\section{Introduction}\label{S: introd}

Let $S=\Bbbk[x_1,\ldots, x_n]$ be a polynomial ring over a field
$\Bbbk$, $I$ a monomial ideal of $S$ and $R=S/I$. If $\J$ is a
subset of the minimal  monomial generating set $\I$ of $I$, we let
$m_\J$ denote the least common multiple of the monomials in $\J$
or the corresponding monomials in $\Bbbk[y_1,\ldots, y_n]$ as
appropriate. The multigraded Poincar\'e series of $R$, $P_R(y,t)$
is  rational, \cite{Ba82}:
 $$P_R(y,t) = \sum_{i=0}^{\infty}  \sum_{{\bf j}\ge 0}
y^{\bf j} {\dim}_k ({\Tor}_i^R(k,k)_{\bf j})
t^i={{\prod_{i=1}^n(1+ty_i)}\over{Q_R(y,t)}}$$ where $Q_R(y,t)\in
Z[y_1,\ldots, y_n][t]$.

The few facts that are known in general about the multigraded
expansion $Q_R(y,t)$ also date back to \cite{Ba82}: the degree of
$Q_R(y,t)$ in $t$ is bounded above by the total degree of $m_\I$,
and the  monomial coefficients of $t$ divide $m_\I$. In this paper
we  discuss the monomial coefficients of $Q_R(y,t)$.

In the first section  of this paper we show that the monomial
coefficients of $t$ in $Q_R(y,t)$ are  least common multiples of
the monomial generators of $I$. We discuss the Koszul homology of
$R$ when $R$ is Golod. We also present a free resolution of $R$
when $I$ is generic.
 In the second section we  discuss the multigraded acyclic closure of $R$ as well as  the  multigraded deviations in terms of the
multigraded homotopy Lie algebra of $R$, and show that the LCM
lattice determines the Poincar\'e series of $R$. This is the
multigraded version of  Theorem 1 of
 \cite{Av02}.

Thanks are due to J.~Backelin and the faculty of the Department of
Mathematics in Stockholm, for useful discussions during my brief
stay in 1996.

\section{The  monomials of $Q_R(y,t)$ }\label{ymonomialsection}

The argument used in the proof of the next proposition has the
flavor of  \cite{Ba82} and uses Golod theory. It uses Lescot's
result, \cite{Le86}, that the Poincar\'e series $M$ over $R$
$$P_R^M(y,t) = \sum_{i=0}^{\infty}  \sum_{{\bf j}\ge 0}
y^{\bf j} {\dim}_k ({\Tor}_i^R(M,k)_{\bf j})
t^i={{f_R^M(y,t)}\over{Q_R(y,t)}}$$ where$f_r^M(y,t))\in
Z[y_1,\ldots, y_n][t]$.

