Erratum to: Transverse lines to surfaces over finite fields

Erratum to: Manuscripta Math. 165, 135–157 (2021). https://doi.org/10.1007/s00229-020-01200-7

In [1, Theorem 4.5], we claim that, given a reduced and geometrically irreducible hypersurface $X = \{F=0\}\subset {\mathbb {P}}^n$ over ${\mathbb {F}}_q$, if it is Frobenius nonclassical, then it is nonreflexive. If we denote by the Gauss map of X, then X is nonreflexive if and only if, at a general point $P\in X$, the pullback of differentials

$$\begin{aligned} d\gamma ^*_P:(\gamma ^*\Omega _{X^*})\otimes {{\mathbb {F}}_q(P)} \longrightarrow \Omega _{X}\otimes {\mathbb {F}}_q(P) \end{aligned}$$

(0.1)

is not injective. In our original argument, we intended to prove this property via [1, Lemma 4.6], which shows that the determinant of the Hessian matrix

$$\begin{aligned} H_F{:}{=}\left( F_{ij} \right) \quad \text {where}\quad F_{ij} = \frac{\partial ^2 F}{\partial X_j\partial X_i} \end{aligned}$$

is zero modulo F. While [1, Lemma 4.6] is correct, this is not sufficient to prove [1, Theorem 4.5]. The subtle error is the following incorrect assertion: a hypersurface X is nonreflexive if and only if the determinant of the Hessian matrix of X vanishes identically on X. Let us explain why this claim fails when $\deg (X)\equiv 1\mod {\text {char}}({\mathbb {F}}_q)$.

Indeed, if we denote $d = \deg (F)$ and $F_i = \partial F/\partial X_i$, then a computation with Euler’s formula shows

$$\begin{aligned} \begin{pmatrix} d(d-1)F &{} (d-1)F_1 &{} \cdots &{} (d-1)F_n \\ (d-1)F_1 &{} F_{11} &{} \cdots &{} F_{n1} \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ (d-1)F_n &{} F_{1n} &{} \cdots &{} F_{nn} \end{pmatrix} = \begin{pmatrix} X_0 &{} X_1 &{} \cdots &{} X_n \\ 0 &{} 1 &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} 1 \end{pmatrix} \cdot H_F\cdot \begin{pmatrix} X_0 &{} 0 &{} \cdots &{} 0 \\ X_1 &{} 1 &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ X_n &{} 0 &{} \cdots &{} 1 \end{pmatrix}\nonumber \\ \end{aligned}$$

(0.2)

from which we see that $\det (H_F)$ is constantly zero whenever $d\equiv 1\mod {\text {char}}({\mathbb {F}}_q)$. Moreover, we actually assume that the Gauss map is finite in our original argument, but we mistakenly ignored this assumption in the statement of [1, Theorem 4.5]. In this erratum, we fix the statement and prove it without using [1, Lemma 4.6].

FormalPara Theorem 0.1

Let $X\subset {\mathbb {P}}^n$ be a reduced and geometrically irreducible Frobenius nonclassical hypersurface over ${\mathbb {F}}_q$. Suppose that $\dim (X) = \dim (X^*)$ (which is satisfied when X is smooth due to Zak’s theorem [2]). Then X is nonreflexive.

Our strategy of proof goes as follows: Let I and $I'$ be the ideal sheaves for X and $X^*$, respectively. Then there is a commutative diagram for sheaves of differentials:

(0.3)

Note that $\gamma = \Gamma |_X$ where $\Gamma$ is the polar map

$$\begin{aligned} \Gamma :{\mathbb {P}}^n\dashrightarrow ({\mathbb {P}}^n)^* : [X_0:\dots :X_n] \longmapsto \left[ \frac{\partial F}{\partial X_0}:\dots :\frac{\partial F}{\partial X_n}\right] . \end{aligned}$$

We have $\gamma ^*\left( \Omega _{({\mathbb {P}}^n)^*}|_{X^*}\right) = \left( \Gamma ^*\Omega _{({\mathbb {P}}^n)^*}\right) |_{X}$ and the vertical arrow in the middle of (0.3) is induced by the pullback of differentials $d\Gamma ^*:\Gamma ^*\Omega _{({\mathbb {P}}^n)^*}\longrightarrow \Omega _{{\mathbb {P}}^n}.$ Let $U\subset {\mathbb {P}}^n$ be an open neighborhood of P where $I(U) = (f)$. In order to prove that (0.1) is not injective, we will prove that the image of the linear map

$$\begin{aligned} d\Gamma ^*_P:(\Gamma ^*\Omega _{({\mathbb {P}}^n)^*})\otimes {{\mathbb {F}}_q(P)} \longrightarrow \Omega _{{\mathbb {P}}^n}\otimes {\mathbb {F}}_q(P) \end{aligned}$$

(0.4)

has dimension at most $n-2 = \dim (X)-1$ modulo df.

