Proof. We show that the map:

\begin{equation*}\operatorname{End}_{\operatorname{g}}(S) \longrightarrow \prod_{i=1}^l \operatorname{End}_k(S_i),\end{equation*}

is a closed immersion and hence $\operatorname{End}_g(S)$ is an affine submonoid. Since S is generated as an algebra by elements in $S_1,\ldots,S_l$, an endomorphism is completely determined by the above restrictions and hence the map is injective. The fact that the map respects composition (and is well-defined) is immediate since we consider only graded endomorphisms. Thus $\operatorname{End}_{\operatorname{g}}(S)$ is a submonoid and it only remains to show that it is a closed subset; that is, cut out by polynomials.

To do this, we write down the corresponding collection of matrices with respect to the basis of each $S_i = S_i' \oplus S_i^{\prime\prime}$ given by monomials of degree α_i:

(*)\begin{equation} \phi \longleftrightarrow \left( \left( \begin{array}{c|c} A_i & 0 \\ \hline B_i & C_i\\ \end{array} \right)\right)_{i=1}^l, \end{equation}

where $B_i \in \operatorname{Hom}_k(S_i' , S_i^{\prime\prime})$. We shall often suppress the brackets in this notation.

The matrices A_i and B_i come from evaluating the single variables in $S_i'$. The C_i come from evaluating monomials in $S_i^{\prime\prime}$ which are products of 2 or more variables in $S_j'$ with j ≠ i and hence C_i is completely determined by A_j and B_j for j ≠ i. We claim that the elements of C_i are polynomials in elements of A_j and B_j for j ≠ i.

Let us prove this claim. Consider monomials $x^D,x^E \in S_i^{\prime\prime}$, where D and E are effective non-prime divisors with class α_i. Both x^D and x^E are elements of the monomial basis of $S_i^{\prime\prime}$ so that for $\phi \in \operatorname{End}_{\operatorname{g}}(S)$:

\begin{equation*}\phi(x^D) = \cdots + c_i^{DE}x^E + \cdots,\end{equation*}

where $c_i^{DE}$ is the corresponding entry in C_i. Then $x^D = x_{\rho_1} \cdots x_{\rho_s}$ is a product of variables allowing duplications with $x_{\rho_i} \not\in S_i'$. Thus,

\begin{equation*}\phi(x_{\rho_1}) \, \cdots \,\phi(x_{\rho_s}) = \cdots + c_i^{DE}x^E + \cdots.\end{equation*}

But each $\phi(x_{\rho_k})$ is a linear combination of monomials with coefficients given by elements of A_j and B_j with j ≠ i. Thus the elements of the C_i are given by polynomials in the elements of $A_{j},B_{j}$ and we are done.

On the other hand, the A_i and B_i are chosen completely arbitrarily. In other words we have a bijection of sets:

\begin{align*} \operatorname{End}_{\operatorname{g}}(S) & \longleftrightarrow \prod_{i=1}^l \operatorname{Hom}_k(S_i',S_i), \\ \phi & \longleftrightarrow \begin{pmatrix} A_i\\ B_i \end{pmatrix}. \end{align*}

The 0 in the top right-hand corner of the matrices (*) comes from the fact that:

\begin{equation*}\phi(S_i^{\prime\prime}) \cap S_i' = 0,\end{equation*}

since $S_0 = k$, and monomials in $S_i^{\prime\prime}$ contain more than one variable.

It remains to remark that $\operatorname{Aut}_{\operatorname{g}}(S)$ is the group of invertible elements in a linear algebraic monoid. It follows from [Reference Putcha29, Corollary 3.26] that $\operatorname{Aut}_{\operatorname{g}}(S)$ is a linear algebraic group.

Proof. It is clear that the matrices above form a unipotent subgroup and we refer the reader to [Reference Cox11, Theorem 4.2] for a proof that it is in fact the unipotent radical. We prove that it is positively graded by the distinguished $\mathbb G_m$.

