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Let $X$ be a connected compact Riemann surface of genus $g$. Choose an oriented symplectic basis $a_1,\dots ,a_g,b_1,\dots ,b_g$ of $H_1(X,\mathbb{Z})$, with $a_i\cdot b_j={\delta }_{ij}$. For a closed complex differential form $\alpha$, write $A_i(\alpha )={\int }_{a_i}\alpha$ and $B_i(\alpha )={\int }_{b_i}\alpha$. The Riemann bilinear relations begin with the identityTo prove it, cut $X$ along the chosen cycles to obtain a fundamental polygon $P$. On this simply connected polygon, a closed differential form has a primitive $F$, with $dF=\alpha$. Stokes theorem gives ${\int }_P\alpha \wedge \beta ={\int }_{\partial P}F\beta$. The two copies of each edge have opposite orientations, and their values of $F$ differ by the period acquired in passing between them. Pairing the copies of the $a_i$ and $b_i$ edges gives, respectively, $-B_i(\alpha )A_i(\beta )$ and $A_i(\alpha )B_i(\beta )$. Summing proves the identity; the convention $a_i\cdot b_i=1$ fixes its sign.

For holomorphic one-forms $\omega ,\eta$, their wedge product is zero because locally both are multiples of $dz$. Thus the first holomorphic Riemann bilinear relations isFor a nonzero holomorphic one-form $\omega =f(z)dz$, the second relation follows by applying the bilinear identity to $\omega ,\overline{\omega }$:In particular, a holomorphic one-form with all $a$-periods zero must vanish. Since $\dim H^0(X,K_X)=g$, the map taking a holomorphic one-form to its $a$-periods is an isomorphism. We can therefore choose normalized holomorphic one-forms ${\omega }_1,\dots ,{\omega }_g$ with ${\int }_{a_i}{\omega }_j={\delta }_{ij}$. Put $B_{ij}={\int }_{b_i}{\omega }_j$. The first relation applied to ${\omega }_j,{\omega }_k$ gives $B_{jk}=B_{kj}$, and the second applied to ${\sum }_j{c}_j{\omega }_j$ gives $2{\overline{c}}^{,t}(\mathrm{Im}B)c>0$ for $c\ne 0$. ConsequentlyThese conclusions make the construction of the Jacobian variety work. Integration identifies the image of $H_1(X,\mathbb{Z})$ in $H^0(X,K_X{)}^{\ast }$ with the period lattice $\Lambda ={\mathbb{Z}}^g+B{\mathbb{Z}}^g$. The real-linear map $(m,n)\mapsto m+Bn$ is invertible: its imaginary part is $Yn$, and $Y$ is invertible. Hence $\Lambda$ is a discrete lattice of full real rank $2g$, andis a compact complex torus, rather than a non-Hausdorff quotient. Changing the symplectic basis changes the displayed coordinates but not this intrinsic quotient.

The Riemann bilinear relations also supply its polarization of a complex torus. Use the Hermitian form $H(z,w)=z^t{Y}^{-1}\overline{w}$, linear in the first argument, and set $E=-\mathrm{Im}H$. Symmetry of $B$ givesThus $E$ is an integral positive Riemann form on a complex torus. Its matrix on the displayed lattice basis is the unimodular symplectic matrix. The corresponding positive holomorphic line bundle, equivalently the one constructed from theta automorphy in Solution 4, gives a principal polarization of a complex torus. By the Kodaira embedding theorem this makes the complex torus projective, hence an abelian variety. The normalization $E=-\mathrm{Im}H$ here is important because $H$ was chosen linear in its first argument. If $g=0$, the spaces of holomorphic one-forms and periods are zero and the Jacobian variety is a point.

Let $v_i$ be the primitive lattice generator of the ray ${\tau }_i$. In the toric divisor class sequence, the first map sends a torus character exponent $m\in M$ to its principal divisor on a toric variety:The second map sends an invariant Weil divisor to its class modulo principal divisors. Completeness makes the rays span $N_{\mathbb{R}}$, so the first map is injective. These are the maps in the given sequence; its exactness need not be reproved.

For $D={\sum }_i{r}_i{F}_i$, letThe torus characters ${\chi }^m$ with $m\in P_D\cap M$ form a basis of $\mathcal{L}(D)=\{f\in \mathbb{C}(X):\mathrm{div}(f)+D\ge 0\}\cup \{0\}$. This works for Weil divisors, not only for Cartier divisors. It is the character basis associated to the lattice polytope of a toric divisor.

For a Cartier divisor, the toric ampleness criterion is as follows. Each maximal cone $\sigma$ has local Cartier data $m_{\sigma }\in M$ satisfying $⟨m_{\sigma },v_i⟩=-r_i$ on its rays. The divisor is ample exactly whenEquivalently the support function is strictly concave, with these linear pieces. For a Q-Cartier Weil divisor use rational $m_{\sigma }$ and clear their denominators; the criterion means that a positive multiple is ample Cartier. An arbitrary Weil divisor need not be Q-Cartier, so this qualification cannot be omitted.

