Projective symmetries of Chebyshev polynomials

Introduction

For an integer $d\geq1$, let $\varphi_d(X)\in\mathbb{Z}[X]$ be the monic Chebyshev polynomial determined by

Thus $\varphi_0=2$, $\varphi_1=X$, and $\varphi_{d+1}=X\varphi_d-\varphi_{d-1}$. The polynomial $\varphi_d$ is also known as the Dickson polynomial $D_d(X,1)$; see [6] for background on Dickson polynomials.

Let $k$ be an algebraically closed field of odd characteristic $p$, and assume that $p\nmid d$. Denote by

the zero set of $\varphi_d$, and put

Since $\varphi_d(-X)=(-1)^d\varphi_d(X)$, the involution $\sigma:X\mapsto-X$ belongs to $H_d$.

For comparison in characteristic zero, Mosunov [8 Thm. 1.3] computed the rational linear automorphism groups of the homogenized Chebyshev forms of the first and second kinds. For the first kind, he obtains a cyclic group of order 2 when the degree is odd and a dihedral group $D_2$ of order 4 when the degree is even; if transformations sending the form to its negative are also allowed, the resulting group is $D_2$ in both cases. This is a different notion from the projective setwise stabilizer considered here, but it gives a useful characteristic-zero comparison.

The main result proven in the present manuscript is as follows.

Denote by $D_d$ the dihedral group of order $2d$. The two exceptional cases are described in Propositions 5.2 and 5.3, namely $H_d\cong D_d$ if $2d=p^r+1$, and $H_d\cong S_3$ if $4d=p^r+1$. We consider the result and the techniques used in the proof interesting by themselves; besides this, they can be used to describe the automorphism groups of hyperelliptic curves defined by $y^2=\varphi_d(x)$ (at least when $d$ is even). In our manuscript [10] we use the knowledge of $H_d$ in every relevant characteristic, to determine the automorphisms of the plane degree $d$ curves defined by $y^d=\varphi_d(x)$ for all pairs $(\operatorname{char}(k),d)$ such that $\operatorname{char}(k)\nmid2d$.

The paper is organized as follows. Section 2 recalls some basic facts about $\varphi_d$, including a short elementary proof of its separability. Section 3 contains two consequences of Lucas’ theorem. In Section 4 the stabilizer in $\operatorname{Aut}(\mathbb{P}^1)$ of the set of zeros of $\varphi_d$ is studied. The main results will be proven in Section 5.

Chebyshev polynomials

A derivative identity and separability

A consequence of the relation (1) defining $\varphi_d$ is the following.

Proof. The evaluation map

is injective. Differentiating (1) with respect to $t$ and multiplying by $t$ gives

Squaring this and applying the identities

and

completes the proof. □

Separability of $\varphi_d(X)\in k[X]$ whenever $\operatorname{char}(k)\nmid2d$ is well known; see for example [3 Thm. 6.1(b)], [5 Thm. 7.2(a)], [9 Lemma 2.5], [1 Rem. 6.2]. The preceding identity implies it as follows.

Proof. Suppose that $\alpha$ were a multiple zero of $\varphi_d$. Then $\varphi_d(\alpha)=\varphi'_d(\alpha)=0$. Considering (2) over $k$, evaluation at $\alpha$ gives

which is impossible because $p$ is odd and $p\nmid2d$. Hence $\varphi_d$ is separable. Since $k$ is algebraically closed and $\deg\varphi_d=d$, the polynomial has exactly $d$ distinct zeros in $k$. □

Coefficients

Write

In particular,

The differential equation

Differentiating (2) in $\mathbb{Z}[X]$ gives

Since $\varphi'_d$ is not the zero polynomial, one obtains

Lucas' theorem and a relation between certain polynomials

Recall the following result by É. Lucas [7].

This will now be used to translate the property $2d\in1+p^{\mathbb{Z}_{\geq1}}$ into a statement about a certain polynomial $P_d(Z)$ in characteristic $p$. For every integer $d>0$ consider

where $k$ is an algebraically closed field of characteristic $p\geq3$.

