^1)\).
Step 3: \(H(P_{(a,b)}) \neq H(P_{(c,d)})\) for \(P_{(c,d)} \notin \mathcal{O} \cup \mathcal{O}'\). The Weierstrass semigroup at the places of \(\mathcal{O}\) shows that the linear series \(|(q+1)P|\) has dimension \(\lfloor m/2 \rfloor + 2\) for every rational place \(P\). Applying 4 to \(Y, \ldots, Y^{\lfloor m/2 \rfloor}\) and using the maximality of the curve gives \[\Bigl\{q+1-\bigl\lfloor\tfrac{m}{2}\bigr\rfloor,\;\ldots,\;q+1\Bigr\} \subset H(P_{(c,d)}),\] so \(H(P_{(c,d)})\) has exactly one nonzero element smaller than \(q+1 - \lfloor m/2 \rfloor\).
\(m\) even. Applying 4 to \(Y^{m/2}\) and \(Y^{m/2-1}\) gives \(q+1-m,\, q+1-(m-2) \in H(P_{(a,b)})\); since \(m > 4\), both are smaller than \(q+1-\lfloor m/2 \rfloor\).
\(m\) odd, \(m \neq 5\). We obtain \(q+1-(m-1),\, q+1-(m-3) \in H(P_{(a,b)})\), both smaller than \(q+1-\lfloor m/2 \rfloor\).
\(m = 5\). Proposition 19 yields a rational function with \(T\)-valuation \(3\) at \(P_{(a,b)}\), giving \[\{q-3,\, q-1,\, q,\, q+1\} \subset H(P_{(a,b)}) \quad \text{and} \quad \{q-2,\, q-1,\, q,\, q+1\} \subset H(P_{(c,d)}).\]
In all cases \(H(P_{(a,b)}) \neq H(P_{(c,d)})\), completing the proof. ◻
Since \(\mathcal{X}_{m,q}\) is a maximal curve over \(\mathbb{F}_{q^2}\), by [11], its automorphism group \({\rm Aut}(\mathcal{X}_{m,q})\) is defined over \(\mathbb{F}_{q^2}\). In particular, the orbit of any \(\mathbb{F}_{q^2}\)-rational place under \({\rm Aut}(\mathcal{X}_{m,q})\) is contained in the set of \(\mathbb{F}_{q^2}\)-rational places. Moreover, places in the same orbit share the same Weierstrass semigroup [1]. Throughout this section we assume \(p \nmid \binom{m}{2+(m \bmod 2)}\), which provides sufficient information on the orbits of rational places to identify the automorphism group with the group given in Proposition 4 or 13, according to whether or not \(m = (q+1)/2\).
This case requires separate treatment: by Remark 16, we have \(H(P_{(a,b)}) = H(P)\) for any rational place with \(2a^{(q+1)/4} + 1 = 0\) and \(P \in \mathcal{O}_0 \cup \mathcal{O}_m\). As a finite number of automorphism groups may be checked by an algorithm, we therefore assume \(q > 31\) for the remainder of this subsection.
Lemma 4. \({\rm Aut}(\mathcal{X}_{4,q})\) has no nontrivial \(p\)-Sylow subgroup.
Proof. The results on Weierstrass semigroups from Sections 4 and 5 imply that \(\mathcal{O}_\infty\) is an orbit under \({\rm Aut}(\mathcal{X}_{4,q})\). By the Orbit-Stabilizer theorem, any nontrivial \(p\)-Sylow subgroup \(S\) partitions \(\mathcal{X}_{4,q}(\mathbb{F}_{q^2})\) into orbits of \(p\)-power size. Since \(p \nmid |\mathcal{O}_\infty| = \frac{q+1}{2}\), the subgroup \(S\) must fix some point of \(\mathcal{O}_\infty\). However, maximal curves have \(p\)-rank zero [12], and every automorphism of order \(p\) of a curve of \(p\)-rank zero has exactly one fixed point [12]. Since \(|\mathcal{O}_\infty| \not\equiv 1 \pmod{p}\), the subgroup \(S\) would have at least two fixed points in \(\mathcal{O}_\infty\), a contradiction. ◻
Although it is a priori possible for a place \(P_{(a,b)}\) with \(2a^{(q+1)/4} + 1 = 0\) to lie in the orbit of \(P_{(a,0)}\), the assumption \(q > 31\) rules this out.
Lemma 5. The orbit of \(P_{(a,0)}\) under \({\rm Aut}(\mathcal{X}_{4,q})\) is \(\mathcal{O}_0 \cup \mathcal{O}_m\).
