Hi,
the Hirzebruch surface $F_n$ admits a deformation for $0\leq m\leq n$ defined by the equation
$$
\mathcal{M}=\{ ([x_0:x_1],[y_0:y_1:y_2],t) \in \mathbb{P}^1 \times \mathbb{P}^2 \times \mathbb{C}| x_0^ny_1-x_1^ny_0 + t x_0^{n-m}x_1^m y_2 =0\}
$$
The central fiber is $F_n$ while a noncentral fiber is isomorphic to $F_{n-2m}$.
Each fiber $\mathcal{M}_t$ is toric. However it seems almost always impossible to find a differentiable action of the torus $T^2$ on $\mathcal M$ that would preserve and act holomorphically and torically on the fibers.
I would like to know whether this is correct and why.
Thanks a lot.
Best Answer
What you say is correct and here is a proof.
First remark is that if there were such an action of $T^2$ holomorphic on the fibers and smooth on $M$, it would be just simply holomorphic on $M$. Indeed, for a fixed toric surface, $T^2$ can act on it holomorphically only in a discreet number of ways, that (provided we fix a base in $H_1(T^2,\mathbb Z))$ are parametrized by $SL(2,\mathbb Z)$. The space $M\setminus M_0$ is a locally holomorphically trivial fibration, so $T^2$ would act on it holomorphically. Now, since the action of $T^2$ on $M$ is continuous it must be holomorphic, because any continuous vector field on a complex manifold, holomorphic outside a set of co-dimension $1$ is holomorphic.
So, to finish we need to show that there is no holomorphic action of $T^2$ on $M$. To do this take first any Kahler metric on $M$ and homogenise it under the action of $T^2$. Now, the action of $T^2$ becomes Hamiltonian, so we have a fiberwise moment map. For each fiber the image of the moment map is a $4$-gon on $\mathbb R^2$ and this $4$-gon varies continuously with the fiber. Here we get a contradiction. Indeed, the polygons corresponding to $F_n$ can not be continuously deformed to those of $F_{n-2m}$ without degenerating them at some point (I assume $n\ne m$ but I guess one can exclude this case as well).
Note that in the last reasoning we could have just taken a product Kahler form on $\mathbb P^1\times \mathbb P^2\times \mathbb C$ and restrict it to $M$. In this case the image of the moment map would not have varied at all (but the above reasoning is more universal since it does not use any information about $M$).