Take positive weights $q=(q_0,\dots ,q_n)$, reduced so their common greatest common divisor is one. A toric definition of weighted projective space usesIf the weights are well-formed, meaning $\gcd (q_0,\dots ,\widehat{q_i},\dots ,q_n)=1$ for every $i$, the $v_i$ are primitive: the quotient $N/\mathbb{Z}{v}_i$ has torsion of order that omitted-weight greatest common divisor. The lattice $N$ itself is free because $q$ is primitive. The positive relation ${\sum }_i{q}_i{v}_i=0$ implies that proper subsets generate pointed cones and that they meet along common faces. Completeness can be seen directly: represent a vector by real coordinates $a_i$ and subtract ${\min }_i(a_i/q_i)$ times $q$. The resulting coordinates are nonnegative and at least one is zero, placing the vector in one of the displayed cones. Their common-face property follows from this unique normalization of each class modulo $\mathbb{R}q$. This gives the fan of the well-formed weighted projective space.

The dual lattice isIts map in the divisor sequence is simply inclusion in ${\mathbb{Z}}^{n+1}$, so the homomorphism $a\mapsto q\cdot a$ identifies the cokernel with $\mathbb{Z}$. It is onto by coprimality of all the weights. ThusChoose integers $a_i$ by the Bezout identity ${\sum }_i{q}_i{a}_i=1$. Then $D={\sum }_i{a}_i{F}_i$ explicitly represents the positive generator $H$.

To verify ampleness in the Weil/Q-Cartier sense, put $\ell =\mathrm{lcm}(q_0,\dots ,q_n)$. On the maximal cone omitting ray $j$, defineThis is integral and has $q\cdot m_j=\ell -\ell =0$. Its pairings with the included rays are $-\ell a_i$, while on the omitted ray they equal $-\ell a_j+\ell /q_j>-\ell a_j$. Consequently $\ell D$ is Cartier and ample by the numerical criterion. This is an explicit class-group generator of a well-formed weighted projective space.

The generator is generally ample Weil, not Cartier. If the source's word “ample” is interpreted as requiring a Cartier divisor, its generator request is false in general: on $\mathbb{P}(1,1,2)$, take $D=F_0$. On the chart omitting the weight-two ray, the Cartier equations force $m=(-1,0,1/2)$, which is not in the integral dual lattice. Twice this datum is integral, so the class-one divisor has Cartier index two and no Cartier divisor generates its class group. With the usual ample-Q-Cartier-Weil interpretation, the construction above proves exactly the requested result. This distinction also explains the use of the divisorial spaces $\mathcal{L}(rD)$ rather than assuming an invertible sheaf $\mathcal{O}(D)$.

Finally let an ample divisor represent the class-group generator. Its class is the positive generator; the negative generator cannot be ample because a positive Cartier multiple has negative degree on the curves where the positive generator has positive degree. After replacing by a linearly equivalent invariant divisor, write it as $D=\sum a_i{F}_i$ with $\sum q_i{a}_i=1$. A character basis element of $\mathcal{L}(rD)$ has exponent $m\in M$ with $m_i+r{a}_i\ge 0$. Set $b_i=m_i+r{a}_i$. Then $b\in {\mathbb{Z}}_{\ge 0}^{n+1}$ and $\sum q_i{b}_i=r$. Conversely every nonnegative exponent vector of weighted degree $r$ gives the unique $m=b-ra\in M$. Thereforeis a bijection of bases in each degree. Adding $m$ and $r$ under multiplication adds these nonnegative exponent vectors, so it preserves products and the unit. A different linearly equivalent representative changes the basis by the corresponding rational-function power and gives the same graded-ring isomorphism. Hence the divisorial section ring is the weighted polynomial ring:The character construction is defined over the ground field $k$, so this is an isomorphism of graded $k$-algebras. It proves the divisorial section ring of a well-formed weighted projective space assertion even when its ample generator is not Cartier.

A singular two-dimensional affine toric variety in the normal fan construction must come from a full-dimensional pointed cone; cones of smaller dimension give smooth charts with a torus factor. Let its primitive boundary generators be $v_0,v_{t+1}$. Choose an oriented lattice basis with $v_0=(1,0)$, and write the other generator as $(a,r)$, $r>0$. Primitivity gives $\gcd (a,r)=1$. Changing the second basis vector by a multiple of the first lets us write the cone asHere $r>1$ because determinant one would give a smooth chart. The sublattice $N^{\prime }=\mathbb{Z}(1,0)+\mathbb{Z}(-q,r)$ has index $r$, and its affine chart is ${\mathbb{C}}^2$. The quotient $N/N^{\prime }$ is cyclic, generated by $(0,1)$, whose coordinates in the two boundary generators are $(q/r,1/r)$. Thus the kernel of the corresponding finite map of tori acts on the two affine coordinates as $(x,y)\mapsto ({\zeta }^{q}x,\zeta y)$. More explicitly, invariant monomials are exactly those $x^i{y}^j$ with $qi+j\equiv 0(\mathrm{mod}r)$, which are the elements of the original dual lattice in the first quadrant. Their ring is $\mathbb{C}[{\sigma }^{\vee }\cap M]$. Hence the singularity is the cyclic quotientThe choice of the two boundary axes is unordered; exchanging them replaces $q$ by its inverse modulo $r$.