Under (*), we have

\begin{equation*}\lambda_g(t) \longleftrightarrow \left( \begin{array}{c|c} t^{-1} I_{n_i} & 0 \\ \hline 0 & Q_i(t) \end{array}\right), \end{equation*}

where

\begin{equation*}Q_i(t) = \operatorname{diag}(t^{-l_1},\ldots,t^{-l_k}),\end{equation*}

are diagonal matrices with $l_j \geq 2$. To see this, consider $x^D = x_{\rho_1} \cdots x_{\rho_l} \in S_i^{\prime\prime}$ again allowing duplications. Then,

\begin{equation*}\lambda_g(t)(x^D) = \lambda_g(t)(x_{\rho_1}) \cdots \lambda_g(t)( x_{\rho_l}) = t^{-l}x^D\end{equation*}

where l has to be greater than 2 since D was a non-prime divisor.

To calculate the weights on the Lie algebra of U consider the conjugation action:

\begin{equation*} \lambda_g(t^{-1}) \left(\begin{array}{cc} I_{n_i} & 0 \\ B_i & C_i \end{array} \right) \lambda_g(t) = \left(\begin{array}{c|c} I_{n_i} & 0 \\ \hline t Q_i(t^{-1})B_i & Q_i(t)C_iQ_i(t) \end{array} \right), \end{equation*}

of an arbitrary element of U by $\lambda_g(t)$. Then the matrix in the bottom left hand corner is given by:

\begin{equation*}\begin{pmatrix} t^{l_1 -1} && \\ & \ddots & \\ & & t^{l_k-1} \end{pmatrix}B_i\end{equation*}

and since each $l_j \geq 2$, the exponents here are strictly positive. This suffices to show that the group is graded unipotent since the matrices B_i describe the Lie algebra.

Proof. Let $X = \mathbb P(a_0 ,\dots a_n) = \operatorname{Proj} S$ where $S = k[x_1, \dots , x_{m} , y_1,\dots , y_{n_l}]$ so that the y_j have the maximum weight $b_l = a_n$. Then $\lambda_{g,N}(\mathbb G_m) \subset G = \operatorname{Aut}_{\operatorname{g}}(S)$ acts on X as follows:

\begin{align*} \lambda_{g,N}(t) \cdot (0: \dots :x_i: \dots : 0) &= (0: \dots :t^{-N}x_i:0: \dots : 0)\\ \lambda_{g,N}(t) \cdot (0: \dots :y_j: \dots : 0) &= (0: \dots: 0 :ty_j: \dots : 0), \end{align*}

for $1\leq i \leq m$ and $1 \leq j \leq n_l$. Let $u \in U \subset G$ be an element of the unipotent radical. By Remark 3.12, u acts on S as follows:

\begin{align*} u \cdot x_i &= x_i + p_i(x_0, \dots , x_{n'})\\ u \cdot y_j &= y_j + q_j(x_0, \dots , x_{n'}), \end{align*}

for $0\leq i \leq n'$ and $0 \leq j \leq n_l$, where $p_i,q_j \in k[x_1,\dots ,x_{m}]$ are weighted homogeneous polynomials (possibly 0) of degree a_i and a_n respectively. Note that $p_i = 0$ for those i such that $a_i = b_1$ is the minimum weight and that p_i and q_j do not contain any factors of y_j, since the y_j all have the same maximal weight. In particular, if $p_i \neq 0$, then $\deg p_i \gt 1$.

Consider the action by conjugation of $\lambda_{g,N}(\mathbb G_m)$ on U, first on the x_i:

\begin{align*} \left(\lambda_{g,N}(t) \cdot u \cdot \lambda_{g,N}(t^{-1}) \right)\cdot x_i &= \left(\lambda_{g,N}(t) \cdot u \right)\cdot t^{N}x_i\\ & = \lambda_{g,N}(t) \cdot (t^{N}x_i + p_i(t^{N}x_0, \dots , t^{N}x_{n'}))\\ & = x_i + t^{-N} p_i(t^{N}x_0, \dots , t^{N}x_{n'}). \end{align*}

Those p_i’s which are non-zero have degree $a_i \gt 1$ and hence if u is a weight vector for the $\lambda_{g,N}(\mathbb G_m)$-action, it has weights $a_iN - N \gt 0$. The argument for the y_i is identical and is omitted.

