Delzant type theorem for torus-equivariantly embedded toric hypersurfaces

Kentaro Yamaguchi Department of Mathematical Sciences, Tokyo Metropolitan University, 1-1 Minami-Ohsawa, Hachioji, Tokyo, 192-0397, Japan [email protected]

Abstract.

In the previous work, we study the closure of a complex subtorus in a toric manifold given by the data of an affine subspace. We call it torus-equivariantly embedded toric manifold when the closure of a complex subtorus is a smooth complex submanifold. In this paper, we clearify the condition for nonsingularity of the closure of a complex subtorus in terms of polytopes. The main result is a generalization of Delzant theorem to the case of torus-equivariantly embedded toric hypersurfaces.

Key words and phrases:

Delzant correspondence, toric Kähler manifold, torus-equivariantly embedding

2020 Mathematics Subject Classification:

Primary 53C40; Secondary 53D20, 14M25

1. Introduction

Symplectic toric manifolds are $2n$ -dimensional symplectic manifolds equipped with the effective Hamiltonian action of an $n$ -dimensional torus. The image of the moment map for the Hamiltonian torus action is the convex hull of the image of the fixed points of the Hamiltonian action [Ati82, GS82]. Due to the work of Delzant [Del88], there is a one-to-one correspondence between symplectic toric manifolds and certain convex polytopes appeared as the moment polytopes of symplectic toric manifolds. The moment polytopes of symplectic toric manifolds are called Delzant polytopes. Moreover, symplectic toric manifolds canonnically admit a Kähler metric, which is called a Guillemin metric [Gui94a, Gui94b, CDG03]. Using the data of a Delzant polytope, the complements of the toric diviors $D$ in the corresponding symplectic toric manifold $X$ can be identified with the complex torus $(\mathbb{C}^{\ast})^{n}$ .

In [Yam24b], we construct the closure $\overline{C(V)}$ of the $k$ -dimensional complex subtorus $C(V)$ in the toric divior complement $X\setminus D\cong(\mathbb{C}^{\ast})^{n}$ from the data of a $k$ -dimensional affine subspace $V\subset\mathfrak{t}^{n}\cong\mathbb{R}^{n}$ . The closure $\overline{C(V)}$ can be expressed by the zero locus of polynomials $f^{\lambda}_{k+1},\ldots,f^{\lambda}_{n}$ for each vertex $\lambda$ of the Delzant polytope, which are determined by the data of the Delzant polytope and the affine subspace. Note that the closure $\overline{C(V)}$ might have singularity at the intersection with the toric diviors $D$ . We show in [Yam24b, Theorem 4.20] that if the closure $\overline{C(V)}$ is a smooth $k$ -dimensional complex submanifold in the symplectic toric manifold $X$ , then the moment polytope for the Hamiltonian $k$ -dimensional torus action on $\overline{C(V)}$ coincides with the moment polytope for the Hamiltonian $k$ -dimensional subtorus action on $X$ . Moreover, the submanifold $\overline{C(V)}$ is a symplectic toric manifold with respect to the $k$ -dimensional torus action on $\overline{C(V)}$ . In particular, if $\overline{C(V)}$ is a complex submanifold, then the moment polytope for the Hamiltonian $k$ -dimensional subtorus action on $X$ is a Delzant polytope. We call the complex submanifold $\overline{C(V)}$ the torus-equivariantly embedded toric manifold.

Even though the moment polytope for the Hamiltonian $k$ -dimensional subtorus action on $X$ is a Delzant polytope as a convex polytope for some affine subspace $V$ , $\overline{C(V)}$ might not be a smooth complex submanifold. For example, let $X=\mathbb{C}P^{2}$ . For any $1$ -dimensional affine subspace $V$ in $\mathfrak{t}^{2}\cong\mathbb{R}^{2}$ , the moment polytope for the Hamiltonian one-dimensional subtorus action associated with $V$ becomes an interval in $(\mathfrak{t}^{1})^{\ast}\cong\mathbb{R}$ , which can be seen as the Delzant polytope of $\mathbb{C}P^{1}$ . However, there are many affine subspaces $V$ such that the closure $\overline{C(V)}$ has a singularity (see [Yam24b, Section 5.1]).

1.1. Main Results

In this paper, we show that a Delzant type correspondence for torus-equivariantly embedded toric manifolds of codimension one, i.e. $k=n-1$ .

Let $V$ be an $(n-1)$ -dimensional affine subspace in $\mathfrak{t}^{n}\cong\mathbb{R}^{n}$ with rational slope, i.e. $\overline{V}\cap\mathbb{Z}^{n}\cong\mathbb{Z}^{n-1}$ , where $\overline{V}$ is the linear part of $V$ . Then, we may take a primitive $\mathbb{Z}$ -basis $p_{1},\ldots,p_{n-1}\in\mathbb{Z}^{n}$ of $\overline{V}\cap\mathbb{Z}^{n}$ and $a\in\mathbb{R}^{n}$ such that $V=\mathbb{R}p_{1}+\cdots+\mathbb{R}p_{n-1}+a$ . We define the injective homomorphism $i_{V}:T^{n-1}\to T^{n}$ by

i_{V}(t_{1},\ldots,t_{n-1}):=\left(\prod_{l=1}^{n-1}t_{l}^{\langle p_{l},e_{1}\rangle},\ldots,\prod_{l=1}^{n-1}t_{l}^{\langle p_{l},e_{n}\rangle}\right),

where $e_{1},\ldots,e_{n}$ are the standard basis of $(\mathfrak{t}^{n})^{\ast}\cong\mathbb{R}^{n}$ . We introduce some notions concerning the pair of a Delzant polytope and an affine subspace.

Definition 1.1.

Let $\Delta$ be a Delzant polytope and $V=\mathbb{R}p_{1}+\cdots+\mathbb{R}p_{n-1}+a$ an $(n-1)$ -dimensional affine subspace in $\mathfrak{t}^{n}$ with rational slope. A vertex $\lambda$ of the polytope $\Delta$ is a good vertex with respect to the pullback $i_{V}^{\ast}:(\mathfrak{t}^{n})^{\ast}\to(\mathfrak{t}^{n-1})^{\ast}$ if the vertex $\lambda$ satisfies the two conditions:

(1)

$i_{V}^{\ast}(\lambda)$ is a vertex of the convex polytope $i_{V}^{\ast}(\Delta)$ ,
(2)

for the direction vectors $v^{\lambda}_{1},\ldots,v^{\lambda}_{n}$ of the edges from the vertex $\lambda$ of the polytope $\Delta$ , the vectors $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n})$ in $(\mathfrak{t}^{n-1})^{\ast}$ are all nonzero.

Definition 1.2.

Let $V=\mathbb{R}p_{1}+\cdots+\mathbb{R}p_{n-1}+a$ be an $(n-1)$ -dimensional affine subspace in $\mathfrak{t}^{n}$ with rational slope. A Delzant polytope $\Delta$ is a good polytope with respect to the map $i_{V}^{\ast}$ if any good vertex $\lambda$ with respect to the map $i_{V}^{\ast}$ satisfies the two conditions:

•

we can choose $n-1$ direction vectors $v^{\lambda}_{1},\ldots,v^{\lambda}_{j-1},v^{\lambda}_{j+1},\ldots,v^{\lambda}_{n}$ such that $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{j-1}),i_{V}^{\ast}(v^{\lambda}_{j+1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n})$ are linearly independent and $i_{V}^{\ast}(\{\sum_{i=1}^{n}a_{i}v^{\lambda}_{i}\mid a_{i}\geq 0\})=\{\sum_{i\neq j}b_{i}i_{V}^{\ast}(v^{\lambda}_{i})\mid b_{i}\geq 0\}$ ,
•

the direction vectors $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{j-1}),i_{V}^{\ast}(v^{\lambda}_{j+1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n})\in\mathbb{Z}^{n-1}$ from the vertex $i_{V}^{\ast}(\lambda)$ of $i_{V}^{\ast}(\Delta)$ form a $\mathbb{Z}$ -basis of $\mathbb{Z}^{n-1}$ .

Note that the first condition in Definition 1.2 are equivalent to the condition that the polytope $i_{V}^{\ast}(\Delta)$ is simple.

The closure $\overline{C(V)}$ can be expressed by the zero locus of a polynomial $f^{\lambda}$ for each vertex $\lambda$ of $\Delta$ . Our first main result is as follows:

Theorem 1.3 (Theorem 3.15).

Let $\Delta$ be a Delzant polytope and $V=\mathbb{R}p_{1}+\cdots+\mathbb{R}p_{n-1}+a$ an $(n-1)$ -dimensional affine subspace in $\mathfrak{t}^{n}$ with rational slope. If the Delzant polytope $\Delta$ is good with respect to the map $i_{V}^{\ast}$ , then the rank of the Jacobian matrix $Df^{\lambda}$ is equal to one at any point of the zero locus of $f^{\lambda}$ for any vertex $\lambda$ of the Delzant polytope $\Delta$ . In particular, $\overline{C(V)}$ is a smooth complex hypersurface in the toric manifold $X$ .

Since complex hypersurfaces $\overline{C(V)}$ are symplectic toric manifolds with respect to the Hamiltonian $T^{n-1}$ -action, we obtain the following:

Corollary 1.4.

If the Delzant polytope $\Delta$ of a toric manifold $X$ is good with respect to the map $i_{V}^{\ast}$ , then the convex polytope $i_{V}^{\ast}(\Delta)$ is a Delzant polytope of the complex hypersurface $\overline{C(V)}$ in $X$ .

We also show the converse of Theorem 1.3.

Theorem 1.5 (Theorem 4.13).

Let $V=\mathbb{R}p_{1}+\cdots+\mathbb{R}p_{n-1}+a$ be an $(n-1)$ -dimensional affine subspace in $\mathfrak{t}^{n}$ with rational slope. If the rank of the Jacobian matrix $Df^{\lambda}$ is equal to one at any point of the zero locus of $f^{\lambda}$ for any vertex $\lambda$ of the Delzant polytope $\Delta$ of the toric manifold $X$ , then the Delzant polytope $\Delta$ is good with respect to the map $i_{V}^{\ast}$ .

From Theorem 1.3 and Theorem 1.5, we obtain the following:

Corollary 1.6.

Let $\Delta$ be a Delzant polytope of a symplectic toric manifold $X$ . Then, $\overline{C(V)}$ is a torus-equivariantly embedded toric hypersurface in $X$ if and only if $\Delta$ is a good Delzant polytope with respect to the map $i_{V}^{\ast}$ .

Corollary 1.6 characterizes the condition when $\overline{C(V)}$ is a smooth complex hypersurface in terms of the combinatorics of Delzant polytopes.

1.2. Sketch of the Proofs

We explain the sketch of the proof of Theorem 1.3. First, in Section 3.1 we show that if the vertex $\lambda$ of the Delzant polytope $\Delta$ does not satisfy the condition (2) in Definition 1.1, then the rank of the Jacobian matrix $Df^{\lambda}$ is equal to one. Next, in Section 3.2 we show that if the vertex $\lambda$ satisfies the condition (2) but does not satisfy the condition (1) in Definition 1.1, then the rank of the Jacobian matrix $Df^{\lambda}$ is equal to one. Finally, in Section 3.3 we consider the case when the vertex $\lambda$ is good with respect to the map $i_{V}^{\ast}$ . The detailed proof of Theorem 1.3 is given in Section 3.4.

The proof of Theorem 1.5 is based on the observation of the conditions of the sets $\mathcal{I}^{+}_{\lambda},\mathcal{I}^{-}_{\lambda}$ , which are defined in Definition 2.3.

1.3. Outline

This paper is organized as follows. In Section 2, we prepare for something about torus-equivariantly embedded toric hypersurfaces. In Section 3, we show Theorem 1.3. In Section 4, we show Theorem 1.5.