The Hirzebruch–Jung resolution is computed by the negative Euclidean algorithm. Put $r_0=r$, $r_1=q$ and recursively setThe nonzero remainders strictly decrease, $0\le r_{i+1}<r_i$, and their greatest common divisor stays one; the process terminates with $r_t=1$, $r_{t+1}=0$. All $b_i\ge 2$, and unwinding the recursion gives the negative continued fractionStart with $v_0=(1,0)$ and $v_1=(0,1)$ and define $v_{i+1}=b_i{v}_i-v_{i-1}$. The determinant of every consecutive pair is one, by induction. Moreoveragain by the same recursion and the initial values $r,q$. Each intermediate ray is therefore inside $\sigma$. To check the ordering explicitly, the first coordinates after $v_1$ are negative, the second coordinates positive, and successive determinants are positive. At the last step the determinant with the endpoint is zero. The vector $v_{t+1}$ is primitive, points in the same direction as $(-q,r)$, and hence equals it. Subdivide by all the intermediate rays $v_1,\dots ,v_t$. Every new cone is regular, and the subdivision has the same support, so it gives a proper smooth resolution.

Each internal ray has a compact invariant curve $E_i\cong {\mathbb{P}}^1$, because the quotient fan at it has the two opposite primitive rays. The relation $v_{i-1}+v_{i+1}=b_i{v}_i$ gives its self-intersection $E_i^2=-b_i$. This can also be checked directly: on the chart for $(v_{i-1},v_i)$ use coordinates $(x,y)$ with $E_i=\{y=0\}$; on the next chart the coordinates are $(u,v)=(x^{b_i}y,x^{-1})$, with $E_i=\{u=0\}$. The normal-coordinate transition is that of ${\mathcal{O}}_{{\mathbb{P}}^1}(-b_i)$, proving the self-intersection. Neighboring exceptional curves meet transversely once and nonneighbors do not meet.

Since $b_i\ge 2$, there is no exceptional minus-one curve. Combinatorially, deleting $v_i$ gives a cone of determinant $b_i>1$. Deleting a longer consecutive block cannot give a regular cone either: determinants of successively more distant rays satisfy the same recursion with coefficients at least two and grow from $0,1$ to integers at least two. Thus no proper smooth coarsening of this subdivision exists. This proves the minimality of the minimal resolution of a cyclic quotient surface singularity; equivalently its exceptional chain has no curve that can be blown down to a smooth point. The number of exceptional curves is the length $t$ of the negative continued fraction.

If $t=1$, then $r/q=b_1$ is an integer, so coprimality gives $q=1$. Conversely $q=1$ gives the single coefficient $[r{]}^{-}$ and one curve. If $t=2$, thenThe numerator and denominator on the right are coprime, so $r=b_1{b}_2-1$ and $q=b_2\ge 2$. Therefore $q>1$ and $q\mid r+1$. Conversely, if these conditions hold, then $b_1=(r+1)/q\ge 2$, $b_2=q\ge 2$ give exactly this two-term expansion. Hence the requested equivalences are

Now consider a complete fan in dimension two with three rays and at most one singularity. Its three two-dimensional cones are the consecutive ray pairs. At least two cones are regular, and they share a ray. After an oriented lattice change, take their consecutive generators to be $v_1=(1,0)$, $v_2=(0,1)$ and $v_3=(-1,b)$: the first cone is a basis, and $\det (v_2,v_3)=1$ forces the first coordinate $-1$. Completeness forces $b<0$, because the third pair must span the missing sector with positive determinant $\det (v_3,v_1)=-b$. Write $b=-r$, $r\ge 1$. The unique positive primitive relation is $v_1+r{v}_2+v_3=0$, so the fan is that of the weighted projective plane $\mathbb{P}(1,r,1)$, or $\mathbb{P}(1,1,r)$ after permuting weights. For $r=1$ this is ${\mathbb{P}}^2$; for $r>1$ its remaining cone has type $X_{r,1}$. This proves the full three-ray classificationThese are the complete three-ray toric surfaces with at most one singular point, including the smooth case.

For four rays and exactly the specified singularity, place the three regular cones consecutively. Normalize the middle one to have generators $(1,0),(0,1)$. Its neighboring regular cones then force the cyclic ray listThe remaining cone has determinant $\det (v_4,v_1)=1-ab=r$, so $ab=1-r$. Thus $a,b$ have opposite signs, and $s=|a|$ is a positive divisor of $r-1$. In the oriented lattice basis $v_4,(0,-1)$,so the singular cone has quotient parameter $a\mathrm{mod}r$. It follows that its parameter is congruent to $\pm s$. If identification with the specified singularity exchanges the two boundary rays, the parameter is inverted; this does not change the criterion, since for $st=r-1$ we have $s^{-1}\equiv -t(\mathrm{mod}r)$, and $t$ is the complementary positive factor. The same identity with a minus sign handles $(-s{)}^{-1}$.