Proof. We write $G = \operatorname{Aut}_{\operatorname{g}}(S) = \prod_{j=1}^l \mathrm{GL}_{n_j} \lt imes U$ and denote $\operatorname{Lie} G$ by:

\begin{equation*}\mathfrak g = \prod_{j=1}^l \mathfrak g\mathfrak l_{n_j} \lt imes \mathfrak u.\end{equation*}

It is clear that $\lambda_{\underline{a}}(\mathbb G_m) \subset \operatorname{Aut}(f)$. To obtain the desired result it suffices to show that the Lie algebras of $\operatorname{Aut}(f)$ and $\lambda_{\underline{a}}(\mathbb G_m)$ agree as sub-Lie algebras of $\mathfrak g$.

The Lie algebra $\mathfrak g$ acts on S by derivation: let $\xi \in \mathfrak g$ and $F \in S$ be arbitrary elements of $\mathfrak g$ and S respectively, then

\begin{equation*}\xi(F) = \sum_{i = 0}^n F_i \, \xi(x_i),\end{equation*}

where $F_i = \frac{\partial F}{\partial x_i}$. Suppose that $\xi \in \operatorname{Lie} (\operatorname{Aut}(f)) \subset \mathfrak g$. Then since f is semi-invariant under the action of $\operatorname{Aut}(f)$, it is also a semi-invariant for the action of $\operatorname{Lie} (\operatorname{Aut}(f))$; that is, $\xi(f) = \tilde{\alpha} f$ for some $\tilde{\alpha} \in k$. The weighted Euler formula tells us that $f = \frac{1}{d}\sum_{i=0}^n a_if_i$ and so,

\begin{equation*}\sum_{i=0}^n f_i (\xi (x_i) - \alpha a_i x_i) = 0,\end{equation*}

where $\alpha = \frac{\tilde{\alpha}}{d}$.

Rearranging, for each i we get an equation:

\begin{equation*}p_if_i = -(p_0f_0 + \cdots +p_{i-1}f_{i-1}+p_{i+1}f_{i+1} + \cdots + p_nf_n),\end{equation*}

where $p_j = \xi(x_j) - \alpha a_j x_j$. Thus $p_i f_i \in(f_0, \ldots ,f_{i-1},f_{i+1}, \ldots , f_n)$.

Since f is quasismooth, its partial derivatives $f_0,\ldots,f_n$ form a regular sequence. Moreover, any permutation of the f_i is a regular sequence. Thus f_i is a non-zero divisor in the ring $S / (f_0, \ldots ,f_{i-1},f_{i+1}, \ldots , f_n)$ and hence $p_i \in (f_0, \ldots ,f_{i-1},f_{i+1}, \ldots , f_n)$. However, $\deg p_i = a_j$ and since we assumed $\deg f \geq \max\{a_j\}+2$, this forces $p_i = 0$ and $\xi(x_i) = \alpha a_i x_i$. Thus α is the only parameter and we have shown that $\operatorname{Lie} (\operatorname{Aut}(f))$ is one dimensional and hence agrees with that of $\operatorname{Lie} \lambda_{\underline{a}}(\mathbb G_m)$.

Moreover, we can see explicitly that:

\begin{equation*}\operatorname{Lie} (\operatorname{Aut}(f)) = \{((\alpha b_j I_{n_j})_{j=1}^l,0) \, | \, \alpha \in k\} \subset \mathfrak g,\end{equation*}

which is precisely the Lie algebra of $\lambda_{\underline{a}}(\mathbb G_m)$.

Proof. Since α is very ample and A corresponds to the monomial basis of S_α, the toric variety $X_A \subset \mathbb P(k^A)^\vee$ is the toric variety X with the embedding defined by α. Thus $X^{\vee, \,\alpha} = X_A^\vee$.

