21.9 Fans, divisors and character sectionsFF296752801A

Proof. Let $Q$ be a face, choose a vertex $v\in Q$, and use the notation of lemma 21.8.2(4). Since $u_1,\ldots ,u_ n$ is the basis dual to $e_1,\ldots ,e_ n$, every $x\in N_{\mathbb R}$ can be written as $x=\sum _ i\alpha _ iu_ i$ with $\alpha _ i=\langle e_ i,x\rangle $, and $\langle m-v,x\rangle =\sum _ it_ i\alpha _ i$ for $m=v+\sum _ it_ ie_ i\in P$. If $v$ minimizes $\langle \cdot ,x\rangle $ on $P$, then $\alpha _ i=\langle (v+e_ i)-v,x\rangle \geq 0$, because $v+e_ i\in P$. If moreover every point of $Q$ minimizes it, and the $i$th facet at $v$ does not contain $Q$, choose $m\in Q$ with $t_ i=\ell _ i(m)>0$; then $\sum _ jt_ j\alpha _ j=0$ with nonnegative terms, so $\alpha _ i=0$. Hence $x$ is a nonnegative combination of the normals of the facets containing $Q$, all of which pass through $v$, that is, $x\in \sigma _ Q$. Conversely, if $x=\sum _\rho c_\rho u_\rho $ with $c_\rho \geq 0$ and all $\rho $ containing $Q$, then $\langle m,x\rangle =\sum _\rho c_\rho (\ell _\rho (m)-a_\rho )$, and each $\ell _\rho $ attains its minimum $0$ on $P$ at every point of $Q$; so does $\langle \cdot ,x\rangle $. This proves the characterization, and it shows that $\sigma _ Q$ is generated by part of the basis $u_1,\ldots ,u_ n$ of $N$.

Every face $Q$ is the intersection of $P$ with the facets containing it. Indeed, if $Q$ is the set where $\langle \cdot ,x\rangle $ is minimal, then $x\in \sigma _ Q$, and writing $x=\sum _\rho c_\rho u_\rho $ as above we see that the minimum is attained exactly where $\ell _\rho =0$ for all $\rho $ with $c_\rho >0$; this set contains the intersection of the facets containing $Q$, which in turn contains $Q$.

The faces of $\sigma _ Q$ are the cones generated by subsets of its generators. Indeed, if $m\in M_{\mathbb R}$ is nonnegative on $\sigma _ Q$ and $x$ is a nonnegative combination of the generators, then $\langle m,x\rangle $ is a sum of nonnegative terms, so the face $\{ x\in \sigma _ Q:\langle m,x\rangle =0\} $ is generated by the generators annihilated by $m$. Conversely, since the generators are part of a basis of $N_{\mathbb R}$, for every subset of them there is an $m$ that vanishes on this subset and equals $1$ on the remaining generators, and $m$ is nonnegative on $\sigma _ Q$. If $J$ is a set of facets containing $v$, the set $Q'=\{ m\in P:\ell _\rho (m)=0\text{ for }\rho \in J\} $ is a face, namely the set where $\sum _{\rho \in J}\langle \cdot ,u_\rho \rangle $ is minimal. It contains $v+se_ i$ for small $s>0$ whenever the $i$th facet at $v$ is not in $J$, so the facets containing $Q'$ are exactly those in $J$, and $\sigma _{Q'}=\operatorname {cone}\{ u_\rho :\rho \in J\} $. Thus every face of $\sigma _ Q$ is in the family.

