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%% Orbifold Riemann--Roch and plurigenera
%% Miles Reid
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\DeclareMathOperator{\Num}{Num}
\DeclareMathOperator{\Proj}{Proj}
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\newcommand{\grow}{\mathrm{grow}}
\newcommand{\orb}{\mathrm{orb}}
\newcommand{\per}{\mathrm{per}}
\newcommand{\tot}{\mathrm{tot}}
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\newcommand{\QED}{\quad QED}
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\newtheorem{rmk}[theorem]{Remark}
\newtheorem{exa}[theorem]{Example}
\begin{document}
\title{Orbifold Riemann--Roch and plurigenera}
\author{Miles Reid}
\date{}
\maketitle
I give a general formula for the Hilbert series of a polarised $n$-dimensional
orbifold (for example, with isolated orbifold points). The result comes from
orbifold RR, and so ultimately from equivariant RR (the Atiyah--Singer
Lefschetz trace formula); however, the formula is organised so that no Chern
or Todd classes appear explicitly, and no Dedekind sums. The formula reduces
much of my work over 20 years to a few lines of computer algebra.
\section{Introduction}
For simplicity, I start with a special case of the general formula, with extra
conditions.
\begin{defn}\label{defn!Porb} \rm A {\em simply polarised orbifold with
isolated orbifold points\/} is a variety $X$ polarised by a sheaf of graded
algebras $\bigoplus\Oh_X(i)$ satisfying the following conditions:
\begin{itemize}
\item $X$ is a projective variety of dimension $n\ge1$ over a field $k$ (e.g.,
$k=\C$), and $\Oh_X(m)$ is an ample invertible sheaf for some $m>0$;
\item $X$ has at most isolated orbifold singularities $\rth(a_1,\dots,a_n)$,
and locally at each point, each $\Oh_X(i)$ is the $i$th eigensheaf of the
$\mu_r$ action.
\end{itemize}
If $n\ge2$, isolated singularities are in codimension $\ge2$, so the $\Oh(i)$
are divisorial sheaves, with $\Oh(i)=\Oh_X(iA)$ for an ample Weil divisor $A$;
in particular, $\Oh(i)$ is then determined by $\Oh(1)$, as the reflexive hull
(or double dual) of $\Oh(1)^{\tensor i}$.
This fails $n=1$; the orbifold curve case is computationally pretty simple, but
some arguments need extra care (see Section~\ref{sec!Cur}). Orbifold behaviour
in codimension~1 can be treated more generally by setting
$\Oh(i)=\Oh(iA):=\Oh([iA])$, with $A=\sum \frac{a_j}{r_j}E_j$ a $\Q$-Weil
divisor. There are in any case theoretical advantages in thinking of
$\bigoplus\Oh(i)$ systematically as a graded structure sheaf.
\end{defn}
Then the ring
\[
R(X)=R(X,\bigoplus\Oh_X(i))=\bigoplus_{i\ge0} H^0(X,\Oh_X(i))
\]
is a finitely generated $k$-algebra, of the form $R(X)=k[x_0,\dots,x_N]/I_X$
with weighted generators $x_i\in H^0(X,\Oh_X(w_i))$ and a weighted homogeneous
ideal $I_X$. The affine variety $\sC_X=\Spec X$ is the {\em weighted cone}
over $X$; the grading induces an action of the multiplicative group $\G_m$ on
$R(X)$ and $\sC_X$ that defines the quotient $X=\Proj R(X)=
(\sC_X\setminus0)/\G_m$. Under the above conditions, $\sC_X$ is nonsingular
outside the origin, and all the orbifold behaviour of $X,\bigoplus\Oh_X(i)$
comes from isolated orbits with cyclic isotropy subgroups $\mu_r\subset\G_m$.
Choosing generators gives $\sC_X\subset\aff^{N+1}$ and
$X\subset\PP(w_0,\dots,w_N)$, where $\aff^{N+1}$ is affine space with
coordinates $x_0,\dots,x_N$ and $\PP(w_0,\dots,w_N)$ is weighted projective
space (w$\PP^N$) with homogeneous coordinates $x_0,\dots,x_N$ of the indicated
weights.
\begin{defn} \rm Write $P_i(R)=\dim_k R_i$ for the dimension of the $i$th
graded piece of a finitely generated graded ring $R=\bigoplus_{i\ge0} R_i$; by
abuse of notation, I call $P_i(X)=P_i(R(X))=h^0(X,\Oh(i))$ the {\em
plurigenera} of $X$. The {\em Hilbert series} of $R(X)$ or of $X$ is the
formal power series
\[
P_X(t)=P_{R(X)}(t)=\sum_{i\ge0} P_i(R)t^i.