Conversely suppose $s\mid r-1$ and $s\equiv \epsilon q(\mathrm{mod}r)$, $\epsilon \in \{1,-1\}$. Choose $a=\epsilon s$ and $b=-\epsilon (r-1)/s$ in the displayed four rays. They are primitive and cyclically ordered, with all four gaps less than a half-plane; the opposite signs of $a,b$ ensure completeness. Three consecutive determinants are one, and the fourth is $r$. Its quotient parameter is $a\equiv q(\mathrm{mod}r)$, so the only singular point has the required type. We have proved the four-ray completion criterion for a cyclic quotient surface singularity:

The smoothness criterion for a toric variety says that every cone must be regular: its primitive ray generators must be part of a lattice basis of $N$. Such a $d$-dimensional cone gives the chart ${\mathbb{A}}^d\times ({\mathbb{C}}^{\ast }{)}^{n-d}$; all charts are smooth precisely under this criterion.

Here is a terminating construction of a smooth fan subdivision. First subdivide compatibly into simplicial polyhedral cones. One method is the barycentric subdivision: choose a rational interior ray in each nonzero cone and perform the star subdivisions in decreasing dimension. Its cones correspond to chains of faces, hence are simplicial. For a simplicial cone with primitive generators $v_i$, let its multiplicity be the index of their integer span in the saturated lattice $N\cap {\mathrm{span}}_{\mathbb{R}}\sigma$.

If a cone is not regular, choose a nonregular face $\tau$ minimal under inclusion. Its half-open fundamental parallelepiped contains a nonzero point of the toric lattice $w={\sum }_i{\alpha }_i{v}_i$, $0\le {\alpha }_i<1$. Every coefficient is positive: otherwise $w$ lies in the span of a regular proper face, whose generators extend to a lattice basis, forcing all the remaining coefficients to be integers and hence zero. Replace $w$ by the primitive vector $w^{\prime }$ on its ray; its coefficients ${\beta }_i$ still satisfy $0<{\beta }_i<1$. Star-subdivide the entire fan at this ray. In each cone containing $\tau$, the new maximal cones replace one of these $v_i$ by $w^{\prime }$, so multilinearity of the determinant makes the new multiplicity ${\beta }_i$ times the old one. It is a positive integer strictly smaller than the old multiplicity. Other cones are unchanged. Select $\tau$ inside a maximal cone of largest multiplicity. The number of maximal cones having that largest multiplicity decreases, and once it reaches zero the largest multiplicity decreases. This lexicographic integer descent terminates at multiplicity one. It proves every finite rational fan has a smooth subdivision, not merely a simplicial one; this is multiplicity descent in toric desingularization.

For the specified threefold, abbreviate the primitive rays byIts boundary two-dimensional faces are $ab,bc,cd,da$. For example their supporting normals, chosen inward, are $(0,1,-1),(1,1,-2),(1,-1,2),(0,0,1)$; each is zero on the named pair and positive on the other two rays. These normals are primitive. Equivalently the greatest common divisor of the two-by-two minors for each ray pair is one, so these faces are regular; the rays themselves are primitive as well. The full cone is three-dimensional with four extremal rays, hence not simplicial and not regular. The orbit-cone correspondence now shows the closed torus-fixed point is its only singular point: every other orbit belongs to the smooth chart of a proper face.

Define the first subdivision by its maximal conesHere a string of letters denotes their nonnegative span; all faces are included. This is the triangulation along diagonal $ac$ of a positive cross-section. Both determinants have absolute value one, so it is smooth. The subdivision has the same support as the original cone, hence its toric morphism is proper and birational and is a resolution. No new rays were added, so it has no exceptional surfaces. Its internal wall $ac$ gives a compact exceptional curve isomorphic to ${\mathbb{P}}^1$, by the two-ray quotient fan at that wall. The only proper coarsening is the original nonregular cone, so this resolution is minimal in the stated fan sense.

For a second resolution, setStart with the other diagonal $bd$. The cone $abd$ is regular, but $bcd$ has determinant of absolute value two. Its primitive interior point $e$ divides it into the three regular cones shown, each with determinant of absolute value one. Therefore ${\Sigma }_2$ is again a smooth resolution and is different from ${\Sigma }_1$.

This fan has exactly one new ray, so its only exceptional surface is $E_e\cong {\mathbb{P}}^2$. To identify it, quotient $N$ by $\mathbb{Z}e$. The three neighboring rays are $\bar{b},\bar{c},\bar{d}$, their sum is zero, and consecutive pairs are bases because $bce,cde,dbe$ are regular. Their quotient fan is the simplex fan of the projective plane. In fact its normal bundle is ${\mathcal{O}}_{{\mathbb{P}}^2}(-2)$: the chart $U_{bcd}$ before subdivision is ${\mathbb{A}}^3/\{\pm 1\}$ since its index-two lattice extension is generated by $(b+c+d)/2$, and the new radial parameter is the square of the ordinary radial parameter of $\mathcal{O}(-1)$. It has the transition factors of $\mathcal{O}(-2)$.

