. This curve has only one singular point at $P = [0:0:1]$ and does not pass through the singular points of $\PP^2_{w}$. Resolving the singularity of the curve with a $(3,2)$-blow up gives rise to one exceptional divisor $E$ with self-intersection number $-1/6$ and $\pi^* \mathcal{C} = \hat{\mathcal{C}} + 6 E$, $K_{\pi} = 4E$. Since $H^0(\PP^2_w, \mathcal{O}_{\PP^2_{w}}(k-6)) = 0$ for $k = 0,\dots,5$, the cokernel of $\pi^{(k)}$ is $$\operatorname{coker}\pi^{(k)} = \frac{\mathcal{O}_{\CC^2,0}}{\{ g \mid \operatorname{mult}_{E} \pi^{*} h > k - 5 \}},$$ whose dimension is $1$ for $k=5$ and $0$ for $k=0,\ldots,4$. According to Theorem \[thm:conucleo\_singular\], $\dim H^1(\tilde{X},\CC) = 2$ and the characteristic polynomial of the monodromy of the covering acting on $H^1(\tilde{X},\CC)$ is $(t-e^{\frac{2\pi i}{6}})(t-e^{\frac{2\pi i 5}{6}}) = t^2 - t + 1$.

Note that the $6$th-cyclic cover of $\PP^2$ ramified along any curve of degree $6$ having just a singular point with the topological type of $x^2+y^3$ does not have any irregularity. Therefore the singular points of the ambient space might affect the irregularity of the covering, even though they have codimension $2$.

A Zariski pair of curves on the weighted projective plane {#sec:zariski-pair}
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Let $\zeta$ be a $3$th-primitive root of unity and for $\lambda = 1, \zeta$, consider the curve $\mathcal{C}_{\lambda} \subset S=\PP^2_{(1,1,3)}$ defined by the polynomial $$G_{\lambda}(x,y,z) = (\lambda y z + x z - x^{a} y^{b})^3 + (y z - x z + x^{a} y^{b})^3
+ (- \lambda y z + \lambda x z + x^{a} y^{b})^3$$ with $a+b=4$ so that it is quasi-homogeneous. As above let $\rho_\lambda: \tilde{X}_\lambda \to S$ be the cyclic branched covering of degree $d=12$ ramifying on $\mathcal{C}_\lambda$. Using $H^1(\tilde{X}_\lambda,\CC)$, which is an invariant of the pair $(S,\mathcal{C}_\lambda)$, we will see that $(S,\mathcal{C}_1)$ and $(S,\mathcal{C}_{\zeta})$ provide a Zariski pair for $(a,b)=(1,3)$. However, one cannot