Proof. Suppose for contradiction that some place \(P_{(a,b)}\) with \(2a^{(q+1)/4} + 1 = 0\) lies in the orbit of \(P_{(a,0)}\) (the Weierstrass semigroup data show this is the only other possibility). The orbit of \(P_{(a,b)}\) under \(G\) (Proposition 4) is \(\mathcal{O}' = \{P_{(\tilde{a},\tilde{b})} : 2\tilde{a}^{(q+1)/4} + 1 = 0\}\), so the orbit of \(P_{(a,0)}\) under the full automorphism group would consist of \(\frac{(q+1)(q+3)}{4}\) places.
By the Orbit-Stabilizer theorem applied to \(G\), we have \(|G_{P_{(a,0)}}| = q+1\). Since \(|{\rm Aut}(\mathcal{X}_{4,q})_{P_{(a,0)}}| \ge |G_{P_{(a,0)}}|\), applying Orbit-Stabilizer to the full automorphism group gives \(|{\rm Aut}(\mathcal{X}_{4,q})| \ge \frac{(q+1)^2(q+3)}{4}\). For \(q > 31\) this exceeds \(84(g-1)\), so [12] forces \(p \mid |{\rm Aut}(\mathcal{X}_{4,q})|\), contradicting Lemma 4. ◻
Proposition 21. Let \(G\) be as in Proposition 4. Then \({\rm Aut}(\mathcal{X}_{4,q}) = G\).
Proof. Suppose \(|{\rm Aut}(\mathcal{X}_{4,q})| > |G|\). By Lemma 5, \(\mathcal{O}_0 \cup \mathcal{O}_m\) is an orbit under \({\rm Aut}(\mathcal{X}_{4,q})\). Applying the Orbit-Stabilizer theorem to both \(G\) and \({\rm Aut}(\mathcal{X}_{4,q})\) shows that there exists \(\gamma \in {\rm Aut}(\mathcal{X}_{4,q})_{P_{(a,0)}} \setminus G_{P_{(a,0)}}\). Let \(C = \langle \sigma \rangle\), where \(\sigma \colon (x,y) \mapsto (x, \delta y)\) and \(\delta\) is a primitive \((q+1)\)-th root of unity. Since \(p \nmid |{\rm Aut}(\mathcal{X}_{4,q})|\), the stabilizer \({\rm Aut}(\mathcal{X}_{4,q})_{P_{(a,0)}}\) is cyclic [12], so \(\gamma\) commutes with \(C\). Hence for every point \(P\) fixed by \(C\), \[\label{sigma-gamma} \sigma^i(\gamma(P)) = \gamma(\sigma^i(P)) = \gamma(P),\tag{6}\] so \(\gamma\) permutes the fixed points of \(C\), namely \(\mathcal{O}_m\). Since \(\mathcal{O}_0 \cup \mathcal{O}_m\) and \(\mathcal{O}_\infty\) are both orbits under \({\rm Aut}(\mathcal{X}_{4,q})\), the automorphism \(\gamma\) preserves \(\mathcal{O}_0\) and \(\mathcal{O}_\infty\), and therefore preserves the divisors of \(x\) and \(y\). It follows that \(\gamma(x,y) = (\mu_1 x, \mu_2 y)\) for some \(\mu_1, \mu_2 \in \mathbb{F}_{q^2}^*\), giving \(\gamma \in G\), a contradiction. ◻
Remark 22. Regarding the remaining values of \(q\), the proprietary software Magma affirms that \(|{\rm Aut}(\mathcal{X}_{4,7})| = 192\), and that \({\rm Aut}(\mathcal{X}_{4,q}) = G\) (as in Proposition 4) for \(q \in \{11, 19, 23, 27, 31\}\).
Under these hypotheses, together with our assumption on the characteristic, the sets \(\mathcal{O}_\infty\) and \(\mathcal{O}_0 \cup \mathcal{O}_m\) are orbits under \({\rm Aut}(\mathcal{X}_{m,q})\).
Lemma 6. \({\rm Aut}(\mathcal{X}_{m,q})\) has no nontrivial \(p\)-Sylow subgroup.
Proof. By the same argument as in Lemma 4, if \({\rm Aut}(\mathcal{X}_{m,q})\) had a nontrivial \(p\)-Sylow subgroup \(S\), its unique fixed point would lie in \(\mathcal{O}_\infty\). Since \(\mathcal{O}_0 \cup \mathcal{O}_m\) is an orbit under the automorphism group, this would force \(p \mid |\mathcal{O}_0 \cup \mathcal{O}_m| = 2\,\frac{q+1}{m}\), a contradiction. ◻
The argument of Proposition 21 now carries over verbatim to give the following.
Proposition 23. Let \(G\) be as in Proposition 4. Then \({\rm Aut}(\mathcal{X}_{m,q}) = G\).
We show that \({\rm Aut}(\mathcal{X}_{m,q}) = G\) also when the characteristic of \(\mathbb{F}_q\) is even. The argument is similar to the previous cases, but requires adjustment since \({\rm Aut}(\mathcal{X}_{m,q})\) now admits a \(2\)-Sylow subgroup.
