Consider a rational normal curve $C\subset\mathbb{P}^d$ of degree $d$, and let $W\subset\mathbb{P}^{d+1}$ be a cone over $C$.
Since $C$ is a toric variety $W$ is toric as well. I would like to ask what is the polytope in $\mathbb{R}^2$ associated to the toric surface $W$.
Thank you very much.
This variety can be obtained by taking the Hirzebruch surface $\mathbb{P}_{\mathbb{P}^1}(\mathcal{O} \oplus \mathcal{O}(n))$, and contracting the section with self intersection $-n$. This may be done by manipulating the Delzant polytope accordingly.
I try to express this using a concrete example and then I hope the idea will be clear.
The Delzant polytope for the Hirzebruch surface $\mathbb{P}_{\mathbb{P}^1}(\mathcal{O} \oplus \mathcal{O}(1)) = Bl_{p}(\mathbb{P^2})$ is the quadrilateral with vertices $(0,0),(0,2),(1,2),(3,0)$. Then contracting the $-1$-curve, we "add in the vertex" along the edge corresponding to the minus one curve, giving the triangle with vertices $(3,0),(0,0),(0,3)$ i.e. the Delzant polytope of $\mathbb{CP}^2$.
This "adding the corner" procedure is the general way of contracting an (invariant) curve with negative self intersection in a toric surface. Take the edge corresponding this curve and continue the two edges adjacent to it until they meet; then add this region to the polytope. I leave you to apply this to the polytopes associated to the other Hirzebruch surfaces.
The final result with give (up to scale) the triangle in $\mathbb{R}^2$ with vertices $(0,0),(0,1)$ and $(n,0)$.