Acknowlegements

The author is grateful to the advisor, Manabu Akaho, for a lot of encouragements and supports. The author would also like to thank Yuichi Kuno and Yasuhito Nakajima for helpful discussions. This work was supported by JST SPRING, Grant Number JPMJSP2156.

2. Reveiw of Torus-equivariantly Embedded Toric Manifolds

In this section, we review the settings in [Yam24b]. We also show some properties of the rank of the Jacobian matrix $Df^{\lambda}$ used in this paper.

2.1. Symplectic Toric Manifolds

In [Del88], symplectic toric manifolds are completely classified by the certain convex polytopes, known as Delzant polytopes:

Definition 2.1.

A convex polytope $\Delta$ in $(\mathfrak{t}^{n})^{\ast}\cong\mathbb{R}^{n}$ is Delzant if $\Delta$ satisfies the following three properties:

•

simple; each vertex has $n$ edges,
•

rational; the direction vectors $v^{\lambda}_{1},\ldots,v^{\lambda}_{n}$ of the edges from any vertex $\lambda\in\Lambda$ can be chosen as integral vectors,
•

smooth; the vectors $v^{\lambda}_{1},\ldots,v^{\lambda}_{n}\in\mathbb{Z}^{n}$ chosen as above form a $\mathbb{Z}$ -basis of $\mathbb{Z}^{n}$ .

Here, $\Lambda$ denotes the set of the vertices of $\Delta$ .

Some literatures (for example, [CDG03]) define Delzant polytopes in terms of the inward pointing normal vectors $u^{\lambda}_{1},\ldots,u^{\lambda}_{n}$ to the $n$ facets sharing each vertex instead of the direction vectors from each vertex. In fact, these two ways to define Delzant polytopes are equivalent because of the following equation:

{{}^{t}[u^{\lambda}_{1}\;\cdots\;u^{\lambda}_{n}]}[v^{\lambda}_{1}\;\cdots\;v^{\lambda}_{n}]=E_{n}

holds for any vertex $\lambda$ of $\Delta$ (see [Yam24b, Lemma 3.10] for example).

Due to the Delzant construction, we obtain the corresponding symplectic toric manifold $X$ from the data of a given Delzant polytope $\Delta$ . Here we give the expression of a system of the inhomogeneous coordinate charts on the corresponding symplectic toric manifold $X$ from the data of the Delzant polytope $\Delta$ . We define two matrices $Q^{\lambda}(=[Q^{\lambda}_{ij}])$ and $D^{\lambda\sigma}(=[d^{\lambda\sigma}_{ij}])$ by

Q^{\lambda}=\left[v^{\lambda}_{1}\;\cdots\;v^{\lambda}_{n}\right],\;D^{\lambda\sigma}=(Q^{\lambda})^{-1}Q^{\sigma}

for any vertices $\lambda,\sigma\in\Lambda$ .

Lemma 2.2.

From the data of a given Delzant polytope $\Delta$ , we can construct an open covering $\{U_{\lambda}\}_{\lambda\in\Lambda}$ of the corresponding toric manifold $X$ and a set of maps $\{\varphi_{\lambda}:U_{\lambda}\to\mathbb{C}^{n}\}_{\lambda\in\Lambda}$ such that

\varphi_{\sigma}\circ\varphi_{\lambda}^{-1}(z^{\lambda})=\left(\prod_{j=1}^{n}(z^{\lambda}_{j})^{d^{\lambda\sigma}_{j1}},\ldots,\prod_{j=1}^{n}(z^{\lambda}_{j})^{d^{\lambda\sigma}_{jn}}\right)

for any vertices $\lambda,\sigma\in\Lambda$ such that $U_{\lambda}\cap U_{\sigma}\neq\emptyset$ .

The detailed construction in terms of the notation used here is given in [Yam24b, Section 2.1 and Section 3.2]. Moreover, the complement of toric diviors in a $2n$ -dimensional symplectic toric manifold can be identified with a complex $n$ -dimensional torus.

2.2. Torus-equivariantly Embedded Toric Hypersurfaces

Let $p_{1},\ldots,p_{n-1}\in\mathbb{Z}^{n}$ be primitive vectors which are linearly independent, and $a\in\mathbb{R}^{n}$ . We consider the $(n-1)$ -dimensional affine subspace $V=\mathbb{R}p_{1}+\cdots+\mathbb{R}p_{n-1}+a$ in $\mathfrak{t}^{n}\cong\mathbb{R}^{n}$ . Write $\overline{V}$ as the linear part of $V$ . Assume that $V$ has a rational slope, i.e. we may assume that $p_{1},\ldots,p_{n-1}\in\mathbb{Z}^{n}$ form a $\mathbb{Z}$ -basis of $\overline{V}\cap\mathbb{Z}^{n}\cong\mathbb{Z}^{n-1}$ . Let $q\in\mathbb{Z}^{n}$ be a primitive basis of the orthogonal subspace to $\overline{V}$ in $\mathfrak{t}^{n}\cong\mathbb{R}^{n}$ .

Through the log-affine coordinate $(\mathbb{C}^{n})^{\ast}\cong\mathfrak{t}^{n}\times T^{n}$ , we define the complex subtorus $C(V)\cong V\times T^{n-1}$ (see [Yam24b, Proposition 4.2] for detail). Recall that the complement of toric diviors in a toric manifold $X$ can be identified with a complex $n$ -dimensional torus $(\mathbb{C}^{\ast})^{n}$ . The closure $\overline{C(V)}$ of $C(V)$ in the toric manifold $X$ is expressed as follows:

Definition 2.3.

The closure $\overline{C(V)}$ of $C(V)$ in the toric manifold $X$ is defined as $\overline{C(V)}=\bigcup_{\lambda\in\Lambda}\varphi_{\lambda}^{-1}(\overline{C_{\lambda}(V)})$ , where $\overline{C_{\lambda}(V)}\subset\varphi_{\lambda}(U_{\lambda})\cong\mathbb{C}^{n}$ is a zero locus of $f^{\lambda}$ . Here, $f^{\lambda}$ are defined by

f^{\lambda}(z^{\lambda})=\prod_{j\in\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}}(z^{\lambda}_{j})^{\langle u^{\lambda}_{j},q\rangle}-e^{\langle a,q\rangle}\prod_{j\in\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}}(z^{\lambda}_{j})^{-\langle u^{\lambda}_{j},q\rangle},

where

	$\displaystyle\mathcal{I}^{+}_{\lambda}=\{i\in\{1,\ldots,n\}\mid\langle u^{\lambda}_{i},q\rangle\geq 0\},$
	$\displaystyle\mathcal{I}^{-}_{\lambda}=\{i\in\{1,\ldots,n\}\mid\langle u^{\lambda}_{i},q\rangle\leq 0\},$
	$\displaystyle\mathcal{I}^{0}_{\lambda}=\{i\in\{1,\ldots,n\}\mid\langle u^{\lambda}_{i},q\rangle=0\}.$

Note that if $\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\emptyset$ , then $\prod_{j\in\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}}(z^{\lambda}_{j})^{\langle u^{\lambda}_{j},q\rangle}=1$ . Similarly, if $\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\emptyset$ , then $\prod_{j\in\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}}(z^{\lambda}_{j})^{-\langle u^{\lambda}_{j},q\rangle}=1$ .

Remark 2.4.

The rank of the Jacobian matrix $Df^{\lambda}$ of $f^{\lambda}$ might be equal to zero at some points in the toric diviors. In particular, $\overline{C(V)}$ might have a singularity at some points in the toric diviors.

Proposition 2.5.

If $\overline{C(V)}$ is a complex submanifold in the toric manifold $X$ , then $\overline{C(V)}$ is toric with respect to the $T^{n-1}$ -action on $\overline{C(V)}$ .

Proof.

We show that the $T^{n-1}$ -action on $\overline{C(V)}$ is effective, Hamiltonian. Define the injective homomorphism $i_{V}:T^{n-1}\to T^{n}$ by

i_{V}(t_{1},\ldots,t_{n-1})=\left(\prod_{l=1}^{n-1}t_{l}^{\langle p_{l},e_{1}\rangle},\ldots,\prod_{l=1}^{n-1}t_{l}^{\langle p_{l},e_{n}\rangle}\right).

Since the toric divior complement $X\setminus D$ can be identified with a complex torus $(\mathbb{C}^{\ast})^{n}$ , we obtain $X\setminus D\cong\mathfrak{t}^{n}\times T^{n}$ . Since $C(V)\cong(\mathbb{C}^{\ast})^{n-1}$ is a complex subtorus in $(\mathbb{C}^{\ast})^{n}\cong X\setminus D$ , we obtain the identification $C(V)\cong V\times i_{V}(T^{n-1})\subset\mathfrak{t}^{n}\times T^{n}\cong X\setminus D$ . Since we defined the $T^{n-1}$ -action on $\overline{C(V)}$ through the map $i_{V}:T^{n-1}\to T^{n}$ and the $T^{n}$ -action on $X$ (see [Yam24b]), we see that the $T^{n-1}$ -action on $\overline{C(V)}$ is effective.

If $\overline{C(V)}$ is a complex submanifold in $X$ , then there exists an embedding $i:\overline{C(V)}\to X$ . Thus, a symplectic form on $\overline{C(V)}$ is $i^{\ast}\omega$ , where $\omega$ is a symplectic form on $X$ . Let $R_{t}:\overline{C(V)}\to\overline{C(V)}$ be a map defined by

R_{t}(x)=x\cdot t,

where $\cdot$ is the $T^{n-1}$ -action on $\overline{C(V)}$ . By the definition of the $T^{n-1}$ -action on $\overline{C(V)}$ , we obtain

R_{t}^{\ast}(i^{\ast}\omega)=i^{\ast}R_{t}^{\ast}\omega=i^{\ast}\omega,

for any $t\in T^{n-1}$ . Thus, the $T^{n-1}$ -action on $\overline{C(V)}$ preserves the symplectic form $i^{\ast}\omega$ on $\overline{C(V)}$ . We show that the map $\overline{\mu}:=i_{V}^{\ast}\circ\mu\circ i:\overline{C(V)}\to(\mathfrak{t}^{n-1})^{\ast}$ is the moment map for the $T^{n-1}$ -action on $\overline{C(V)}$ , where $\mu:X\to(\mathfrak{t}^{n})^{\ast}$ is the moment map for the $T^{n}$ -action on $X$ . By the definition of the $T^{n-1}$ -action on $\overline{C(V)}$ , the map $\overline{\mu}$ is equivariant with respect to the $T^{n-1}$ -action on $\overline{C(V)}$ . For any point $p\in\overline{C(V)}$ and $X\in\mathfrak{t}^{n-1}$ , we have

	$\displaystyle\overline{\mu}_{X}(p)$	$\displaystyle=\langle i_{V}^{\ast}\mu(i(p)),X\rangle$
		$\displaystyle=\langle\mu(i(p)),(i_{V})_{\ast}X\rangle$
		$\displaystyle=\langle\mu(i(p)),X\rangle$
		$\displaystyle=(\mu_{X}\circ i)(p).$

From this calculation, we obtain

	$\displaystyle d\overline{\mu}_{X}$	$\displaystyle=d(\mu_{X}\circ i)$
		$\displaystyle=di^{\ast}\mu_{X}$
		$\displaystyle=i^{\ast}d\mu_{X}$
		$\displaystyle=i^{\ast}\omega(X^{\sharp},-).$

Therefore, the map $\overline{\mu}$ is the moment map for the $T^{n-1}$ -action on $\overline{C(V)}$ equipped with the symplectic form $i^{\ast}\omega$ . ∎

Note that the above discussion is valid for any codimension of $\overline{C(V)}$ .