\]
It is well known that this is a rational function $\Num(t)/\prod(1-t^{w_j})$
whose denominator corresponds to the generators of $R$.
The main point of this paper is to calculate $P_X(t)$ under extra conditions.
Surprisingly, it turns out in many cases to be easier to calculate the
generating function $P_X(t)$ than the individual $P_i(X)$. \end{defn}
\begin{defn}\label{defn!pGor} \rm I say that $X,\bigoplus\Oh_X(i)$ is {\em
projectively Gorenstein} if $R(X)$ is a Gorenstein graded ring. It is known
(compare \cite{GW} and \cite{W}) that this is equivalent to the following
cohomological conditions:
\begin{itemize}
\item $H^j(X,\Oh_X(i))=0$ for all $j\ne0,\dim X$ and all $i$;
\item the $i$th graded piece of $R(X)$ is the complete linear system
$H^0(X,\Oh_X(i))$ (this is a projective normality assumption, that I have
already imposed implicitly in defining $R(X)$);
\item the orbifold canonical sheaf of $X$ is of the form $\Oh_X(k_X)$ for some
integer $k_X$.
\end{itemize}
The final condition is stated here for $n\ge2$, when $\om_X$ is a divisorial
sheaf. More care is needed to handle orbifold behavior in codimension~1:
namely, rather than a single dualising sheaf $\om_X$, we need the graded
dualising sheaf of the graded structure sheaf $\bigoplus\Oh_X(i)$. The case
$n=1$ (orbifold curves) can be treated in terms of fractional divisors
$A=\sum\frac{a_i}{r_i}P_i$ and the orbifold canonical class
$K_{C,\orb}=K_C+\sum\frac{r_i-1}{r_i}P_i$; see \ref{}, \cite{De} and \cite{W}.
\end{defn}
\begin{defn}[Mukai \cite{Mu}] \label{defn!coind} \rm Let $X$ be a projectively
Gorenstein simply polarised orbifold with isolated orbifold points as above.
The {\em coindex} of $X$ is defined by $c=k_X+n+1$ where $k_X$ is as in
Definition~\ref{defn!pGor} and $n=\dim X$. \end{defn}
\begin{rmk}\label{rmk!init} \rm This definition is well known for nonsingular
projectively Gorenstein varieties. The coindex is invariant under passing to a
hyperplane section (of weighted degree~1), since the canonical class increases
by~1 (by the adjunction formula), while the dimension decreases by~1. Clearly,
\begin{itemize} \item $\PP^n$ has coindex~0; \item a quadric hypersurface has
coindex~1; \item an elliptic curve, del Pezzo surface or Fano 3-fold of
index~2 has coindex~2; \item a canonical curve, K3 surface or Fano 3-fold has
coindex~3. \item a surface of general type or Calabi--Yau 3-fold has coindex~4
\end{itemize}
See Remark~\ref{rmk!initP} below for their Hilbert series. \end{rmk}
The following theorem for nonsingular varieties is my model:
\begin{theorem} Let $X$ be a nonsingular projectively Gorenstein variety. Then
its Hilbert series is of the form
\begin{equation}
P_X(t)=\frac{\Num(t)}{(1-t)^{n+1}},
\label{eq!Num}
\end{equation}
where $\Num(t)$ is a symmetric polynomial of degree $c=\coindex X$.
\end{theorem}
\paragraph{Proof} This is an elementary consequence of Hirzebruch
Riemann--Roch, plus the vanishing assumption in Definition~\ref{defn!pGor} and
Serre duality.
