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This part proves $(a)\Rightarrow (b)$. If $M\ne 0$ were a finite-dimensional unital module, its action would give a homomorphismThe kernel is a two-sided ideal and is proper because the identity acts as the identity on $M$. A simple ring therefore makes the homomorphism injective. This is impossible: $A_n(k)$ has the infinitely many linearly independent ordered monomials established in part (a), whereas the endomorphism algebra has dimension $({\dim }_{k}M{)}^2$. Thus simplicity excludes every nonzero finite-dimensional module.

There is also a direct characteristic-zero check: the relation $[D_1,x_1]=1$ acts as a matrix identity. Taking its trace gives $0={\dim }_{k}M$ in $k$. In characteristic zero this forces $M=0$. This agrees with the Weyl algebra has no nonzero finite-dimensional modules criterion, but the injectivity argument proves the requested implication without presupposing characteristic zero.

The Bernstein inequality for Weyl algebra modules requires characteristic zero: for a nonzero finitely generated left or right module $M$ over $A_n(k)$,Equivalently, its Bernstein growth dimension of a Weyl algebra module is at least $n$. Without the characteristic-zero hypothesis this is false, as the module in part (c) has dimension zero in the growth sense. The proof below works over any characteristic-zero field, not just an algebraically closed one.

For a left module, choose a nonzero finite-dimensional generating subspace $M_0$ and put $M_j=F_j{A}_n{M}_0$, where $F_j$ is the Bernstein filtration. Ordered monomials giveMultiplication defines the linear action mapWe prove this map injective by induction on $j$, which is the essential faithful finite-step action of a Weyl algebra argument. For $j=0$, an element is a scalar, and a scalar killing $M_0\ne 0$ is zero. Suppose $a\in F_j$ kills $M_j$. For any generator $z$ and $m\in M_{j-1}$, both $azm$ and $zam$ vanish because $zm\in M_j$ and $m\in M_j$. Thus $[a,z]\in F_{j-1}$ kills $M_{j-1}$, so induction gives $[a,z]=0$ for every generator. By the explicit commutator formulas in part (a), commuting with all $D_i$ forces all positive $x_i$ exponents of $a$ to vanish, and commuting with all $x_i$ forces all positive $D_i$ exponents to vanish. Characteristic zero is used to cancel the nonzero integer exponents. Hence $a$ is scalar and, because it kills $M_0$, is zero. This proves injectivity.

Taking dimensions yieldsThe left side is asymptotic to $j^{2n}/(2n)!$. Consequently ${\dim }_k{M}_{2j}\ge C{j}^n$ for all sufficiently large $j$ and some $C>0$. Taking logarithms and the limsup in the definition of Gelfand–Kirillov dimension of a module proves the claimed lower bound. For a good Bernstein filtration, the associated graded module is finite over $k[x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n]$; its cumulative Hilbert function is eventually polynomial, and the same comparison says its degree is at least $n$.

For right modules, use the opposite ring isomorphism $A_n(k)\cong A_n(k{)}^{\mathrm{op}}$ sending $x_i$ to $x_i$ and $D_i$ to $-D_i$. It respects the Bernstein filtration, so the identical estimate applies. The bound is sharp: the polynomial module $k[x_1,\dots ,x_n]$, generated by $1$, has $M_j$ equal to the polynomials of degree at most $j$, of dimension $\left(\genfrac{}{}{0ex}{}{j+n}{n}\right)$. Its Gelfand–Kirillov dimension of a module is exactly $n$.

The monomials $1,x,\dots ,x^n$ are linearly independent and span, so $\boxed{{\dim }_{\mathbb{C}}V=n+1}$. Use the factorial monomial basis for polynomial differentiation $p_j(x)=x^j/j!$, $0\le j\le n$. Direct differentiation gives $e_i(p_j)={\delta }_{ij}$. HenceIndeed every $\varphi \in V$ has the Taylor expansion $\varphi ={\sum }_{j=0}^n{e}_j(\varphi )p_j$; evaluation on the $p_j$ proves independence of the functionals. Derivative orders include zero. If one uses a convention excluding zero from $\mathbb{N}$, $e_0$ must explicitly be included, since all positive-order derivatives annihilate constants and cannot span the dual space.