Proof. $\Leftarrow$: Given is that $2d=1+p^r$ for some $r\geq1$. Applying Theorem 3.1 with $N=2d$ and $M=2j\leq2d$ yields $P_d(Z)=1+Z^{(1+p^r)/2}$. Choosing $a\in k$ of order dividing $(1+p^r)/2$ and $a\ne1$ completes the proof (note that the assumption that $k$ is algebraically closed, or at least contains an algebraic closure of $\mathbb{F}_p$, is used here to guarantee that $k$ contains the required roots of unity, for every $r\geq1$).

$\Rightarrow$: Consider the base $p$ expansion $2d=\sum N_i p^i$ with $0\leq N_i<p$ for all $i$. We start by showing that $N_i\in\{0,1\}$ for all $i$, and $N_0=1$. Indeed, were $N_i\geq2$ for some $i\geq0$ then Theorem 3.1 applied to $N=2d$ and $M=2p^i$ shows $\binom{2d}{2p^i}\ne0$ in $k$, so $a^{p^i}=1$ which results in the contradiction $a=1$. Thus every base $p$ digit of $2d$ is $0$ or $1$. Moreover, since $p\nmid2d$ implies $N_0\ne0$, one concludes $N_0=1$.

Next, we show that besides $N_0$, exactly one more base $p$ digit of $2d$ equals $1$. As $2d$ is even and $p$ is odd, the number of base $p$ digits $N_r$ equal to $1$, is even. And were $N_r=N_s=1$ with $0<r<s$, then each of $1+p^r$, $1+p^s$ and $p^r+p^s$ are even and $<2d$. Theorem 3.1 shows that the coefficients $\binom{2d}{1+p^r}$, $\binom{2d}{1+p^s}$ and $\binom{2d}{p^r+p^s}$ in $P_d\in k[Z]$ are nonzero, hence $a^{(1+p^r)/2}=a^{(1+p^s)/2}=a^{(p^r+p^s)/2}=1$. This implies

a contradiction. Hence exactly one higher base $p$ digit of $2d$ equals $1$ and $2d=1+p^r$. □

Proof. The hypothesis says that

Since $d>1$ and $p\nmid d$, a nontrivial $d$-th root of unity $a\in k^\times$ exists. Then $P_d(aZ)=P_d(Z)$, and Lemma 3.2 shows $2d=1+p^r$ for some $r\geq1$. □

Relating $\varphi_d$ and $P_d$. The relevance of the polynomials $P_d$ towards our result, starts as follows.

Proof. Introduce an indeterminate $R$ and substitute $Z=R^2$. Put

which implies $T+T^{-1}=2\frac{1+R^2}{1-R^2}$. By (1) it follows that

finishing the proof.

Finite projective stabilizers

In this section $k$ denotes an algebraically closed field of characteristic $p\ge0$, and $d\ge3$ is an integer. Throughout, it is assumed that $p\nmid2d$. By Lemma 2.2, these conditions imply that the set $Z_d$ of zeros of the Chebyshev polynomial $\varphi_d$ in $k$ satisfies $\#Z_d=d\ge3$. We study the group

Restriction to $Z_d$ gives a homomorphism

Elements in its kernel fix every point of $Z_d$, hence at least three distinct points of $\mathbb{P}^1(k)$. Thus the homomorphism is injective and $H_d$ is a finite group. As already observed in Section 1, the equality $\varphi_d(-X)=(-1)^d\varphi_d(X)$ shows that

is an element of order 2 in $H_d$.

Section 6.5.2]) results in

Proof. Assume that $\sigma$ is the unique involution of $H_d$. For every $g \in H_d$, the conjugate $g\sigma g^{-1}$ is again an involution, hence it equals $\sigma$. Thus

the centralizer in $\operatorname{Aut}(\mathbb{P}^1)$ of $\sigma$. This centralizer consists of the maps

Every map of the second type is an involution, so by assumption is not in $H_d$. Hence every $g \in H_d$ has the form $X \mapsto aX$.