It remains to see that $X_A^\vee = \nabla_A$. To see this, we consider the map $\widetilde{\Phi}_A : (k^*)^{n+1} \to k^A$ as local parameters on the torus in the cone $Y_A \subset (k^A)^\vee$ over X_A. Note that since α is very ample, Y_A is an affine toric variety and that $Y_A - \{0\} \to X_A$ is a toric morphism. We claim that $X_A^\vee = \nabla_A$.

For some $f \in (k^A)^\vee$, if $\mathbf T_yY_A \subset V(f) \subset (k^A)$ for some $y \in T_{Y_A}$, where $T_{Y_A}$ is the torus in Y_A, then $f \in Y_A^\vee$. However, if it holds that $\mathbf T_yY_A \subset V(f) \subset (k^A)$ for some torus point $y \in T_Y$, then $f \in \nabla_A^\circ$ (see Definition 4.1), since the torus $T_{Y_A} \subset Y_A$ is contained in the smooth locus of _A.

Thus we have identified $\nabla_A^\circ$ with a non-empty open (and hence dense) subset of $X^{\vee}_A$ given by hyperplanes containing the tangent space to points on the torus. Note that $\nabla_A^\circ \subset \nabla_A$ is a dense subset by definition. Since both $\nabla_A$ and $X_A^\vee$ are irreducible hypersurfaces in $\mathbb P(k^A)$, we must have $X_A^\vee = \nabla_A$.

Proof. We can describe the fibre explicitly:

\begin{align*} p_1^{-1}(x_0) & = \big\{(x_0,[f]) \, \, | \, \, \frac{\partial f}{\partial x_i}(x_0) = 0 \, \, \, 0 \leq i \leq r \big\} \\ & = \bigcap_{i=0}^r \big\{(x_0 , [f]) \, \, | \, \, \frac{\partial f}{\partial x_i}(x_0) = 0 \big\}.\\ \end{align*}

Each of the sets $\{(x_0 , [f]) \, \, | \, \, \frac{\partial f}{\partial x_i}(x_0) = 0\}$ is the vanishing of a linear polynomial in the coefficients of the polynomial f. It follows that $p_1^{-1}(x_0)$ is the intersection of hyperplanes and thus a linear subspace.

For $g \in G$, we have that:

\begin{equation*}g \cdot p_1^{-1}(x_0) = p_1^{-1}(g\cdot x_0),\end{equation*}

as $g \cdot (x_0,[f]) = (g \cdot x_0 , g\cdot [f])$. In particular, they are all linear subspaces of the same dimension.

Proof. Since $B: = G \cdot T = \bigcup_{g \in G} g \cdot T \subset X$ is an open subset of X, it is an integral noetherian scheme. Then

\begin{equation*}B\times |\alpha| = |\alpha|_B \subset |\alpha|_X = X \times |\alpha|,\end{equation*}

is open and $W\subset |\alpha|_X$ is closed, so $W' \subset |\alpha|_B$ is a closed subscheme. To see that all fibres over points in B have the same Hilbert polynomial, we observe that, since the torus acts transitively on itself, $G \cdot x = G \cdot T = B$ for all $x \in T$. So applying Lemma 4.7, we have that all the fibres over B are linear subspaces of the same dimension and thus have the same Hilbert polynomial. Hence we can apply [Reference Hartshorne20, Theorem III.9.9] and conclude that $p_1|_{W'}$ is a flat morphism.

Proof. Consider the map $p_1|_{W'}:W' \to G \cdot T$. Since X is irreducible and $G \cdot T$ is open, $G \cdot T$ is also irreducible. By Corollary 4.8, $p_1|_{W'}$ is flat and hence open. By Lemma 4.7, every fibre is isomorphic to the same projective space, and hence all fibres are irreducible.

Hence we can apply [Reference Grothendieck19, IV.2, Corollaire 2.3.5 (iii)] to $p_1|_{W'}$ and conclude that Wʹ is irreducible.