For minimality, the ray $e$ cannot be deleted in a smooth coarsening. Deleting an interior ray in this triangulation merges all three incident cones into $bcd$, which is nonregular, or merges further into the original nonsimplicial cone. If $e$ is retained, deleting a wall between its three incident cones gives a nonconvex union unless all three are merged, again giving $bcd$. The remaining internal wall $bd$ separates $abd$ and $bde$; their union has four extremal rays $a,b,d,e$, since $a+e=b+d$, and is not regular. Thus no proper coarsening is smooth. These are the two minimal toric resolutions of a four-ray threefold cone.

The factorization can be made completely explicit. Starting with ${\Sigma }_1$, add $e=a+c$ in the wall $ac$, producing a smooth fan ${\Sigma }_A$ with maximal cones $abe,bce,ade,cde$. Next addin the wall $ae$. The resulting common fan has maximal conesStarting instead with ${\Sigma }_2$, the star subdivision at $w=b+d$ in the wall $bd$ gives exactly the same six cones. Each insertion is the sum of the generators of a smooth two-dimensional cone. Such a star subdivision is the toric blowup along an orbit closure of its smooth invariant curve: in local coordinates the center has ideal $(x_1,x_2)$, and the two blowup charts with ratios $x_2/x_1$ and $x_1/x_2$ give the two new cones. We have the zigzag of blowup morphismsThe required birational map is the composite of the inverses of the first two morphisms with the last. This supplies individual smooth toric blowups, rather than appealing to an abstract factorization theorem.

Write $C={\sigma }^{\vee }$ for the dual cone, $S=C\cap M$, and $U_{\sigma }=\mathrm{Spec}\mathbb{C}[S]$ for the affine toric variety. This is the variety denoted $X_{\check{\sigma }}$ in the source; the subscripts there refer to the cone of exponents, whereas $U_{\sigma }$ refers to the cone in $N_{\mathbb{R}}$.

For a face of a polyhedral cone $\tau ⪯\sigma$, define ${\tau }^{\ast }=C\cap {\tau }^{\perp }$. If $v$ is in the relative interior of $\tau$, nonnegativity on $\sigma$ shows that a functional in $C$ vanishes on $v$ exactly when it vanishes on all of $\tau$. Hence ${\tau }^{\ast }=C\cap v^{\perp }$ is a face of $C$. Conversely a face $\gamma$ of $C$ gives ${\gamma }^{\ast }=\sigma \cap {\gamma }^{\perp }$. Every polyhedral face is exposed: the supporting functionals for its containing facets can be added, since they are nonnegative. If $\tau =\sigma \cap {\mu }^{\perp }$, then $\mu \in {\tau }^{\ast }$, soThe same argument applied to $C$ proves $({\gamma }^{\ast }{)}^{\ast }=\gamma$, using the closed-cone identity $C^{\vee }=\sigma$. That identity follows from separation of a point outside a closed convex cone. Finally, enlarging $\tau$ makes its annihilator smaller. Thus these maps are inverse order-reversing bijections, proving face duality for polyhedral cones even when $\sigma$ is not full-dimensional.

For the given integral exposing functional $\mu$, we claimOne inclusion is immediate since $\mu$ vanishes on $\tau$. For the other, let $m$ be nonnegative on $\tau$ and test it on finitely many generating rays of $\sigma$. Where $\mu$ vanishes, those rays belong to $\tau$, so $m$ is already nonnegative. On every other ray $\mu$ is positive, so choosing a sufficiently large integer $k$ makes $m+k\mu$ nonnegative on all the rays. This proves the claim. The corresponding semigroup algebra identity isThus the face chart is an open subvariety.

For a face $\gamma ⪯C$, the span of monomials ${\chi }^m$ with $m\notin \gamma$ is an ideal: if $m+n\in \gamma$ with $m,n\in C$, the face property forces both into $\gamma$. Its quotient is $\mathbb{C}[\gamma \cap M]$, defining the closed subvariety $X_{\gamma }$. If $\mu \notin \gamma$, this subvariety lies in $V({\chi }^{\mu })$.

Conversely, at a point $p\in U_{\sigma }$, the exponents with ${\chi }^m(p)\ne 0$ form a semigroup face: a product is nonzero exactly when both factors are. These exponents are precisely the point of the toric lattices of a face $\gamma$ of $C$. To see the geometric assertion, choose finitely many monoid generators by the Gordan lemma. The nonzero generators span a cone. If that cone were not a face, a rational nonnegative relation would express a sum of its generators using an omitted generator with a positive coefficient. Clearing denominators and evaluating the corresponding monomials would equate a nonzero product to zero. Conversely, any point of the toric lattice in their cone has a positive integral multiple expressible using them, so its monomial must also be nonzero. This proves the monomial support face of a toric point assertion. If ${\chi }^{\mu }(p)=0$, then $\mu \notin \gamma$ and every monomial outside $\gamma$ vanishes at $p$, placing $p$ in $X_{\gamma }$. Therefore, as reduced closed subvarieties,The assertion concerns the closed set with its reduced structure; a monomial's principal zero subscheme can have multiplicities.