Every $x$ attains its minimum on $P$ on some face $Q$, and then $x\in \sigma _ Q$ by the characterization; hence the cones cover $N_{\mathbb R}$. Let $Q_1,Q_2$ be faces, and let $Q$ be the intersection of $P$ with all facets containing both. It is a face (if no facet contains both, $Q=P$; otherwise it is the set where the sum of the corresponding functionals is minimal), and every face containing $Q_1\cup Q_2$ contains $Q$, since it is the intersection of $P$ with the facets containing it. Since $Q\supset Q_1\cup Q_2$, it is the smallest face containing $Q_1\cup Q_2$. By the characterization, $x\in \sigma _{Q_1}\cap \sigma _{Q_2}$ if and only if the face where $\langle \cdot ,x\rangle $ is minimal contains $Q_1\cup Q_2$, that is, if and only if it contains $Q$; so $\sigma _{Q_1}\cap \sigma _{Q_2}=\sigma _ Q$. The generators of $\sigma _ Q$, the normals of the facets containing $Q_1\cup Q_2$, form a subset of the generators of $\sigma _{Q_ k}$, so $\sigma _ Q$ is a face of $\sigma _{Q_1}$ and of $\sigma _{Q_2}$.

For the last assertion, let $|x|=1$ and choose a vertex $v$ minimizing $\langle \cdot ,x\rangle $ on $P$. The coefficients $\alpha _ i=\langle e_ i,x\rangle $ above are nonnegative and satisfy $\alpha _ i\leq |e_ i|$. Their sum $\lambda $ is positive, because $x\neq 0$, and at most $\sum _ i|e_ i|$. Let $\Lambda $ be the maximum of these finite sums over all vertices. Dividing by $\lambda $ and omitting the zero coefficients gives the required expression, with all $u_{\rho _ i}$ in the cone $\sigma _{\{ v\} }$. In dimension zero the fan consists of the zero cone, and there is no unit vector. $\square$

Proof. Fix a vertex $v$ and use the chart $U_ v\cong \mathbb C^ n$ of theorem 21.8.3. For $\tau \in T$ we have $w_ i(\tau \cdot z)=\chi ^{e_ i}(\tau )w_ i(z)$, and $\tau \mapsto (\chi ^{e_ i}(\tau ))_ i$ maps $T$ onto $(\mathbb C^*)^ n$; so $T$ acts on $U_ v$ by independent scalings of the coordinates. The orbits in $U_ v$ are therefore the sets

and $O_{v,I}$ has dimension $n-|I|$. Let $Q_{v,I}=\{ m\in P:\ell _ i(m)=0\text{ for }i\in I\} $. It is a face, namely the set where $\sum _{i\in I}\langle \cdot ,u_ i\rangle $ is minimal on $P$. It contains $v$ and the points $v+se_ j$ for $j\notin I$ and small $s>0$, and it lies in $v+\operatorname {span}\{ e_ j:j\notin I\} $, because $m-v=\sum _ i\ell _ i(m)e_ i$; so $\dim Q_{v,I}=n-|I|$. Since $Z_ m/Z_ v=\prod _ iw_ i^{\ell _ i(m)}$, a homogeneous coordinate $Z_ m$, $m\in A$, is nonzero on $O_{v,I}$ if and only if $\ell _ i(m)=0$ for all $i\in I$, that is, if and only if $m\in Q_{v,I}$.

The charts $U_ v$ are $T$-invariant, because the action of $T$ multiplies $Z_ v$ by $\chi ^ v(\tau )\neq 0$. Hence every orbit $O$ in $X_ P$ meets, and therefore lies in, some chart $U_ v$, so it is one of the sets $O_{v,I}$. The set of $m\in A$ with $Z_ m\neq 0$ on $O$ does not depend on the chart, and it equals $Q_{v,I}\cap A$; since a face is the convex hull of its vertices, which are lattice points, $Q_{v,I}$ is the convex hull of this set. Thus each orbit $O$ determines a face $Q(O)$, with $\dim Q(O)=\dim O$. Distinct orbits determine distinct faces: if $Q(O)=Q(O')=Q$ and $v$ is a vertex of $Q$, then $Z_ v\neq 0$ on both orbits, so both lie in $U_ v$, where an orbit is determined by the set of indices $i$ with $v+e_ i\notin Q$. Every face $Q$ arises: choose a vertex $v\in Q$ and let $I$ be the set of indices of the facets at $v$ that contain $Q$; then $Q_{v,I}=Q$, because $Q$ is the intersection of $P$ with the facets containing it (proposition 21.9.1), and these facets all pass through $v$. We write $O_ Q$ for the orbit of the face $Q$; in particular $O_ P=\iota (T)$.

