# Any three plane conics has a line meeting them in a configuration of points with an order 5 symmetry

Edit: In light of Jason's answer, the below asserted statement must be wrong, which means that the intersection theory argument alluded to in the background section must be wrong. What follows is the question as originally stated:

Statement: Let $C$ be union of 3 smooth conics in $\mathbb{P}^2$ meeting each other in nodes. Then there exists a line $L\cong \mathbb{P}^1$ such that the configuration of 6 points $x_1,…,x_6\in L$ obtained by intersecting $L$ with $C$ has an automorphism of order 5.

The above fact is surprising (to me at least) since by a dimension count, one cannot get an arbitrary configuration of six points and the configuration of points with an order 5 automorphism is unique. The fact seems to follow from some work a student of mine is doing involving genus 2 K3 surfaces (more below), but the statement seems so classical, that it feels like there should be a direct elementary proof.

Question: Is there a "classical" proof of the above statement?

In fact, the statement seems to hold for any (possibly reducible) sextic curve $C$ with singularities no worse than nodes as long as (1) there is at least one node, and (2) there are no components which are lines.

Background:

Given a sextic plane curve $C$ with at worst nodal singularities, we get a K3 surface by taking the double cover of the plane branched over $C$ and then resolving the ordinary double point singularities. The double covers of the lines in the plane give rise to a two dimensional linear system of genus 2 curves on the K3 which give a rational map $\mathbb{P}^2--\to \overline{M}_2$. One can compute the class of the closure of the image (call it $D$) in the chow ring of $\overline{M}_2$ by intersecting it with test curves in $\overline{M}_2$ (this is tricky so it is possible that we are making mistakes here) and one gets a formula for $D$ as a linear combination of boundary divisors. If $C$ is smooth, the linear combination has integer coefficients, but under the assumption that there is at least one node and no lines in $C$, the coefficients have 5's in the denominator. I think the only way that can happen is if $D$ meets the (unique I think) genus 2 curve with an order 5 automorphism. This gives rise to the statement about the points on the line in $\mathbb{P}^2$.

Question 2: Does the statement hold when $C$ contains lines? For example if $C$ is the union of 6 disjoint lines?

The above proof breaks down in this case.

• Did you try to find such a line explicitly for "random" nodal $C$? This could corroborate or disprove an argument by intersection theory so "tricky [that] it is possible we are marking mistakes". – Noam D. Elkies Aug 8 '15 at 13:43
• @Noam. My attempt to find such lines explicitly led me to the statement about conics which seemed to be the easiest place to corroborate or disprove my argument. Given Jason's argument, it looks like there must be mistakes in my $\overline{M}_2$ argument. – Jim Bryan Aug 8 '15 at 14:59