The vectors $q,v^{\lambda}_{1},\ldots,v^{\lambda}_{n},u^{\lambda}_{1},\ldots,u^{\lambda}_{n}$ used here have the following relation:

Lemma 2.6 ([Yam24b, Lemma 4.14]).

\sum_{j=1}^{n}\langle u^{\lambda}_{j},q\rangle i_{V}^{\ast}(v^{\lambda}_{j})=0

holds.

In this case, the rank of the Jacobian matrix $Df^{\lambda}$ is independent of the choice of $a\in\mathbb{R}^{n}$ of the affine subspace $V=\mathbb{R}p_{1}+\cdots+\mathbb{R}p_{n-1}+a$ [Yam24a, Proposition 2.12], i.e. we may assume that $a=0$ .

2.3. Properties of the Rank of the Jacobian Matrices

In this section, we prepare for the proof of the main results.

Lemma 2.7.

Let $\lambda$ be a vertex of $\Delta$ .

•

If $\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\emptyset$ , then $\overline{C_{\lambda}(V)}\cap\{z^{\lambda}_{j}=0\}=\emptyset$ for any $j\in\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}$ .
•

If $\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\emptyset$ , then $\overline{C_{\lambda}(V)}\cap\{z^{\lambda}_{j}=0\}=\emptyset$ for any $j\in\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}$ .

Proof.

We assume that $\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\emptyset$ . Then, since the defining equation $f^{\lambda}(z^{\lambda})=0$ for $\overline{C_{\lambda}(V)}$ is

f^{\lambda}(z^{\lambda})=1-\prod_{j\in\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}}(z^{\lambda}_{j})^{-\langle u^{\lambda}_{j},q\rangle},

we obtain that $\overline{C_{\lambda}(V)}\cap\{z^{\lambda}_{j}=0\}=\emptyset$ for any $j\in\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}$ .

Similarly, if we assume that $\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\emptyset$ , then we obtain $\overline{C_{\lambda}(V)}\cap\{z^{\lambda}_{j}=0\}=\emptyset$ for any $j\in\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}$ . ∎

Lemma 2.8.

If $\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\emptyset$ (or $\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\emptyset$ ), then $\overline{C_{\lambda}(V)}$ is nonsingular.

Proof.

Assume that $\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\emptyset$ . From the proof of Lemma 2.7, the defining equation $f^{\lambda}(z^{\lambda})=0$ for $\overline{C_{\lambda}(V)}$ is

f^{\lambda}(z^{\lambda})=1-\prod_{j\in\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}}(z^{\lambda}_{j})^{-\langle u^{\lambda}_{j},q\rangle}.

Since we obtain

\frac{\partial f^{\lambda}}{\partial z^{\lambda}_{i}}=\begin{cases*}\langle u^{\lambda}_{i},q\rangle\prod_{j\in\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}}(z^{\lambda}_{j})^{-\langle u^{\lambda}_{j},q\rangle-\delta_{ij}}&if $i\in\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}$,\\ 0&otherwise,\end{cases*}

the points where the rank of the Jacobian matrix $Df^{\lambda}$ of $f^{\lambda}$ is equal to zero should be in $\overline{C_{\lambda}(V)}\cap(\bigcup_{j\in\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}}\{z^{\lambda}_{j}=0\})$ . However, since $\overline{C_{\lambda}(V)}\cap(\bigcup_{j\in\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}}\{z^{\lambda}_{j}=0\})=\emptyset$ from Lemma 2.7, the rank of $Df^{\lambda}$ is equal to one, i.e. $\overline{C_{\lambda}(V)}$ does not have a singular point.

Similarly, we assume that $\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\emptyset$ , then we obtain that the rank of $Df^{\lambda}$ is equal to one. ∎

Lemma 2.9.

Assume that

\mathcal{I}^{+}_{\lambda}=\{1,\ldots,n-1\},\;\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\{n\}

\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\{n\},\;\mathcal{I}^{-}_{\lambda}=\{1,\ldots,n-1\}.

If $\langle u^{\lambda}_{n},q\rangle=\pm 1$ , then $\overline{C_{\lambda}(V)}$ is nonsingular.

Proof.

Assume that $\mathcal{I}^{+}_{\lambda}=\{1,\ldots,n-1\},\;\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\{n\}$ . From the defining equation for $\overline{C_{\lambda}(V)}$ , we obtain that

\frac{\partial f^{\lambda}}{\partial z^{\lambda}_{i}}=\begin{cases*}\langle u^{\lambda}_{i},q\rangle\prod_{j\in\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}}(z^{\lambda}_{j})^{\langle u^{\lambda}_{j},q\rangle-\delta_{ij}}&if $i=1,\ldots,n-1$,\\ \langle u^{\lambda}_{n},q\rangle(z^{\lambda}_{n})^{-\langle u^{\lambda}_{n},q\rangle-1}&if $i=n$.\end{cases*}

If $\langle u^{\lambda}_{n},q\rangle=-1$ , then

\frac{\partial f^{\lambda}}{\partial z^{\lambda}_{n}}=-1.

Therefore, the rank of the Jacobian matrix $Df^{\lambda}$ is equal to one, i.e. $\overline{C_{\lambda}(V)}$ is nonsingular. ∎

3. Delzant Polytopes of Torus-equivariantly Embedded Toric Hypersurfaces

In this section, we show the first main result (Theorem 3.15).

Since $i_{V}^{\ast}:(\mathfrak{t}^{n})^{\ast}\to(\mathfrak{t}^{n-1})^{\ast}$ is surjective, we have the following:

Lemma 3.1.

Let $v_{1},\ldots,v_{n}\in\mathbb{R}^{n}\cong(\mathfrak{t}^{n})^{\ast}$ be linearly independent. There exist $j_{1},\ldots,j_{n-1}\in\{1,\ldots,n\}$ such that $i_{V}^{\ast}(v_{j_{1}}),\ldots,i_{V}^{\ast}(v_{j_{n-1}})\in(\mathfrak{t}^{n-1})^{\ast}\cong\mathbb{R}^{n-1}$ are linearly independent.

Hereafter, we assume that $j_{1}=1,\ldots,j_{n-1}=n-1$ otherwise specified.

3.1. Case of $\mathcal{J}_{\lambda}\neq\emptyset$

Let $\lambda$ be a vertex of the Delzant polytope $\Delta$ . Define $\mathcal{J}_{\lambda}=\{i\mid\langle p_{1},v^{\lambda}_{i}\rangle=\cdots=\langle p_{n-1},v^{\lambda}_{i}\rangle=0\}$ . In this section, we show that if the vertex $\lambda$ of $\Delta$ does not satisfy the condition (2) in Definition 1.1, then $\overline{C_{\lambda}(V)}$ is nonsingular.

Lemma 3.2.

The vertex $\lambda$ of $\Delta$ does not satisfy the condition (2) in Definition 1.1 if and only if $\mathcal{J}_{\lambda}\neq\emptyset$ .

Proof.

If the vertex $\lambda$ of $\Delta$ does not satisfy the condition (2) in Definition 1.1, then there exists a direction vector $v^{\lambda}_{j}$ such that the vector $i_{V}^{\ast}(v^{\lambda}_{j})\in(\mathfrak{t}^{n-1})^{\ast}$ is zero. Note that the pullback $i_{V}^{\ast}:(\mathfrak{t}^{n})^{\ast}\to(\mathfrak{t}^{n-1})^{\ast}$ is given by

i_{V}^{\ast}(\xi)=\left(\langle p_{1},\xi\rangle,\ldots,\langle p_{n-1},\xi\rangle\right).

Since $i_{V}^{\ast}(v^{\lambda}_{j})=(0,\ldots,0)$ , we obtain $\langle p_{1},v^{\lambda}_{j}\rangle=0,\ldots,\langle p_{n-1},v^{\lambda}_{j}\rangle=0$ , i.e. $j\in\mathcal{J}_{\lambda}$ . In particular, $\mathcal{J}_{\lambda}\neq\emptyset$ .

If $\mathcal{J}_{\lambda}\neq\emptyset$ , then there exists an element $j_{0}\in\mathcal{J}_{\lambda}$ . By the definition of the set $\mathcal{J}_{\lambda}$ , we have $\langle p_{1},v^{\lambda}_{j_{0}}\rangle=0,\ldots,\langle p_{n-1},v^{\lambda}_{j_{0}}\rangle=0$ , i.e. the vector $i_{V}^{\ast}(v^{\lambda}_{j_{0}})$ is zero. ∎

In particular, $\mathcal{J}_{\lambda}\neq\emptyset$ is equivalent to $|\mathcal{J}_{\lambda}|=1$ .

Lemma 3.3.

If $\mathcal{J}_{\lambda}\neq\emptyset$ , then $\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\emptyset$ or $\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\emptyset$ .

Proof.

Let $j_{0}\in\mathcal{J}_{\lambda}$ . By the definition of $\mathcal{J}_{\lambda}$ , we obtain $v^{\lambda}_{j_{0}}\in V^{\perp}=\mathbb{R}q$ . Since the vectors $v^{\lambda}_{j_{0}},q$ are nonzero, there exists $b\in\mathbb{R}\setminus\{0\}$ such that $q=bv^{\lambda}_{j_{0}}$ . Since $\langle u^{\lambda}_{i},v^{\lambda}_{j}\rangle=\delta_{ij}$ (see [Yam24b, Lemma 3.10]), we have

\langle u^{\lambda}_{i},q\rangle=\langle u^{\lambda}_{i},bv^{\lambda}_{j_{0}}\rangle=b\langle u^{\lambda}_{i},v^{\lambda}_{j_{0}}\rangle=b\delta_{ij_{0}}

for any $i=1,\ldots,n$ . If $b>0$ , then $i\in\mathcal{I}^{+}_{\lambda}$ for any $i$ , i.e. $\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\emptyset$ . If $b<0$ , then $i\in\mathcal{I}^{-}_{\lambda}$ for any $i$ , i.e. $\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\emptyset$ . ∎

From Lemma 3.3 and Lemma 2.8, we obtain the following:

Corollary 3.4.

If $\mathcal{J}_{\lambda}\neq\emptyset$ , then $\overline{C_{\lambda}(V)}$ is nonsingular.

For a vertex $\lambda$ of $\Delta$ , we define the cone $\mathcal{C}_{\lambda}:=\{\sum_{i=1}^{n}a_{i}v^{\lambda}_{i}\mid a_{i}\geq 0\}$ .

Proposition 3.5.

If $\mathcal{J}_{\lambda}\neq\emptyset$ , then $i_{V}^{\ast}(\lambda)$ is a vertex of the convex polytope $i_{V}^{\ast}(\Delta)$ .

Proof.

From Lemma 3.1, we may assume that the vectors $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n-1})$ are linearly independent. Then, $n\in\mathcal{J}_{\lambda}$ , i.e. $i_{V}^{\ast}(v^{\lambda}_{n})=0\in(\mathfrak{t}^{n-1})^{\ast}$ .

If there does not exist any nontrivial subspace in the cone $i_{V}^{\ast}(\mathcal{C}_{\lambda})$ , then $i_{V}^{\ast}(\lambda)$ is a vertex in the polytope $i_{V}^{\ast}(\Delta)$ . Let $W^{\lambda}_{V}$ be a subspace contained in the cone $i_{V}^{\ast}(\mathcal{C}_{\lambda})$ . For any element $x\in W^{\lambda}_{V}\subset i_{V}^{\ast}(\mathcal{C}_{\lambda})$ , there exist $r_{1},\ldots,r_{n}\geq 0$ such that

x=\sum_{j=1}^{n}r_{j}i_{V}^{\ast}(v^{\lambda}_{j})=\sum_{j=1}^{n-1}r_{j}i_{V}^{\ast}(v^{\lambda}_{j}).