1 Proof of the theorem

Let $[Y_0:\dots :Y_n]$ be homogeneous coordinates for $({\mathbb {P}}^n)^*$ so that the polar map $\Gamma$ can be written as $Y_i = F_i$, and let $y_i = Y_i/Y_0$ be the affine coordinates for the chart $\{Y_0\ne 0\}$. Assume without loss of generality that the point $P\in X$ belongs to the open subset

$$\begin{aligned} U{:}{=}\{X_0\ne 0\}\cap \{F_0\ne 0\}\cap \left( \bigcup _{i=1}^{n} \{F_i\ne 0\}\right) \subset {\mathbb {P}}^n. \end{aligned}$$

If we write $x_i{:}{=}X_i/X_0$ and let $f_i = f_i(x_1,\ldots ,x_n)$ be the dehomogenization of $F_i$ with respect to $X_0$, then $\Gamma |_U$ can be expressed as $y_i = f_i/f_0 = F_i/F_0$. In this setting, the map of differentials $d\Gamma ^*|_U$ sends each $dy_i$ to

$$\begin{aligned} dy_i= & {} \sum _{j=1}^n \frac{\partial (f_i/f_0)}{\partial x_j} dx_j = \sum _{j=1}^n\left( \frac{(\partial f_i/\partial x_j)f_0 - f_i(\partial f_0/\partial x_j) }{f_0^2} \right) dx_j\\= & {} \sum _{j=1}^n\left( \frac{F_{ij}F_0 - F_iF_{0j} }{F_0^2} \right) dx_j \end{aligned}$$

This linear map corresponds to the square matrix $M_F/F_0$ where $M_F$ is given by

$$\begin{aligned} M_F{:}{=}\begin{pmatrix} F_{11}-\frac{F_1}{F_0}F_{01} &{} F_{21}-\frac{F_2}{F_0}F_{01} &{} \cdots &{} F_{n1}-\frac{F_n}{F_0}F_{01}\\ F_{12}-\frac{F_1}{F_0}F_{02} &{} F_{22}-\frac{F_2}{F_0}F_{02} &{} \cdots &{} F_{n2}-\frac{F_n}{F_0}F_{02}\\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ F_{1n}-\frac{F_1}{F_0}F_{0n} &{} F_{2n}-\frac{F_2}{F_0}F_{0n} &{} \cdots &{} F_{nn}-\frac{F_n}{F_0}F_{0n} \end{pmatrix}. \end{aligned}$$

Now we extend the above matrix to the following one

$$\begin{aligned} H''_F{:}{=}\begin{pmatrix} 0 &{} F_1 &{} F_2 &{} \cdots &{} F_n\\ F_1 &{} F_{11}-\frac{F_1}{F_0}F_{01} &{} F_{21}-\frac{F_2}{F_0}F_{01} &{} \cdots &{} F_{n1}-\frac{F_n}{F_0}F_{01}\\ F_2 &{} F_{12}-\frac{F_1}{F_0}F_{02} &{} F_{22}-\frac{F_2}{F_0}F_{02} &{} \cdots &{} F_{n2}-\frac{F_n}{F_0}F_{02}\\ \vdots &{} \vdots &{} \vdots &{} \ddots &{} \vdots \\ F_n &{} F_{1n}-\frac{F_1}{F_0}F_{0n} &{} F_{2n}-\frac{F_2}{F_0}F_{0n} &{} \cdots &{} F_{nn}-\frac{F_n}{F_0}F_{0n} \end{pmatrix}. \end{aligned}$$

Lemma 1.1

Let $(0,a_1,\ldots ,a_n)$ be a nonzero vector where $a_i\in \overline{{\mathbb {F}}_q}$. If

$$\begin{aligned} (0,a_1,\ldots ,a_n)\cdot H''_F(P) = 0 \qquad \text {where}\qquad P\in U, \end{aligned}$$

then the column space of $M_F(P)$ has dimension $\le n-2$ modulo $(F_1(P),\ldots ,F_n(P))^t$.

