Let $\pi:\mathcal{C} \to \Delta$ be a family of projective curves of genus $g \ge 2$ over the unit disc $\Delta$, smooth over the punctured disc $\Delta\backslash \{0\}$ and central fiber $\pi^{-1}(0)$ is an irreducible nodal curve with exactly one node. For $t$ close to $0$, denote by $\delta_t$ the vanishing cycle on $H_1(\mathcal{C}_t,\mathbb{Z})$, where $\mathcal{C}_t:=\pi^{-1}(t)$. I am looking for conditions on the central fiber $\pi^{-1}(0)$ such that there exists a $1$-cycle $\gamma \in H_1(\mathcal{C}_t,\mathbb{Z})$ satisfying: $\gamma.\delta_t=1$. Any idea/reference will be most welcome.

If I understand correctly, this is true if $\pi$ is a degeneration of elliptic curves.

  • $\begingroup$ There exist simple closed curves $a,b$ on $C_t$ whose homology classes (also denoted $a,b$) satisfy $a\cdot b = 1$. Since the central fiber is irreducible with exactly one node, the vanishing cycle $\delta_t$ is a non-separating simple closed curve on $C_t$; it follows from the classification of surfaces that there is an orientation-preserving homeomorphism $f$ of $C_t$ such that $f(b) = \delta_t$. You may then take $\gamma = f(a)$. $\endgroup$ – Tony Aug 3 '18 at 14:47

Because the symplectic form on $H_1(C_t, \mathbb Z)$ is a perfect pairing, it suffices to check that there is a group homomorphism $H_1(C_t,\mathbb Z) \to \mathbb Z$ that sends $ \gamma$ to $1$, which follows if $\gamma$ is not divisible by any $n>1$ in $H_1(C_t,\mathbb Z)$.

Because $\gamma$ is defined as the generator of the kernel of $H_1(C_t,\mathbb Z) \to H_1(C_0,\mathbb Z)$, $\gamma$ is indivisible as long as $H_1(C_0,\mathbb Z)$ is nontorsion, which follows from it being an irreducible nodal curve.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.