Since $W^{\lambda}_{V}$ is a linear space, $(-x)\in W^{\lambda}_{V}$ . There exist $s_{1},\ldots,s_{n}\geq 0$ such that

-x=\sum_{j=1}^{n}s_{j}i_{V}^{\ast}(v^{\lambda}_{j})=\sum_{j=1}^{n-1}s_{j}i_{V}^{\ast}(v^{\lambda}_{j}).

Let $R_{j}:=r_{j}+s_{j}$ for $j=1,\ldots,n$ . Since $x+(-x)=0$ , we obtain

0=\sum_{j=1}^{n-1}R_{j}i_{V}^{\ast}(v^{\lambda}_{j}).

Since we assume that $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n-1})$ are linearly independent, we obtain that $R_{1}=\cdots=R_{n-1}=0$ . Since $r_{1},\ldots,r_{n-1},s_{1},\ldots,s_{n-1}\geq 0$ , we obtain $r_{1}=\cdots=r_{n-1}=s_{1}=\cdots=s_{n-1}=0$ , which means that $x=0$ . Therefore, we obtain $W^{\lambda}_{V}=\{0\}$ , i.e. there is no nontrivial subspace in the cone $i_{V}^{\ast}(\mathcal{C}_{\lambda})$ . The point $i_{V}^{\ast}(\lambda)$ is a vertex of the convex polytope $i_{V}^{\ast}(\Delta)$ . ∎

This proposition tells us that if the vertex $\lambda$ of $\Delta$ does not satisfy the condition (2) in Definition 1.1, then the vertex $\lambda$ satisfies the condition (1).

Proposition 3.6.

If $\mathcal{J}_{\lambda}\neq\emptyset$ , then the vectors $i_{V}^{\ast}(v^{\lambda}_{j_{1}}),\ldots,i_{V}^{\ast}(v^{\lambda}_{j_{n-1}})$ for $j_{1},\ldots,j_{n-1}\notin\mathcal{J}_{\lambda}$ are the direction vectors from the vertex $i_{V}^{\ast}(\lambda)$ of the polytope $i_{V}^{\ast}(\Delta)$ .

Proof.

Let $j\in\mathcal{J}_{\lambda}$ , i.e. $i_{V}^{\ast}(v^{\lambda}_{j})=0$ . From Lemma 3.1, the $(n-1)$ vectors $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{j-1}),i_{V}^{\ast}(v^{\lambda}_{j+1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n})$ are linearly independent.

For the rest of the proof, we show that

i_{V}^{\ast}\left(\left\{\sum_{i=1}^{n}a_{i}v^{\lambda}_{i}\mid a_{i}\geq 0\right\}\right)=\left\{\sum_{i\neq j}b_{i}i_{V}^{\ast}(v^{\lambda}_{i})\mid b_{i}\geq 0\right\}.

Since it is clear that the right hand side is a subset of the left hand side, we show the other inclusion. For any $a_{1},\ldots,a_{n}\geq 0$ , we obtain

i_{V}^{\ast}\left(\sum_{i=1}^{n}a_{i}v^{\lambda}_{i}\right)=\sum_{i=1}^{n}a_{i}i_{V}^{\ast}(v^{\lambda}_{i})=\sum_{i\neq j}a_{i}i_{V}^{\ast}(v^{\lambda}_{i}).

Therefore, $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{j-1}),i_{V}^{\ast}(v^{\lambda}_{j+1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n})$ are the direction vectors from the vertex $i_{V}^{\ast}(\lambda)$ . ∎

3.2. Case of $i_{V}^{\ast}(\lambda)$ is not a vertex

Let $\lambda$ be a vertex of the Delzant polytope $\Delta$ . In this section, we show that if the vertex $\lambda$ of $\Delta$ does not satisfy the condition (1) in Definition 1.1, i.e. there is a nontrivial subspace $W^{\lambda}_{V}$ in the cone $i_{V}^{\ast}(\mathcal{C}_{\lambda})$ , then $\overline{C_{\lambda}(V)}$ is nonsingular. From Proposition 3.5, this vertex $\lambda$ satisfies the condition (2), i.e. $\mathcal{J}_{\lambda}=\emptyset$ .

Let $x$ be a nonzero element in the nontrivial subspace $W^{\lambda}_{V}$ . Since $x\in W^{\lambda}_{V}$ , we obtain $(-x)\in W^{\lambda}_{V}$ . Since the subspace $W^{\lambda}_{V}$ is in the cone $i_{V}^{\ast}(\mathcal{C}_{\lambda})$ , there exist $r_{1},\ldots,r_{n},s_{1},\ldots,s_{n}\geq 0$ such that

x=\sum_{j=1}^{n}r_{j}i_{V}^{\ast}(v^{\lambda}_{j}),\;(-x)=\sum_{j=1}^{n}s_{j}i_{V}^{\ast}(v^{\lambda}_{j}).

Let $R_{j}=r_{j}+s_{j}$ for $j=1,\ldots,n$ . Then, we obtain

(3.1)

0=x+(-x)=\sum_{j=1}^{n}(r_{j}+s_{j})i_{V}^{\ast}(v^{\lambda}_{j})=\sum_{j=1}^{n}R_{j}i_{V}^{\ast}(v^{\lambda}_{j}).

From Lemma 3.1, we may assume that the vectors $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n-1})$ are linearly independent.

Lemma 3.7.

Assume that the vectors $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n-1})$ are linearly independent. If $i_{V}^{\ast}(\lambda)$ is not a vertex of the polytope $i_{V}^{\ast}(\Delta)$ , then $R_{n}\neq 0$ in Equation 3.1.

Proof.

Assume on the contrary that $R_{n}=0$ . Then, from Equation 3.1 we obtain

0=\sum_{j=1}^{n-1}R_{j}i_{V}^{\ast}(v^{\lambda}_{j}).

Since the vectors $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n-1})$ are linearly independent, we obtain $R_{1}=\cdots=R_{n-1}=0$ . Since $R_{j}=r_{j}+s_{j}$ and $r_{j},s_{j}\geq 0$ , we obtain $r_{j}=s_{j}=0$ for any $j=1,\ldots,n$ , i.e. $x=0$ . This is contradiction to the assumption that $x$ is a nonzero element in the subspace $W^{\lambda}_{V}$ . ∎

Lemma 3.8.

Proof.

From Lemma 3.7, $R_{n}\neq 0$ . Then, Equation 3.1 follows

i_{V}^{\ast}(v^{\lambda}_{n})=-\frac{1}{R_{n}}\sum_{j=1}^{n-1}R_{j}i_{V}^{\ast}(v^{\lambda}_{j}).

Since from Lemma 2.6 we calculate

	$\displaystyle\sum_{j=1}^{n-1}\langle u^{\lambda}_{j},q\rangle i_{V}^{\ast}(v^{\lambda}_{j})$	$\displaystyle=-\langle u^{\lambda}_{n},q\rangle i_{V}^{\ast}(v^{\lambda}_{n})$
		$\displaystyle=\langle u^{\lambda}_{n},q\rangle\frac{1}{R_{n}}\sum_{j=1}^{n-1}R_{j}i_{V}^{\ast}(v^{\lambda}_{j})$
		$\displaystyle=\sum_{j=1}^{n-1}\langle u^{\lambda}_{n},q\rangle\frac{R_{j}}{R_{n}}i_{V}^{\ast}(v^{\lambda}_{j}),$

we obtain

\sum_{j=1}^{n-1}\left(\langle u^{\lambda}_{j},q\rangle-\langle u^{\lambda}_{n},q\rangle\frac{R_{j}}{R_{n}}\right)i_{V}^{\ast}(v^{\lambda}_{j})=0.

Since we assume that the vectors $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n-1})$ are linearly independent, we obtain

(3.2)

\langle u^{\lambda}_{1},q\rangle=\langle u^{\lambda}_{n},q\rangle\frac{R_{1}}{R_{n}},\ldots,\langle u^{\lambda}_{n-1},q\rangle=\langle u^{\lambda}_{n},q\rangle\frac{R_{n-1}}{R_{n}}.

Since $R_{n}\neq 0$ means that $R_{n}>0$ , we say that

(3.3)

\frac{R_{1}}{R_{n}}\geq 0,\ldots,\frac{R_{n-1}}{R_{n}}\geq 0.

Assume on the contrary that $\langle u^{\lambda}_{n},q\rangle=0$ . Then, from Equation 3.2, we obtain $\langle u^{\lambda}_{1},q\rangle=\cdots=\langle u^{\lambda}_{n-1},q\rangle=0$ . Since $u^{\lambda}_{1},\ldots,u^{\lambda}_{n}$ form a $\mathbb{Z}$ -basis of $\mathbb{Z}^{n}$ , if $\langle u^{\lambda}_{1},q\rangle=\cdots=\langle u^{\lambda}_{n},q\rangle=0$ , then $q=0$ . This is contradiction to the assumption that $q$ is a basis of the orthogonal subspace $V^{\perp}$ to the $(n-1)$ -dimensional subspace $V$ in $\mathfrak{t}^{n}\cong\mathbb{R}^{n}$ . Therefore, $\langle u^{\lambda}_{n},q\rangle\neq 0$ . ∎

Lemma 3.9.

If $i_{V}^{\ast}(\lambda)$ is not a vertex of the polytope $i_{V}^{\ast}(\Delta)$ , then $\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\emptyset$ or $\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\emptyset$ .

Proof.

From Lemma 3.1, we may assume that the vectors $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n-1})$ are linearly independent.

By Lemma 3.8, we obtain $\langle u^{\lambda}_{n},q\rangle\neq 0$ . Moreover, since Equation 3.2 and Equation 3.3, if $n\in\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}$ , then $1,\ldots,n-1\in\mathcal{I}^{+}_{\lambda}$ , i.e. $\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}=\emptyset$ . Similarly, if $n\in\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}$ , then $1,\ldots,n-1\in\mathcal{I}^{-}_{\lambda}$ , i.e. $\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}=\emptyset$ . ∎

From Lemma 3.9 and Lemma 2.8, we obtain the following:

Corollary 3.10.

If $i_{V}^{\ast}(\lambda)$ is not a vertex of the polytope $i_{V}^{\ast}(\Delta)$ , then $\overline{C_{\lambda}(V)}$ is nonsingular.

3.3. Case of $\mathcal{J}_{\lambda}=\emptyset$ and $i_{V}^{\ast}(\lambda)$ is a vertex

Let $\lambda$ be a vertex of the Delzant polytope $\Delta$ . In this section, we consider the case when the vertex $\lambda$ of $\Delta$ is a good vertex with respect to the map $i_{V}^{\ast}:(\mathfrak{t}^{n})^{\ast}\to(\mathfrak{t}^{n-1})^{\ast}$ in the sense of Definition 1.1. In this case, we suppose the following:

Assumption 3.11.

The vectors $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n-1})\in\mathbb{Z}^{n}$ are linearly independent. Moreover, we have

i_{V}^{\ast}\left(\left\{\sum_{i=1}^{n}a_{i}v^{\lambda}_{i}\mid a_{i}\geq 0\right\}\right)=\left\{\sum_{j=1}^{n-1}b_{j}i_{V}^{\ast}(v^{\lambda}_{j})\mid b_{j}\geq 0\right\}.

Assumption 3.11 implies that $i_{V}^{\ast}(\lambda)$ is a vertex of the polytope $i_{V}^{\ast}(\Delta)$ . Note that if the polytope $i_{V}^{\ast}(\Delta)$ is simple, then Assumption 3.11 holds.

Lemma 3.12.

Under Assumption 3.11, if the vertex $\lambda$ of $\Delta$ is a good vertex with respect to the map $i_{V}^{\ast}:(\mathfrak{t}^{n})^{\ast}\to(\mathfrak{t}^{n-1})^{\ast}$ , then we obtain

\mathcal{I}^{+}_{\lambda}=\{1,\ldots,n-1\},\;\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\{n\}

\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\{n\},\;\mathcal{I}^{-}_{\lambda}=\{1,\ldots,n-1\}.