Differentiation satisfies $D{p}_0=0$ and $D{p}_j=p_{j-1}$ for $j\ge 1$. With column vectors of coefficients, its matrix $A$ has entries $A_{ij}=1$ if $j=i+1$ and zero otherwise, for indices $0\le i,j\le n$:The dual map is $D^{\ast }(\ell )=\ell \circ D$, not a conjugate-linear adjoint. Thus $D^{\ast }{e}_i=e_{i+1}$ for $i<n$ and $D^{\ast }{e}_n=0$. Therefore $\boxed{[D^{\ast }]=[D{]}^T}$ in the corresponding dual basis. For $n=0$ both matrices are the one-by-one zero matrix.

For nonnegative integers $m,n$, Fourier orthogonality givesEvery polynomial is a finite linear combination of these monomials, and distinct monomials are linearly independent. ThereforeThis is an algebraic basis: no infinite linear combinations are needed in the polynomial vector space.

Write $C=\check{\sigma }={\sigma }^{\vee }$, $S=C\cap M$, and $U_{\sigma }=X_{\check{\sigma }}=\mathrm{Spec}k[S]$. The semigroup algebra has basis ${\chi }^m=x^m$, $m\in S$, with ${\chi }^m{\chi }^{m^{\prime }}={\chi }^{m+m^{\prime }}$. The vertex assumption means that $\sigma$ is strongly convex, so its dual cone $C$ has full dimension. By the Gordan lemma, $S$ is finitely generated: choose integral toric cone generators and subtract integer parts of their nonnegative coefficients; the remainder lies in a bounded parallelepiped with only finitely many lattice points. This argument also applies when $C$ has lineality space, by including generators in both directions there.

Choose an integral interior point $u\in C$. For every $m\in M$, both $tu$ and $tu+m$ belong to $S$ for all sufficiently large integers $t$. Thus $S-S=M$. Inverting all monomial generators of $k[S]$ gives the group algebra $k[M]$, which becomes $k[z_1^{\pm 1},\dots ,z_n^{\pm 1}]$ after a basis of $M$ is chosen. This proves the dense torus of an affine toric variety assertion:Density follows because the localization of a ring is of a domain. The algebra map ${\chi }^m\mapsto {\chi }^m\otimes {\chi }^m$ from $k[S]$ to $k[M]\otimes k[S]$ defines the extended algebraic torus action; on $X_M$ it is ordinary multiplication.

For a face of a polyhedral cone $\nu ⪯C$, set $I_{\nu }=({\chi }^m:m\in S\setminus \nu )$. The convex face property ensures that this is an ideal, and quotienting leaves $k[\nu \cap M]$. Hence $X_{\nu }=\mathrm{Spec}k[\nu \cap M]$ is a closed subvariety of $U_{\sigma }$. Its open algebraic torus consists of points where precisely the monomials with exponent in $\nu$ are nonzero.

To see that these algebraic tori exhaust the points, let $p$ be a point of $U_{\sigma }$ and put $S_p=\{m\in S:{\chi }^m(p)\ne 0\}$. Multiplicativity gives $m+m^{\prime }\in S_p$ if and only if both $m,m^{\prime }\in S_p$. Thus $S_p$ is a semigroup face. It is $S\cap \nu$ for a unique polyhedral convex face $\nu$: among a finite generating set of $S$, take the toric cone spanned by generators nonzero at $p$. A relation expressing a sum with a generator outside this toric cone as a combination of generators inside would, after clearing rational denominators, contradict the displayed multiplicativity property. Therefore the toric cone is a convex face. Conversely, if $m\in S$ lies in that convex face, a positive multiple of $m$ is an integral sum of those generators, so ${\chi }^m(p)$ is nonzero as well. This also proves uniqueness.

The point $p$ lies in $X_M$ exactly when its convex face is all of $C$. ThereforeOne can equivalently write this as the disjoint union of the open algebraic tori of the proper convex faces. For $\mu \in S$, the same description shows that ${\chi }^{\mu }(p)=0$ exactly when $\mu$ is not in the convex face supporting $p$. On every $X_{\nu }$ with $\mu \notin \nu$ the monomial vanishes identically. Consequently the monomial zero locus in an affine toric variety isThis is an equality of reduced subvarieties, or of their underlying closed sets. The principal subscheme itself may have multiplicity, as $x^2=0$ already shows on ${\mathbb{A}}^1$.