Proof. Note that by the definition of $\nabla$

\begin{equation*}\nabla_A = \overline{\{[f] \in |\alpha| \, \, | \, \, \exists x \in T\, \text{such that } \frac{\partial f}{\partial x_i}(x) = 0\, \text{for all } i\} } = \overline{p_2(p_1^{-1}(T))},\end{equation*}

where $T \subset X$ is the torus. Then since $ T \subset G \cdot T$, it holds that $p_1^{-1}(T) \subset p_1^{-1}(G \cdot T)$. Thus

\begin{equation*}\overline{p_2(p_1^{-1}(T))} \subset \overline{p_2(p_1^{-1}(G \cdot T))}.\end{equation*}

Applying the definition of Wʹ and the observation that $\overline{p_2(p_1^{-1}(T))} =\nabla_A$, we conclude:

\begin{equation*}\nabla_A \subset \overline{p_2(W')}.\end{equation*}

Then since Wʹ is irreducible, $\overline{p_2(W')}$ is irreducible. Also note that $\operatorname{codim} p_2(W') \geq 1$, since the quasismooth locus in $|\alpha|$ is open and Wʹ is disjoint from $|\alpha|^{\text{QS}}$. Hence $\overline{p_2(W')}$ is an irreducible closed subvariety of codimension 1. Then as $\nabla_A$ is an irreducible subvariety of codimension 1, we conclude that:

\begin{equation*}\nabla_A = \overline{p_2(W')},\end{equation*}

which completes the first part of the theorem.

Now we prove that $\nabla_A$ is G-invariant. Note that both maps p ₁ and p ₂ are G-equivariant since they are restrictions of projections. Then as $G \cdot T$ is G-invariant it follows that $W' = p_1^{-1} (G \cdot T)$ is G-invariant and thus $\nabla_A = \overline{p_2(W')}$ is also G-invariant.

Proof. By definition, $\nabla_A = \textbf{V}(\Delta_A) \subset |\alpha|$. The automorphism group G acts on $H^0(|\alpha|,\mathcal O_{|\alpha|}(\deg \Delta_A))$. Since $\nabla_A$ is G-invariant, for every $g \in G$ we have that $\textbf{V}(g \cdot \Delta_A) = \textbf{V}(\Delta_A)$. Thus $g \cdot \Delta_A = \chi(g) \Delta_A$ for some $\chi(g) \in k^*$. It follows from the group action laws that $\chi(g' g) = \chi(g')\chi(g)$ and thus $\chi : g \mapsto \chi(g)$ is a character. This proves the result.

Notation 5.1.

We give the parameter space:

\begin{equation*}\mathcal Y_d = \operatorname{Div}_X^d = \mathbb P(k[x_1, \dots , y_{n_l}]_d),\end{equation*}

the following coordinates of the coefficients of the monomials: $(u_0:\cdots : u_{L} : v_0 : \cdots : v_M) \in \mathcal Y_d$, where the v_j correspond to monomials in the y_j and all have the same total degree, and the u_i are the coefficients of monomials containing an x_i for some $1 \leq i \leq n'$. The integer M is defined by:

\begin{equation*}M = \begin{pmatrix} n_l + d_n \\ d_n \end{pmatrix}, \end{equation*}

where $d_n = \frac{d}{a_n}$ and the integer L is computed in terms of the $a_i's$, but its exact value is not required for the subsequent discussion.

Proof. As noted above, $\lambda_{g,N}$ acts on monomials containing only variables y_i with weight $-d_n$. Suppose that $\underline{x}^I \in V = k[x_1, \dots , y_{n_l}]_d$ is another monomial containing at least one x_i variable. Then $\lambda_{g,N}(t) \cdot \underline{x}^I = t^{r(N,I)}\underline{x}^I$ and since $\lambda_{g,N}(t) \cdot x_i = t^N x_i$ we have that $r(N,I) \gt -d_n$. Hence $V_{\min} = k[y_1, \dots , y_{n_l}]_d$ and thus

\begin{equation*}Z_{\min} = \mathbb P(V_{\min})=\{(0: \cdots : 0 : v_0 : \cdots : v_M) \, \, | \, \, \exists j: \, \, v_j \neq 0 \},\end{equation*}

using Notation 5.1.