Proof.

From Assumption 3.11, there exist $S_{1},\ldots,S_{n-1}\in\mathbb{R}_{\geq 0}$ such that

(3.4)

i_{V}^{\ast}(v^{\lambda}_{n})=\sum_{j=1}^{n-1}S_{j}i_{V}^{\ast}(v^{\lambda}_{j}).

From Lemma 2.6, we calculate

	$\displaystyle 0$	$\displaystyle=\sum_{j=1}^{n}\langle u^{\lambda}_{j},q\rangle i_{V}^{\ast}(v^{\lambda}_{j})$
		$\displaystyle=\sum_{j=1}^{n-1}\langle u^{\lambda}_{j},q\rangle i_{V}^{\ast}(v^{\lambda}_{j})+\langle u^{\lambda}_{n},q\rangle i_{V}^{\ast}(v^{\lambda}_{n})$
		$\displaystyle=\sum_{j=1}^{n-1}\left(\langle u^{\lambda}_{j},q\rangle+S_{j}\langle u^{\lambda}_{n},q\rangle\right)i_{V}^{\ast}(v^{\lambda}_{j}).$

Since $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n-1})$ are linearly independent, we obtain

(3.5)

\langle u^{\lambda}_{1},q\rangle=-S_{1}\langle u^{\lambda}_{n},q\rangle,\ldots,\langle u^{\lambda}_{n-1},q\rangle=-S_{n-1}\langle u^{\lambda}_{n},q\rangle.

Note that if $S_{j}=0$ , then $\langle u^{\lambda}_{j},q\rangle=0$ .

Assume on the contrary that $\langle u^{\lambda}_{n},q\rangle=0$ . If $\langle u^{\lambda}_{n},q\rangle=0$ , then we obtain

(3.6)

\langle u^{\lambda}_{1},q\rangle=\cdots=\langle u^{\lambda}_{n-1},q\rangle=0

from Equation 3.5. Since the vectors $u^{\lambda}_{1},\ldots,u^{\lambda}_{n}$ form a $\mathbb{Z}$ -basis of $\mathbb{Z}^{n}$ , Equation 3.6 means that $q=0$ , i.e. $V^{\perp}=\{0\}$ . This is contradiction to the assumption that $V^{\perp}$ is the orthogonal subspace to the $(n-1)$ -dimensional subspace $V$ in $\mathfrak{t}^{n}\cong\mathbb{R}^{n}$ . Therefore, $\langle u^{\lambda}_{n},q\rangle\neq 0$ .

Since $\langle u^{\lambda}_{n},q\rangle\neq 0$ , we obtain $n\not\in\mathcal{I}^{0}_{\lambda}$ , i.e. $n\in\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}$ or $n\in\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}$ . From Equation 3.5, if $n\in\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}$ , then $1,\ldots,n-1\in\mathcal{I}^{-}_{\lambda}$ . Similarly, if $n\in\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}$ , then $1,\ldots,n-1\in\mathcal{I}^{+}_{\lambda}$ . ∎

By the definition of the pullback $i_{V}^{\ast}:(\mathfrak{t}^{n})^{\ast}\to(\mathfrak{t}^{n-1})^{\ast}$ , it is clear that the vectors $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n})$ are integral vectors.

Lemma 3.13.

Under Assumption 3.11, if the vectors $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n-1})$ form a $\mathbb{Z}$ -basis of $\mathbb{Z}^{n-1}$ , then $\langle u^{\lambda}_{n},q\rangle=\pm 1$ .

Proof.

Since the vectors $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n-1})\in\mathbb{Z}^{n-1}$ form a $\mathbb{Z}$ -basis of $\mathbb{Z}^{n-1}$ and $i_{V}^{\ast}(v^{\lambda}_{n})\in\mathbb{Z}^{n-1}$ , there exist $Q_{1},\ldots,Q_{n-1}\in\mathbb{Z}$ such that

i_{V}^{\ast}(v^{\lambda}_{n})=\sum_{j=1}^{n-1}Q_{j}i_{V}^{\ast}(v^{\lambda}_{j}).

Moreover, since $i_{V}^{\ast}(v^{\lambda}_{n})\in i_{V}^{\ast}(\mathcal{C}_{\lambda})$ from Assumption 3.11, $i_{V}^{\ast}(v^{\lambda}_{n})$ can be expressed as the form in Equation 3.4. From these equations, we obtain

\sum_{j=1}^{n-1}\left(Q_{j}-S_{j}\right)i_{V}^{\ast}(v^{\lambda}_{j})=0.

Since the vectors $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n-1})$ are linearly independent,

(3.7)

Q_{1}=S_{1},\ldots,Q_{n-1}=S_{n-1}.

Since $S_{1},\ldots,S_{n-1}\geq 0$ , we have $Q_{1},\ldots,Q_{n-1}\geq 0$ . Thus, we obtain $Q_{1},\ldots,Q_{n-1}\in\mathbb{Z}_{\geq 0}$ .

From Equation 3.5 and Equation 3.7, we obtain

(3.8)

\langle u^{\lambda}_{1},q\rangle=-Q_{1}\langle u^{\lambda}_{n},q\rangle,\ldots,\langle u^{\lambda}_{n-1},q\rangle=-Q_{n-1}\langle u^{\lambda}_{n},q\rangle.

From Equation 3.8, we obtain

{{}^{t}\!\left[\begin{matrix}u^{\lambda}_{1}&\cdots&u^{\lambda}_{n}\end{matrix}\right]}q=\left[\begin{matrix}\langle u^{\lambda}_{1},q\rangle\\ \vdots\\ \langle u^{\lambda}_{n},q\rangle\end{matrix}\right]=\langle u^{\lambda}_{n},q\rangle\left[\begin{matrix}-Q_{1}\\ \vdots\\ -Q_{n-1}\\ 1\end{matrix}\right]

and since ${{}^{t}[u^{\lambda}_{1}\;\cdots\;u^{\lambda}_{n}]}[v^{\lambda}_{1}\;\cdots\;v^{\lambda}_{n}]=E_{n}$ , we obtain

q=\langle u^{\lambda}_{n},q\rangle\left[\begin{matrix}v^{\lambda}_{1}&\cdots&v^{\lambda}_{n}\end{matrix}\right]\left[\begin{matrix}-Q_{1}\\ \vdots\\ -Q_{n-1}\\ 1\end{matrix}\right].

Since the vector $q\in\mathbb{Z}^{n}$ is primitive, we obtain $\langle u^{\lambda}_{n},q\rangle=\pm 1$ . ∎

From Lemma 2.9, Lemma 3.12 and Lemma 3.13, we obtain the following:

Corollary 3.14.

Assume that the vertex $\lambda$ of the Delzant polytope $\Delta$ is a good vertex with respect to the map $i_{V}^{\ast}:(\mathfrak{t}^{n})^{\ast}\to(\mathfrak{t}^{n-1})^{\ast}$ . Under Assumption 3.11, if there exist distinct $j_{1},\ldots,j_{n-1}\in\{1,\ldots,n\}$ such that the vectors $i_{V}^{\ast}(v^{\lambda}_{j_{1}}),\ldots,i_{V}^{\ast}(v^{\lambda}_{j_{n-1}})\in\mathbb{Z}^{n-1}$ form a $\mathbb{Z}$ -basis of $\mathbb{Z}^{n-1}$ , then $\overline{C_{\lambda}(V)}$ is nonsingular.

3.4. The Proof of the first Main Theorem

In this section, we give a proof of the first main result.

Theorem 3.15.

If the Delzant polytope $\Delta$ is good with respect to the map $i_{V}^{\ast}$ , then the rank of the Jacobian matrix $Df^{\lambda}$ is equal to one at any point for any vertex $\lambda$ of $\Delta$ . In particular, $\overline{C(V)}$ is a smooth complex hypersurface in the toric manifold $X$ corresponding to $\Delta$ .

Proof.

It is sufficient to consider each vertex of the Delznt polytope $\Delta$ .

If a vertex $\lambda$ of $\Delta$ is not a good vertex with respect to the map $i_{V}^{\ast}$ , then the vertex $\lambda$ of $\Delta$ does not satisfy the condition (1) or (2) in Definition 1.1. If $\lambda$ does not satisfy the condition (2), i.e. $\mathcal{J}_{\lambda}\neq\emptyset$ , then Corollary 3.4 tells us that $\overline{C_{\lambda}(V)}$ is nonsingular. If $\lambda$ does not satisfy the condition (1), then Corollary 3.10 tells us that $\overline{C_{\lambda}(V)}$ is nonsingular. Thus, if a vertex $\lambda$ of $\Delta$ is not a good vertex with respect to the map $i_{V}^{\ast}$ , then $\overline{C_{\lambda}(V)}$ is nonsingular.

Since we assume that the Delzant polytope $\Delta$ is good with respect to the map $i_{V}^{\ast}$ , if a vertex $\lambda$ of $\Delta$ is a good vertex with respect to the map $i_{V}^{\ast}$ , then Assumption 3.11 holds for such $\lambda$ and there exist distinct $j_{1},\ldots,j_{n-1}\in\{1,\ldots,n\}$ such that the vectors $i_{V}^{\ast}(v^{\lambda}_{j_{1}}),\ldots,i_{V}^{\ast}(v^{\lambda}_{j_{n-1}})\in\mathbb{Z}^{n-1}$ form a $\mathbb{Z}$ -basis of $\mathbb{Z}^{n-1}$ . In Corollary 3.14, we show that $\overline{C_{\lambda}(V)}$ is nonsingular in this case.

Therefore, if $\Delta$ is good with respect to the map $i_{V}^{\ast}$ , then $\overline{C_{\lambda}(V)}$ is nonsingular for any vertex $\lambda$ of $\Delta$ , i.e. $\overline{C(V)}$ is a smooth complex hypersurface in the toric manifold $X$ . ∎

4. From Torus-equivariantly Embedded Toric Hypersurfaces to Delzant Polytopes

In this section, we show the second main result (Theorem 4.13).

Assume that the rank of the Jacobian matrix $Df^{\lambda}$ is equal to one at any point of the zero locus of $f^{\lambda}$ for any vertex $\lambda$ of the Delzant polytope, i.e. $\overline{C(V)}$ is nonsingular. In this case, as we show in Proposition 2.5, $\overline{C(V)}$ is toric with respect to the Hamiltonian $T^{n-1}$ -action defined in [Yam24b] with the moment map $\overline{\mu}=i_{V}^{\ast}\circ\mu\mid_{\overline{C(V)}}$ . Moreover, we have the following:

Theorem 4.1 ([Yam24b, Theorem 4.20]).

Assume that $\overline{C(V)}$ is a complex submanifold in the toric manifold $X$ . Then, we obtain

\overline{\mu}(\overline{C(V)})=i_{V}^{\ast}(\Delta),

where $\Delta$ is the Delzant polytope of the toric manifold $X$ .

Recall that if $\overline{C(V)}$ is a complex submanifold, then $\overline{C(V)}$ is a symplectic toric manifold with respect to the $T^{n-1}$ -action on $\overline{C(V)}$ (Proposition 2.5). By the Delzant correspondence, the moment polytope of the submanifold $\overline{C(V)}$ should be Delzant, i.e. we obtain the following:

Corollary 4.2.

Assume that $\overline{C(V)}$ is a complex submanifold in the toric manifold $X$ . Then, the polytope $i_{V}^{\ast}(\Delta)$ is a Delzant polytope.

In particular, the polytope $i_{V}^{\ast}(\Delta)$ is simple, i.e. Assumption 3.11 holds under the assumption that $\overline{C(V)}$ is a smooth complex hypersurface.

To show the second main result, we have to consider the case when the vertex $\lambda$ of $\Delta$ is good with respect to the map $i_{V}^{\ast}$ .