\begin{proposition}\label{lcmcoefficients}
Let $Q_R(y,t)=1+\sum (\sum c_{\boldsymbol{j}} y^{\boldsymbol{j}})t^i$. Then $y^{\boldsymbol{j}}$ is
equal to a least common multiple of a subset of the minimal
monomial generating set of $I$.
\end{proposition}
\begin{proof} By a standard argument, \cite{Ba82},  it is enough to prove  Proposition \ref{lcmcoefficients}
when the ideal $I$ is generated by squarefree monomials. In this
case the  $y$ monomials of $Q_R(y,t)$ are squarefree, \cite{Ba82}.
We do induction on the number of variables $n$ that divide the
monomials of  the minimal generating set of $I$, the case $n=1$
being trivial. Consider a term of $Q_R(y,t)$ whose coefficient in
$y$ does not involve $y_i$. We separate the generators of $I$ that
involve $x_i$. Let $I=L_i+ x_i J_i$ where  $L_i, J_i\subset
\Bbbk[x_1,\ldots,\hat{x_i}, \ldots, x_{n-1}]$ and let
$A=\Bbbk[x_1,\ldots, \hat{x_i}, \ldots, x_{n-1}]/L_i$ and
$A'=\Bbbk[x_1,\ldots, \hat{x_i}, \ldots, x_{n-1}]/(L_i+J_i)$.
Therefore $R\cong A[x_i]/x_i J_i$ and
 $A[x_i]\mapright{} R$ is Golod,   which implies that
$$P_R(y,t)={P_{A[x_i]}\over (1-t(P^R_{A[x_i]}-1))}=
{(1+ty_i)P_A\over (1-t^2P_{A[x_i]}^{x_iJ_i} ) }= {(1+ty_i)P_A
\over (1-t^2y_i P_A^{J_i})}=$$
$$={(1+ty_i)P_A \over (1-t y_i(P_A^{A'}-1))}=
{\prod (1+ty_i)\over Q_A+ty_iQ_A-ty_if_A^{A'}}.$$ The $y$
monomials that do not involve $y_i$ are terms of $Q_A$, so by
induction they have the desired form. Suppose now that a monomial
of $Q_R(y,t)$ involves all $y_i$: it is the product of all the
variables $y_i$ and is  equal to the least common multiple of all
the generators in $I$.
\end{proof}

In \cite{ChRe95} we noticed that the terms of $Q_R(y,t)$ when $I$
is a complete or almost complete intersection come from the set
$\{ (-1)^{l_\J}t^{(|\J|+l_\J)}m_\J\}$. Here $\J$ ranges over the
subsets of the minimal monomial generating set of $I$ and $l_\J$
is the number of connected components of the graph we get when we
connect the elements of $\J$ by an edge if they have a variable in
common. In \cite{ChRe95} we also confirmed that this is the case
when the Taylor resolution \cite{Ta60}, \cite{Ei97} of $S/I$ over
$S$   is minimal.

In Remark \ref{golod-denom-rem} we confirm that the same holds for
the monomial ideals $I$ such that $R/I$ is Golod. First we recall
and expand a comment by Fr\"oberg, \cite{Fr79}.

\begin{remark} Let $R=S/I$ be Golod and $K_\bullet$ be the Koszul
complex on the variables $\bar{x_i}$ of $R$. Then $H(K_\bullet)$
is isomorphic to $\Bbbk[X_{A_1},\ldots, X_{A_r}]/(X_{A_1},\ldots,
X_{A_r})^2$ where $A_i$ is a subset of the minimal monomial
generating set of $I$ and the degree of $X_{A_i}$ equals the
cardinality of $A_i$ .

Indeed since $K_\bullet$ admits trivial Massey operations the
product of two elements of $H_{\ge 1}(K_\bullet)$ is zero,
\cite{GuLe69}, \cite{Av98}. Moreover $H(K_\bullet)$ is always of
the form $\Bbbk[X_1, \ldots, X_r]/L$: $L$ is generated by
quadratics, the $X_i's$ form a minimal multiplicative generating
set for $H(K_\bullet)$, and $X_i X_j=(-1)^{|X_i| |X_j|} X_i X_j$
where $|X_i|$ is the degree of $X_i$. Let $F_\bullet$ be the
Taylor complex on the minimal monomial generators of $I$ over $S$,
\cite{Ta60} and let $E_\bullet$ be the minimal multigraded
resolution of $S/I$ over $Q$. $E_\bullet$ is a direct summand of
$F_\bullet$. It follows that the multidegrees of  the generators
of $E_i$ are among the multidegrees of the generators of $F_i$.
Therefore they are the multidegrees of  least common multiples of
subsets $A_j$ of $\I$. It follows that $H(K\bullet)=H(\Bbbk\otimes
E_\bullet)= \Bbbk[X_{A_1},\ldots, X_{A_r}]/L$. Since $X_{A_j}
X_{A_t}=0$,  $X_{A_j} X_{A_t} \in L$.
\end{remark}