Notation 5.7.

For the rest of this section, we fix N > 0 and $\hat{U} = \lambda_{g,N} \lt imes U$.

Proof. We begin by observing that

\begin{equation*}\mathcal Y - \mathcal Y_{\min}^0 = \{(u_0 : \cdots : u_{M'} : 0 : \cdots : 0)\}.\end{equation*}

Take some $f \in \mathcal Y - \mathcal Y_{\min}^0$. We know that f contains no monomials made up of only the y_i. Thus $(0 : \dots : 0 : 1) \in X$ will be a common zero for all $\frac{\partial f}{\partial x_i}$ and $\frac{\partial f}{\partial y_i}$ since d is a Cartier degree (as monomials of the form $y_{n_l}^{d_n-1}x_i$ can never be homogeneous of degree d). It follows that f is not quasismooth and hence

\begin{equation*} \mathcal Y^{\text{QS}} \subset \mathcal Y_{\min}^0.\end{equation*}

Suppose that $f \in Z_{\min}$. Then f is a polynomial in the y_i and so $(1:0: \cdots : 0) \in X$ will be a common zero for all $\frac{\partial f}{\partial x_i}$ and $\frac{\partial f}{\partial y_i}$. Thus f is not quasismooth and we have that:

\begin{equation*}Z_{\min} \subset \mathcal Y^{\text{NQS}}.\end{equation*}

Since $\mathcal Y^{\text{NQS}}$ is a G-invariant subset, it follows that:

\begin{equation*}U \cdot Z_{\min} \subset \mathcal Y^{\text{NQS}}\end{equation*}

and so we conclude that:

\begin{equation*} \mathcal Y^{\text{QS}} \subset \mathcal Y_{\min}^0 - U \cdot Z_{\min}.\end{equation*}

Proof. Let $G = \operatorname{Aut}_{\operatorname{g}}(S)$; then a general automorphism in the unipotent radical $\phi \in U \subset G$ is given by:

\begin{equation*} \phi : \left\{\begin{array}{lr} x_i \longmapsto x_i \\ y_j \longmapsto y_j+ p_{\phi,j}(x_1, \dots , x_{n}), \end{array}\right. \end{equation*}

for $1 \leq i \leq n$ and $1 \leq j \leq m$ and with $p_{\phi,j} \in k[x_1, \dots , x_{n}]_r$. Composing two such elements $\phi,\psi \in U$ gives

\begin{equation*} \phi \circ \psi : \left\{\begin{array}{lr} x_i \longmapsto x_i \\ y_j \longmapsto y_j+ p_{\phi,j}(x_1, \dots , x_{n})\!\!\!\!&+ \,\, p_{\psi,j}(x_1, \dots , x_{n}). \end{array}\right. \end{equation*}

It follows that any two automorphisms commute and hence U is abelian and thus

\begin{equation*}U \simeq \mathbb G_a^L.\end{equation*}

Let us prove the second statement. Let $d' = \frac{d}{b}$ and recall that $Z_{\min} = \mathbb P(k[y_1, \dots , y_m]_{d'})$. Take some $[f] \in Z_{\min}^{\text{ss},R}$ and $\phi \in \operatorname{Stab} _U([f])$. Assume that ϕ in non-trivial.

Since the $\mathrm{GL}_n$ acts trivially on $Z_{\min}$ it follows that $Z_{\min}^{\text{ss},R} = Z_{\min}^{\text{ss},\mathrm{GL}_m}$. Note that this only holds for semistability. Then considering the standard maximal torus in $\mathrm{GL}_m$, the reductive Hilbert–Mumford criteria imply that the barycentre of the section polytope of $\mathcal O_{Z_{\min}}(d')$ is contained in the Newton polytope of $[f]$. Since $d \gt b+1$, it follows that each variable y_j appears in $[f]$ and thus $\phi \cdot [f]$ must contain at least one x_i. Hence $\phi \cdot [f] \neq [f]$ and we have a contradiction.