Such $g$ preserves $\mathcal{Z}_d$, hence the monic and separable polynomials $\varphi_d(X)$ and $a^{-d}\varphi_d(aX)$ are equal, hence

Comparing the coefficients of $X^{d-2}$ and using (4) gives $a^2 = 1$, because $p \nmid d$. Therefore $H_d = \langle\sigma\rangle$.

Assume now that $H_d \supsetneq\langle\sigma\rangle$. By Lemma 4.2, $H_d$ contains an involution $\tau\ne\sigma$. Put

Then $\mu\in H_d$ and

Note that $\mu\ne id$ since $\sigma\ne\tau$, and $\mu\ne\sigma$ since $\tau\ne id$.

Proof. Write $u := \mu(\infty) \in\mathbb{P}^1$. If $u = \infty$, then $\mu(X) = aX + b$ for certain $a,b \in k$. Since $\tau(X) = -aX-b$ is an involution, we have $a^2 = 1$ and $(a-1)b = 0$. If $a=-1$, then $b=0$ because the characteristic is not 2, and hence $\tau= id$, a contradiction. Thus $a=1$. Hence $\mu(X) = X+b$.

Since $\mu\in H_d$, the polynomials $\varphi_d(X+b)$ and $\varphi_d(X)$ are monic of degree $d$ and separable, with the same zero set. Hence the two polynomials are equal. Comparing the coefficients of $X^{d-1}$ gives $db=0$, implying $b=0$, a contradiction. Thus $u \ne\infty$, i.e., $u \in k$.

The equality $\sigma\mu\sigma= \mu^{-1}$ and the fact that $\mu^{-1}(u) = \infty$ show that $\mu(-u) = \infty$. Hence $\mu$ is given as

for some $s \in k^\times$. From $\mu(\mu(-u)) = u$ it follows that $\mu$ is an involution if and only if $u = 0$.

The root identity

Let $\mu\in H_d$ be as before, given as in (7). Since $\mu^{-1}$ preserves $\mathcal{Z}_d$, we have $\varphi_d(u) \ne0$: indeed, the proof of Lemma 4.3 shows $\mu^{-1}(u) = \infty\notin\mathcal{Z}_d$. Write the $d$ zeros of $\varphi_d$ as $\alpha_1,\ldots,\alpha_d$. Since $\mu^{-1}$ permutes them and $\mu^{-1}(\alpha_i)=-u-\frac{s}{u-\alpha_i}$, one computes

i.e.,

A logarithmic-differential identity

Set

Thus $\omega=-\frac{1}{2}\mathrm{d}\log(X^2-4)$. The differential $\omega$ has simple poles at $2$ and $-2$, both of residue $-1/2$, and a simple pole at $\infty$ of residue $1$.

Proof. Using (8), put

Let $\beta$ be a zero of $\varphi_d$. By Lemma 2.2, $\varphi_d'(\beta)\ne0$. Moreover $\beta\ne\pm2$, since $\varphi_d(2)=2$ and $\varphi_d(-2)=2(-1)^d$. Also, $\beta\ne-u$ because $\mu$ sends $-u$ to $\infty$ whereas $H_d$ preserves $\mathbb{Z}_d\ni\beta$. The same remarks apply with $\beta$ replaced by $\mu(\beta)$.

Evaluating the differential equation (5) at a zero $\alpha$ of $\varphi_d$ gives

At $X=\beta$, differentiation of $H_\mu(X)=(X+u)^d\varphi_d(\mu(X))$ gives

As $H_\mu=\varphi_d(u)\varphi_d$ and $\mu'(X)=-s/(X+u)^2$ and $\mu''(X)/\mu'(X)=-2/(X+u)$, from (12) one obtains

Put $t=X+u$, $B=ut+s$, and $Q=B^2-4t^2$. Since

we have

Consequently,

The numerator has degree at most 4. For every $\beta\in\mathcal{Z}_d$, the denominator is nonzero: indeed, $\beta\ne\pm2$, $\beta\ne-u$, and $\mu(\beta) \in\mathcal{Z}_d$ implies $\mu(\beta) \ne\pm2$, so $Q(\beta) \ne0$. By (13), the numerator therefore vanishes at all $d \ge5$ distinct points of $\mathcal{Z}_d$. Hence it is identically zero, proving (10).

