The tangent line at a marked point on $M_{0,n}/\mathfrak S_{n-1}$, after a table of Grothendieck

Introduction

Let $k$ be a field of characteristic zero and $n \ge3$. Let $M_{0,n}$ be the moduli space of $n$ distinct ordered points $x_1,\ldots,x_n$ on $\mathbb{P}^1$, up to $\operatorname{PGL}_2$. For each $i$ the tangent lines $T_{x_i}\mathbb{P}^1$ form a line bundle $L_i$ on $M_{0,n}$, and we put $L=L_1\otimes\cdots\otimes L_n$. The symmetric group $\mathfrak{S}_n$ acts on $M_{0,n}$ by renumbering the points and acts compatibly on $L$; the subgroup $\mathfrak{S}_{n-1}$ fixing the last point acts on $L_n$. This defines classes

and the question is their order: the least $e\ge1$ for which $L^{\otimes e}$, or $L_n^{\otimes e}$, has a nowhere-vanishing section compatible with the action.

The source

The University of Montpellier holds the mathematical papers that Alexandre Grothendieck left to Jean Malgoire, and publishes their facsimiles. Page 70 of its folder 146, Groupoïdes de Teichmüller, has a table which gives, for $n=4,5,6$, the best exponent that makes $L$ and the $L_i$ trivial compatibly with the symmetric group. Its first row gives $(n-1)(n-2)=6,12,20$. In the column $n=4$ it gives 3 for $L$ and 6 for $L_i$, and both are right. The columns $n=5$ and $n=6$ hold only queried values: $12?$ for both bundles when $n=5$, and $20?$ for $L$ and $60?$ for $L_i$ when $n=6$. Below the table, a computation for $n=5$ ends with $(L^{\otimes4})^3\simeq\mathcal{O}$, that is, $L^{\otimes12}$ is trivial. Page 72 builds sections indexed by the three-element subsets of the points. For $n=4$ it writes $L^{\otimes6}\simeq1$, which is true and weaker than the order 3. For $n=5$ it writes $L^{\otimes6}\simeq1$ again (our transcription marks the words before the formula as an uncertain reading); this is false, since by Theorem 1 $L$ then has order 4. The theorem completes the table. For $n=5$ the queried 12 is right for $L_5$, but it is not minimal for $L$, whose order is

4. For $n=6$ the orders are 5 and 20, not 20 and 60. In every column the order of $L_n$ is the number in the first row.

	$n=4$		$n=5$		$n=6$
	$L$	$L_n$	$L$	$L_n$	$L$	$L_n$
Page 70	3	6	12?	12?	20?	60?
Theorem 1	3	6	4	12	5	20

What was known

The stack $[M_{0,n}/S_n]$ is the stack of smooth binary forms of degree $n$ up to scalars, modulo $\operatorname{PGL}_2$. Arsie and Vistoli computed the Picard groups of stacks of smooth cyclic covers of projective spaces from such presentations by forms [1 Th. 5.1]. For $n=2g+2$ with $g\geq2$, Gorchinskiy and Viviani identify $[M_{0,n}/S_n]$ with the stack $\mathcal{D}_{2g+2}$ of $\mathbb{P}^1$-bundles with an étale divisor of degree $2g+2$, and find $\operatorname{Pic}\mathcal{D}_{2g+2}\cong\mathbb{Z}/2(n-1)$ [5 Prop. 3.4, Th. 3.6]. The order of $[L]$ follows from these methods in a few lines (Remark 3), but we have not found it stated. For $n=2g+2$ and $g\geq2$, Edidin and Hu compute the integral Chow ring of $[M_{0,2g+2}/S_{2g+1}]$, which they present as binary forms with one marked root, as $\mathbb{Z}[l]/(2g(2g+1)l)$ [3 Th. 1.3], building on Pernice’s description of the stack of 1-pointed hyperelliptic curves [8]. So $\operatorname{Pic}[M_{0,n}/S_{n-1}]$ is cyclic of order $(n-1)(n-2)$ for even $n\geq6$. Their generator is written in terms of characters of a torus [3 Th. 5.8], and its identification with $L_n$ is not stated. For $n=4$, sending an elliptic curve $E$ to $E/\{\pm1\}\cong\mathbb{P}^1$, with the origin as last point, makes $\mathcal{M}_{1,1}$ a $\mu_2$-gerbe over $[M_{0,4}/S_3]$. Pulling back along it is injective on Picard groups, and its image consists of the classes on which $\mu_2$ acts trivially. Since $\operatorname{Pic}\mathcal{M}_{1,1}=\mathbb{Z}/12$ is generated by the Hodge bundle $\lambda$, on which $-1$ acts by $-1$ [7, 4], the image is $2\mathbb{Z}/12\cong\mathbb{Z}/6$, and $L_4$ pulls back to $T_0E^{\otimes2}\cong\lambda^{-2}$, of order 6. For odd $n$ we found no printed statement. D. Edidin confirms that their paper treats only the even case and that he knows no reference for the odd one [2]. The author asked the question on MathOverflow (question 516243) and posted the argument of Section 2 as an answer.

