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Let $f:\mathbb{C}^2\to \mathbb{C}$ be an isolated plane curve singularity. Consider the versal deformation space $\mathbb{C}^\mu$ parameterizing deformations $f_\lambda$ for $\lambda \in \mathbb C^\mu$. This has discriminant set $D \subset \mathbb C^\mu$: by construction $\lambda \in D$ if and only if $f_\lambda^{-1}(0)$ is singular.

On $\mathbb C^2$ we have the standard volume form $dx \wedge dy$, and for any $f_\lambda$, we can consider the associated ``Gelfand-Leray form'' $$ \omega_\lambda = \frac{dx}{(f_\lambda)_y} = - \frac{dy}{(f_\lambda)_x}, $$ which determines a (nowhere-vanishing) holomorphic $1$-form on the level surface $f_\lambda^{-1}(0)$ for any $\lambda \not \in D$.

One can study the area of the Gelfand-Leray form as a function of $\lambda$. Define for $\lambda \not \in D$: $$ A(\lambda) = \int_{f^{-1}_\lambda(0)} \omega_\lambda \wedge \overline{\omega_\lambda}. $$

Under reasonable circumstances (e.g. for Brieskorn-Pham singularities; I'm not sure how generally this holds), the area is finite.

I'm interested in understanding what happens to $A(\lambda)$ as $\lambda$ tends towards $D$. When the limiting value $\lambda_0$ parametrizes a nodal curve, for instance, I'm pretty sure you can see that $A(\lambda)$ tends to infinity; I can see this same behavior explicitly in some other carefully-constructed examples. My main question is whether this must happen for all degenerations to singular curves.

Consider any family $\lambda(t)$ for $t \in [0,1)$ with $\lambda(t) \in D$ if and only if $t = 0$. As $t \to 0$, must $A(\lambda(t)) \to \infty$?

I would ideally hope that this question has already been answered in the literature, but I can't find any discussion of this in the book of Arnol'd--Gusein-Zade--Varchenko.

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