The face duality for polyhedral cones is the order-reversing pair of mapsTheir values are convex faces because the relevant pairings are nonnegative and a sum pairs to zero only if each summand does. More explicitly, finitely many nonnegative exposing functionals can be summed to expose the indicated intersection. For a convex face $\tau$, choose $m_0\in C$ exposing it: $\tau =\sigma \cap m_0^{\perp }$. Then $m_0\in C\cap {\tau }^{\perp }$, so $\sigma \cap (C\cap {\tau }^{\perp }{)}^{\perp }\subseteq \tau$, while the reverse inclusion is immediate. Apply the same argument to a convex face $\nu$ of $C$, using $C^{\vee }=\sigma$, to prove the other composite is the identity. Inclusion reversal follows directly from annihilators. This proves the bijection, not just the existence of the maps.

For $\tau ⪯\sigma$, write $\nu =C\cap {\tau }^{\perp }$ and $L=M\cap {\tau }^{\perp }$. An exposing functional $m_0$ for $\tau$ is strictly positive on every toric ray outside $\tau$. Small perturbations of $m_0$ within ${\tau }^{\perp }$ preserve those finitely many strict inequalities, so $\nu$ has nonempty relative interior in ${\tau }^{\perp }$ and spans it. Therefore $\nu \cap M$ generates $L$, by the interior-point argument used above within that subspace. Thus the open algebraic torus of $X_{\nu }$ is $\mathrm{Spec}k[L]=X_{{\tau }^{\perp }}$. Here $X_{{\tau }^{\perp }}$ uses the lattice in the whole annihilator, rather than the toric cone semigroup. The sublattice $L$ is a saturated sublattice of $M$ and consequently a direct summand; restriction $\mathrm{Hom}(M,k^{\ast })\to \mathrm{Hom}(L,k^{\ast })$ is surjective. The extended algebraic torus action is multiplication on this open algebraic torus and hence transitive. All other monomials vanish there. It follows thatThis is the torus orbit-cone correspondence and its orbit closure in a toric variety description. Under convex face duality, $F_{\gamma }={\bigcup }_{\eta ⪰\gamma }O(\eta )$.

Finally, choose integral $m_{\tau }\in C$ exposing $\tau$. Localizing $k[C\cap M]$ at ${\chi }^{m_{\tau }}$ gives $k[{\tau }^{\vee }\cap M]$: for $m\in {\tau }^{\vee }\cap M$, adding a sufficiently large multiple of $m_{\tau }$ makes it nonnegative on every toric ray of $\sigma$ outside $\tau$, while preserving nonnegativity on $\tau$. Hence $X_{\check{\tau }}=U_{\tau }=D({\chi }^{m_{\tau }})$ is the usual affine open chart inside $U_{\sigma }$. An orbit $O(\gamma )$ belongs to it exactly when $\gamma ⪯\tau$. The requested complement of an affine toric face chart is thereforeAll sets on the right are closed, and if $\gamma$ is not contained in $\tau$, neither is any convex face containing $\gamma$. This justifies replacing the excluded orbits by their whole closures.

First suppose $\Delta$ has full dimension in $M_{\mathbb{R}}$. Form the toric cone over it in $M_{\mathbb{R}}\oplus \mathbb{R}$ and the graded semigroup algebrawhere degree zero contains only $k$. The lattice semigroup is finitely generated by the Gordan lemma. The projective toric variety of a lattice polytope is ${\mathbb{P}}_{\Delta }=\mathrm{Proj}{R}_{\Delta }$. Using all lattice points in all dilates avoids any assumption that the degree-one lattice points generate the ring.

For each vertex $v$ of $\Delta$, the degree-zero part after inverting ${\chi }^{v}t$ isIndeed its exponents are $m-lv$ with $m\in l\Delta$. Conversely, every lattice point of the tangent toric cone has this form for sufficiently large integral $l$. These charts cover the Proj construction: each point of a dilate is a rational convex combination of vertices, so some power of its monomial is a product of vertex monomials. Their dual cones are the vertex toric cones of the inward normal fan of a polytope. The normal toric fan is complete because every linear functional on $\Delta$ attains its minimum on a convex face. This describes ${\mathbb{P}}_{\Delta }$ as a complete toric variety. Translating $\Delta$ by $m_0\in M$ gives the graded isomorphism $(m,l)\mapsto (m+l{m}_0,l)$, so containing the origin is unnecessary. If $\Delta$ is not full-dimensional, translate a lattice vertex to zero and use $M\cap {\mathrm{span}}_{\mathbb{R}}(\Delta -\Delta )$; the construction is then complete of dimension $\dim \Delta$.