Proof. The first statement follows directly from the $\hat{U}$-Theorem (Theorem 2.13) and the second statement follows from the fact that a geometric quotient is an open map (since it is a topological quotient) and that Proposition 5.8 we that $ \mathcal Y^{\text{QS}} \subset \mathcal Y^{s,\hat{U}}$ is an open subset.

Proof. Consider the geometric quotient map $q:\mathbb P^n \rightarrow X$ defined by the injective map of graded rings $q^* : k[x_0,\dots , x_n] \rightarrow k[\overline{x}_0, \dots \overline{x}_n]$ sending $x_i \mapsto \overline{x}_i^{a_i}$. Let $\widetilde{Y} = q^{-1}(Y) \subset \mathbb P^n$, then $\widetilde{Y}$ is the hypersurface defined by the polynomial $f(x_0^{a_0}, \dots , x_n^{a_n})$ and is of degree $d \geq \max\{a_0, \dots , a_n\} + 2 \geq 3$. As Y is quasismooth, we have that $\widetilde{Y}$ is smooth (see for example [Reference Batyrev and Cox8, Proposition 3.5]). This implies that the barycentre of the section polytope of $\mathbb P^n$ is contained in the interior of the Newton polytope of $\widetilde{Y}$ by the reductive moduli problem for hypersurfaces of degree $d \geq 3$ [Reference Mumford, Fogarty and Kirwan27, Proposition 4.2].

Let $P = \operatorname{Conv}(de_0, \dots , de_n)$ denote the section polytope of $\mathbb P^n$ and $P_{X} = \operatorname{Conv}(d_0e_0, \dots , d_ne_n)$ denote the section polytope of X, where $d_i = \frac{d}{a_i}$. Then the respective barycentres are given by $B = \frac{1}{n+1}\sum_{i=1}^nde_i$ and $B_X = \frac{1}{n+1}\sum_{i=0}^nd_ie_i$.

Consider the map $\varphi : \mathbb Q^{n+1} \rightarrow \mathbb Q^{n+1}$ given by $\varphi(e_i) = \frac{1}{a_i}e_i$ for $i = 0, \dots , n$, where the e_i are the unit vectors. Note that $\varphi(B) = B_X$. Under this linear map, the Newton polytope of $\widetilde{Y}$ is sent to the Newton polytope of Y. Moreover, $B \in NP(\widetilde{Y})^{\circ}$ implies that $B_X \in NP(Y)^{\circ}$.

Proof. We drop the linearisation from the notation. Note that it suffices to prove that $ \mathcal Y^{\text{QS}} \subset \mathcal Y^{\text{s},T}$, since by the non-reductive Hilbert–Mumford criterion of Theorem 2.19 we get $ \mathcal Y^{\text{QS}} \subset \mathcal Y^{\text{s},G}$. Indeed, as $ \mathcal Y^{\text{QS}}$ is G-invariant, we have that $ \mathcal Y^{\text{QS}} \subset g \cdot \mathcal Y^{\text{s},T}$ and hence

\begin{equation*} \mathcal Y^{\text{QS}} \subset \bigcap_{g \in G} g \cdot \mathcal Y^{\text{s},T} = \mathcal Y^{\text{s},G},\end{equation*}

by Theorem 2.19.

Let us prove that $ \mathcal Y^{\text{QS}} \subset \mathcal Y^{\text{s},T}$. Suppose that $Y \subset X$ is a quasismooth hypersurface of degree d. Then by Lemma 5.17, the polytope $\operatorname{NP}(Y)$ contains the origin O. Thus by the Hilbert–Mumford criterion of Theorem 2.18 quasismooth hypersurfaces are T-stable for the twisted linearisation $\mathcal O(1)$.