In detail, RR implies that $\chi(\Oh_X(i))$ is a polynomial of degree $n$ in
$i$:
\[
\chi(\Oh_X(i)) = \chi(\Oh_X(iD)) = \int \ch(iD)\cdot\Td(X)
\]
where $D=c_1(\Oh_X(1))$. By the vanishing assumption there is no intermediate
cohomology, and by Serre duality $h^n(\Oh(i))=h^0(\Oh(k_X-i))$, so that
\begin{align}
\chi(\Oh_X(i)) & = h^0(X,\Oh(i)) + (-1)^n h^n(X,\Oh(i)) \notag \\[4pt]
& = P_i(X) + (-1)^n P_{k_X-i}(X).
\end{align}
In particular $P_i(X)$ is a polynomial in $i$ of degree $n$ for $i\ge k_X+1$.
(Note that when $k_X<0$, this says that $H^0(\Oh_X(i))=H^n(\Oh_X(i))=0$ for
$k_X+1*k_X$, it makes sense to
divide the formal sum up as a formal power series $P_X(t)$ in positive powers
of $t$ plus $(-1)^nt^{k_X}P_X(1/t)$, which is $t^{k_X}$ times a power series
in negative powers of $t$. On the other hand, multiplying the formal sum
formally by $(1-t)^{n+1}$ gives zero, since $\chi(\Oh_X(i))$ is a polynomial
of degree $n$ in $i$. This proves (\ref{eq!fn}).
(\ref{eq!fn}) implies that the numerator of $P(X)$ is a symmetric polynomial,
and completes the proof. \QED
\begin{rmk}\label{rmk!initP} \rm For low values of $c$, the numerator
$\Num(t)$ in (\ref{eq!Num}) is
\[
\begin{array}{|c|l|}
\hline
\hbox{coindex} & \Num(t) \\
\hline
\hline
c=0 & 1 \\
c=1 & 1+t \\
c=2 & 1+dt+t^2 \\
c=3 & 1+(g-2)t+(g-2)t^2+t^3 \\
c=4 & 1+at+bt^2+at^3+t^4 \\
\hline
\end{array}
\]
This proves the Hilbert series converse of Remark~\ref{rmk!init}. The form of
these polynomials is familiar: for example, a regular surface of general type
has $c=4$, with $a=p_g-3$ and $b=K^2-(2p_g-4)$. The formula itself works
perfectly well even if $p_g=0$, so $a=-3$.
My convention is to take the $[c/2]$ coefficients of $\Num(t)$ as the basic
invariants of $X$. One effect is that we study the Hilbert series in terms of
plurigenera themselves; relating the initial plurigenera to the global
topological invariants of $X$ (the Todd classes, the terms in
$\int\ch(iA)\Td(X)$ of Hirzebruch RR) becomes a secondary issue.
The classic case is when $R(X)$ admits a regular sequence $x_0,\dots,x_n$ of
elements of degree~1; geometrically, this holds if and only if $|\Oh_X(1)|$ is
a free linear system. Then $R(X)$ is a free graded module over the polynomial
ring $k[x_0,\dots,x_n]$, and its generators are in one-to-one correspondence
with a $k$-vector space basis of the Artinian quotient ring
$R(X)/(x_0,\dots,x_n)$; the Hilbert numerator of $R(X)$ is the Hilbert series
of this Artinian quotient. Thus passing to the numerator of
$P_X(t)=\frac{\Num(X)}{(1-t)^{n+1}}$ has the effect of normalising $X$ to
dimension~$-1$. \end{rmk}
\subsection{Isolated orbifold singularities}
Now let $k_X$ be projectively Gorenstein with isolated orbifold singularities,
with $k_X$ and $c$ as in Definitions~\ref{defn!pGor}--\ref{defn!coind}. The
ingredients in the plurigenus formula are as follows:
\begin{itemize}
\item the dimension $n$;
\item the canonical weight $k_X$;
\item the coindex $c=k_X+n+1$;
\item the first $[c/2]$ plurigenera $P_i(X)$ for $i=1,\dots,[c/2]$;
\item a basket $\sB$ of isolated orbifold points
$\sB=\{\rth(a_1,\dots,a_n)\}$, with $r\ge2$ and $a_1,\dots,a_n\in[1,r-1]$
coprime to $r$.