For the simplex in question put $Q={\prod }_i{q}_i$. Its inequalities are $x_i\ge 0$ and ${\sum }_i{q}_i{x}_i\le Q$, so its inward normal toric rays are ${\mathbb{R}}_{+}{e}_i$ and ${\mathbb{R}}_{+}(-q_1,\dots ,-q_n)$. Here the $e_i$ in the dual space denote the standard dual basis. Apply the real linear map $A(e_i)=e_i/q_i$. It carries these toric rays and their convex face incidences to the standard simplex fan, with toric rays $e_1,\dots ,e_n,-{\sum }_i{e}_i$. The transported lattice isWith $\gcd (q_1,\dots ,q_n)=1$, the primitive vectors of this toric fan in $N^{\prime }$ are $v_i=e_i/q_i$ and $v_0=-{\sum }_i{e}_i$. The latter is primitive because $\lambda {\sum }_i{e}_i\in N^{\prime }$ requires each $\lambda q_i$ to be integral, whose smallest positive solution is $\lambda =1$. Their unique primitive relation isThus mutual coprimality beyond the displayed common gcd condition is not needed.

For a general complete toric variety, introduce one variable $z_{\rho }$ for each toric ray. Its toric homogeneous coordinate ring, or Cox ring, isThe toric divisor class sequence is $0\to M\to {\mathbb{Z}}^{\Sigma (1)}\to \mathrm{Cl}(X_{\Sigma })\to 0$, where the first map is $m\mapsto (⟨m,v_{\rho }⟩{)}_{\rho }$. Thus the graded piece $S_{\alpha }$ has basis the monomials $\prod z_{\rho }^{b_{\rho }}$ with $b_{\rho }\ge 0$ and $\sum b_{\rho }[F_{\rho }]=\alpha$. For an invariant divisor $D=\sum a_{\rho }{F}_{\rho }$, monomials in degree $[D]$ correspond to algebraic torus characters ${\chi }^m$ through $b_{\rho }=a_{\rho }+⟨m,v_{\rho }⟩\ge 0$. This also identifies that graded piece with the global sections of $\mathcal{O}(D)$.

In our example $M^{\prime }=(N^{\prime }{)}^{\ast }={⨁}_i{q}_i\mathbb{Z}{e}_i^{\ast }$, and its map to ${\mathbb{Z}}^{n+1}$ sends ${\sum }_i{q}_i{b}_i{e}_i^{\ast }$ to $(-{\sum }_i{q}_i{b}_i,b_1,\dots ,b_n)$. The quotient is therefore $\mathbb{Z}$, via $(c_0,\dots ,c_n)\mapsto c_0+{\sum }_i{q}_i{c}_i$. In particular $\mathrm{deg}{z}_0=1$ and $\mathrm{deg}{z}_i=q_i$, giving the explicit graded isomorphismThe target has the grading inherited from ordinary total degree in $k[Y_0,\dots ,Y_n]$. Its displayed generators are algebraically independent, which proves injectivity as well as surjectivity onto that subring.

To obtain the quotient, the toric irrelevant ideal is generated by ${\prod }_{\rho \notin \sigma (1)}{z}_{\rho }$. Here each maximal toric cone omits exactly one toric ray, so it is $(z_0,\dots ,z_n)$, and the open set is ${\mathbb{A}}^{n+1}\setminus \{0\}$. The diagonalizable algebraic group with algebraic torus character group $\mathrm{Cl}(X)=\mathbb{Z}$ is $k^{\ast }$. Its action on the Cox coordinates is dictated by their degrees, namelyHere the $y_i$ name the independent Cox coordinates; the graded-subring realization above does not change their weights. More concretely, the quotient chart where $y_i\ne 0$ has coordinate ring $(S[y_i^{-1}]{)}_0$, the invariant ring for this action. These charts glue to $\mathrm{Proj}S$ and to the specified toric fan charts. Positive weights make each orbit closed inside the punctured affine space, and in characteristic zero the finite stabilizers cause no failure of this geometric quotient. This is the weighted projective space construction.