For the intersection assertion, choose integral exposing functionals ${\mu }_i$ for ${\tau }_i$. Both are nonnegative on $\sigma$, so ${\tau }_1\cap {\tau }_2=\sigma \cap ({\mu }_1+{\mu }_2{)}^{\perp }$. Localization then givesThis uses the dual of the intersection, as printed in the PDF; the TeX's perpendicular symbol in that position is not the correct transcription.

For a general toric fan, the points with a fixed monomial support face form a torus orbit. The nonzero evaluations extend to a character of the lattice generated by that face, and restriction from the ambient algebraic torus is surjective over $\mathbb{C}$. Thus the affine calculation yields the orbit-cone correspondence: $U_{\sigma }$ contains exactly the orbits $O(\nu )$ for faces $\nu ⪯\sigma$. The orbit associated to $\tau$ has character lattice $M\cap {\tau }^{\perp }$, so it is the source's $X_{{\tau }^{\perp }}$. The same monomial ideal calculation, chart by chart, shows that its closure isIndeed, inside a chart containing $\tau$, the closure has exponent face ${\sigma }^{\vee }\cap {\tau }^{\perp }$, and its smaller support faces correspond exactly to cones containing $\tau$.

Fix $\sigma \in \Sigma$, as required for its affine chart. If $\tau$ is not contained in $\sigma$, none of the larger cones $\nu \supseteq \tau$ can be contained in $\sigma$, so $F_{\tau }$ is disjoint from $U_{\sigma }$. Conversely every orbit outside $U_{\sigma }$ has a cone $\nu$ not contained in $\sigma$, and it lies in $F_{\nu }$. Consequently the global complement is exactly

Let $G$ be a finite undirected $d$-regular graph on $N$ vertices, with adjacency matrix $A$ and normalized random walk matrix $P=A/d$. Define edge expansion and normalized conductance byAn expander family has orders tending to infinity, uniformly bounded maximum degree, and a uniform positive lower bound on $h$. For a regular graph the constant vector is an eigenvector of $P$ with eigenvalue one. Write its eigenvalues as $1={\mu }_1\ge {\mu }_2\ge \cdots \ge {\mu }_N\ge -1$, and let $\delta =1-{\mu }_2$. This spectral gap measures the energetic cost of departing from a constant function.

For every mean-zero vector $f$, the spectral theorem and the graph's energy identity giveApply this to $f=1_S-|S|1/N$. Its norm squared is $|S|(1-|S|/N)$ and its energy is $e(S,S^c)/d$. ThereforeThus a uniform spectral gap gives uniform edge expansion when $d$ is fixed. This also bounds exterior vertex expansion: each exterior vertex accounts for at most $d$ boundary edges, so $|N(S)\setminus S|\ge e(S,S^c)/d$.

For the converse, take a mean-zero eigenvector $f$ for ${\mu }_2$ and choose its sign so that its positive support has at most $N/2$ vertices. Put $g=f^{+}$, which is nonzero. Truncation decreases energy in the following precise sense: for any two real numbers $a,b$,Summing over edges yields $⟨g,(I-P)g⟩\le ⟨g,(I-P)f⟩=\delta \Vert g{\Vert }_2^2$. For the threshold sets $S_t=\{v:g(v{)}^2>t\}$, every nonempty set lies in that small positive support. Integrating its boundary count gives the layer-cake identityThe Cauchy-Schwarz inequality bounds the left side bybecause the first sum is at most $d\delta \Vert g{\Vert }_2^2$ and the second at most $2d\Vert g{\Vert }_2^2$. Cancelling proves the Cheeger inequality in both directions:Hence bounded-degree regular families have uniform edge expansion exactly when their second-largest normalized eigenvalue stays uniformly below one.

Absolute eigenvalues give additional information. Put $\rho ={\max }_{i\ge 2}|{\mu }_i|$. Centering the indicators of $S,T$ and applying the operator-norm bound on the mean-zero subspace proves the expander mixing lemma:Here $e(S,T)=1_S^{T}A{1}_T$, so internal edges are counted as oriented incidences. A small $\rho$ makes edge counts resemble uniformly distributed edges. It also gives rapid random walk mixing: starting at a vertex, $P^t$ contracts the mean-zero component of its point mass by at most ${\rho }^t$ in Euclidean norm; Cauchy-Schwarz inequality then gives total variation distance at most $\frac{1}{2}\sqrt{N}{\rho }^t$ from the uniform distribution. A bipartite graph has eigenvalue $-1$ and its non-lazy walk alternates sides, even if it has good edge expansion. The lazy walk $(I+P)/2$ has nonconstant eigenvalues between zero and $1-\delta /2$, resolving this periodicity. It is important to distinguish the gap $1-{\mu }_2$ controlling cuts from the absolute bound controlling non-lazy mixing.

Here is a fully specified deterministic construction of an infinite family. Its degree is large but fixed. SetChoose the lexicographically first ordered list $a_1,\dots ,a_d\in W$ satisfyingThis finite deterministic search terminates. To prove existence, choose the list uniformly and independently. For each nonzero $u$, its signs are independent unbiased signs; the exponential-moment estimate $\cosh t\le e^{t^2/2}$ and Markov inequality give an exceptional probability at most $2e^{-d/200}$. A union bound gives total failure probability at mostThus at least one qualifying list exists. Fix it once and for all.