\begin{remark}\label{golod-denom-rem}
Thus monomial Golod rings are special Golod rings. Let $R$ be a
Golod ring. From our previous remark and using the fact that
$m_A\neq m_{A_1}\cdots m_{A_t}$ for any partition of $A$ where
$X_A$ and $X_{A_j}$ are among the generators of $H(K_\bullet)$,we
see that $A$ is connected and $l_A=1$. Let $H_i(K)_{\bf j}$ denote
the subspace of $H_i(K)$ of multidegree $\bf j$. Thus the terms of
the denominator of the Poincar\'e series of $R$,
$Q_R(y,t)=1-\sum_i \sum_{\bf j} \dim_\Bbbk H_i(K)_{\bf j} y^{\bf
j} t^{i+1}$ of the Golod ring $R$ come from the set $\{
(-1)t^{(|\J|+1)}m_\J\}$ where $\J$ ranges over the subsets of the
minimal monomial generating set of $I$.
\end{remark}

Generic ideals and their minimal resolutions were first introduced
in \cite{BaPeSt98} and generalized in \cite{MiStYa00}. Let $I$ be
a generic ideal. If the multidegrees of  two minimal monomial
generators of $I$ are equal for some variable then there is a
third monomial generator of $I$ whose multidegree is strictly
smaller than the multidegree of the least common multiple of the
other two. The minimal resolution $E_\bullet$ of $S/I$ over $S$ is
determined by the Scarf complex of $\D_I$. The latter  consists of
the subsets $\J$ of the index set of $I$, $\Phi$, such that
$m_{\J}\neq m_{\L}$ for any other subset $\L$ of $\Phi$. We note
that $E_\bullet$ has the structure of a DG algebra,
\cite{BaPeSt98}. The Eagon resolution of $\Bbbk$ over $R=S/I$ was
discussed in \cite{GuLe69}. Below we give the generators for the
Eagon resolution when $I$ is generic.

\begin{theorem}
\label{res-thm} Let $R=S/I$ where $I$  is a generic ideal of $S$,
let $K_\bullet$ be the Koszul complex on the $x_i$ over $R$,
$X_0=0$, and  $X_i$ be the free $R$-module with generators $T_{\rm
L}$ where $|\L|=i$ and $\L$ is in $\Delta_I$,  the Scarf complex
of $I$. Then  $(Y_\bullet, d_\bullet)$ is a free resolution of
$\Bbbk$ over $R$, where
$$Y_i=\sum_{j+t_1+\cdots+t_l+l=i}
                  K_j \otimes
          X_{t_1}\otimes\cdots \otimes X_{t_l},$$
and
\begin{flushleft}
$d(T_\J)=(m_\J/x_\J ) e_\J,$
\end{flushleft}
\begin{flushleft}
$d(T_{\I_1}\otimes\cdots\otimes T_{\I_l})=
 d(T_{\I_1} \otimes \cdots\otimes T_{\I_{l-1}})
\otimes T_{\I_l}
  +$
\end{flushleft}
\begin{flushright}
$(-1)^{\displaystyle{\sum_{i=1}^{l-1}
 (|\I_i|+1)}}
{\displaystyle T_{\I_1}\otimes\cdots\otimes
  T_{\I_{l-2}}\otimes \delta(\I_{l-1},\I_l})(T_{\I_{l-1}}\wedge T_{\I_l}),$
\end{flushright}
where $\delta(\I_{l-1},\I_l)={{ m_{\I_{l-1}} m_{\I_l}\over
m_{\I_{l-1}\cup \I_l} }} $ if $\I_{l-1}\cup\I_l\in \Delta_I$ and 0
otherwise
\begin{flushleft}
$d(e_\L \otimes T_{\I_1}\otimes\cdots\otimes
T_{\I_l})d(e_\L)\otimes T_{\I_1}\otimes\cdots\otimes T_{\I_l}+
(-1)^{|\L|}e_\L\otimes d(T_{\I_1}\otimes\cdots\otimes
 T_{\I_l}).$
\end{flushleft}
\end{theorem}

\begin{proof} The theorem follows immediately from \cite{Iy97}, Theorem 1.2 and the preceeding comments.
\end{proof}