Lemma 7. Every nontrivial \(2\)-Sylow subgroup \(S\) of \({\rm Aut}(\mathcal{X}_{m,q})\) has order \(2\).
Proof. When \(p = 2\) we necessarily have \(m\) odd, so \(\mathcal{O}_\infty\) and \(\mathcal{O}_0 \cup \mathcal{O}_m\) are orbits under \({\rm Aut}(\mathcal{X}_{m,q})\). Since \(|\mathcal{O}_\infty| = \frac{q+1}{m}\) is odd, the unique fixed point of \(S\) lies in \(\mathcal{O}_\infty\), so \(S\) acts freely on \(\mathcal{O}_0 \cup \mathcal{O}_m\), giving \(|S| \mid |\mathcal{O}_0 \cup \mathcal{O}_m| = 2\,\frac{q+1}{m}\). Since \(\frac{q+1}{m}\) is odd, we have \(4 \nmid |\mathcal{O}_0 \cup \mathcal{O}_m|\), and hence \(|S| = 2\). ◻
Proposition 24. Let \(G\) be as in Proposition 4. Then \({\rm Aut}(\mathcal{X}_{m,q}) = G\).
Proof. Suppose \(|{\rm Aut}(\mathcal{X}_{m,q})| > |G|\) and let \(\gamma \in {\rm Aut}(\mathcal{X}_{m,q})_{P_{(a,0)}} \setminus G_{P_{(a,0)}}\). Since every nontrivial \(2\)-Sylow subgroup has order \(2\) and \(2 \mid |G|\), we may assume \(\gamma\) has odd order. Let \(C = \langle \sigma \rangle\) be as in Proposition 21. By [12] the tame part of \({\rm Aut}(\mathcal{X}_{m,q})\) is cyclic, so \(\gamma\) commutes with \(C\), and the remainder of the proof is identical to that of Proposition 21. ◻
As noted at the end of Section 4, \(\mathcal{O}\) is contained in an orbit under \({\rm Aut}(\mathcal{X}_{(q+1)/2,q})\). The single case \((m, q) = (4, 7)\) may be checked by an algorithm, so we assume \(q \ge 11\). By the general-case discussion of the preceding section, \(\mathcal{O}\) is itself an orbit under this group.
Proposition 25. Let \(G\) be as in Proposition 13. If \(q \equiv 3 \pmod{4}\) and \(q \ge 11\), then \({\rm Aut}(\mathcal{X}_{(q+1)/2,q}) = G\).
Proof. We first show \(p \nmid |{\rm Aut}(\mathcal{X}_{(q+1)/2,q})|\) by an argument analogous to the case \(m < \frac{q+1}{2}\). By Proposition 20, the set \(\mathcal{O}' = \{P_{(a,b)} : 2a^2 + 1 = 0\}\) is an orbit under \({\rm Aut}(\mathcal{X}_{(q+1)/2,q})\) of cardinality \(2(q+1) \equiv 2 \pmod{p}\). Thus a nontrivial \(p\)-Sylow subgroup of \({\rm Aut}(\mathcal{X}_{(q+1)/2,q})\) would have at least two fixed points in \(\mathcal{O}'\), contradicting [12].
The rest follows as in Proposition 21. Assuming \(|{\rm Aut}(\mathcal{X}_{(q+1)/2,q})| > |G|\), we obtain \(\gamma\) acting on \(\mathcal{O}_m\). Since \(\mathcal{O}_0\) and \(\mathcal{O}_\infty\) are orbits under \(\langle \sigma \rangle\) (where \(\sigma \colon (x,y) \mapsto (x, \delta y)\) with \(\delta\) a primitive \((q+1)\)-th root of unity) and \(\mathcal{O}\) is an orbit under \({\rm Aut}(\mathcal{X}_{(q+1)/2,q})\), an analysis analogous to 6 shows that \(\gamma\) either fixes or interchanges \(\mathcal{O}_0\) and \(\mathcal{O}_\infty\). Since these sets have different cardinalities, \(\gamma\) fixes each of \(\mathcal{O}_0\), \(\mathcal{O}_\infty\), and \(\mathcal{O}_m\), and the argument of Proposition 21 gives \(\gamma \in G\), a contradiction. ◻
João Paulo Guardieiro was partially supported by FAPESP (Brazil), grant no./19443-4. Yuri da Silva was partially supported by CAPES (Brazil) – Finance Code 001. Saeed Tafazolian was partially supported by CNPq grant no./2025-4, FAEPEX grant no./25, and FAPESP grant no./00923-6. The authors thank Daniel Cariello for his assistance with the rank computation for the matrix in 5 .