The closure $\overline{O_ Q}$ is $T$-invariant, hence a union of orbits, and it lies in the closed set $\{ Z_ m=0\text{ for }m\in A\setminus Q\} $. If $O_{Q'}\subset \overline{O_ Q}$, then $Q'\cap A\subset Q$, and so $Q'\subset Q$. Conversely, let $Q'\subset Q$ be faces, let $v$ be a vertex of $Q'$, and let $I\subset I'$ be the index sets of $Q$ and $Q'$ at $v$. A point $w\in O_{Q'}=O_{v,I'}$ is the limit as $\varepsilon \downarrow 0$ of the points obtained from $w$ by replacing the coordinates $w_ i=0$, $i\in I'\setminus I$, by $\varepsilon $, and these points lie in $O_{v,I}=O_ Q$. Hence $O_{Q'}\subset \overline{O_ Q}$.

Let $\rho $ be a facet and $D_\rho =\overline{O_\rho }$. If $v\in \rho $ and $\rho $ is the $i$th facet at $v$, the orbits in $U_ v$ indexed by faces contained in $\rho $ are the sets $O_{v,I}$ with $i\in I$, so $D_\rho \cap U_ v=\{ w_ i=0\} $; if $v\notin \rho $, then $D_\rho \cap U_ v=\varnothing $, because $Z_ v=0$ on $D_\rho $. Hence $D_\rho $ is a $T$-invariant closed submanifold of codimension one. Every proper face is contained in a facet, being the intersection of the facets containing it; hence $X_ P\setminus \iota (T)$, the union of the orbits of the proper faces, is the union of the divisors $D_\rho $.

Each $D_\rho $ is irreducible, that is, not the union of two analytic subsets different from $D_\rho $. First, $D_\rho $ is connected, being the closure of the orbit $O_\rho $: in a chart $U_ v$ in which $\rho $ is the $i$th facet at $v$, the coordinates $w_ j$, $j\neq i$, identify $O_\rho =O_{v,\{ i\} }$ with the connected set $(\mathbb C^*)^{n-1}$. Next, let $Z\subsetneq D_\rho $ be an analytic subset. The set of interior points of $Z$ in $D_\rho $ is open, and it is closed in $D_\rho $. Indeed, if $p$ lies in its closure, then $p\in Z$, and on a connected coordinate neighborhood of $p$ in $D_\rho $ the local holomorphic functions defining $Z$ vanish on a nonempty open set, hence identically by the identity theorem, so that $p$ is an interior point. As $D_\rho $ is connected and $Z\neq D_\rho $, the set $Z$ has empty interior in $D_\rho $. If $D_\rho =Z_1\cup Z_2$ with analytic subsets $Z_1,Z_2\neq D_\rho $, then the nonempty open set $D_\rho \setminus Z_1$ would be contained in $Z_2$, which has empty interior.

Let $m\in M$. Since $m=\sum _ j\langle m,u_ j\rangle e_ j$, the character $\chi ^ m$ is the Laurent monomial $\prod _ jw_ j^{\langle m,u_ j\rangle }$ in the coordinates of $U_ v$. Hence it extends to a meromorphic function on each $U_ v$, and these extensions agree on the dense set $\iota (T)$, hence on overlaps. Near each point of the dense orbit $O_\rho \cap U_ v$ of $D_\rho \cap U_ v=\{ w_ i=0\} $, it is $w_ i^{\langle m,u_ i\rangle }$ times a nowhere vanishing holomorphic function; so its order along $D_\rho $, the difference of the orders of numerator and denominator in the sense of definition 6.12.3, is $\langle m,u_ i\rangle =\langle m,u_\rho \rangle $. In particular, $\chi ^ m$ is holomorphic on $U_ v\cong \mathbb C^ n$ if and only if all the exponents $\langle m,u_ j\rangle $ are nonnegative, that is, if and only if $\langle m,x\rangle \geq 0$ for every $x$ in $\sigma _{\{ v\} }=\operatorname {cone}(u_1,\ldots ,u_ n)$. $\square$