Let $H$ be the Cayley graph on $W$ whose $j$th neighbor of $x$ is $x+a_j$. It is an undirected $d$-regular port-labeled graph, allowing repeated edges and loops in this intermediate construction. Its characters $x\mapsto (-1{)}^{u\cdot x}$ form an orthogonal eigenbasis, with the displayed averages as normalized eigenvalues. Therefore $\rho (H)\le 1/10$. Each loop counts as one transition port; degrees here mean row sums of the adjacency matrix.

We next define the zig-zag product and prove the bound used for iteration. A port labeling of a $D_0$-regular undirected graph $F$ gives an involution ${\mathrm{Rot}}_F(v,a)=(w,b)$, recording the endpoint and reverse port. If a $d$-regular graph $H_0$ has vertex set $[D_0]$, the product has vertices $(v,a)\in V(F)\times [D_0]$. For each ordered pair of small-graph ports, move from $a$ to $a^{\prime }$ in $H_0$, follow the big edge ${\mathrm{Rot}}_F(v,a^{\prime })=(w,b^{\prime })$, and move from $b^{\prime }$ to $b$ in $H_0$. The endpoint is $(w,b)$; reversing the steps reverses the two small-graph port labels. The product is undirected, has $|V(F)|D_0$ vertices, and has degree $d^2$.

Write $B=I\otimes P_{H_0}$ and let $R$ be the permutation matrix of ${\mathrm{Rot}}_F$. The normalized product operator is $BRB$. For a mean-zero unit vector, decompose $f=u+v$ into its cloud-constant part and its cloud-orthogonal part. Then $Bu=u$, $\Vert Bv\Vert \le \rho (H_0)\Vert v\Vert$, and $\Vert R\Vert =1$. The compression of $R$ to cloud-constant vectors is exactly $P_F$, so $|⟨u,Ru⟩|\le \rho (F)\Vert u{\Vert }^2$. Expanding the quadratic form and applying Cauchy-Schwarz inequality givesThe operator is symmetric, so its mean-zero operator norm is bounded by this same quantity. No unproved product spectral estimate is needed.

Start with the loop-inclusive complete graph $G_0$ on $D$ vertices, whose normalized matrix is $J/D$ and whose nonconstant eigenvalues are all zero. Given a $D$-regular $G_t$, its two-step graph $G_t^2$ has degree $D^2$ and normalized operator $P_{G_t}^2$. Label a two-step edge by its ordered pair of ports; reversal reverses the path and its reverse ports. These $D^2$ labels are the vertex set of the fixed graph $H$. DefineIt has degree $d^2=D$, and its number of vertices is $N_{t+1}=N_t{D}^2$, hence $N_t=D^{2t+1}$. By the proved bound,Starting from zero, induction gives $\rho (G_t)\le 1/2$, since $(1/2{)}^2+21/100=46/100<1/2$. The cut-energy argument therefore gives multiplicity-counted edge expansion at least $D/4$ for every $G_t$.

To obtain ordinary simple graphs, delete loops and merge parallel edges. The maximum degree is at most $D$, and any one unordered vertex pair had multiplicity at most $D$. Thus every cut retains at least $1/D$ of its multiplicity-counted boundary, givingThese simple graphs form an expander family. Every choice is deterministic: the finite base search is fixed independently of $t$, and the recursive edge rules determine all subsequent adjacency lists. Once that base is fixed, adjacency lists can be generated successively using the previous lists, with a constant amount of work per output edge. This is squaring and zig-zag iteration yields bounded-degree expanders. The spectral argument supplies the uniform expansion guarantee and explains why squaring improves the eigenvalues while the product restores a constant degree.

Write $[A{]}^{\omega }$ for the infinite subsets of an infinite $A\subseteq \mathbb{N}$. A Ramsey set of infinite subsets is a family $E\subseteq [\mathbb{N}{]}^{\omega }$ such that each infinite ground set $A$ has an infinite subset $B$ withThe convention asking only for one such homogeneous cone is weaker; the proof below establishes the displayed property and, in fact, the stronger completely Ramsey set property.

For a finite stem $s$ and an infinite reservoir $A$ lying above every element of $s$, putThese sets form the basic open neighbourhoods of the Ellentuck topology, or star topology. Here $s$ is an initial segment of every member of $[s,A]$. A star-Borel set means a member of the sigma-algebra generated by these open sets. A family is completely Ramsey if every $[s,A]$ has a homogeneous refinement $[s,B]$ with $B\subseteq A$ infinite and the same stem $s$. We prove this for every star-Borel set, beginning with open sets.

Fix a star-open family $O$ and a basic neighbourhood $[s,A]$. An infinite reservoir $T$ above a finite stem $t$ accepts $t$ if $[t,T]\subseteq O$. It rejects $t$ if no infinite subset of $T$ accepts $t$. A reservoir decides $t$ if it accepts or rejects it. Any reservoir has a refinement deciding a prescribed stem: take an accepting refinement if there is one, and otherwise the current reservoir rejects. Both kinds of decision persist under infinite thinning.