Let $u,v,w$ be the consecutive primitive toric ray vectors around the compact invariant curve $F=F_v$. Smoothness gives oriented determinants $\det (u,v)=\det (v,w)=1$, hence $u+w=bv$ for an integer $b$. Choose an algebraic torus character $m$ with $⟨m,u⟩=0$ and $⟨m,v⟩=1$. The principal divisor on a toric variety has coefficient one at $F$, coefficient $b$ at $F_w$, and zero at $F_u$. The other invariant curves do not meet $F$, while each adjacent one meets it transversely once. Intersecting this principal divisor with the proper curve $F$ gives $0=F^2+b$. This proves the self-intersection formula for a toric surface divisor even when the ambient surface is not complete.

If $F^2=-1$, then $b=1$ and $v=u+w$. The vectors $u,w$ form a lattice basis since $\det (u,w)=1$, and their toric cone is the union of the two toric cones adjacent to $v$. Deleting $v$ is consequently a smooth toric fan coarsening. Its inverse is precisely the toric blowup at a torus-fixed point, obtained by inserting the sum of the two primitive basis vectors. Conversely, that insertion gives $u+w=v$, and hence algebraic self-intersection $-1$. Thus the toric contraction of a minus-one curve satisfiesFor the cyclic quotient description, first take a strongly convex toric cone of dimension two. Write its primitive boundary vectors as $a,b$ and put $N_0=\mathbb{Z}a+\mathbb{Z}b$, $r=[N:N_0]$. Since $a$ is primitive, extend it to a basis of $N$ and write $b=ua+rc$ with $\gcd (u,r)=1$. Thus $N/N_0$ is cyclic of order $r$. In the lattice $N_0$, the toric cone is smooth and its toric variety is ${\mathbb{A}}^2$. Passing to the larger lattice $N$ takes the finite quotient whose invariant monomials are precisely those in the smaller dual lattice $M=N^{\ast }\subset N_0^{\ast }$. Relative to the two boundary coordinates, $c=(b-ua)/r$, so a generator acts with weights $(-u,1)$ modulo $r$. Replacing that generator by a suitable power makes the first weight one and the second some $q$ with $0<q<r$ and $\gcd (q,r)=1$. Therefore, for the singular case,The cyclic quotient surface singularity can equivalently be described by the first-quadrant toric cone inIndeed its coordinate ring is $k[x^i{y}^j:i,j\ge 0,i+qj\equiv 0(\mathrm{mod}r)]=k[x,y{]}^{{\mu }_r}$. Characteristic zero ensures the stated ordinary finite-group quotient interpretation.

The restriction to singular toric cones of dimension two is necessary for $r>1$. A smooth toric cone has $r=1$ and gives ${\mathbb{A}}^2$ with the trivial quotient. A toric cone of dimension one gives ${\mathbb{A}}^1\times k^{\ast }$, and the zero toric cone gives $(k^{\ast }{)}^2$. These latter surfaces cannot be a quotient of the displayed kind, since they have nonconstant invertible regular functions, whereas $k[x,y{]}^{{\mu }_r}$ has only constant units. Thus the blanket affine-surface wording requires these separate cases.

For the singular case, compute the negative continued fraction by the ceiling form of the Euclidean algorithm. Start with $a_0=r$, $a_1=q$, and, while $a_i>0$, setThen $0\le a_{i+1}<a_i$, so the process terminates at $a_{s+1}=0$. The gcd is unchanged at each step, hence $a_s=1$. Since $a_{i-1}>a_i>0$, every $b_i\ge 2$. Rearranging the recurrence givesNow define lattice vectorsTo verify this Hirzebruch–Jung resolution algorithm, write $v_i=(p_i,a_i)/r$ with $p_0=0$, $p_1=1$ and $p_{i+1}=b_i{p}_i-p_{i-1}$. Induction gives $a_i\equiv q{p}_i(\mathrm{mod}r)$, so $v_i\in L$. It also gives $p_i{a}_{i+1}-a_i{p}_{i+1}=-r$. Thus every adjacent pair has absolute determinant $1/r$, the covolume of $L$, and is a basis of $L$. The $p_i$ increase strictly, while the $a_i$ decrease to zero, so these toric rays occur in order inside the first quadrant. At termination, $a_s=1$ and the determinant identity gives $p_{s+1}=r$, hence $v_{s+1}=(1,0)$. All boundary and inserted vectors are primitive because they occur in lattice bases.

