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.

and the basesuch 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. $\endgroup$ – Allen Knutson Jul 4 '11 at 17:37