The stack as polynomials modulo the affine group

Put $m=n-1\geq2$. Since $\operatorname{PGL}_2$ acts transitively on $\mathbb{P}^1$, we may send the last point to $\infty$; its stabiliser is the affine group $\operatorname{Aff}=\mathbb{G}_a\rtimes\mathbb{G}_m$, acting on $\mathbb{A}^1$ by $x\mapsto ax+b$, and the other $m$ points, unordered, are the roots of a monic squarefree polynomial of degree $m$. Taking the slice $x_n=\infty$ gives an isomorphism of stacks

where $U_m\subset\mathbb{A}^m$ is the open set of monic squarefree polynomials $x^m+c_1x^{m-1}+\cdots+c_m$, on which $\operatorname{Aff}$ acts by substitution (Figure 1).

The translations $\mathbb{G}_a$ act freely, and every orbit meets the closed set $c_1=0$ exactly once, since $m$ is invertible in $k$: replacing $f(x)$ by $f(x-c_1/m)$ centres the roots. So

where $t\in\mathbb{G}_m$ acts by $x\mapsto tx$, that is, on the coefficient $c_i$ with weight $i$. The complement of $V_m$ in $\mathbb{A}^{m-1}$ is the discriminant hypersurface $\operatorname{disc}=0$. As a function of the roots, $\operatorname{disc}=\prod_{i<j}(r_i-r_j)^2$ is homogeneous of degree $m(m-1)$, so it is a semi-invariant of weight $m(m-1)$. It is irreducible in $k[c_2,\ldots,c_m]$. Indeed, the discriminant of the generic monic polynomial is irreducible in $k[c_1,\ldots,c_m]$ and invariant under translation of the roots. Since $m$ is invertible, $k[c_1,\ldots,c_m]=k[c_1,c'_2,\ldots,c'_m]$, where the $c'_j$ are the coefficients of the centred polynomial $f(x-c_1/m)$, and the discriminant is the same polynomial in $c'_2,\ldots,c'_m$, so it stays irreducible there.

Proof. Since $V_m$ is open in an affine space, $\operatorname{Pic} V_m=0$, so a $\mathbb{G}_m$-linearised line bundle on $V_m$ is the trivial bundle with a linearisation. A linearisation is a unit $\phi(t,x)$ of $\mathcal{O}(\mathbb{G}_m \times V_m)=\mathcal{O}(V_m)[t^{\pm1}]$ satisfying the cocycle rule. As $\operatorname{disc}$ is irreducible, the units of $\mathcal{O}(V_m)$ are the $c\operatorname{disc}^{j}$ with $c\in k^\times$, so $\phi=c\operatorname{disc}(x)^j t^a$, and $\phi(1,x)=1$ forces $c=1$ and $j=0$. So every linearisation is a character $t\mapsto t^a$. Two characters $t^a$ and $t^b$ give isomorphic bundles exactly when there is a unit $u$ on $V_m$ with $u(t\cdot x)=t^{b-a}u(x)$, that is, a semi-invariant unit of weight $b-a$. The units $c\operatorname{disc}^{j}$ have weights $jm(m-1)$. Hence $\operatorname{Pic}[V_m/\mathbb{G}_m]=\operatorname{Pic}^{\mathbb{G}_m}(V_m)\cong\mathbb{Z}/m(m-1)$, generated by the character of weight $1$.

Proof of Theorem 1. In the model $[U_m/\operatorname{Aff}]$ the last point is $\infty$. In the coordinate $u=1/x$ at $\infty$, the map $x\mapsto ax+b$ is $u\mapsto u/(a+bu)$, whose derivative at $u=0$ is $a^{-1}$; so $\mathbb{G}_m$ acts on $T_\infty\mathbb{P}^1$ through the character of weight $-1$, and $[L_n]$ corresponds to $-1\in\mathbb{Z}/m(m-1)$. It generates the group, which proves (1).

