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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}^1$$\mathbb{P}^{n-1}$. OnNote that onece this is done, the other hand$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 curvedivisor $E$ gives a boundary face $F$ of the newly obtained moment polytope $\Delta_M$. The blowing up process is alwaysthen realized as replacing a toric fixed point with the moment polytope -1 curve$\Delta_E$ of $\mathbb{P}^{n-1}$. Note that $\Delta_E=F\subset\Delta_M$.

As an illustrative example, so locallyconsider $\mathbb{C}^n$ blowing up at the symplectic manifold looks likeorigin. The result is the total space $M$ of $\mathscr{O}(-1)\rightarrow\mathbb{P}^1$$\mathscr{O}(-1)\rightarrow\mathbb{P}^{n-1}$, which is also a non-compact toric symplectic manifoldvariety. The fixed point $0\in\mathbb{C}^n$ is now replaced by a simplex under the moment map, and itsthis simplex is exactly the moment polytope of $P_M$ can be described explicitly$\mathbb{P}^{n-1}$. ThusIn fact, blowing up a smooth point will give you the moment polytopesame local geometry, and the normal bundle of your toric symplectic manifold after blow up looks likethe exceptional divisor in the whole space is always $P_M$$\mathscr{O}(-1)$, which explainsso the reasongeneral case is in fact not more complicated.

You must restrict to the case of blowing up at a fixed point of the torus action, otherwise the manifold is no longer toric. Naively one just replaces the corner with a $\mathbb{P}^1$. On the other hand, the exceptional curve is always a -1 curve, so locally the symplectic manifold looks like the total space $M$ of $\mathscr{O}(-1)\rightarrow\mathbb{P}^1$, which is also a toric symplectic manifold, and its moment polytope $P_M$ can be described explicitly. Thus the moment polytope of your toric symplectic manifold after blow up looks like $P_M$, which explains the reason.

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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You must restrict to the case of blowing up at a fixed point of the torus action, otherwise the manifold is no longer toric. Naively one just replaces the corner with a $\mathbb{P}^1$. On the other hand, the exceptional curve is always a -1 curve, so locally the symplectic manifold looks like the total space $M$ of $\mathscr{O}(-1)\rightarrow\mathbb{P}^1$, which is also a toric symplectic manifold, and its moment polytope $P_M$ can be described explicitly. Thus the moment polytope of your toric symplectic manifold after blow up looks like $P_M$, which explains the reason.