# Monodromy group of the generic plane curve

Let's work over $$\mathbb{C}$$. The degree $$d$$ curves in $$\mathbb{P}^2_{\mathbb{C}}$$ are parameterized by a projective space $$|\mathcal{O}_{\mathbb{P}^2}(d)|$$. Let $$U_d\subset |\mathcal{O}_{\mathbb{P}^2}(d)|$$ be the locus parameterizing smooth curves. Let $$u\in U_{1,d}$$ be a closed point, let $$X_u$$ be the corresponding plane curve. For any $$m\in\mathbb{Z}$$, the fundamental group $$\pi_1(U_{1,d},u)$$ acts on $$\mathrm{H}^1(X_u,\mathbb{Z})$$. The monodrmoy action preserves the intersection form. Let's denote the image of $$\pi_1(U_{1,d},u)$$ in $$\mathrm{Sp}_{2g}(\mathbb{Z})$$ by $$\Gamma_{1,d}$$.

$$\bullet$$When $$d$$ is even, by Theorem 4(i)(https://math.unice.fr/~beauvill/pubs/mono.pdf), the group $$\Gamma_{1,d}=\mathrm{Sp}_{2g}(\mathbb{Z})$$.

$$\bullet$$When $$d$$ is odd, Theorem 4(ii) there claimed that there exists a $$\mathbb{Z}/2\mathbb{Z}$$-quadratic form $$q_X$$ on $$\mathrm{H}^1(X,\mathbb{Z}/2\mathbb{Z})\to\mathbb{Z}/2\mathbb{Z}$$ such that $$\Gamma_{1,d}\subset\mathrm{Sp}_{2g}(\mathbb{Z})$$ is the stabilizer of $$q_X$$. (I think this $$q_X$$ comes from classification of purely algebraic structure called vanishing lattices'', and I am not sure what it is in explicit cases..)

$$\textbf{Question}$$: (1) Is this $$q_X$$ the one associated with the theta characteristic $$\mathcal{O}_{\mathbb{P}^2}(\frac{d-3}{2})|_{X_u}$$? (I think the answer is yes, as explained by abx in the comment.)

(2) If so, are there some explicit expression of generators of the group $$\Gamma_{1,d}=\mathrm{Stab}_{\mathrm{Sp}_{2g}(\mathbb{Z})}(q_X)$$? (Ok, I found it Reference on generators of subgroups of symplectic groups, the generators are given in Mumford Tata lectures on theta I, P208)

(3)The goal was to check $$\Gamma_{1,d}$$ has no common eigenvector in $$\mathrm{H}^1(X,\mathbb{Z}/m\mathbb{Z})$$ for every $$m\in\mathbb{Z}$$. It can be checked by generators displayed in (2), but I think this is a much weaker statement than determining the structure of $$\Gamma_{1,d}$$, is there a simpler way to check this?

[There is a quadratic form given by theta characteristic $$\mathcal{N}:=\mathcal{O}_{\mathbb{P}^2}(\frac{d-3}{2})|_{X_u}$$: We identify $$\mathrm{H}^1(X,\mathbb{Z}/2\mathbb{Z})=\mathrm{H}^2(X,\mu_2)=\mathrm{Pic}^0(X)[2]$$, then for any $$2$$-torsion line bundle $$\mathcal{L}$$, we assign the number $$q(\mathcal{L})=h^0(\mathcal{L\otimes N})-h^0(\mathcal{L})$$ in $$\mathbb{Z}/2\mathbb{Z}$$. I think $$\Gamma_{1,d}$$ stabilizes this $$q$$, but I am not sure does this mean $$\Gamma_{1,d}$$ is exactly the stabilizer of this $$q$$?]

• The theorem you quote implies that there is a unique $q$ preserved by the monodromy, so the form you describe is equal to $q_X$, and $\Gamma_{1,d}$ is exactly its stabilizer. – abx May 16 at 7:32
• @abx Thanks, but I didn't quite get the point.. how is the uniqueness implied? – Qixiao May 16 at 7:41
• Do we know the fixed quadratic forms of $\Gamma_{1,d}$ is $1$-dimensional? – Qixiao May 16 at 7:48
• Yes. Two such quadratic forms differ by a linear form, and there is no nonzero linear form invariant under $\operatorname{O}(q)$. – abx May 16 at 7:56
• @abx Thanks, I think by taking dual, this is equivalent to show $SpO(q):=Sp\cap O(q)$ has no common eigenvector in $\mathrm{H}^1(X,\mathbb{Z}/2\mathbb{Z})$? I am not sure how this can be seen directly without knowing $q$? – Qixiao May 16 at 8:05