Subdivide the quadrant by $v_1,\dots ,v_s$. Every resulting toric cone is smooth by the smoothness criterion for a toric variety, and the fan subdivision gives a proper birational toric morphism, hence a resolution. Each interior toric ray has a complete one-dimensional star, so its exceptional curve of a surface resolution is ${\mathbb{P}}^1$. The recurrence yieldsThis proves minimality: no exceptional curve of a surface resolution can be contracted to a smooth point by the Castelnuovo contraction criterion. There is also a direct toric fan check. For fixed $i$, the determinants of $v_i$ with $v_{i+h}$, measured in units of the lattice covolume, satisfy the same recurrence with coefficients at least two and initial values zero and one. They are at least $h$. Omitting any intervening toric ray therefore leaves a toric cone of determinant at least two, so no nontrivial coarsening is smooth. Thus this is the minimal resolution of a cyclic quotient surface singularity.

If all exceptional algebraic self-intersections are $-2$, all $b_i=2$ and induction gives $[2,\dots ,2{]}^{-}=(s+1)/s$. Since $r,q$ are coprime, this forces $r=s+1$, $q=s$. Conversely $r/(r-1)$ has exactly $r-1$ coefficients equal to two. ThereforeFor the final toric fan, the minimal smooth fan subdivision of the toric cone from $(0,1)$ to $(r,1)$ inserts the toric rays $(j,1)$ for $1\le j<r$. Adjacent determinants have absolute value one, and $(j-1,1)+(j+1,1)=2(j,1)$, so its inserted curves have algebraic self-intersection $-2$ and it is the minimal fan subdivision just described. The second original toric cone, from $(r,1)$ to $(1,0)$, is already smooth.

Write $v_j=(j,1)$ and $e=(1,0)$. In the resolved toric fan, the rightmost interior toric ray obeys $v_{r-1}+e=v_r$, so its curve has algebraic self-intersection $-1$ and can be blown down to a smooth point. After deleting it, the same relation $v_{r-2}+e=v_{r-1}$ applies to the new rightmost interior toric ray. Continue deleting $v_r,v_{r-1},\dots ,v_1$ in that order. Each deletion is the inverse of a toric blowup at a torus-fixed point. The last toric fan has only the quadrant toric cone and defines ${\mathbb{A}}^2$. ConsequentlyAll maps induce the identity on the dense algebraic torus, so this composite is the same morphism as $X_{{\Sigma }^{\prime }}\to X_{\Sigma }\to {\mathbb{A}}^2$. The intermediate smooth toric fans need not include the singular toric fan $\Sigma$.

The dense algebraic torus has coordinate ring $k[M]$, a Laurent polynomial ring and hence a unique factorization domain. Its divisor class group is therefore zero. Restrict an arbitrary Weil divisor $D$ to the algebraic torus and choose a rational function on an algebraic variety $f$ whose divisor there is $D{|}_T$. Then $D-\mathrm{div}(f)$ is supported on the algebraic torus boundary. The prime divisors in that boundary are precisely the $F_j$ corresponding to the toric rays, by the torus orbit-cone correspondence. ConsequentlySince the variety is smooth, every Weil divisor is Cartier. This gives invariant representatives for all divisor classes, without assuming that the original divisor was invariant.

Now use such a representative. The associated vector space isThe principal divisor on a toric variety formula gives $\mathrm{div}({\chi }^m)={\sum }_j⟨m,e_j⟩F_j$. Define the lattice polytope of a toric divisorFor a maximal toric cone $\sigma$, its basic toric ray vectors are a basis of $N$, so there is a unique integral $m_{\sigma }$ with $⟨m_{\sigma },e_j⟩=-a_j$ on its toric rays. The algebraic torus character ${\chi }^{m_{\sigma }}$ is a local frame for $\mathcal{O}(D)$ on $U_{\sigma }$. Every global section restricts on the algebraic torus to a finite Laurent polynomial ${\sum }_m{c}_m{\chi }^m$. Relative to this frame, regularity on $U_{\sigma }$ means ${\sum }_m{c}_m{\chi }^{m-m_{\sigma }}\in k[{\sigma }^{\vee }\cap M]$. Since distinct algebraic torus characters are independent, this holds exactly when each exponent with nonzero coefficient satisfies the inequalities for the toric rays of $\sigma$. Taking all charts provesCompleteness makes $P_D$ bounded when nonempty: a nonzero recession vector would be nonnegative on every toric ray of a complete toric fan, and therefore on all of $N_{\mathbb{R}}$, which is impossible. The displayed space is consequently finite-dimensional.