Proof. Let $c \in H^0(\mathcal Y,\mathcal O(1)) = (k[x_1, \dots , x_n ,y]_d)^\vee$ be the section corresponding to the coefficient of the monomial $y^{d'}$. By Example 4.15, we have that $ \mathcal Y^{\text{QS}} = \mathcal Y_{c \cdot\Delta_A} \subset \mathcal Y_c$ and hence $ \mathcal Y^{\text{QS}}$ is an affine variety. In this case, $Z_{\min} = \{(0: \cdots : 0 : c) \, \, | \, \, c \neq 0\}$ is a point, and so by Remark 2.14 the quotient:

\begin{equation*}q_U : \mathcal Y_{\min}^0 \longrightarrow \mathcal Y_{\min}^0/U,\end{equation*}

from Corollary 5.16 is a trivial U-bundle. Hence $\mathcal Y_{\min}^0/U$ is affine by [Reference Asok and Doran3, Theorem 3.14]. Thus $ \mathcal Y^{\text{QS}} \to \mathcal Y^{\text{QS}} /U$ is a trivial bundle and $Q = \mathcal Y^{\text{QS}}/U$ is an affine variety.

Consider the action of $R = G /U $ on Q. Since $ \mathcal Y^{\text{QS}}$ is affine, Lemma 2.4 implies that Q admits a geometric quotient by R if and only if all the G-orbits are closed in $ \mathcal Y^{\text{QS}}$. Then as $d \geq r+2$, Theorem 3.15 implies that all stabiliser groups are finite giving that the action on $ \mathcal Y^{\text{QS}}$ is closed. Hence we have a geometric quotient:

\begin{equation*} \mathcal Y^{\text{QS}} / G = Q / R.\end{equation*}

Since Q is an affine variety and $Q/R$ is a reductive quotient, we conclude that $ \mathcal Y^{\text{QS}} / G$ is a projective over affine variety.

Proof. Since Y is a proper algebraic scheme, by [Reference Matsumura and Oort26, Lemma 3.4], we can identify the Lie algebra of the automorphism group of Y with the vector space $H^0(Y,\mathcal T_Y)$ of global vector fields on Y, where $\mathcal T_Y$ is the tangent sheaf. If we show that $ H^0(Y,\mathcal T_Y) = 0$, then we can conclude that $\dim \operatorname{Aut}(Y) = 0$ and since $\operatorname{Stab} _{G}(Y) \subset \operatorname{Aut}(Y)$ we have that $\dim \operatorname{Stab} _G(Y) = 0$.

Let us prove that $H^0(Y,\mathcal T_Y) = 0$. Let $N = \dim Y = n+m -1$; then by Serre duality:

\begin{equation*}H^0(Y,\mathcal T_Y) \simeq H^N(Y, \Omega_Y \otimes \omega_Y)^\vee,\end{equation*}

where ω_Y is the canonical line bundle of Y. Let $\mathcal O_Y(1,1)$ be the restriction of $\mathcal O_X(1,1) = \mathcal O_{\mathbb P^n}(1) \boxtimes \mathcal O_{\mathbb P^m}(1)$ to Y. By the adjunction formula we have that $\omega_Y \cong \mathcal O_Y(d-n-1,e-m-1)$ and hence by our assumption we have that ω_Y is very ample. Then by Kodiara-Nakano vanishing ([Reference Esnault and Viehweg17, Theorem 1.3]) we have that $H^N(Y,\Omega_Y(d-n-1,e-m-1)) = 0$. Hence

\begin{equation*}H^0(Y,\mathcal T_Y) \simeq H^N(Y,\Omega_Y(d-n-1,e-m-1))^\vee = 0.\end{equation*}

We conclude that $\dim \operatorname{Aut}(Y) = 0$.

Proof. First, note that the discriminant $\Delta_A$ is a true invariant for the G-action since G has no non-trivial characters. Then by Theorem 4.10, we have that:

\begin{equation*}\mathcal Y^{\text{SM}} = \mathcal Y_{\Delta_A},\end{equation*}

since G acts transitively on X and hence $\mathcal Y^{\text{SM}} \subset \mathcal Y^{\text{ss}}$. Finally, if $d \gt n+1$ and $e \gt m+1$, by Proposition 5.23, we have that the stabiliser for every point in $\mathcal Y^{\text{SM}}$ is finite and hence all the orbits are closed. It follows that $\mathcal Y^{\text{SM}} \subset \mathcal Y^{\text{s}}$.