Proof. (1) Since $u_1,\ldots ,u_ n$ is the basis of $N$ dual to $e_1,\ldots ,e_ n$, every $m\in M$ is $m=\sum _ j\langle m,u_ j\rangle e_ j$ with integer coefficients, and $\sigma _{\{ v\} }=\operatorname {cone}(u_1,\ldots ,u_ n)$. Hence $\langle m,\cdot \rangle \geq 0$ on $\sigma _{\{ v\} }$ if and only if all coefficients $\langle m,u_ j\rangle $ are nonnegative. The second assertion follows from the last assertion of proposition 21.9.2, since $\chi ^ m=\prod _ jw_ j^{\langle m,u_ j\rangle }$ on $U_ v$.

(2) By theorem 21.8.3, $Z_{v'}/Z_ v=\prod _ jw_ j^{b_ j(v')}$ on $U_ v$, with $b_ j(v')=\ell _ j(v')=\langle v'-v,u_ j\rangle \geq 0$; this holomorphic function is the extension of $\chi ^{v'-v}$ to $U_ v$. Since $U_ v\cap U_{v'}=U_ v\cap \{ Z_{v'}\neq 0\} $, the first assertion follows. On $U_ v\cap U_{v'}$ we have $w'_ i=Z_{v'+e_ i'}/Z_{v'}=w^{b(v'+e_ i')}/w^{b(v')}$, a quotient of monomials whose denominator does not vanish there, and $b_ j(v'+e'_ i)-b_ j(v')=\langle e'_ i,u_ j\rangle $. A negative exponent $\langle e'_ i,u_ j\rangle $ forces $b_ j(v')>0$, so the corresponding factor $w_ j$ divides the denominator and does not vanish on $U_ v\cap U_{v'}$.

(3) By proposition 21.9.1, applied to the faces $\{ v\} $ and $\{ v'\} $, the set $Q$ is the smallest face containing $v$ and $v'$, and $\sigma _{\{ v\} }\cap \sigma _{\{ v'\} }=\sigma _ Q$. The facets containing $Q$ are the facets $\rho _ j$ through $v$ with $\ell _ j(v')=0$. If $x=\sum _ j\alpha _ ju_ j\in \sigma _{\{ v\} }$ with $\alpha _ j\geq 0$, then $\langle v'-v,x\rangle =\sum _ j\alpha _ j\ell _ j(v')$ is a sum of nonnegative terms, so it vanishes if and only if $\alpha _ j=0$ whenever $\ell _ j(v')>0$, that is, if and only if $x\in \operatorname {cone}\{ u_ j:\ell _ j(v')=0\} =\sigma _ Q$. $\square$

Proof. Fix $j$ and the variables other than $z_ j$. The resulting function $g(\zeta )$ is holomorphic on $\mathbb C^*$, and for $k\in \mathbb Z$ the integral $\frac1{2\pi }\int _0^{2\pi }g(\varrho \mathrm e^{\mathrm i\theta })\varrho ^{-k}\mathrm e^{-\mathrm ik\theta }\, \mathrm d\theta =\frac1{2\pi \mathrm i}\int _{|\zeta |=\varrho }g(\zeta )\zeta ^{-k-1}\, \mathrm d\zeta $ does not depend on $\varrho >0$, by Cauchy's theorem on the annulus between two such circles. Integrating over the other angles, by Fubini's theorem, we see that $c_ m(r)$ does not depend on $r_ j$; hence it does not depend on $r$. The definition gives the Cauchy estimates $|c_ m|\leq C(r)r^{-m}$, where $C(r)$ is the maximum of $|f|$ on the torus $\{ |z_ j|=r_ j\} $. Let $K\subset (\mathbb C^*)^ n$ be compact, and choose $0<a\leq b$ with $a\leq |z_ j|\leq b$ on $K$ for all $j$. Given $m$, take $r_ j=2b$ if $m_ j\geq 0$ and $r_ j=a/2$ if $m_ j<0$. Then, for $z\in K$,

where $C_ K$ is the largest of the numbers $C(r)$ for the $2^ n$ tori with $r_ j\in \{ a/2,2b\} $. Since $\sum _{m\in \mathbb Z^ n}2^{-(|m_1|+\cdots +|m_ n|)}=3^ n$, the series converges normally on $K$.