\item[] Each $\rth(a_1,\dots,a_n)\in\sB$ is required to satisfy
\[
k_X+\sum_{j=1}^n a_j \equiv 0 \quad\hbox{mod $r$.}
\]
\end{itemize}
Using these ingredients, I now cook up an {\em initial term} $P_{I,X}(t)$, and
for each point $\rth(a_1,\dots,a_n)\in\sB$ an {\em orbifold contribution}
$P_{\tot}(\rth(a_1,\dots,a_n),k_X)$, so that
\[
P_X(t)=P_{I,X}(t)+\sum_{\sB}P_{\tot}(\rth(a_1,\dots,a_n),k_X).
\]
In turn, $P_{\tot}$ is the sum of a {\em periodic term} $P_{\per}$ and a {\em
growing term} $P_{\grow}$ (see Theorem~\ref{th!main}). Each of these terms is
computed by a simple recipe.
\paragraph{Initial term} The initial term $P_{I,X}(t)$ is
\begin{equation}
P_{I,X}(t)=\frac{A(t)}{(1-t)^{n+1}}\,,
\end{equation}
where $A(t)$ is a symmetric polynomial of degree $c$ with integer
coefficients, uniquely determined by the condition that the formal power
series $P_{I,X}$ has the given $P_i(X)$ as coefficient of $t^i$ up to
$i=[c/2]$. The recipe is:
\begin{enumerate}
\renewcommand{\labelenumi}{(\arabic{enumi})}
\item set $A_0=\sum_{i=0}^{[c/2]} P_it^i$ (with $P_0=1$);
\item set $P'_i=$ coefficient of $t^i$ in $A_1=(1-t)^{n+1}A$ for
$i=0,\dots,[c/2]$ ;
\item set $A(t)=\sum_{i=0}^c P''_it^i$ where $P''_i=P'_i$ or $P'_{c-i}$.
\end{enumerate}
\begin{exa} \rm $n=3$, $c=5$, $P_1=3$, $P_2=7$, $A_0=1+3t+7t^2$; then
$A_1=(1-t)^4A_0=1-t+t^2+\cdots$, so that $P_I(t):=
\frac{1-t+t^2+t^3-t^4+t^5}{(1-t)^4}$;
Here one checks that $P_I(t)=\frac{1+2t^3+t^6}{(1-t)^3(1-t^2)}$ is the Hilbert
series of the nonsingular canonical 3-fold $X(6,6)\subset\PP(1,1,1,2,3,3)$.
\end{exa}
\paragraph{Periodic term} The orbifold contribution of each
$\rth(a_1,\dots,a_n)\in\sB$ is a sum of two pieces: first the periodic part
\begin{equation}
P_{\per}(t)=\frac{N_{\per}}{1-t^r}, \quad\hbox{where}\quad
N_{\per}=\sum_{i=1}^{r-1}b_it^i \quad\hbox{with $b_i\in\Q$}.
\end{equation}
Here $N_{\per}(\rth(a_1,\dots,a_n))$ is defined as the inverse mod
$1+t+\cdots+t^{r-1}$ of $\prod(1-t^{a_i})$ (see Lemma~\ref{lem!Nper} for
details and recipes). This term is fractional and periodic (the denominator
$1-t^r$ just repeats the terms from $1$ to $r-1$ with period $r$). It records
the deviation of $P_i(X)$ from being a polynomial in $i$.
\paragraph{Growing term} The growing term depends on the orbifold point
$\rth(a_1,\dots,a_n)$ and on the global canonical weight $k_X$ or the
coindex $c=k_X+n+1$. It is
\begin{equation}
P_{\grow}(\rth(a_1,\dots,a_n),k)=\frac{B(t)}{(1-t)^{n+1}}\,,
\end{equation}
where $B$ is a Laurent polynomial with support in $[[c/2]+1,[c/2]+r-1]$ such
that the formal power series $P_{\per}+P_{\grow}$ is zero up to degree $[c/2]$.
