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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.

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I think the right thing to look for is a $T^2$ action on the total space and the base such that one circle acts on each fiber, and the other circle acts nontrivially on the base, and on the central fiber. So at least you get the $T^2$ action on that fiber. –  Allen Knutson Jul 4 '11 at 17:37
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2 Answers

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 $M$).

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Thanks a lot Dima ! I think you also have to arrange so that the Kaehler class does not depend on the fibers. Otherwise, the moment map would vary. –  Yann Jul 3 '11 at 5:01
    
Note, that in fact this proof works as well in the symplectic category, not only in the algebro-geometric caterogry –  Dmitri Jul 3 '11 at 17:40
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This is more of a comment than an alternative solution to Dmitri's solution (unfortunately I am again unable to comment; I still don't understand the rules here). Yann can also prove his assertion by infinitesimal deformation theory. This has the feature that it works over an arbitrary base field (even in positive characteristic). The first-order deformations of a smooth, toric variety $X$ are the same as vectors $v$ in $H^1(X,T_X)$. For the torus $T$ acting on $X$, the tangent sheaf $T_X$ is $T$-linearized. For an algebraic subgroup $S$ of $T$, the action of $S$ on $X$ extends to the first-order deformation if and only if $v$ is $S$-invariant.

By the description of the tangent sheaf $T_X$ in Section 4.3 of Fulton's book, $H^1(X,T_X)$ equals the direct sum over all irreducible $T$-divisors $D$ of $H^1(D,N_{D/X})$, where $N_{D/X}$ is the normal sheaf of $D$ in $X$, i.e., $\mathcal{O}_X(D)|_D$. For such a $D$, there is a one-parameter subgroup $S_D$ of $T$ which fixes $D$ pointwise, not just setwise (this can be checked on a $T$-invariant open affine which intersects $D$, hence follows from the special case that $X$ equals $\mathbb{A}^m$ cross a torus). The invertible sheaf $N_{D/X}$ is $T$-linearized, hence $S_D$-linearized. But $S_D$ acts trivially on $D$, hence this $S_D$-linearization is just scaling by a character of $S_D$. Again by looking at a $T$-invariant open affine, one can see that this character is an embedding, i.e., it has trivial kernel. This $S_D$-linearization gives an action of $S_D$ on $H^1(D,N_{D/X})$ which is scaling by the same character (or perhaps the dual character, depending on your sign conventions). In particular, for a first-order deformation of $X$ corresponding to a vector $v$ in $\oplus_D H^1( D, N_{D/X} )$, if the projection of $v$ into the factor $H^1(D,N_{D/X})$ is nonzero, then $v$ is not $S_D$-invariant. Thus the action of $S_D$ on $X$ does not extend to the first-order deformation, in fact the action of no nontrivial subgroup of $S_D$ extends.

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