We show by induction on $n$ that its sum is $f$. For $n=1$ this is the Laurent theorem of one-variable theory, with the coefficients given by the integral above. Let $n\geq 2$, and write $z=(z',z_ n)$ with $z'\in (\mathbb C^*)^{n-1}$. For fixed $z'$, the one-variable theorem gives $f(z',z_ n)=\sum _ ka_ k(z')z_ n^ k$ with $a_ k(z')=\frac1{2\pi }\int _0^{2\pi }f(z',\mathrm e^{\mathrm i\theta })\mathrm e^{-\mathrm ik\theta }\, \mathrm d\theta $. Differentiation under the integral sign shows that $a_ k$ is smooth and satisfies the Cauchy–Riemann equations, so it is holomorphic on $(\mathbb C^*)^{n-1}$, and by Fubini's theorem its Laurent coefficients are the numbers $c_{(m',k)}$, $m'\in \mathbb Z^{n-1}$. By the induction hypothesis, $a_ k(z')=\sum _{m'}c_{(m',k)}z'^{m'}$, so $f(z)=\sum _ k\sum _{m'}c_{(m',k)}z'^{m'}z_ n^ k$. This double series converges absolutely, by the previous paragraph, so it may be summed as $\sum _ mc_ mz^ m$.

For the uniqueness, integrate the normally convergent series $\sum _{m''}c'_{m''}r^{m''}\mathrm e^{\mathrm i\langle m'',\theta \rangle }=f(r_1\mathrm e^{\mathrm i\theta _1},\ldots )$ term by term against $(2\pi )^{-n}r^{-m}\mathrm e^{-\mathrm i\langle m,\theta \rangle }\, \mathrm d\theta $; since $\int _0^{2\pi }\mathrm e^{\mathrm il\theta }\, \mathrm d\theta =0$ for every integer $l\neq 0$, only the term $m''=m$ survives, and we obtain $c'_ m=c_ m(r)=c_ m$. Finally, if $f$ is holomorphic on $\mathbb C^ n$, its Taylor series at the origin converges normally on every polydisc by (5.2.1), in particular on every torus, so by uniqueness the coefficients $c_ m$ are the Taylor coefficients for $m\in \mathbb Z_{\geq 0}^ n$ and vanish for all other $m$. $\square$

Proof. For vertices $v$ and $v'$, the coordinate sections satisfy $s_ v/s_{v'}=Z_ v/Z_{v'}=\chi ^{v-v'}$ on $\iota (T)$. Hence the section $s_0=\chi ^{-v}s_ v$ of $L$ over $\iota (T)$ does not depend on the vertex $v$. In the chart $U_ v$, its coefficient $\chi ^{-v}$ with respect to the frame $s_ v$ is a Laurent monomial in $w$, so $s_0$ extends to a meromorphic section on each $U_ v$; the extensions agree on overlaps, because they agree on the dense set $\iota (T)$. By proposition 21.9.2, if $v$ is a vertex of the facet $\rho $, then $\operatorname {ord}_{D_\rho }s_0=\operatorname {ord}_{D_\rho }\chi ^{-v}=-\langle v,u_\rho \rangle =a_\rho $, since $\ell _\rho (v)=0$. As $s_0$ has no zeros or poles on $\iota (T)$, its divisor is $D_ P$. Similarly, the meromorphic function $f_ v=\chi ^{-v}$ on $U_ v$ has divisor $\sum _{\rho \ni v}a_\rho (D_\rho \cap U_ v)$, which is the restriction of $D_ P$ to $U_ v$, because $D_\rho \cap U_ v=\varnothing $ for $v\notin \rho $. The line bundle $\mathcal O_{X_ P}(D_ P)$ of the divisor therefore has over $U_ v$ the local frame $1/f_ v$, and its transition functions, in the sense of definition 6.14.9, are $f_{v'}/f_ v=\chi ^{v-v'}$. These are the transition functions of $L$ in the frames $s_ v$, since $s_ v=\chi ^{v-v'}s_{v'}$. Hence the map sending $s_ v$ to $1/f_ v$ for every vertex $v$ is an isomorphism $L\cong \mathcal O_{X_ P}(D_ P)$.