The result takes the form
\[
P(X)=P_{I,X}(t)+\sum_{\sB} P_{\tot}(\rth(a_1,\dots,a_n),k_X)
\]
where the
$P_{I,X}(t)$ is the rational function with denominator $(1-t)^{n+1}$ and
numerator the symmetric polynomial of degree $c$ which is uniquely determined
by the condition that $P_{I,X}(t)$ {\em initial term}
\begin{defn}
\end{defn}
\begin{defn}
\end{defn}
\section{Some lemmas}
I calculate the contributions
\[
P_{\tot}(\rth(a_1,\dots,a_n),k) = P_{\per}(\rth(a_1,\dots,a_n)) + P_{grow}(\rth(a_1,\dots,a_n),k)
\]
for an isolated quotient singularity $\rth(a_1,\dots,a_n)$ on a polarised
variety with $K=kA$.
\begin{lemma} Consider a polynomial
\[
B = \sum_{i=1}^{r-1} b_it^i \in\Q[t],
\]
and suppose given $r,n\in\N$ and an interval $J=[d+1,\dots,d+r-1]$ of
$r-1$ consecutive integers. Then there exists a unique Laurent polynomial
\[
A = \sum_{j\in J} \al_jt^j \in\Q[t,t\1]
\]
supported in $J$ such that $A-(1-t)^nB$ is divisible by $1+t+\cdots+t^{r-1}$:
\[
A-(1-t)^nB =\frac{1-t^r}{1-t}L
\]
with $L$ a Laurent polynomial.
\end{lemma}
It follows that
\[
\frac{B}{1-t^r}+\frac{L}{(1-t)^{n+1}} = \frac{A}{(1-t)^n(1-t^r)}.
\]
Later I write $P_{\per}=\frac{B}{1-t^r}$, $P_{grow}=\frac{L}{(1-t)^{n+1}}$, so
that $P_{\tot}=P_{\per}+P_{grow}=\frac{A}{(1-t)^n(1-t^r)}$.
\paragraph{Proof} The quotient ring
\[
V = \Q[t]/(1+t+\cdots+ t^{r-1})
\]
is an $(r-1)$-dimensional vector space based by $1,t,\dots,t^{r-2}$. However,
$t$ maps to an invertible element of $V$, so that also
\[
V =\Q[t,t\1]/(1+t+\cdots+ t^{r-1}),
\]
and the $r-1$ elements $t^j$ for $j\in J$ form another basis of $V$.
Therefore the class of $(1-t)^nB$ modulo the ideal of $\Q[t,t^{-1}]$ generated
by $1+t+\cdots+ t^{r-1}$ can be written in a unique way as a linear combination of
$t^j$ for $j\in J$. \QED
\begin{defn} \rm The numerator
\[
N_{\per}(\rth(a_1,\dots,a_n)) = \sum_{i=1}^{r-1} b_it^i
\]
of $P_{\per}$ is defined as the inverse modulo $1+t+\cdots+t^{r-1}$ of $\prod_{j=1}^n
(1-t^{a_j})$, and $P_{\per}$ itself is defined by
\[
P_{\per} = \frac{N_{\per}}{1-t^r}.
\]
\end{defn}
\begin{lemma}\label{lem!Nper} (1) When $n=1$,
\[
N_{\per}(\rth(a))=\frac1r\sum_{i=1}^{r-1}-\overline{bi}\,t^i,
\]
where as usual
\begin{align*}
& \hbox{$b$ is the inverse of $a$ modulo $r$, and} \\
& \hbox{in $\overline{bi}$, the bar denotes smallest residue mod $r$.}
\end{align*}
(2) For general $n$,
\[
N_{\per}(\rth(a_1,\dots, a_n)) = \prod_{j=1}^n N_{\per}(\rth(a_j)) \quad
\mathrm{mod}\ 1+t+\cdots+t^{r-1}.
\]
\end{lemma}
\paragraph{Proof of (1)} Consider
\[
(1-t^a)\left(\sum_{i=1}^{r-1} -\overline{bi}\,t^i\right) =
\sum_{i=1}^{r-1}-\overline{bi}\,t^i + \sum_{i=1}^{r-1}\overline{bi}\,t^{a+i}
\]
Working modulo $1+t+\cdots+t^{r-1}$ allows me to substitute $t^r=1$, and to
subtract a scalar multiple of $1+t+\cdots+t^{r-1}$ or of $t+t^2+\cdots+t^r$.
On applying these rules, the right-hand side evaluates to $r$.
For example, $r=7$ and $a=2$; then $b=4$ and we consider
\[
-4t - t^2 - 5t^3 - 2t^4 - 6t^5 - 3t^6;
\]
multiplying by $1-t^2$ gives
\[
\begin{matrix}
-4t&-t^2&-5t^3&-2t^4&-6t^5&-3t^6\\
&&+4t^3&+t^4&+5t^5&+2t^6&+6t^7&+3t^8;
\end{matrix}
\]
according to my rules, I can replace $3t^8$ by $3t$ and $6t^7$ by
$6=(7 - 1)$, and the whole sums to
\[
7-(1+t+t^2+\cdots+t^6)=7.