As an immediate corollary we get the characterization of Golod
rings when $I$ is a generic ideal, which generalizes Proposition
2.4 of \cite{Ga00}.

\begin{proposition} Let $I$ be a generic ideal. $S/I$ is Golod if and only if  $m_{\I_{1}} m_{\I_2}\neq m_{\I_{1}\cup \I_2} $ whenever  $\I_{1}\cup\I_2\in \Delta_I$.
\end{proposition}

\section{Multigraded deviations and acyclic closures}\label{multdeviations}

In this section we introduce the multigraded acyclic closure
$R\<X{{\rm >}}$ of $\Bbbk$ over $R$. For the construction of the
usual acyclic closure, we refer to Construction 6.3.1,
\cite{Av98}. To this end we choose $x_{\boldsymbol{j}}$, the $\G$-variables
of $X_n$  so that $cls(\theta(x_{\boldsymbol{j}}))$ has multidegree $\boldsymbol{j}$ and
$\{ cls(\theta(x_{\boldsymbol{j}}))\}$ is a  minimal multigraded generating
set  of $H_{n-1}(R\<X_{\le n-1}{{\rm >}})$.  We define the
multidegree of $x_{\boldsymbol{j}}$ to be $\boldsymbol{j}$ and we let $X_{n,\boldsymbol{j}}$
consist of all $ x\in X_n$ whose  multidegree  is $ \boldsymbol{j}$. We will
show that the cardinality of $X_{n,\boldsymbol{j}}$ appears as an exponent in
a product decomposition of the multigraded Poincar\'e series.

First we state and prove the multigraded version of Remark 7.1.1,
\cite{Av98}.

\begin{proposition} For each formal multigraded power series with integer coefficients
$$P(y,t)=1+\sum_{i=1}^\infty
(\sum_{\boldsymbol{j}} a_{i, \boldsymbol{j}}y^{\boldsymbol{j}}) t^i$$ where for each $i$,
$a_{i,\boldsymbol{j}}=0$ for all but finitely many values of $\boldsymbol{j}$,
 there exist uniquely defined integers
$e_{n,{\boldsymbol{j}}}\in Z$ such that
$$P(y,t)= {{\prod_{i=1}^\infty \prod_{\boldsymbol{j}}(1+y^{\boldsymbol{j}} t^{2i-1})^{e_{2i-1,{ \boldsymbol{j}}}} }\over
    {\prod_{i=1}^\infty \prod_{\boldsymbol{j}}(1-y^{\boldsymbol{j}} t^{2i})^{e_{2i,{\boldsymbol{j}} }} }}
$$
and the product converges in the (t)-adic topology of the ring
$Z[y,t]$.
\end{proposition}

\begin{proof}
We set $P_0(t)=1$ and assume by induction that $P(y,t)\equiv
P_{n-1}(y,t) $ modulo $ t^{n+1}$. If $P(y,t)-P_{n-1}(y,t)\equiv
\sum e_{n,{\boldsymbol{j}}}y^{\boldsymbol{j}} t^n (\modulo\ t^{n+1})$, then  we set
$P_n(y,t)=P_{n-1}(y,t) \prod_{\boldsymbol{j}} (1+y^{\boldsymbol{j}} t^n)^{e_{n,\boldsymbol{j}}}$ if
$n$ is odd and
 $P_n(y,t)=P_{n-1}(y,t) / \prod_{\boldsymbol{j}} (1-y^{\boldsymbol{j}} t^n)^{e_{n,\boldsymbol{j}}}$ if $n$ is even.
Then it is clear that $P(t)\equiv P_n(t) (\modulo\ t^{n+1})$ and
the other assertions follow as well.
\end{proof}