Hereafter, we assume that the vertex $\lambda$ of $\Delta$ is good with respect to the map $i_{V}^{\ast}$ . If $\overline{C(V)}$ is a complex hypersurface, then from Lemma 3.12, we may further assume that $\mathcal{I}^{+}_{\lambda}=\{1,\ldots,n-1\}$ and $\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\{n\}$ under the Assumption 3.11.

Lemma 4.3.

Assume that the vertex $\lambda$ of $\Delta$ is good with respect to the map $i_{V}^{\ast}$ . Under the Assumption 3.11, we obtain $\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}\neq\emptyset$ and $\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}\neq\emptyset$ .

Proof.

From Lemma 3.12, we may assume that $\mathcal{I}^{+}_{\lambda}=\{1,\ldots,n-1\}$ and $\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\{n\}$ .

Assume on the contrary that $\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\emptyset$ , i.e. $\langle u^{\lambda}_{1},q\rangle=\cdots=\langle u^{\lambda}_{n-1},q\rangle=0$ . From Lemma 2.6, we obtain

0=\sum_{j=1}^{n}\langle u^{\lambda}_{j},q\rangle i_{V}^{\ast}(v^{\lambda}_{j})=\langle u^{\lambda}_{n},q\rangle i_{V}^{\ast}(v^{\lambda}_{n}).

Since $\langle u^{\lambda}_{n},q\rangle\neq 0$ , we obtain $i_{V}^{\ast}(v^{\lambda}_{n})=0$ , which is contradiction to the assumption that the vertex $\lambda$ of $\Delta$ is good with respect to the map $i_{V}^{\ast}$ . ∎

Lemma 4.4.

Assume that the vertex $\lambda$ of $\Delta$ is good with respect to the map $i_{V}^{\ast}$ and $\mathcal{I}^{+}_{\lambda}=\{1,\ldots,n-1\}$ and $\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\{n\}$ . If $\overline{C(V)}$ is nonsingular, then we obtain at least one of the following:

(1)

there exists $i_{0}\in\mathcal{I}^{+}_{\lambda}$ such that $\langle u^{\lambda}_{i_{0}},q\rangle=1$ and $\langle u^{\lambda}_{j},q\rangle=0$ for any $j\in\mathcal{I}^{+}_{\lambda}\setminus\{i_{0}\}$ ,
(2)

$\langle u^{\lambda}_{n},q\rangle=-1$ .

Proof.

From the defining equation for $\overline{C_{\lambda}(V)}$ , we obtain that

f^{\lambda}(z^{\lambda})=\prod_{j\in\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}}(z^{\lambda}_{j})^{\langle u^{\lambda}_{j},q\rangle}-\prod_{j\in\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}}(z^{\lambda}_{j})^{-\langle u^{\lambda}_{j},q\rangle}.

Note that $\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}\neq\emptyset$ and $\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}\neq\emptyset$ from Lemma 4.3. Assume that $\mathcal{I}^{+}_{\lambda}=\{1,\ldots,n-1\},\;\mathcal{I}^{-}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\{n\}$ . From this equation, we obtain

\frac{\partial f^{\lambda}}{\partial z^{\lambda}_{i}}=\begin{cases*}\langle u^{\lambda}_{i},q\rangle\prod_{j\in\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}}(z^{\lambda}_{j})^{\langle u^{\lambda}_{j},q\rangle-\delta_{ij}}&if $i=1,\ldots,n-1$,\\ \langle u^{\lambda}_{n},q\rangle(z^{\lambda}_{n})^{-\langle u^{\lambda}_{n},q\rangle-1}&if $i=n$.\end{cases*}

If $\overline{C(V)}$ is nonsingular, then there exists $i_{0}\in\{1,\ldots,n\}$ such that

\frac{\partial f^{\lambda}}{\partial z^{\lambda}_{i_{0}}}(o)\neq 0

at the origin $o\in\varphi_{\lambda}(U_{\lambda})\cap\overline{C_{\lambda}(V)}$ . If such $i_{0}\in\mathcal{I}^{+}_{\lambda}$ , then we obtain $\langle u^{\lambda}_{j},q\rangle-\delta_{i_{0}j}=0$ for $j=1,\ldots,n-1$ , i.e. $\langle u^{\lambda}_{i_{0}},q\rangle=1$ and $\langle u^{\lambda}_{j},q\rangle=0$ for any $j\in\mathcal{I}^{+}_{\lambda}\setminus\{i_{0}\}$ . If such $i_{0}\notin\mathcal{I}^{+}_{\lambda}$ , i.e. $i_{0}=n$ , then $\langle u^{\lambda}_{n},q\rangle=-1$ . ∎

Note that we obtain the similar result to Lemma 4.4 when we assume that $\mathcal{I}^{+}_{\lambda}\setminus\mathcal{I}^{0}_{\lambda}=\{n\}$ and $\mathcal{I}^{-}_{\lambda}=\{1,\ldots,n-1\}$ .

4.1. The first case in Lemma 4.4

We consider the first case in Lemma 4.4. Hereafter, we assume that $\langle u^{\lambda}_{1},q\rangle=1$ and $\langle u^{\lambda}_{2},q\rangle=\cdots=\langle u^{\lambda}_{n-1},q\rangle=0$ for simplicity when we consider the first case in Lemma 4.4.

Lemma 4.5.

Assume that the vertex $\lambda$ of $\Delta$ is good with respect to the map $i_{V}^{\ast}$ . Moreover, assume that $\langle u^{\lambda}_{1},q\rangle=1$ and $\langle u^{\lambda}_{2},q\rangle=\cdots=\langle u^{\lambda}_{n-1},q\rangle=0$ . Then, we obtain

(4.1)

i_{V}^{\ast}(v^{\lambda}_{1})=-\langle u^{\lambda}_{n},q\rangle i_{V}^{\ast}(v^{\lambda}_{n}).

Proof.

Since we assume that $\langle u^{\lambda}_{1},q\rangle=1$ and $\langle u^{\lambda}_{2},q\rangle=\cdots=\langle u^{\lambda}_{n-1},q\rangle=0$ , we obtain

\sum_{j=1}^{n}\langle u^{\lambda}_{j},q\rangle i_{V}^{\ast}(v^{\lambda}_{j})=i_{V}^{\ast}(v^{\lambda}_{1})+\langle u^{\lambda}_{n},q\rangle i_{V}^{\ast}(v^{\lambda}_{n}).

From Lemma 2.6, the left hand side of this equation is equal to zero. Therefore, we obtain Equation 4.1. ∎

Remark 4.6.

If $\lambda$ is good, Assumption 3.11 holds, and $\overline{C_{\lambda}(V)}$ is nonsingular, then we have $i_{V}^{\ast}(v^{\lambda}_{1})=-\langle u^{\lambda}_{n},q\rangle i_{V}^{\ast}(v^{\lambda}_{n})$ with $-\langle u^{\lambda}_{n},q\rangle\in\mathbb{Z}_{>0}$ . Therefore, we may use the vector $i_{V}^{\ast}(v^{\lambda}_{n})$ instead of $i_{V}^{\ast}(v^{\lambda}_{1})$ in Assumption 3.11, i.e. we may assume the following:

•

the vectors $i_{V}^{\ast}(v^{\lambda}_{2}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n})$ are linearly independent,

•

we have

i_{V}^{\ast}\left(\left\{\sum_{i=1}^{n}a_{i}v^{\lambda}_{i}\mid a_{i}\geq 0\right\}\right)=\left\{\sum_{j=2}^{n}b_{j}i_{V}^{\ast}(v^{\lambda}_{j})\mid b_{j}\geq 0\right\}.

Lemma 4.7.

Assume the same as in Lemma 4.5. Then, we obtain

\det[p_{1}\;\cdots\;p_{n-1}\;q]=\pm\langle q,q\rangle\det[i_{V}^{\ast}(v^{\lambda}_{2})\;\cdots\;i_{V}^{\ast}(v^{\lambda}_{n})].

Proof.

Since from Lemma 4.5, we obtain

	$\displaystyle\left[\begin{matrix}{{}^{t}p_{1}}\\ \vdots\\ {{}^{t}p_{n-1}}\\ {{}^{t}q}\end{matrix}\right]\left[v^{\lambda}_{1}\;\cdots\;v^{\lambda}_{n}\right]$	$\displaystyle=\left[\begin{matrix}\langle p_{1},v^{\lambda}_{1}\rangle&\cdots&\langle p_{1},v^{\lambda}_{n}\rangle\\ \vdots&\ddots&\vdots\\ \langle p_{n-1},v^{\lambda}_{1}\rangle&\cdots&\langle p_{n-1},v^{\lambda}_{n}\rangle\\ \langle q,v^{\lambda}_{1}\rangle&\cdots&\langle q,v^{\lambda}_{n}\rangle\end{matrix}\right]$
		$\displaystyle=\left[\begin{matrix}i_{V}^{\ast}(v^{\lambda}_{1})&\cdots&i_{V}^{\ast}(v^{\lambda}_{n})\\ \langle q,v^{\lambda}_{1}\rangle&\cdots&\langle q,v^{\lambda}_{n}\rangle\end{matrix}\right]$
		$\displaystyle=\left[\begin{matrix}-\langle u^{\lambda}_{n},q\rangle i_{V}^{\ast}(v^{\lambda}_{n})&i_{V}^{\ast}(v^{\lambda}_{2})&\cdots&i_{V}^{\ast}(v^{\lambda}_{n})\\ \langle q,v^{\lambda}_{1}\rangle&\langle q,v^{\lambda}_{2}\rangle&\cdots&\langle q,v^{\lambda}_{n}\rangle\end{matrix}\right],$

we obtain by using the cofactor expansion that

	$\displaystyle\det[p_{1}\;\cdots\;p_{n-1}\;q]\det[v^{\lambda}_{1}\;\cdots\;v^{\lambda}_{n}]$
	$\displaystyle=\det{{}^{t}[p_{1}\;\cdots\;p_{n-1}\;q]}\det[v^{\lambda}_{1}\;\cdots\;v^{\lambda}_{n}]$
	$\displaystyle=\det\left[\begin{matrix}-\langle u^{\lambda}_{n},q\rangle i_{V}^{\ast}(v^{\lambda}_{n})&i_{V}^{\ast}(v^{\lambda}_{2})&\cdots&i_{V}^{\ast}(v^{\lambda}_{n})\\ \langle q,v^{\lambda}_{1}\rangle&\langle q,v^{\lambda}_{2}\rangle&\cdots&\langle q,v^{\lambda}_{n}\rangle\end{matrix}\right]$
	$\displaystyle=\det\left[\begin{matrix}0&i_{V}^{\ast}(v^{\lambda}_{2})&\cdots&i_{V}^{\ast}(v^{\lambda}_{n})\\ \langle q,v^{\lambda}_{1}\rangle+\langle q,v^{\lambda}_{n}\rangle\langle u^{\lambda}_{n},q\rangle&\langle q,v^{\lambda}_{2}\rangle&\cdots&\langle q,v^{\lambda}_{n}\rangle\end{matrix}\right]$
	$\displaystyle=\pm\left(\langle q,v^{\lambda}_{1}\rangle+\langle q,v^{\lambda}_{n}\rangle\langle u^{\lambda}_{n},q\rangle\right)\det[i_{V}^{\ast}(v^{\lambda}_{2})\;\cdots\;i_{V}^{\ast}(v^{\lambda}_{n})].$

Since we assume $\langle u^{\lambda}_{1},q\rangle=1$ and $\langle u^{\lambda}_{2},q\rangle=\cdots=\langle u^{\lambda}_{n-1},q\rangle=0$ , we obtain

\langle q,v^{\lambda}_{1}\rangle+\langle q,v^{\lambda}_{n}\rangle\langle u^{\lambda}_{n},q\rangle=\sum_{j=1}^{n}\langle q,v^{\lambda}_{j}\rangle\langle u^{\lambda}_{j},q\rangle=\langle q,q\rangle.