\]
In the general case, for clarity, break up the second sum as a sum over $a+i\le r-1$
and another over $a+i\ge r$, and change the dummy index from $i$ to $j=a+i$.
Then the second sum is
\begin{multline}
\sum_{j=a+1}^{r-1}\overline{b(j-a)}t^j
+\sum_{j=r}^{r+a-1}\overline{b(j-a)}t^{j-r} \\
=\sum_{j=a+1}^{r-1}(\overline{bj}-1)t^j
+(r-1)t^r+\sum_{j=r+1}^{r+a-1}(\overline{bj}-1)t^j.
\end{multline}
In fact, the coefficient of $t^j$ is $\overline{bj}-1$ whenever
$\overline{bj}\ne0$. The only exception is the coefficient of $t^r$, which has
$\overline{b(j-a)}=r-1$. Now in each term with $j\ge r$, substitute
$t^r\mapsto1$ to get (as in the example)
\[
(1-t^a)\sum-\overline{bi}t^i=r-(1+t+\cdots+t^{r-1})=r.
\]
(The missing term $i=a$ in the second sum has $\overline{ab}=1$, so its
coefficient is zero.) \QED
\section{Curves}\label{sec!Cur}
although some considerations may be confusing at first
sight
\begin{thebibliography}{GW}
\bibitem[De]{De} Michel Demazure, Anneaux gradu\'es normaux, in Introduction
\`a la th\'eorie des singularit\'es, II, Hermann, Paris (1988) pp.~35--68
\bibitem[GW]{GW} S. Goto and K-i Watanabe, On graded rings. I, J. Math.
Soc. Japan {\bf30} (1978) 179--213
\bibitem[Mu]{Mu} MUKAI Shigeru, Biregular classification of Fano 3-folds and Fano
manifolds of coindex~3, Proc. Nat. Acad. Sci. U.S.A. {\bf86} (1989) 3000--3002
\bibitem[W]{W} WATANABE Kei-ichi, Some remarks concerning Demazure's
construction of normal graded rings, Nagoya Math. J. {\bf 83} (1981)
203--211
\end{thebibliography}
\end{document}
========================
Start with \rth(a) and define b = a^-1 mod r. Define
Scores (b/r) = i in ZZ such that bi/r scores
= i in ZZ such that [bi/r] > [b(i-1)/r].
This is obviously invariant under translation i -> r + i.
It also the reflection symmetry
i <-> -a+1-i and i <-> r-a+1-i.
While slightly less obvious, this boils down to
bi + r - bi == 0 mod r
so that bi scores if and only if b(-i+1) scores. The
function NoScores(b/r) has the same symmetries.
Now for any k with a+k == 0 mod r, I define Pnum(\rth(a),k)
to be
either a segment of Scores(b/r)
or a segment of NoScores(b/r)
(depending on parity) that is
a symmetric Laurent polynomial of degree k+r+1
that is zero in degree up to [c/2] where c = k+2.
so that Ptot = Pnum/(1-t)*(1-t^r) is symmetric of degree k.
For exa 1/7(2), b = 4, Scores (4/7) =
.. + t^-3 + t^-1 + 1 + t^2 + t^4 + t^6 + t^7 + t^9 + ..
(you only need to write down one segment of this mod r)
For k = -2, k+r+1 eq 6, I could take the segment
1+t^2+t^4+t^6 for Scores(4/7), but that has nonzero coeff
in degree zero = c/2, so I must do
-t-t^3-t^5.
For ex P_I = 1/(1-t)^2 and Ptot = (-t-t^3-t^5)/(1-t)*(1-t^7)
gives (1+t^2+t^4+t^6)/(1-t)*(1-t^7), and this corresponds to
the genuine example C_8 in PP(1,2,7).
Sim for k = 5, t^4+t^6+t^7+t^9 is symmetric of deg 13, and
is zero in degree up to c/2 = 3.