I believe this is incorrect. Let $V$ be a $3$-dimensional vector space, so that $\mathbb{P}V$ is $\mathbb{P}^2$. Let $\mathbb{P}(V^\vee)$ be the dual projective space parameterizing lines $L\subset \mathbb{P}V$. Let $\mathbb{P}(S)\subset \mathbb{P} V \times \mathbb{P}(V^\vee)$ denote the universal line, and let $S$ denote the locally free sheaf on $\mathbb{P}(V^\vee)$ that is a subsheaf of $V\otimes\mathcal{O}_{\mathbb{P}(V^\vee)}$ and whose associated projective bundle is $\mathbb{P}(S)$. Let $A$ denote $\mathbb{P}(\text{Sym}^2(S^\vee))$, the parameter space for pairs $([L],[D])$ of a line $L\subset \mathbb{P} V$ and an effective, degree $2$ Cartier divisor $D\subset L$. Finally, denote by $$B \subset A \times \mathbb{P}(\text{Sym}^2(V^\vee))$$ the closed subscheme parameterizing triples $([L],[D],[C])$ of a line $L\subset \mathbb{P} V$, an effective, degree $2$ Cartier divisor $D\subset L$, and an effective, degree $2$ Cartier divisor $C\subset \mathbb{P} V$ (i.e., a plane conic), such that $D$ is contained in $C$. Since the restriction linear transformation, $$\text{Sym}^2(V^\vee) \to H^0(L,\mathcal{O}_L(2)),$$ is surjective, the projection morphism, $$\text{pr}_A : B \to A,$$ is a projective bundle of fiber dimension $3$, i.e., a $\mathbb{P}^3$-bundle.
Now denote by $A_3$ the self-fiber product $$A_3 = A\times_{\mathbb{P}(V^\vee)} A \times_{\mathbb{P}(V^\vee)} A,$$ parameterizing data $([L],[D_1],[D_2],[D_3])$ of a line $L\subset \mathbb{P} V$, and a triple of effective, degree $2$ Cartier divisors $D_i \subset L$. Similarly, denote by $B_3$ the self-fiber product $$B_3 = B\times_{\mathbb{P}(V^\vee)} B \times_{\mathbb{P}(V^\vee)} B,$$ parameterizing data $([L],[D_1],[D_2],[D_3],[C_1],[C_2],[C_3])$ of a line $L\subset \mathbb{P} V$, a triple of effective, degree $2$ Cartier divisors $D_i \subset L$, and a triple of effective, degree $2$ Cartier divisors $C_i\subset \mathbb{P} V$ (i.e., plane conics) such that $D_i \subset C_i$. Since $\text{pr}_A$ is a $\mathbb{P}^3$-bundle, also the projection $$\text{pr}_{A_3}:B_3 \to A_3, \ ([L],[D_1],[D_2],[D_3],[C_1],[C_2],[C_3]) \mapsto ([L],[D_1],[D_2],[D_3]),$$ is a $\mathbb{P}^3\times \mathbb{P}^3\times \mathbb{P}^3$-bundle. In particular, $\text{pr}_{A_3}$ is faithfully flat of fiber dimension $9$.
Now, consider the locally closed subset $I\subset A_3$ parameterizing data $([L],[D_1],[D_2],[D_3])$ such that the degree $6$ Cartier divisor $E=D_1+D_2+D_3 \subset L$ is invariant under a cyclic subgroup $G\subset \text{Aut}(L)$ of order $5$. Consider the dimension of $I$. First of all, $\mathbb{P}(V^\vee)$ has dimension $2$. For fixed $[L]$ in $\mathbb{P}(V^\vee)$, every cyclic subgroup $G$ of $\text{Aut}(L) \cong \textbf{PGL}_2$ lifts to $\text{Aut}(L,\mathcal{O}(1)) \cong \textbf{SL}_2$, i.e., $G$ is diagonalizable. The eigenspaces are two distinct points of $L$, say $p_0$ and $p_\infty$. Thus, the parameter space for such subgroups $G$ is just $L\times L \setminus \Delta$, which is $2$-dimensional. Finally, for fixed $L$ and fixed $G$, there is a $1$-dimensional parameter space for $G$-invariant, degree $6$ Cartier divisors $E\subset L$, namely the $7$ isolated points $m\underline{p}_0 + n\underline{p}_\infty$ for $(m,n)\in \mathbb{Z}_{\geq 0}\times \mathbb{Z}_{\geq 0}$ with $m+n=6$ together with the two $\mathbb{A}^1$s parameterizing $\underline{p}_0 + \underline{G\cdot q}$, resp. $\underline{p}_\infty + \underline{G\cdot q}$ for $q\in L\setminus\{p_0, p_\infty\}$. Thus $I$ has dimension $2+2+1 = 5$. Since $\text{pr}_{A_3}$ is faithfully flat of fiber dimension $9$, $\text{pr}_{A_3}^{-1}(I)$ has dimension $5+9 = 14$.
Finally, consider the parameter space $$J =\mathbb{P}(\text{Sym}^2(V^\vee)) \times \mathbb{P}(\text{Sym}^2(V^\vee)) \times \mathbb{P}(\text{Sym}^2(V^\vee))$$ of triples $([C_1],[C_2],[C_3])$ of effective, degree $2$ Cartier divisors $C_i\subset \mathbb{P} V$, i.e., plane conics. Of course $J$ is isomorphic to $\mathbb{P}^5 \times \mathbb{P}^5 \times \mathbb{P}^5$, which has dimension $15$. Since $\text{pr}_{A_3}^{-1}(I)$ has dimension $14$, which is strictly less than $\text{dim}(J)$, the projection morphism, $$\text{pr}_J : \text{pr}_{A_3}^{-1}(I) \to J,$$ cannot be dominant. Thus, the image is not Zariski dense. The Zariski closure of the image is a proper closed subvariety $K\subset J$. For every $([C_1],[C_2],[C_3])$ parameterized by a point of the dense, Zariski open subset $J\setminus K$ of $J$, there exists no $([L],[D_1],[D_2],[D_3])$ in $I$ such that $D_i\subset C_i$ for $i=1,2,3$.
• Thanks Jason. So it looks like there must be a mistake in the $\overline{M}_2$ intersection theory argument then. The family of genus 2 curves on the K3 is nodal except for an isolated set of curves each of which has a tacnode singularity. I have to resolve the map $\mathbb{P}^2-\to\overline{M}_2$ to get the divisor $D$. Do we agree that I should then (given your argument) find that $D$ is an integer linear combination of boundary divisors (or at least not have 5s in the denominator)? – Jim Bryan Aug 8 '15 at 15:12
• @Jason $-$ CW-ing your own answer is usually a mechanism for disclaiming individual credit for work done by others or jointly with others. Here it looks like there's no reason for you not to be individually recognized. – Noam D. Elkies Aug 8 '15 at 18:17