Proof

The hypothesis implies that

$$\begin{aligned} (a_1,\ldots ,a_n)\cdot M_F(P) = 0 \qquad \text {and}\qquad (a_1,\ldots ,a_n)\cdot (F_1(P),\ldots ,F_n(P))^t = 0. \end{aligned}$$

The first equation implies that of ${\text {rk}}(M_F(P))\le n-1$. If ${\text {rk}}(M_F(P))\le n-2$, the proof is done. If ${\text {rk}}(M_F(P)) = n-1$, the second equation above implies that $(F_1(P),\ldots ,F_n(P))^t$ belongs to the column space of $M_F(P)$, which proves the claim. $\square$

The matrix $H''_F$ is related to the Hessian matrix $H_F$ in the following way: If we restrict the matrix on the left hand side of (0.2) to $X = \{F=0\}$ and then divide its first row and first column by $(d-1)$, we will get

$$\begin{aligned} H'_F{:}{=}\begin{pmatrix} 0 &{} F_1 &{}\cdots &{} F_n\\ F_1 &{} F_{11} &{} \cdots &{} F_{n1}\\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ F_n &{} F_{1n} &{} \cdots &{} F_{nn}\\ \end{pmatrix} = \begin{pmatrix} 1 &{} 0 &{} \cdots &{} 0\\ \frac{F_{01}}{F_0} &{} 1 &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ \frac{F_{0n}}{F_0} &{} 0 &{} \cdots &{} 1 \end{pmatrix}\cdot H''_F. \end{aligned}$$

(1.1)

However, we cannot divide by $d-1$ in the field when $d\equiv 1 \mod p$. We see that the matrix $H''_F$ is the correct replacement for the usual Hessian in positive characteristic, especially when $d\equiv 1\mod p$.

On the other hand, X is Frobenius nonclassical means that there exists a polynomial R that satisfies

$$\begin{aligned} FR = X_0^q F_0+X_1^qF_1+\cdots +X_n^q F_n. \end{aligned}$$

(1.2)

Lemma 1.2

Let $P = [1:X_1:\dots :X_n]\in X\cap U$. Then

$$\begin{aligned} \left( 1-d+R, X_1-X_1^q, X_2-X_2^q, \ldots , X_n-X_n^q\right) \cdot H'_F(P) = 0. \end{aligned}$$

Proof

Subtracting Euler’s formula $dF = X_0F_0+\cdots + X_nF_n$ by (1.2) gives

$$\begin{aligned} (d-R)F=(X_0-X_0^q)F_0+(X_1-X_1^q)F_1+\cdots + (X_n-X_n^q)F_n. \end{aligned}$$

(1.3)

Taking partial derivatives of both sides with respect to $X_i$ followed by a rearrangement gives

$$\begin{aligned} -R_iF= & {} (1-d+R)F_i +(X_0-X_0^q)F_{0i} \nonumber \\{} & {} + (X_1-X_1^q)F_{1i} + \cdots + (X_n-X_n^q)F_{ni} \end{aligned}$$

(1.4)

Then the statement follows by a straightforward computation with (1.3), (1.4), the hypothesis that $X_0 = 1$, and the fact that $F(P)=0$. $\square$

Proof of Theorem 0.1

Pick a general $P = [1:X_1:\dots :X_n]\in X\cap U$. First we compute

$$\begin{aligned}&\left( 1-d+R, X_1-X_1^q, X_2-X_2^q, \ldots , X_n-X_n^q\right) \cdot \begin{pmatrix} 1 &{} 0 &{} \cdots &{} 0\\ \frac{F_{01}}{F_0} &{} 1 &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ \frac{F_{0n}}{F_0} &{} 0 &{} \cdots &{} 1 \end{pmatrix} \\&\quad = \left( (1-d+R)+\sum _{j=1}^n\left( \frac{F_{0j}}{F_0}(X_j-X_j^q)\right) , X_1-X_1^q, \ldots , X_n-X_n^q \right) \\&\quad \overset{(1.4)}{=} \left( 0, X_1-X_1^q, \ldots , X_n-X_n^q \right) \end{aligned}$$

Lemma 1.2 and relation (1.1) implies that $\left( 0, X_1-X_1^q, \ldots , X_n-X_n^q \right) \cdot H''_F(P) = 0,$ which implies that $\left( X_1-X_1^q, \ldots , X_n-X_n^q \right) \cdot M_F(P) = 0.$ By applying Lemma 1.1, we conclude that the image of $d\gamma ^*_P$ has dimension $\le n-2$ modulo $\sum _{i=1}^nF_i(P)dx_i = df$, thus it cannot be injective. This shows that the Gauss map $\gamma$ is inseparable, whence X is nonreflexive. $\square$