The point is that the formula should cover things like
PP(2,7) for which k = -9, c = -7, so that P_I = 0, and
the contributions Ptot(1/7(2),-9) and Ptot(1/2(1),-9) have
to cancel out in deg < 0, and give 1/(1-t^2)(1-t^7) in
deg >= 0. This points the way to what we have to do to
reclaim Kaori's results in this language.
=========================
Case dim 1
Pper and Pgrow are simple and conceptual;
they form a foundation for the case of isolated n-dimensional singularities;
The orbifold point $\rth(a)$ at P of curve corresponds to a rational divisor
$(b/r)P$ where $b = a\1$ mod $r$.
Pper is the "small change" lost in rounding down: the rational divisor
$(bi/r)P$ has fractional part ${bi/r}P$.
Rational functions have integral poles, so the RR space \sL(iA) or divisorial
sheaf \Oh_C(iA) corresponding to the fractional divisor iA (defined by the
usual valuation condition $\div f + iA \ge0$) is actually \sL([iA]) or
\Oh_C([iA]). On a curve, if A = (b/r)P then when we apply RR to iA, we lost
the fractional part {bi/r} in loose change. This loss contributes
\[
Pper(\rth(a)=\sum_{i\ge0} -{bi/r}t^i
\]
to the Hilbert series. Since the sum is periodic the sum over $i\ge0$ equals
$\frac1{1-t^r}$ times the sum over an interval $i=1,\dots,r-1$ (the
contribution is zero when $i$ is a multiple of $r$, so don't worry about the
ends of the intervals).
Lemma Pper(\rth(a)=\sum_{i\ge0} -{bi/r}t^i is the inverse of (1-t^a) mod
(1-t^r)/(1-t).
It is important to understand what this contribution Pper(\rth(a) is (and is
not). The polynomial RR formula on a curve is
\[
\hbox{RR}(C,iA)=1-g+i\deg A
\]
with $\deg A\in H^2(C,\Q)=\Q$. The right-hand side is a polynomial function
of $i$. To relate it to $\chi(\Oh_C(iA))$, I must add the contribution
$-{bi/r}$ to account for the passage from the fractional divisor $iA$ to
the first Chern class of $\Oh_C([iA])$.
Start with \rth(a) and define b = a^-1 mod r. Define
Scores (b/r) = i in ZZ such that bi/r scores
= i in ZZ such that [bi/r] > [b(i-1)/r].
This is obviously invariant under translation i -> r + i.
It also the reflection symmetry
i <-> -a+1-i and i <-> r-a+1-i.
While slightly less obvious, this boils down to
bi + r - bi == 0 mod r
so that bi scores if and only if b(-i+1) scores. The
function NoScores(b/r) has the same symmetries.
Now for any k with a+k == 0 mod r, I define Pnum(\rth(a),k)
to be
either a segment of Scores(b/r)
or a segment of NoScores(b/r)
(depending on parity) that is
a symmetric Laurent polynomial of degree k+r+1
that is zero in degree up to [c/2] where c = k+2.
so that Ptot = Pnum/(1-t)*(1-t^r) is symmetric of degree k.
For exa 1/7(2), b = 4, Scores (4/7) =
.. + t^-3 + t^-1 + 1 + t^2 + t^4 + t^6 + t^7 + t^9 + ..
(you only need to write down one segment of this mod r)
For k = -2, k+r+1 eq 6, I could take the segment
1+t^2+t^4+t^6 for Scores(4/7), but that has nonzero coeff
in degree zero = c/2, so I must do
-t-t^3-t^5.
For ex P_I = 1/(1-t)^2 and Ptot = (-t-t^3-t^5)/(1-t)*(1-t^7)
gives (1+t^2+t^4+t^6)/(1-t)*(1-t^7), and this corresponds to
the genuine example C_8 in PP(1,2,7).
Sim for k = 5, t^4+t^6+t^7+t^9 is symmetric of deg 13, and
is zero in degree up to c/2 = 3.
The point is that the formula should cover things like
PP(2,7) for which k = -9, c = -7, so that P_I = 0, and
the contributions Ptot(1/7(2),-9) and Ptot(1/2(1),-9) have
to cancel out in deg < 0, and give 1/(1-t^2)(1-t^7) in
deg >= 0. This points the way to what we have to do to
reclaim Kaori's results in this language.
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