Thus, we obtain

	$\displaystyle\pm\left(\langle q,v^{\lambda}_{1}\rangle+\langle q,v^{\lambda}_{n}\rangle\langle u^{\lambda}_{n},q\rangle\right)\det[i_{V}^{\ast}(v^{\lambda}_{2})\;\cdots\;i_{V}^{\ast}(v^{\lambda}_{n})]$
	$\displaystyle=\pm\langle q,q\rangle\det[i_{V}^{\ast}(v^{\lambda}_{2})\;\cdots\;i_{V}^{\ast}(v^{\lambda}_{n})].$

Since $v^{\lambda}_{1},\ldots,v^{\lambda}_{n}$ form a basis of $\mathbb{Z}^{n}$ , $\det[v^{\lambda}_{1}\;\cdots\;v^{\lambda}_{n}]=\pm 1$ , and

\det[p_{1}\;\cdots\;p_{n-1}\;q]=\pm\langle q,q\rangle\det[i_{V}^{\ast}(v^{\lambda}_{2})\;\cdots\;i_{V}^{\ast}(v^{\lambda}_{n})].

∎

4.2. The second case in Lemma 4.4

Next, we consider the second case in Lemma 4.4, i.e. we assume that $\langle u^{\lambda}_{n},q\rangle=-1$ and $\mathcal{I}^{+}_{\lambda}=\{1,\ldots,n-1\}$ .

Lemma 4.8.

Assume that the vertex $\lambda$ of $\Delta$ is good with respect to the map $i_{V}^{\ast}$ and that $\langle u^{\lambda}_{n},q\rangle=-1$ and $\mathcal{I}^{+}_{\lambda}=\{1,\ldots,n-1\}$ . Then, we obtain

(4.2)

i_{V}^{\ast}(v^{\lambda}_{n})=\sum_{j=1}^{n-1}\langle u^{\lambda}_{j},q\rangle i_{V}^{\ast}(v^{\lambda}_{j}).

Proof.

From Lemma 2.6, we obtain

\displaystyle 0=\sum_{j=1}^{n}\langle u^{\lambda}_{j},q\rangle i_{V}^{\ast}(v^{\lambda}_{j})=\sum_{j=1}^{n-1}\langle u^{\lambda}_{j},q\rangle i_{V}^{\ast}(v^{\lambda}_{j})-i_{V}^{\ast}(v^{\lambda}_{n}).

Thus, we obtain Equation 4.2. ∎

Lemma 4.9.

Assume that $\langle u^{\lambda}_{n},q\rangle=-1$ and $\mathcal{I}^{+}_{\lambda}=\{1,\ldots,n-1\}$ . Then, we obtain

\det[p_{1}\;\cdots\;p_{n-1}\;q]=\pm\langle q,q\rangle\det[i_{V}^{\ast}(v^{\lambda}_{1})\;\cdots\;i_{V}^{\ast}(v^{\lambda}_{n-1})].

Proof.

Since we see

\displaystyle\left[\begin{matrix}{{}^{t}p_{1}}\\ \vdots\\ {{}^{t}p_{n-1}}\\ {{}^{t}q}\end{matrix}\right]\left[v^{\lambda}_{1}\;\cdots\;v^{\lambda}_{n}\right]=\left[\begin{matrix}i_{V}^{\ast}(v^{\lambda}_{1})&\cdots&i_{V}^{\ast}(v^{\lambda}_{n})\\ \langle q,v^{\lambda}_{1}\rangle&\cdots&\langle q,v^{\lambda}_{n}\rangle\end{matrix}\right],

we obtain by using the cofactor expansion that

	$\displaystyle\det[p_{1}\;\cdots\;p_{n-1}\;q]\det[v^{\lambda}_{1}\;\cdots\;v^{\lambda}_{n}]$
	$\displaystyle=\det\left[\begin{matrix}i_{V}^{\ast}(v^{\lambda}_{1})&\cdots&i_{V}^{\ast}(v^{\lambda}_{n})\\ \langle q,v^{\lambda}_{1}\rangle&\cdots&\langle q,v^{\lambda}_{n}\rangle\end{matrix}\right]$
	$\displaystyle=\det\left[\begin{matrix}i_{V}^{\ast}(v^{\lambda}_{1})&\cdots&i_{V}^{\ast}(v^{\lambda}_{n-1})&\sum_{j=1}^{n-1}\langle u^{\lambda}_{j},q\rangle i_{V}^{\ast}(v^{\lambda}_{j})\\ \langle q,v^{\lambda}_{1}\rangle&\cdots&\langle q,v^{\lambda}_{n-1}\rangle&\langle q,v^{\lambda}_{n}\rangle\end{matrix}\right]$
	$\displaystyle=\det\left[\begin{matrix}i_{V}^{\ast}(v^{\lambda}_{1})&\cdots&i_{V}^{\ast}(v^{\lambda}_{n-1})&0\\ \langle q,v^{\lambda}_{1}\rangle&\cdots&\langle q,v^{\lambda}_{n-1}\rangle&\langle q,v^{\lambda}_{n}\rangle-\sum_{j=1}^{n-1}\langle q,v^{\lambda}_{j}\rangle\langle u^{\lambda}_{j},q\rangle\end{matrix}\right]$
	$\displaystyle=\left(\langle q,v^{\lambda}_{n}\rangle-\sum_{j=1}^{n-1}\langle q,v^{\lambda}_{j}\rangle\langle u^{\lambda}_{j},q\rangle\right)\det[i_{V}^{\ast}(v^{\lambda}_{1})\;\cdots\;i_{V}^{\ast}(v^{\lambda}_{n-1})].$

Since we assume $\langle u^{\lambda}_{n},q\rangle=-1$ , we obtain

\langle q,v^{\lambda}_{n}\rangle-\sum_{j=1}^{n-1}\langle q,v^{\lambda}_{j}\rangle\langle u^{\lambda}_{j},q\rangle=-\sum_{j=1}^{n}\langle q,v^{\lambda}_{j}\rangle\langle u^{\lambda}_{j},q\rangle=-\langle q,q\rangle.

Thus, we obtain

	$\displaystyle\left(\langle q,v^{\lambda}_{n}\rangle-\sum_{j=1}^{n-1}\langle q,v^{\lambda}_{j}\rangle\langle u^{\lambda}_{j},q\rangle\right)\det[i_{V}^{\ast}(v^{\lambda}_{1})\;\cdots\;i_{V}^{\ast}(v^{\lambda}_{n-1})]$
	$\displaystyle=-\langle q,q\rangle\det[i_{V}^{\ast}(v^{\lambda}_{1})\;\cdots\;i_{V}^{\ast}(v^{\lambda}_{n-1})].$

Since $v^{\lambda}_{1},\ldots,v^{\lambda}_{n}$ form a basis of $\mathbb{Z}^{n}$ , $\det[v^{\lambda}_{1}\;\cdots\;v^{\lambda}_{n}]=\pm 1$ , and

\det[p_{1}\;\cdots\;p_{n-1}\;q]=\pm\langle q,q\rangle\det[i_{V}^{\ast}(v^{\lambda}_{1})\;\cdots\;i_{V}^{\ast}(v^{\lambda}_{n-1})].

∎

4.3. The Prood of the second Main Theorem

By calculating the determinant of the Gram matrix of $[p_{1}\;\cdots\;p_{n-1}\;q]$ , we obtain the following:

Lemma 4.10.

For $V=\mathbb{R}p_{1}+\cdots+\mathbb{R}p_{n-1}$ and $V^{\perp}=\mathbb{R}q$ , we obtain

\det[p_{1}\;\cdots\;p_{n-1}\;q]=\pm\left(\langle q,q\rangle\det\left[\begin{matrix}\langle p_{1},p_{1}\rangle&\cdots&\langle p_{1},p_{n-1}\rangle\\ \vdots&\ddots&\vdots\\ \langle p_{n-1},p_{1}\rangle&\cdots&\langle p_{n-1},p_{n-1}\rangle\end{matrix}\right]\right)^{\frac{1}{2}}.

Proof.

Since the Gram matrix of $[p_{1}\;\cdots\;p_{n-1}\;q]$ is

	$\displaystyle\left[\begin{matrix}{{}^{t}p_{1}}\\ \vdots\\ {{}^{t}p_{n-1}}\\ {{}^{t}q}\end{matrix}\right][p_{1}\;\cdots\;p_{n-1}\;q]$	$\displaystyle=\left[\begin{matrix}\langle p_{1},p_{1}\rangle&\cdots&\langle p_{1},p_{n-1}\rangle&\langle p_{1},q\rangle\\ \vdots&\ddots&\vdots&\vdots\\ \langle p_{n-1},p_{1}\rangle&\cdots&\langle p_{n-1},p_{n-1}\rangle&\langle p_{1},q\rangle\\ \langle q,p_{1}\rangle&\cdots&\langle q,p_{n-1}\rangle&\langle q,q\rangle\end{matrix}\right]$
		$\displaystyle=\left[\begin{matrix}\langle p_{1},p_{1}\rangle&\cdots&\langle p_{1},p_{n-1}\rangle&0\\ \vdots&\ddots&\vdots&\vdots\\ \langle p_{n-1},p_{1}\rangle&\cdots&\langle p_{n-1},p_{n-1}\rangle&0\\ 0&\cdots&0&\langle q,q\rangle\end{matrix}\right],$

we obtain

	$\displaystyle(\det[p_{1}\;\cdots\;p_{n-1}\;q])^{2}$	$\displaystyle=\det\left(\left[\begin{matrix}{{}^{t}p_{1}}\\ \vdots\\ {{}^{t}p_{n-1}}\\ {{}^{t}q}\end{matrix}\right][p_{1}\;\cdots\;p_{n-1}\;q]\right)$
		$\displaystyle=\det\left[\begin{matrix}\langle p_{1},p_{1}\rangle&\cdots&\langle p_{1},p_{n-1}\rangle&0\\ \vdots&\ddots&\vdots&\vdots\\ \langle p_{n-1},p_{1}\rangle&\cdots&\langle p_{n-1},p_{n-1}\rangle&0\\ 0&\cdots&0&\langle q,q\rangle\end{matrix}\right]$
		$\displaystyle=\langle q,q\rangle\det\left[\begin{matrix}\langle p_{1},p_{1}\rangle&\cdots&\langle p_{1},p_{n-1}\rangle\\ \vdots&\ddots&\vdots\\ \langle p_{n-1},p_{1}\rangle&\cdots&\langle p_{n-1},p_{n-1}\rangle\end{matrix}\right].$

Since Gram matrices are positive definite, i.e. their determinant are nonnegative, we obtain the desired result. ∎

In our case, we obtain more about the determinant of Gram matrices. Though the following proposition is found in books [Mar03, Proposition 1.9.8] and [Ebe13, Proposition 1.2], we give a proof in order to make this paper self-contained.

Proposition 4.11.

Assume that $p_{1},\ldots,p_{n-1}\in\mathbb{Z}^{n}$ form a $\mathbb{Z}$ -basis of $V\cap\mathbb{Z}^{n}\cong\mathbb{Z}^{n-1}$ . Let $q\in\mathbb{Z}^{n}$ be a primitive basis of the orthogonal subspace to $V$ in $\mathbb{R}^{n}$ . For $V=\mathbb{R}p_{1}+\cdots+\mathbb{R}p_{n-1}$ and $V^{\perp}=\mathbb{R}q$ , we obtain

\det\left[\begin{matrix}\langle p_{1},p_{1}\rangle&\cdots&\langle p_{1},p_{n-1}\rangle\\ \vdots&\ddots&\vdots\\ \langle p_{n-1},p_{1}\rangle&\cdots&\langle p_{n-1},p_{n-1}\rangle\end{matrix}\right]=\langle q,q\rangle.