First refine $A$ to decide $s$. If the decision is acceptance, the required homogeneous neighbourhood has already been found. Suppose it is rejection. We use deciding all finite stems by fusion to selectfrom nested infinite reservoirs. After selecting $b_n$, successively thin the remaining reservoir to decide every stem $s\cup a$ with $a\subseteq \{b_1,\dots ,b_n\}$. There are only finitely many such $a$, so every stage can be completed. Choose $b_{n+1}$ from that thinned reservoir and repeat. Let $B=\{b_1,b_2,\dots \}$. For every finite $a\subseteq B$, the taildecides $s\cup a$; for $a=\varnothing$, use all of $B$. Indeed, the stem was decided at the stage when its largest point was selected, and its final tail is a subset of the reservoir retained at that stage. The empty stem remains rejected by the initial decision.

The crucial finite-obstruction fact is that a rejected stem has only finitely many accepting one-point extensions. More precisely, suppose $B/a$ rejects $s\cup a$. If infinitely many $b\in B/a$ had $B/b$ accepting $s\cup a\cup \{b\}$, collect those $b$ into an infinite $C$. Any $X\in [C{]}^{\omega }$ has a least point $b$, and its remaining tail lies in $B/b$. Acceptance at $b$ would then imply $s\cup a\cup X\in O$. Thus $C$ would accept $s\cup a$, contradicting rejection. This proves the finitely many accepting extensions of a rejected stem assertion with the reservoirs specified.

Thin $B$ a second time. Having selected a finite prefix of $C$, maintain that every stem $s\cup a$ with $a$ a subset of that prefix is rejected relative to its tail in $B$. Each such stem has finitely many forbidden next points leading to acceptance. Choose the next point of $C$ above the prefix and outside the union of these finitely many finite forbidden sets. The deciding property of $B$ then makes every newly created stem rejected. Starting with the rejected empty extension gives an infinite $C\subseteq B$ all of whose finite extensions of $s$ are rejected.

We claim $[s,C]\cap O=\varnothing$. Otherwise take $Y\in [s,C]\cap O$. Star-openness supplies a basic neighbourhood inside $O$ containing $Y$. Extend its finite stem along $Y$ until it contains $s$; the new stem has the form $s\cup a$ for a finite initial segment $a$ of $Y\setminus s$. The remaining infinite tail of $Y$ is a subset of $B/a$ and accepts that stem, contradicting rejection. This proves the fusion proof for open Ellentuck sets: every star-open set is completely Ramsey.

Next call $N$ completely Ramsey-null if every $[s,A]$ has a same-stem refinement disjoint from $N$. Every star-nowhere dense set is completely Ramsey-null. Indeed, the complement $D$ of its star-closure is star-open and dense. Apply the just-proved open-set result inside $[s,A]$. The homogeneous alternative disjoint from $D$ would be a nonempty open set contained in the closure of a nowhere-dense set, which is impossible. Hence some $[s,B]$ lies entirely in $D$ and avoids $N$.

Countable unions of completely Ramsey-null sets are again completely Ramsey-null. Here is the full Ellentuck meagre-set fusion lemma. Given $N_1,N_2,\dots$ and $[s,A]$, construct selected prefixes $P_j=\{b_1,\dots ,b_j\}$ and nested unused infinite reservoirs. At stage $j$, refine the reservoir, still lying above $\max P_j$, to arrangeEach constraint can be imposed by complete Ramsey-nullness, with all earlier constraints preserved by thinning; there are finitely many constraints. Choose $b_{j+1}\in T_j$ and retain the tail above it for the next stage. Start with $P_0=\varnothing$ and the reservoir $A$.

Let $B=\{b_1,b_2,\dots \}$. If $Y\in [s,B]$ and $i$ is fixed, take any $j\ge i-1$ and put $a=(Y\setminus s)\cap P_j$. Every remaining point of $Y\setminus s$ lies in $T_j$, so $Y\in [s\cup a,T_j]$ and $Y\notin N_i$. ThereforeThis construction handles all subsets of a selected prefix, not merely the full prefix: members of $[s,B]$ may skip arbitrarily many selected points. Since a meagre set is a countable union of nowhere-dense sets, every star-meagre set is completely Ramsey-null.

It remains to pass from star-open sets to star-Borel sets. Let $\mathcal{C}$ consist of the families $E$ expressible aswhere $O$ is star-open and $M$ is star-meagre. We verify directly that $\mathcal{C}$ is a sigma-algebra. It contains every open set. Countable unions remain in $\mathcal{C}$, becauseand a countable union of meagre sets is meagre. For complements, replace $O^c$ by the open set $\mathrm{int}(O^c)$. The omitted part is $\overline{O}\setminus O$, the boundary of the open set $O$, which is closed and nowhere dense: a nonempty open subset of $\overline{O}$ must meet $O$, so it cannot lie in this boundary. Thusa meagre set. Subsets of meagre sets are meagre as well, since taking a subset preserves nowhere denseness. These observations prove the claimed sigma-algebra property. Hence every star-Borel set belongs to $\mathcal{C}$.