This is a problem about the holomorphic fibration on a complex manifold.
Does there exist a holomorphic fibration of genus two over $\mathbb{CP}^{1}$ with 7 nodal singularities?
If you are aware of such a construction, could you please outline the construction or provide a reference for it?
Thanks
Let $S$ be the product $C\times \mathbb{P}^1$, with $C$ a genus two curve. Take points $p_1,\dots,p_7\in S$ in seven distinct fibers. Now blow-up $S$ in the points $p_1,\dots,p_7$. Then the induced fibration has 7 singular fibers, and each fiber is of the form $C\cup E$ where $E$ is a smooth rational curve intersecting $C$ in one point.
$\def\PP{\mathbb{P}}$Here is a solution without $-1$ curves. Indeed, my family is semi-stable or, if you want the total space to be smooth, then some fibers are $-2$ curves.
My family lives inside $\PP^1 \times \PP^1 \times \PP^1$ with coordinates $(x,y,t)$; the map to $\PP^1$ is projection onto the $t$ coordinate. Start with two $(1,1)$ curves $p(x,t)$ and $q(x,t)$ in the $(x,t)$ variables which meet transversely. For a concrete example, I'll use $xt=2$ and $(3-x)(3-t)=2$ meeting at $(1,2)$ and $(2,1)$. Now, inductively, find $t_1$, $x_1$, $t_2$, $x_2$, ..., $t_4$, $x_4$, $t_5$ such that $p(x_i, t_i)=0$ and $q(x_i, t_{i+1})=0$. As a concrete example, we can take $$11/10, 20/11, 17/13, 26/17, 41/25, 50/41, 137/73, 146/137, 521/265$$
Our equation is $$y^2 = p(x,t) q(x,t) \prod_{i=1}^4 (x-x_i).$$ This will be nodal when $t$ is the projection of one of the two intersection points of $p(x,t)=q(x,t)=0$, (in the example, $1$ or $2$) or when $t$ is one of the five $t_i$ (in the example, $11/10$, $17/13$, $41/25$, $137/73$, $521/265$). The five nodal curves of the latter type have two nodes, so their normalization is genus zero; the two nodal curevs of the first type have one node so their normalization is genus one.
The total space of this family has $A_1$ singularities at the intersections of the factors of $p(x,t) q(x,t) \prod_{i=1}^4 (x-x_i)$. If you blow them up, they'll turn into $-2$ curves.
There are lots of variations of this construction. You could, instead, use a $(2,2)$ curve $r(x,t)$ with a single node, and arrange the four lines $(x-x_i)$ to meet $r(x,t)$ at only four distinct $t$-coordinates. An easy way to do this is to take $r(x,t)$ to use only even power of $x$, and choose $x_2=-x_1$, $x_4=-x_3$. You will then get nodes at these four $t$-coordinates, at the projection of the node of $r(x,t)$, and at the two ramification points of $r(x,t) \to \mathbb{P}^1$.
I thought that you could get down to six points by choosing $t_1$ such that $t_1=t_5$ in the first construction, but the only solutions are the degenerate cases $t_1=t_2=t_3=t_4=t_5 = (\mbox{1 or 2 in the example})$. I don't have a strong opinion as to whether six is possible (without $-1$ curves).
Indeed, I wouldn't be surprised if there were a solution where we take a double cover of a smooth $(2,6)$ curve $R$ in $\PP^1 \times \PP^1$, chosen such that the degree $6$ projection $R \to \PP^1 \times \PP^1$ has ramification profile $((2,2,2)^6, \ (2,2,1,1))$. I don't know how you'd build such a thing, though. If a cover of $\PP^1$ exists with this ramification profile, then we can prove it by writing down the monodromy in the group $S_6$. But I don't know how to force the resulting curve to be hyperelliptic, so that it will have a second, degree $2$ map, to $\PP^1$. Thinking about parameter counting, being hyperelliptic is a condition of codimension $g-2$, so $3$ for a genus $5$ curve. I have $4$ parameters available in choosing my $7$ branch points down in $\PP^1$. So it seems possible. If we pull this off, there will be no $-2$ curves.
Here is a partial result. Let $X$ be a smooth projective surface admitting a genus-2 fibration $\pi : X \to \mathbb{P}^1$ whose singular fibers are all irreducible and nodal. Then $\pi_{\ast}\omega_{X|\mathbb{P}^1}$ is a rank-2 vector bundle on $\mathbb{P}^1$, and we have a morphism $f : X \to \mathbb{P}(\pi_{\ast}\omega_{X|\mathbb{P}^1})$ of $\mathbb{P}^1-$varieties which is a branched double covering. The fact that the singular fibers of $\pi$ are all nodal implies that the branch curve $B \subset \mathbb{P}(\pi_{\ast}\omega_{X|\mathbb{P}^1})$ of $f$ is smooth. Applying Riemann-Hurwitz to each component of $B$, we see that if $\pi$ has an odd number of singular fibers which are all irreducible and nodal, then at least one of them has 2 or more nodes.
