On the global monodromy of a Lefschetz fibration arising from the Fermat surface of degree 4 (Q616641)

Let \(P(x_0,x_1,x_2,x_3)=x_0^4+x_1^4+x_2^4+x_3^4\) be the Fermat equation and let \(X=\left\{[x_0:x_1:x_2:x_3]\in\mathbb{P}^3(\mathbb{C}) \mid P=0 \right\}\) be the Fermat surface of degree \(4\) and \(X^\lor \) be its dual variety. Fix a line \(L\) (1-dimensional projective line) in the dual projective space such that it intersects \(X^\lor\) transversely. For \(\alpha=\{\alpha_0x_0+\alpha_1x_1+\alpha_2x_2+\alpha_3x_3=0\}\in L\), let \(F_\alpha\) be the curve \(X\cap\alpha\) and we consider a fibration on \(L\) with fibers \(F_\alpha\) . In this paper, the author calculates the topological structure of this fibration, including global monodromy. (Partially he uses computer.) There are 36 Lefschetz degenerations on this fibration and he determined the vanishing cycles of all of 36 singular fibers. We may think that because the degree of the surface is \(4\), it is not so hard to calculate (by hand) any topological datae. (We can get all solutions of any equation of degree \(4\) by hand, using Ferrari's formula.) But indeed a general fiber is a genus \(3\) Riemann surface \(\Gamma_3\), and the global monodromy is a map from the fundamental group of \(L\setminus\{\text{sigular loci}\}\) to the mapping class group of \(\Gamma_3\). Anyhow we need to `trace the movement' of solutions of the defining equation (with a parameter \(\alpha\)) by using the computer, and it might turn out to be a mathematically tough problem.