At a finite root $r$, the derivative of $x\mapsto ax+b$ is $a$, so each $T_{r_i}\mathbb{P}^1$ has weight $+1$, and the pull-back of $L$ to $[M_{0,n}/\mathfrak{S}_{n-1}]$ has weight $m-1=n-2$. Its order in $\mathbb{Z}/(n-1)(n-2)$ is $n-1$, so the order of $[L]$ is a multiple of $n-1$. Conversely, write the points as vectors $v_i\in k^2\setminus\{0\}$ and $[v_i v_j]=\det(v_i,v_j)$. Since $T_{\langle v\rangle}\mathbb{P}^1\cong\operatorname{Hom}(\langle v\rangle,k^2/\langle v\rangle)\cong\langle v\rangle^{\otimes-2}\otimes\det k^2$, and since the centre of $\operatorname{GL}_2$ acts trivially on $\langle v\rangle^{\otimes-2}\otimes\det k^2$, descent along the $\mathbb{G}_m^n$-torsor of vectors over configurations shows that trivialising $L^{\otimes(n-1)}$ compatibly with the action amounts to giving a nowhere-vanishing function of the $v_i$, homogeneous of degree $2(n-1)$ in each $v_i$, transforming under $g\in\operatorname{GL}_2$ by $\det(g)^{n(n-1)}$ and invariant under renumbering. The function

is such a function: each $v_i$ occurs in $2(n-1)$ brackets, it vanishes nowhere on $M_{0,n}$, and the right-hand side is invariant under $\mathfrak{S}_n$. This proves (2).

An elementary version

The argument can be phrased without stacks, in the vector model used above. In the same way, a trivialisation of $L_n^{\otimes e}$ compatible with $\mathfrak{S}_{n-1}$ is a nowhere-vanishing function $F(v_1,\ldots,v_n)$ which is homogeneous of degree $2e$ in $v_n$ and of degree $0$ in the other $v_i$, transforms by $\det(g)^e$ under $\mathrm{GL}_2$, and is symmetric in $v_1,\ldots,v_{n-1}$.

Upper bound. The function

has these properties for $e=(n-1)(n-2)$.

Lower bound. A trivialisation over $k$ gives one over an algebraic closure, so we may assume that $k$ contains the roots of unity used below. Let $\omega$ be a primitive $(n-1)$-th root of unity, and consider the configuration with $x_n=0$ and $x_1,\ldots,x_{n-1}$ the $(n-1)$-th roots of unity. Write $0=\langle(0,1)\rangle$ and the other points as $\langle(z,1)\rangle$. The matrix $g=\operatorname{diag}(\omega,1)$ fixes $(0,1)$ and permutes the vectors $(z,1)$, so $F(gv)=F(v)$ by symmetry, while $F(gv)=\det(g)^eF(v)=\omega^eF(v)$. Hence $\omega^e=1$. With $\zeta$ a primitive $(n-2)$-th root of unity and the configuration formed by $x_n=0$, $\infty$ and the $(n-2)$-th roots of unity, the same argument with $\operatorname{diag}(\zeta,1)$ gives $\zeta^e=1$. This matrix also multiplies the vector $(1,0)$ of $\infty$ by $\zeta$, which $F$ does not see, since it has degree $0$ in that variable. As $n-1$ and $n-2$ are coprime, $(n-1)(n-2)$ divides $e$ (Figure 2).

Both directions, for every $n$, and the corresponding statements for $L$, are formalised in Lean 4 with Mathlib, over any field containing the relevant roots of unity, with $F$ allowed to be any function on the points over that field (theorems trivL_iff and trivLi_iff of Grothendieck.Folder146, in the repositories Commutator-IO/grothendieck-project and Commutator-IO/grothendieck-project-palomar; they belong to the Lean appendix of [6]). The dictionary between equivariant line bundles on $M_{0,n}$ and such functions is not formalised, that is, the identification of the equivariant Picard group with the first cohomology of $\mathfrak{S}$ with values in the units; Mathlib does not have the spaces $M_{0,n}$.

Statement on the use of artificial intelligence

The transcription of folder 146, the search of the literature, the general formula, its proofs, the Lean formalisation and a first draft of this text were produced with language models (Claude Opus 5 and Opus 5.5, Anthropic), under the author’s direction. The author checked the proofs and the references, and takes full responsibility for the content. The formal proofs are checked mechanically by Lean.

Acknowledgements

The author thanks Dan Edidin for his answer about the odd case. The facsimiles are published by the University of Montpellier, grothendieck.umontpellier.fr.

References

Commutator, Paris, France

Email address: [email protected]