$\square$

Remark 4.5. The differential form (11) makes the local information transparent. Since $\mu(-u)=\infty$ and $\mu$ is a Möbius transformation, residues are preserved under pullback. Indeed, if $t$ is a local parameter at a point $P$ and $z=ct+O(t^2)$, $c\ne0$, is a local parameter at $\mu(P)$, then the coefficient of $dt/t$ is unchanged. Hence

Thus the total residue of the right-hand side at $X=-u$ is $2d-1$. Furthermore, whenever $\mu^{-1}(2)$ or $\mu^{-1}(-2)$ is finite, it is a genuine simple pole of $\mu^*\omega$, with residue $-1/2$. Neither of these points equals $-u$, since $\mu(-u)=\infty$. Hence these poles cannot be cancelled by the term $2(d-1)\,dX/(X+u)$, whose only possible finite pole is at $X=-u$.

Proof. Introduce $Y$ with $Y^2=X+2$ and put

Then $T+T^{-1}=h(X)$, and the defining identity for $\varphi_d$ gives

Now substitute

Since $Y+2=(Z+Z^{-1})^2$ and $Y-2=(Z-Z^{-1})^2$, we obtain

On the other hand,

The substitution $k[X]\hookrightarrow k[Z,Z^{-1}]$, $X\mapsto Z^4+Z^{-4}$, is injective. Therefore $F_d=\varphi_d$ if and only if

Expanding the left-hand-side shows that this is equivalent to

Applying Lemma 3.3 with $2d$ in place of $d$, we obtain $4d=1+p^r$ for some $r\ge1$.

Conversely, if $4d=1+q$ with $q=p^{r}$, then one computes

showing $F_d=\varphi_d$.

The case $u=0$.

Proposition 4.7. Assume that $k$ has characteristic $p$, that $d\geq5$ and $p\nmid2d$. If $u=0$, then $2d=1$ in $k$ and $s=\pm4$ and $H_d$ contains a nontrivial element fixing both $2$ and $-2$.

Proof. When $u=0$, the differential $\omega$ is regular at $X=0$. By Remark 4.5, the residue of the right-hand-side of (11) at $X=0$ is $2d-1$. Hence $2d=1$ in $k$. Both terms on the right in (11) have at most simple poles at $0$, so this residue cancellation makes the singularity at $0$ removable. Note that $u=0$ means that $\mu$ is given by $\mu(X)=s/X$. The remaining finite poles of $\mu^{*}\omega$ are the genuine simple poles

each of residue $-1/2$. The left-hand-side $-X\mathrm{d}X/(X^{2}-4)$ has finite poles precisely at $2$ and $-2$. Therefore $\{s/2,-s/2\}=\{2,-2\}$, so $s=\pm4$. Hence one of $\mu$ and $\sigma\mu$ equals $X\mapsto4/X$, which is nontrivial and fixes $2$ and $-2$.

The endpoint case $u^{2}=4$.

Since $\mu^{-1}(X)=-u+s/(X-u)$, replacing $\mu$ by $\mu^{-1}$ changes $u$ to $-u$ and leaves $s$ unchanged. Thus we may assume $u=2$.

Proposition 4.8. Assume that $k$ has characteristic $p$ and $d\geq5$ and $p\nmid2d$, and suppose $u^{2}=4$. Then, after replacing $\mu$ by $\mu^{-1}$ if necessary, $4d=1$, $s=-16$ in $k$, so $\mu^{-1}(X)=(2X+12)/(2-X)$. In $\mathbb{Z}$, it follows that $4d=1+p^{r}$ for some $r\geq1$.

