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Can someone please explain why blow up in a Symplectic toric manifold corresponds to chopping off a corner in the Delzant polytope?

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You must restrict to the case of blowing up at a fixed point of the torus action, otherwise the manifold is no longer toric, and the remainings are nonsense. Naively one just replaces the corner with a $\mathbb{P}^{n-1}$. Note that onece this is done, the $T^n$-action will have degenerate orbits when acting on this exceptional divisor $E\cong\mathbb{P}^{n-1}$. The generic orbits are easily seen to be isomorphic to $T^{n-1}$, that's why under the moment map of the effective $T^n$-action, the exceptional divisor $E$ gives a boundary face $F$ of the newly obtained moment polytope $\Delta_M$. The blowing up process is then realized as replacing a toric fixed point with the moment polytope $\Delta_E$ of $\mathbb{P}^{n-1}$. Note that $\Delta_E=F\subset\Delta_M$.

As an illustrative example, consider $\mathbb{C}^n$ blowing up at the origin. The result is the total space of $\mathscr{O}(-1)\rightarrow\mathbb{P}^{n-1}$, which is a non-compact toric variety. The fixed point $0\in\mathbb{C}^n$ is now replaced by a simplex under the moment map, and this simplex is exactly the moment polytope of $\mathbb{P}^{n-1}$. In fact, blowing up a smooth point will give you the same local geometry, and the normal bundle of the exceptional divisor in the whole space is always $\mathscr{O}(-1)$, so the general case is in fact not more complicated.

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This is Homework 22 in Ana Cannas da Silva's Lectures on Symplectic Geometry (2001). It's also described in her articles Symplectic toric manifolds (2003, p.124) and Symplectic Geometry (2006, p.172).

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