At the torus-fixed point $p_{\sigma }$ in the smooth chart $U_{\sigma }\cong {\mathbb{A}}^n$, a section ${\chi }^m$ has local monomial ${\chi }^{m-m_{\sigma }}$. Its value is nonzero exactly when this monomial is constant, namely when $m=m_{\sigma }$. Thus if the complete linear system of a divisor has no base point, some section is nonzero at $p_{\sigma }$, forcing $m_{\sigma }\in P_D$. Conversely, if $m_{\sigma }\in P_D$ for every maximal toric cone, its algebraic torus character is a global section whose local expression on $U_{\sigma }$ is one. It therefore vanishes nowhere on that whole chart, and these charts cover $X$. This proves the toric basepoint-free criterion:The analogous toric ampleness criterion, stated without proof, replaces the inequalities off $\sigma$ by strict ones:Equivalently, the normal fan of a polytope $P_D$ is exactly $\Sigma$.

For the final fan subdivision, take $n\ge 2$ so that $\tilde{e}$ is a new toric ray, as the construction requires. The vectors $e_1,\dots ,e_n$ are a lattice basis, so their sum is primitive. Replacing any one basis vector by this sum gives another basis. Thus the star subdivision gives a smooth toric blowup at a torus-fixed point $\pi :X_{{\Sigma }^{\prime }}\to X_{\Sigma }$, with algebraic exceptional divisor $E=\tilde{F}$. On the old chart, a local equation of $D$ is ${\chi }^{-m_{\sigma }}$, whose order on the new toric ray is $-⟨m_{\sigma },\tilde{e}⟩={\sum }_{i=1}^n{a}_i=a$. All old toric ray coefficients are unchanged. ThereforeWe prove ampleness of $D_c=c{D}^{\prime }-E$ directly by inequalities, giving the ample divisor after blowing up a torus-fixed point result.

Let $u_1,\dots ,u_n$ be the basis of $M$ dual to $e_1,\dots ,e_n$. On the new maximal toric cone ${\sigma }_i$, which omits $e_i$, the required local algebraic torus character isIndeed its pairings on $e_j$, $j\ne i$, are $-c{a}_j$, while its pairing on $\tilde{e}$ is $-ca+1$, exactly the negative of the new coefficient $ca-1$. Its inequality on the omitted old toric ray has strict gap one:For any toric ray $e_j$ outside the original toric cone, put ${\delta }_j=⟨m_{\sigma },e_j⟩+a_j$. The original ampleness gives positive integers ${\delta }_j$, and the new strict gaps arefor all sufficiently large $c$.

For any unchanged maximal toric cone $\eta \ne \sigma$, the local algebraic torus character is $c{m}_{\eta }$. All strict inequalities on old toric rays follow from those for $D$. Its inequality on the new toric ray has gapEach summand is nonnegative, and at least one is positive because $\eta \ne \sigma$ cannot contain all its toric rays. Hence ${ϵ}_{\eta }$ is a positive integer, making this gap positive for $c\ge 2$. There are only finitely many remaining inequalities. For example, any integer satisfyingworks, with the inner maximum omitted when its index set is empty. The toric ampleness criterion now givesThe new-ray assumption is essential to the printed construction. In dimension one, $\tilde{e}=e_1$ and the fan subdivision is unchanged, so the two purported distinct divisors coincide. If the displayed definition of $D^{\prime }$ is applied literally on ${\mathbb{P}}^1$ with coefficients $a_1=-2$, $a_2=3$, then $D$ has degree one but $D^{\prime }$ has degree $-1$, contradicting the conclusion. The argument above applies to the intended blowup in dimension at least two.

Take the right coaction of the coordinate Hopf algebra of quantum SL2 on the quantum plane to beIt is extended multiplicatively and sends $1$ to $1\otimes 1$. The matrix comultiplication and counit verify the coaction identities on $x,y$. To check that it descends to the quantum plane, expandThe last bracket vanishes because $bc=cb$ and $da-ad=(q-q^{-1})bc$.

For the requested calculation, $ca=qac$ and $yx=qxy$ giveMultiply by $\delta (y)$, put all monomials into the order $x^r{y}^s$, and use $bc=cb$. The answer isA left-coaction convention transposes where the matrix coefficients occur; the displayed answer fixes the right coaction throughout.