Proof. Take $u=2$. The differential $\omega$ has residue $-1/2$ at $X=-2$. By Remark 4.5, the residue of the right-hand side of (11) at $-2$ is $2d-1$. In $k$ this implies $2d-1=-1/2$, hence $4d=1$.

Since $\mu(\infty)=u=2$, the inverse image of the pole $2$ of $\omega$ is $\infty$ and $\mu^{-1}(-2)$ is finite and is a genuine simple pole of $\mu^{*}\omega$ of residue $-1/2$. It cannot equal $-2$ because $\mu(-2)=\infty$. Thus comparison with the two finite poles $\{2,-2\}$ of $\omega$ forces $\mu^{-1}(-2)=2$, or equivalently $2+s/4=\mu(2)=-2$, implying $s=-16$. Therefore

Applying (8) to $\mu^{-1}$ gives

Since $(2-X)^{d}=(-1)^{d}(X-2)^{d}$, this is equivalent to

Lemma 4.6 therefore yields $4d=1+p^{r}$ for some $r\geq1$.

The generic case.

Proposition 4.9. Assume that $k$ has characteristic $p$ and $d\geq5$ and $p\nmid2d$ and $u\notin\{0,\pm2\}$. Then $2d=1$, $s=4-u^{2}$ in $k$. In particular, $\mu$ fixes both $2$ and $-2$.

Proof. Since $u^{2}\neq4$, the differential $\omega$ is regular at $X=-u$. By Remark 4.5, the residue of the right-hand side of (11) is $2d-1$. Hence $2d=1$ in $k$.

The two terms on the right in (11) have at most simple poles at $-u$, so the singularity there is removable. Moreover $\mu(\infty)=u\notin\{2,-2\}$, so both $\mu^{-1}(2)$ and $\mu^{-1}(-2)$ are finite. By Remark 4.5 they are distinct, genuine simple poles of $\mu^*\omega$, each with residue $-1/2$. Comparison with the finite poles of $\omega$ gives

If $\mu(2)=-2$ then $\mu(-2)=2$ and $(u+2)^2=-s=(u-2)^2$. Thus $8u=0$, impossible because the characteristic is odd and $u\ne0$. Hence $\mu$ fixes both $2$ and $-2$, implying $s=4-u^2$.

Proof of the main theorem

Proof of Theorem 1.1. Recall that $\sigma:X\mapsto-X$ belongs to $H_d$, and assume that $H_d\supsetneq\langle\sigma\rangle$. By Lemma 4.2, $H_d$ contains an involution $\tau\ne\sigma$; put $\mu:=\sigma\tau$ so that $\mu\ne id$ and $\sigma\mu\sigma=\mu^{-1}$. By Lemma 4.3, there exist $u\in k$ and $s\in k^\times$ such that $\mu(X)=u+s/(X+u)$. We distinguish three cases.

• If $u=0$, then Proposition 4.7 shows that $H_d$ contains a nonidentity element fixing both $2$ and $-2$. Hence Proposition 3.5 gives $2d=1+p^r$ for some $r\ge1$.

• If $u^2=4$, Proposition 4.8 shows $4d=1+p^r$ for some $r\ge1$.

• Finally, suppose that $u\ne0$ and $u^2\ne4$. Then Proposition 4.9 shows $2d=1$, $s=4-u^2$ in $k$, and $\mu\ne id$ fixes both $2$ and $-2$. Thus Proposition 3.5 gives $2d=1+p^r$ for some $r\ge1$.

Therefore

We now prove the converse. If $2d=p^r+1$, Proposition 3.5 produces a nonidentity element of $H_d$ fixing $2$ and $-2$. Since such an element is not in $\langle\sigma\rangle$, one concludes $H_d\supsetneq\langle\sigma\rangle$.

If $4d=p^r+1$, Lemma 4.6 implies that $\mu$ defined by $\mu(X):=(2X+12)/(2-X)$ is in $H_d$. Hence also in this case $H_d\supsetneq\langle\sigma\rangle$.