Proof.

By the assumption, there exists a vector $p_{n}\in\mathbb{Z}^{n}$ such that $p_{1},\ldots,p_{n-1},p_{n}$ form a $\mathbb{Z}$ -basis of $\mathbb{Z}^{n}$ . Since we have

	$\displaystyle\left[\begin{matrix}{{}^{t}p_{1}}\\ \vdots\\ {{}^{t}p_{n-1}}\\ {{}^{t}p_{n}}\end{matrix}\right]\left[p_{1}\;\cdots\;p_{n-1}\;q\right]$	$\displaystyle=\left[\begin{matrix}\langle p_{1},p_{1}\rangle&\cdots&\langle p_{1},p_{n-1}\rangle&\langle p_{1},q\rangle\\ \vdots&\ddots&\vdots&\vdots\\ \langle p_{n-1},p_{1}\rangle&\cdots&\langle p_{n-1},p_{n-1}\rangle&\langle p_{n-1},q\rangle\\ \langle p_{n},p_{1}\rangle&\cdots&\langle p_{n},p_{n-1}\rangle&\langle p_{n},q\rangle\end{matrix}\right]$
		$\displaystyle=\left[\begin{matrix}\langle p_{1},p_{1}\rangle&\cdots&\langle p_{1},p_{n-1}\rangle&0\\ \vdots&\ddots&\vdots&\vdots\\ \langle p_{n-1},p_{1}\rangle&\cdots&\langle p_{n-1},p_{n-1}\rangle&0\\ \langle p_{n},p_{1}\rangle&\cdots&\langle p_{n},p_{n-1}\rangle&\langle p_{n},q\rangle\end{matrix}\right],$

we obtain by the cofactor expansion that

	$\displaystyle\det[p_{1}\;\cdots\;p_{n-1}\;p_{n}]\det[p_{1}\;\cdots\;p_{n-1}\;q]$
	$\displaystyle=\det\left[\begin{matrix}\langle p_{1},p_{1}\rangle&\cdots&\langle p_{1},p_{n-1}\rangle&0\\ \vdots&\ddots&\vdots&\vdots\\ \langle p_{n-1},p_{1}\rangle&\cdots&\langle p_{n-1},p_{n-1}\rangle&0\\ \langle p_{n},p_{1}\rangle&\cdots&\langle p_{n},p_{n-1}\rangle&\langle p_{n},q\rangle\end{matrix}\right]$
	$\displaystyle=\langle p_{n},q\rangle\det\left[\begin{matrix}\langle p_{1},p_{1}\rangle&\cdots&\langle p_{1},p_{n-1}\rangle\\ \vdots&\ddots&\vdots\\ \langle p_{n-1},p_{1}\rangle&\cdots&\langle p_{n-1},p_{n-1}\rangle\end{matrix}\right].$

Since $p_{1},\ldots,p_{n}$ form a $\mathbb{Z}$ -basis of $\mathbb{Z}^{n}$ , we obtain $\det[p_{1}\;\cdots\;p_{n-1}\;p_{n}]=\pm 1$ . From Lemma 4.10, we obtain

(4.3)

\langle p_{n},q\rangle^{2}\det\left[\begin{matrix}\langle p_{1},p_{1}\rangle&\cdots&\langle p_{1},p_{n-1}\rangle\\ \vdots&\ddots&\vdots\\ \langle p_{n-1},p_{1}\rangle&\cdots&\langle p_{n-1},p_{n-1}\rangle\end{matrix}\right]=\langle q,q\rangle.

Since we have

\left[\begin{matrix}\langle p_{1},q\rangle\\ \vdots\\ \langle p_{n-1},q\rangle\\ \langle p_{n},q\rangle\end{matrix}\right]=\left[\begin{matrix}{{}^{t}p_{1}}\\ \vdots\\ {{}^{t}p_{n-1}}\\ {{}^{t}p_{n}}\end{matrix}\right]q,

and $p_{1},\ldots,p_{n}$ form a $\mathbb{Z}$ -basis of $\mathbb{Z}^{n}$ , we obtain

	$\displaystyle q$	$\displaystyle=\left[\begin{matrix}{{}^{t}p_{1}}\\ \vdots\\ {{}^{t}p_{n-1}}\\ {{}^{t}p_{n}}\end{matrix}\right]^{-1}\left[\begin{matrix}\langle p_{1},q\rangle\\ \vdots\\ \langle p_{n-1},q\rangle\\ \langle p_{n},q\rangle\end{matrix}\right]$
		$\displaystyle=\left[\begin{matrix}{{}^{t}p_{1}}\\ \vdots\\ {{}^{t}p_{n-1}}\\ {{}^{t}p_{n}}\end{matrix}\right]^{-1}\left[\begin{matrix}0\\ \vdots\\ 0\\ \langle p_{n},q\rangle\end{matrix}\right]$
		$\displaystyle=\langle p_{n},q\rangle\left[\begin{matrix}{{}^{t}p_{1}}\\ \vdots\\ {{}^{t}p_{n-1}}\\ {{}^{t}p_{n}}\end{matrix}\right]^{-1}\left[\begin{matrix}0\\ \vdots\\ 0\\ 1\end{matrix}\right].$

Since we assume that the vector $q$ is primitive, we obtain $\langle p_{n},q\rangle=\pm 1$ . Therefore, from Equation 4.3 we obtain the desired equation. ∎

From Lemma 4.10 and Proposition 4.11, we obtain the following:

Corollary 4.12.

Assume that $p_{1},\ldots,p_{n-1}\in\mathbb{Z}^{n}$ form a $\mathbb{Z}$ -basis of $V\cap\mathbb{Z}^{n}\cong\mathbb{Z}^{n-1}$ . Assume that the vector $q\in\mathbb{Z}^{n}$ is a primitive basis of the orthogonal subspace to $V$ in $\mathbb{R}^{n}$ . For $V=\mathbb{R}p_{1}+\cdots+\mathbb{R}p_{n-1}$ and $V^{\perp}=\mathbb{R}q$ , we obtain

\det[p_{1}\;\cdots\;p_{n-1}\;q]=\pm\langle q,q\rangle.

We use this fact to show that the second main result:

Theorem 4.13.

If the rank of the Jacobian matrix $Df^{\lambda}$ is equal to one at any point of the zero locus of $f^{\lambda}$ for any vertex $\lambda$ of the Delzant polytope $\Delta$ , i.e. $\overline{C(V)}$ is nonsingular, then $\Delta$ is good with respect to the map $i_{V}^{\ast}$ .

Proof.

It is sufficient to consider each vertex $\lambda$ of the Delznt polytope $\Delta$ .

From Lemma 3.12 and Lemma 4.4, if the vertex $\lambda$ of $\Delta$ is good with respect to the map $i_{V}^{\ast}$ , then we obtain the two cases:

(1)

there exists $i_{0}\in\mathcal{I}^{+}_{\lambda}$ such that $\langle u^{\lambda}_{i_{0}},q\rangle=1$ and $\langle u^{\lambda}_{j},q\rangle=0$ for any $j\in\mathcal{I}^{+}_{\lambda}\setminus\{i_{0}\}$ ,
(2)

$\langle u^{\lambda}_{n},q\rangle=-1$ .

We first consider the first case. In this case, we can assume further that $\langle u^{\lambda}_{1},q\rangle=1$ and $\langle u^{\lambda}_{2},q\rangle=\cdots=\langle u^{\lambda}_{n-1},q\rangle=0$ . Under this assumption, Remark 4.6 shows that the vectors $i_{V}^{\ast}(v^{\lambda}_{2}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n})$ are linearly independent and

i_{V}^{\ast}\left(\left\{\sum_{i=1}^{n}a_{i}v^{\lambda}_{i}\mid a_{i}\geq 0\right\}\right)=\left\{\sum_{j=2}^{n}b_{j}i_{V}^{\ast}(v^{\lambda}_{j})\mid b_{j}\geq 0\right\}.

From Lemma 4.7 and Corollary 4.12, we obtain

\det[i_{V}^{\ast}(v^{\lambda}_{2})\;\cdots i_{V}^{\ast}(v^{\lambda}_{n})]=\pm 1,

which shows that the vectors $i_{V}^{\ast}(v^{\lambda}_{2}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n})$ form a $\mathbb{Z}$ -basis of $\mathbb{Z}^{n-1}$ .

We next consider the second case. In this case, from Lemma 4.9 and Corollary 4.12, we obtain

\det[i_{V}^{\ast}(v^{\lambda}_{1})\;\cdots i_{V}^{\ast}(v^{\lambda}_{n-1})]=\pm 1,

which shows that the vectors $i_{V}^{\ast}(v^{\lambda}_{1}),\ldots,i_{V}^{\ast}(v^{\lambda}_{n-1})$ form a $\mathbb{Z}$ -basis of $\mathbb{Z}^{n-1}$ .

Therefore, $\Delta$ is good with respect to the map $i_{V}^{\ast}$ . ∎

References

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Delzant type theorem for torus-equivariantly embedded toric hypersurfaces

Abstract.

Key words and phrases:

2020 Mathematics Subject Classification:

1. Introduction

1.1. Main Results

Definition 1.1.

Definition 1.2.

Theorem 1.3 (Theorem 3.15).

Corollary 1.4.

Theorem 1.5 (Theorem 4.13).

Corollary 1.6.

1.2. Sketch of the Proofs

1.3. Outline

Acknowlegements

2. Reveiw of Torus-equivariantly Embedded Toric Manifolds

2.1. Symplectic Toric Manifolds

Definition 2.1.

Lemma 2.2.

2.2. Torus-equivariantly Embedded Toric Hypersurfaces

Definition 2.3.

Remark 2.4.

Proposition 2.5.

Proof.

Lemma 2.6 ([Yam24b, Lemma 4.14]).

2.3. Properties of the Rank of the Jacobian Matrices

Lemma 2.7.

Proof.

Lemma 2.8.

Proof.

Lemma 2.9.

Proof.

3. Delzant Polytopes of Torus-equivariantly Embedded Toric Hypersurfaces

Lemma 3.1.

3.1. Case of 𝒥λ≠∅\mathcal{J}_{\lambda}\neq\emptyset

Lemma 3.2.

Proof.

Lemma 3.3.

Proof.

Corollary 3.4.

Proposition 3.5.

Proof.

Proposition 3.6.

Proof.

3.2. Case of iV∗​(λ)i_{V}^{\ast}(\lambda) is not a vertex

Lemma 3.7.

Proof.

Lemma 3.8.

Proof.

Lemma 3.9.

Proof.

Corollary 3.10.

3.3. Case of 𝒥λ=∅\mathcal{J}_{\lambda}=\emptyset and iV∗​(λ)i_{V}^{\ast}(\lambda) is a vertex

Assumption 3.11.

Lemma 3.12.

Proof.

Lemma 3.13.

Proof.

Corollary 3.14.

3.4. The Proof of the first Main Theorem

Theorem 3.15.

Proof.

4. From Torus-equivariantly Embedded Toric Hypersurfaces to Delzant Polytopes

Theorem 4.1 ([Yam24b, Theorem 4.20]).

Corollary 4.2.

Lemma 4.3.

Proof.

Lemma 4.4.

Proof.

4.1. The first case in Lemma 4.4

Lemma 4.5.

Proof.

Remark 4.6.

Lemma 4.7.

Proof.

4.2. The second case in Lemma 4.4

Lemma 4.8.

Proof.

Lemma 4.9.

Proof.

3.1. Case of $\mathcal{J}_{\lambda}\neq\emptyset$

3.2. Case of $i_{V}^{\ast}(\lambda)$ is not a vertex

3.3. Case of $\mathcal{J}_{\lambda}=\emptyset$ and $i_{V}^{\ast}(\lambda)$ is a vertex