Assume $(X,\omega)$ is a compact real $2n$-dimensional symplectic manifold with a Hamiltonian torus action corresponding to the moment map $\mu:X\to \mathfrak{t}^*\cong \mathbb{R}^n$. In this situation Atiyah-Guillemin-Sternberg convexity theorem states that $\mu(X)=\Delta$ is a rational convex polytope. Suppose $\xi\in\mathfrak{t}$ generates a circle subgroup $S^1_\xi\subset \mathbb{T}^k$. We further assume that $\xi$ is prime in the sense that $\xi\neq m\xi'$, $m>1$, for another element with the same property.

The action of $S^1_\xi$ is Hamiltonian with moment map $h_\xi=\left\langle \mu,\xi\right\rangle$ and it commutes with the action of $\mathbb{T}^k$. However, the $S^1_\xi$-action may not be free. For a regular value $\epsilon\in \mathbb{R}$ of $h_\xi$, if the $S^1_\xi$-action is free, the ordinary symplectic cut of Lerman[Remark 1.5] along $$V_\epsilon= \{ x\in X: h_\xi(x)=\epsilon \}\subset X$$ produces two symplectic manifolds $(X_+,\omega_+,\mu_+)$ and $(X_-,\omega_-,\mu_-)$ with induced $\mathbb{T}^k$-actions and moment polytopes

$$ \Delta_+=\Delta\cap\{ p\in \mathfrak{t}^*: \left\langle p,\xi\right\rangle\geq \epsilon\},\quad \Delta_-=\Delta\cap\{ p\in \mathfrak{t}^*: \left\langle p,\xi\right\rangle\leq \epsilon\},$$ respectively.

Now, I think: for generic $\epsilon$, the $S^1_\xi$-action is free if and only if the resulting polytopes are smooth, i.e. along each boundary strata, the normal vectors to the intersecting codimension-one faces are part of an integral basis for $\mathfrak{t}$.

Question: Where can I find a reference for this claim? It should have been stated somewhere.

Assume $(X,\omega)$ is a compact real $2n$-dimensional symplectic manifold with a Hamiltonian torus action corresponding to the moment map $\mu:X\to \mathfrak{t}^*\cong \mathbb{R}^k$. In this situation Atiyah-Guillemin-Sternberg convexity theorem states that $\mu(X)=\Delta$ is a rational convex polytope. Suppose $\xi\in\mathfrak{t}$ generates a circle subgroup $S^1_\xi\subset \mathbb{T}^k$. We further assume that $\xi$ is prime in the sense that $\xi\neq m\xi'$, $m>1$, for another element with the same property.

The action of $S^1_\xi$ is Hamiltonian with moment map $h_\xi=\left\langle \mu,\xi\right\rangle$ and it commutes with the action of $\mathbb{T}^k$. However, the $S^1_\xi$-action may not be free. For a regular value $\epsilon\in \mathbb{R}$ of $h_\xi$, if the $S^1_\xi$-action is free, the ordinary symplectic cut of Lerman[Remark 1.5] along $$V_\epsilon= \{ x\in X: h_\xi(x)=\epsilon \}\subset X$$ produces two symplectic manifolds $(X_+,\omega_+,\mu_+)$ and $(X_-,\omega_-,\mu_-)$ with induced $\mathbb{T}^k$-actions and moment polytopes

$$ \Delta_+=\Delta\cap\{ p\in \mathfrak{t}^*: \left\langle p,\xi\right\rangle\geq \epsilon\},\quad \Delta_-=\Delta\cap\{ p\in \mathfrak{t}^*: \left\langle p,\xi\right\rangle\leq \epsilon\},$$ respectively.

Now, I think: for generic $\epsilon$, the $S^1_\xi$-action is free if and only if the resulting polytopes are smooth, i.e. along each boundary strata, the normal vectors to the intersecting codimension-one faces are part of an integral basis for $\mathfrak{t}$.

Question: Where can I find a reference for this claim? It should have been stated somewhere.