This completes the proof.

Remark 5.1. The same argument also gives the characteristic-zero case. If $k$ is algebraically closed of characteristic zero and $d>4$, then $H_d=\langle\sigma\rangle$. Indeed, a second involution would lead, according as $u=0$, $u^2=4$, or $u\ne0$ and $u^2\ne4, to the equality \$2d=1$ or $4d=1$ in $k$, which is impossible in characteristic zero.

Denote by $D_d$ the dihedral group of order $2d$.

Proposition 5.2. Assume that $k$ has characteristic $p$ and $d\ge5$ and $2d=p^r+1$ for some $r\ge1$. Then $H_d\cong D_d$.

Proof. Put $q=p^r$. By Lemma 3.4 and the proof of Proposition 3.5, the coordinate $Z=(X-2)/(X+2)$ identifies $\mathcal{Z}_d$ with the zero set of $1+Z^d$. Choose $\eta\in k^\times$ with $\eta^d=-1$. After the further scaling $Z\mapsto\eta^{-1}Z$, one obtains that $H_d$ is conjugate to

The group $K$ contains

and the involution $\iota:Z\mapsto Z^{-1}$. Together they generate a subgroup $\cong D_d$ of $K$. We now show that $K$ is exactly this dihedral subgroup.

First, $p\nmid\#K$, since otherwise an element $g\in K$ of order $p$ exists. Such an element has exactly one fixed point in $\mathbb{P}^1(k)$; consequently its action on the $d$-element set $\mu_d$ has orbits of length $p$ in addition to at most one fixed point. Thus $d\equiv0$ or $1\pmod p$. On the other hand $2d=q+1$ gives $2d\equiv1\pmod p$, contradicting the above. Hence $p\nmid\#K$.

By the classification (originally due to L.E. Dickson) of finite subgroups of $\operatorname{PGL}_2$ [4 Thm. A.8], the noncyclic group $K$ is either dihedral or one of $A_4,S_4,A_5$. Since $K$ contains an element of order $d\ge5$, the groups $A_4$ and $S_4$ are impossible. If $K\cong A_5$, the existence of an element of order $d\ge5$ in $K$ forces $d=5$. But then $q=2d-1=9$, so $p=3$, contradicting $p\nmid|A_5|$. Hence $K$ is dihedral, say of order $2n$.

Let $C_n$ be its cyclic subgroup of index two. Since every element outside $C_n$ has order $2$ and $R$ contains an element of order $d\ge5$, we have $R\le C_n$. Thus $C_n$ centralizes $R$. A generator $Z\mapsto\zeta Z_久久爱 of $R$ has $\zeta\ne\pm1$, and its centralizer in $\operatorname{PGL}_2(k)$ consists of the scalings $Z \mapsto aZ$. Therefore every element of $C_n$ is a scaling. A scaling preserving $\mu_d$ must satisfy $a^d=1$, so $C_n=R$. Hence $n=d$ and $K \cong D_d$, finishing the proof. □

Acknowledgments

The first author was partially supported by the Brazilian funding agencies CNPq (grant 302774/2025-4), FAPESP (grant 2024/00923-6) and FAEPEX (grant 3485/25).

Declaration on the use of generative AI

The problem addressed in this paper arose from the authors’ earlier work on automorphisms of Chebyshev plane curves. The first author used Claude Opus 5.5 as an aid in developing the argument presented in Lemma 4.4. The authors subsequently reviewed and revised the argument and take full responsibility for the mathematical content of the paper.

Departamento de Matemática - Instituto de Matemática, Estatística e Computação Científica (IMECC) - Universidade Estadual de Campinas (UNICAMP), Rua Sérgio Buarque de Holanda, 651, Cidade Universitária, Zeferino Vaz, Campinas, SP 13083-859, Brazil

Email address: [email protected]

Bernoulli Institute for Mathematics, Computer Science, and Artificial Intelligence, Nijenborgh 9, 9747 AG Groningen, the Netherlands

Email address: [email protected]