Every ordered monomial in the degree-$n$ homogeneous component has the form $x^{n-j}{y}^j$, $0\le j\le n$. The formulas in part (a) showEach nonzero image still has total degree $n$; $K^{-1}$ does too. Since $E,F,K,K^{-1}$ generate the quantum enveloping algebra of sl2, the degree-$n$ component is a submodule, with basis $x^n,x^{n-1}y,\dots ,y^n$ and dimension $n+1$.

Use the Polynomial representation of the Heisenberg Lie algebra on $V=\mathbb{C}[x]$:The product rule gives $[d/dx,x]f=f$, verifying the required bracket, and the identity commutes with both operators.

For irreducibility, let $W$ be a nonzero invariant subspace and choose a nonzero polynomial of degree $d$ in it. Its $d$th derivative is a nonzero constant, so $1\in W$. Repeated multiplication by $x$ then puts every monomial in $W$, giving $W=\mathbb{C}[x]$.

For faithfulness, suppose $a\rho (p)+b\rho (q)+dI=0$. Applying this operator to $1$ gives $bx+d=0$, so $b=d=0$. Applying what remains to $x$ gives $a=0$. The representation is therefore faithful and irreducible. It is necessarily infinite-dimensional, by the finite-dimensional classification just proved.

Use the natural dihedral group generators $\rho z=e^{2\pi i/n}z$ and $mz=\bar{z}$, and orient the unfolding parameter so the trivial equilibrium loses stability as $\mu$ increases through zero. Rotation equivariance requires a monomial $z^p{\bar{z}}^q$ in $\dot{z}$ to have $p-q\equiv 1(\mathrm{mod}n)$, while reflection makes its coefficient real. These constraints give the dihedral steady-state normal form. The reflection axes have angles $\theta =k\pi /n$ and one-dimensional fixed-point subspaces of a group action; their equilibria are the branches guaranteed by the equivariant branching lemma.

For $n=3$, a generic quadratic anisotropy dominates:In polar coordinates, $\dot{r}=\mu r+b{r}^2\cos 3\theta +O(r^3)$ and $\dot{\theta }=-br\sin 3\theta +O(r^2)$. At a nonzero equilibrium, $\chi =\cos 3\theta =\pm 1$ and $r=-\mu /(b\chi )+O({\mu }^2)>0$. There are three symmetry-related equilibria on each parameter side, on opposite half-axes as the sign of $\mu$ changes. The radial and angular eigenvalues are $-\mu +O({\mu }^2)$ and $3\mu +O({\mu }^2)$, respectively. Hence all three nonzero branches are saddles in the full plane, even though radial stability exchanges at the crossing. The origin is a sink for $\mu <0$ and a source for $\mu >0$. The reflection normaliser has no residual sign reversal in this odd dihedral case, so these are transcritical-like branches, not symmetry-forced pitchforks.

For $n=4$, both relevant nonlinearities are cubic:There are two conjugacy classes of reflection isotropy: the coordinate axes, with $\chi =\cos 4\theta =1$, and the diagonals, with $\chi =-1$. Generically $b\ne 0$ and $a\pm b\ne 0$. Each class has a pitchfork bifurcation, with four symmetry-related equilibria when its squared radius is positive:These formulas give all generic existence and stability cases. If $a<-|b|$, both classes emerge for $\mu >0$; the class with $b\chi >0$ consists of stable nodes and the other of saddles. If $a>|b|$, both classes exist for $\mu <0$; the class with $b\chi >0$ consists of saddles and the other of sources. If $|a|<|b|$, one class exists on each side and both consist of saddles. Equality $a=\pm b$ is nongeneric and requires higher-order terms. Unlike the odd dihedral cases, the residual normaliser quotient contains a sign reversal, enforcing exact paired pitchfork branches.

For $n=5$, cubic radial saturation occurs before the first angular anisotropy:For generic $a,b\ne 0$, ten equilibria lie on the reflection half-axes whenever $-\mu /a>0$, divided into two distinct five-point group orbits. Their radii and eigenvalues areIf $a<0$, they emerge for $\mu >0$: the five with $b\chi >0$ are stable nodes and the other five are saddles. If $a>0$, they exist for $\mu <0$: the five with $b\chi >0$ are saddles and the others are sources. The leading amplitude scaling is pitchfork-like, but the quartic term distinguishes opposite half-axes and generally gives different corrections to their radii; there is no exact sign-reversal symmetry on an individual fixed line. Thus $D_3$ has quadratic saddle branches, $D_4$ has two cubic axial branch types, and $D_5$ has cubic radial branches whose angular stability splits at quartic order.