Second Hirzebruch surface as Delzant space associated to trapezoid I am trying to understand how the second Hirzebruch surface arises as the Delzant space associated to the trapezoid $\Delta \in (\mathbb{R}^2)^\ast$ given by the vertices $(0,0) , (1,0), (1-a,a), (0,a)$. 
Applying the Delzant construction to the trapezoid, for the associated Delzant space I obtain, with $N \simeq \mathbb{T}^2$, $z = (z_1, \ldots, z_4) \in \mathbb{C}^4$,
$$
M_\Delta = \{(z_1, \ldots, z_4) \text{ } | \text{ }  \vert z_1 \vert^2 + \vert z_2 \vert^2 + \vert z_3 \vert^2 = 1, \vert z_2 \vert^2 + \vert z_4 \vert^2 = a \} / N,
$$ 
where the action of $\mathbb{T}^2$ on $\mathbb{C}^4$ is given by 
$$
(\alpha, \beta) \cdot z = (\alpha z_1, \alpha\beta z_2, \alpha z_3, \beta z_4) \\ \mu(z) =  \tfrac{1}{2}(\vert z_1 \vert^2 + \vert z_2 \vert^2 + \vert z_3 \vert^2, \vert z_2 \vert^2 + \vert z_4 \vert^2) + (1, a).
$$
Now, first off, I have trouble understanding, why $M_\Delta$ is the same as the set 
$$
M_1 = \{(z_1, \ldots, z_4) \text{ } | \text{ } (z_1, z_3) \neq 0, (z_2, z_4) \neq 0\} / (\mathbb{C}^\times)^2\}
$$
and then second, how this is a Hirzebruch surface and how one could show that. (I have not studied Hirzebruch surfaces thoroughly, I just stumbled upon them trying to understand this example for a Delzant polytope and space and very briefly as the symplectic blow up).
As this is my first question, I hope I gave enough information concerning the context and hope my question is precise enough. 
Thanks!
 A: For the first part, it follows since you extend the torus action to a $(\mathbb{C}^*)^2$ action on $M_1$ by rescaling the coordinates. In other words,
you see $\mathbb{C}^*\times \mathbb{C}^*$ acting on $M_1$ by 
the same formula you gave for $(\alpha, \beta) \cdot z$, but now $\alpha, \beta$ are in $\mathbb{C}^*$ instead of in $S^1$.
My answer to the second question might be going around in circles from what you understand:
First, as a smooth manifold, the Hirzebruch surface can be understood as either the non-trivial $\mathbb{C}P^1$ bundle over $\mathbb{C}P^1$, or, equivalently, as $\mathbb{C}P^2$ blown up once.
A symplectic structure (up to deformation) is described by the cohomology class represented by the symplectic form, which, in the picture of $\mathbb{C}P^2 \# \overline {\mathbb{C}P^2}$ is described by the area of the line and of the exceptional divisor (with the condition that the line has greater area than the divisor).
The polytope you describe is obtained by cutting off the triangle with vertices $(1,0)$, $(0,1)$ and $(0,0)$. This triangle is $\mathbb{C}P^2$ with its standard symplectic structure. Normalize the line to have area $1$. Now, cutting the top of the triangle off corresponds to taking a blow up. The area of the exceptional class is now $a$.
Very concretely, if you consider the triangle (corresponding to $\mathbb{C}P^2$) and then look at the pre-image of the line $y=a$, you see that you obtain a sphere $S^3$. If you look at the induced foliation on $S^3$ obtained by looking at the kernel of $\omega$ restricted to the sphere, you will see that this is the Hopf fibration. The quotient by this $S^1$ action is then symplectic. Furthermore, this $S^1$ is a subgroup of the initial torus acting on the ambient manifold. I think it should be easy to check at this point that the resulting moment polytope is the one you describe.
If you want more details, look for a proof that an equivariant blow up (or symplectic cut) creates a moment polytope that is obtained by cutting off the corner of the old polytope. This is discussed, for instance, in Ana Cannas da Silva's book, available online here (or in a paper copy -- I recommend getting it):
http://www.math.ist.utl.pt/~acannas/Books/toric.pdf
see page 37 of the book (page 42 of the pdf file).