Let $s$ be a holomorphic section of $L^ k$ and write $s=fs_0^ k$ on $\iota (T)$, with $f$ holomorphic on $T\cong (\mathbb C^*)^ n$. By lemma 21.9.4, applied in the coordinates of a basis of $M$, we have $f=\sum _{m\in M}c_ m\chi ^ m$, with normal convergence on compact subsets of $T$. The coefficients $c_ m$ do not depend on the basis: a change of basis replaces each character by the corresponding monomial in the new coordinates and preserves normal convergence on compact sets, so the uniqueness assertion of the lemma applies. Now fix a vertex $v$. The coefficient $g_ v$ of $s$ with respect to the frame $s_ v^ k$ is a holomorphic function on $U_ v\cong \mathbb C^ n$, and on $\iota (T)$

Since $u_1,\ldots ,u_ n$ is the basis of $N$ dual to the basis $e_1,\ldots ,e_ n$ of $M$, the map $m\mapsto (\langle m-kv,u_ i\rangle )_ i$ is a bijection of $M$ onto $\mathbb Z^ n$, so this is the Laurent expansion of $g_ v$ in the coordinates $w$. By the last assertion of lemma 21.9.4, $c_ m=0$ unless $\langle m-kv,u_ i\rangle \geq 0$ for every facet $i$ at $v$. Since $\langle v,u_\rho \rangle =-a_\rho $ for $v\in \rho $, this condition reads $\langle m,u_\rho \rangle +ka_\rho \geq 0$. Every facet contains a vertex, so for $k>0$ the conditions at all vertices say exactly that $m/k\in P$, that is, $m\in kP$. For $k=0$ they say that $\langle m,u_\rho \rangle \geq 0$ for every $\rho $; then $p+sm\in P$ for $p\in P$ and all $s\geq 0$, so $m=0$ by boundedness. Thus only the finitely many $m\in kP\cap M$ occur. Hence $s=\sum _{m\in kP\cap M}c_ m\chi ^ ms_0^ k$ on the dense set $\iota (T)$, and therefore on $X_ P$, once we know that the sections on the right are holomorphic. Indeed, for $m\in kP\cap M$ all exponents $\langle m-kv,u_ i\rangle $ are nonnegative, so $\chi ^ ms_0^ k$ extends holomorphically to every chart. Finally, the uniqueness of Laurent coefficients shows that the sections $\chi ^ ms_0^ k$ are linearly independent.

The action of $T$ on $\mathbb C^ A$ by $\tau \cdot (Z_ m)=(\chi ^ m(\tau )Z_ m)$ induces actions on the tautological line bundle (example 10.6.6), whose fiber over $[Z]$ is the line $\mathbb CZ$, on its dual $\mathcal O(1)$, on $L^ k$, and on $H^0(X_ P,L^ k)$ by $(\tau \cdot s)(x)=\tau \cdot s(\tau ^{-1}\cdot x)$. It multiplies the linear form $Z_ v$, and hence $s_ v$, by $\chi ^{-v}(\tau )$, and it maps the function $\chi ^ v$ to $\chi ^ v(\tau ^{-1}\, \cdot \, )=\chi ^{-v}(\tau )\chi ^ v$; therefore it fixes $s_0$ and multiplies $\chi ^ ms_0^ k$ by $\chi ^{-m}(\tau )$. So each summand is a line on which $T$ acts through a character. Finally, $h^0(X_ P,L^ k)=\# (kP\cap M)$, and the limit follows from theorem 21.7.3. In dimension zero there is one section, and the volume is one. $\square$