 We define the {\it multigraded} $(n,{\boldsymbol{j}})$ {\it deviation} of $R$, denoted by
$\epsilon_{n,\boldsymbol{j}}$, to be the exponent $e_{n, {\boldsymbol{j}}}$ in the product
decomposition of the Poincar\' e series of $R$.  As in Theorem
7.1.3, \cite{Av98} it follows that
\begin{proposition} Let $I$ be a monomial ideal of $S$, $R=S/I$ and
$R\<X{\rm >}$ be the multigraded acyclic closure of $\Bbbk$ over
$R$. Then
$$P_R(y,t)= {{\prod_{i=1}^\infty \prod_{\boldsymbol{j}}(1+y^{\boldsymbol{j}} t^{2i-1})^{\card(X_{2i-1,{ \boldsymbol{j}}})} }\over
    {\prod_{i=1}^\infty \prod_{\boldsymbol{j}}(1-y^{\boldsymbol{j}} t^{2i})^{\card(X_{2i,{\boldsymbol{j}} }) }  }}
$$
and $\card X_{n,\boldsymbol{j}}=\epsilon_{n,\boldsymbol{j}}(R)$.
\end{proposition}

Since $R\< X{\rm >}$ is multigraded, it follows that the homotopy
Lie algebra of $R$, $\pi^\bullet(R)=H Der_R^\gamma(R\< X{{\rm >}},
R\< X{{\rm >}})$ has a multigraded structure. Moreover $\pi^n(R)
\cong \Hom_{\Bbbk}({\Bbbk}X_n, \Bbbk)$, (Theorem 10.2.1,
\cite{Av98}), and $\pi^n_{\boldsymbol{j}}(R) \cong
\Hom_\Bbbk({\Bbbk}X_{n,{-\boldsymbol{j}}}, \Bbbk)$ while $\rank_\Bbbk
\pi^n_{\boldsymbol{j}}(R)=\epsilon_{n,-\boldsymbol{j}}$.

\begin{remark}
We let $K_\bullet$ denote the Koszul complex on $x_1,\ldots, x_n$
over $S$ and $T_\bullet$ the Taylor complex on the minimal
generators of $I$ over $S$. The complexes $K_\bullet$, $T_\bullet$
have the structure of a DG $\Gamma$ algebra by Lemma 9,
\cite{Av02}; they are also multigraded  and
 $H_\bullet (T\otimes_S K)_{\boldsymbol{j}}$
$\cong H_\bullet(T\otimes_S \Bbbk)_{\boldsymbol{j}}\cong H_\bullet
(R\otimes_S K)_{\boldsymbol{j}}$, where $\boldsymbol{j} \in L_I$. In Lemma 11 of
\cite{Av02}, it is shown that $\pi^{\ge 2}(R)\cong \pi^\bullet
(R\otimes_S K)$.  We remark that $R\otimes_S K$ is a multigraded
DG $\Gamma$ algebra, and one can choose a multigraded DG $\Gamma$
algebra $U$ so that we have the factorization $R\otimes_S
K\mapright{} U\mapright{}\Bbbk$ with the properties of Lemma 11
and Remark 10 of \cite{Av02}: $\pi(R\otimes_S K)$ is the graded
$\Bbbk$ dual of  the residue of $H_\bullet(U\otimes_{R\otimes_S K}
\Bbbk)$ modulo multigraded relations and is multigraded. The
homomorphisms of Lemma 11, \cite{Av02} are multigraded and
$\pi^{l+2}_{\boldsymbol{j}}(R)\cong \pi^{l}_{\boldsymbol{j}} (R\otimes_S K)$. Finally we
have $\pi^{l}_{\boldsymbol{j}} (T\otimes_S K)\cong \pi^{l}_{\boldsymbol{j}} (T\otimes_S
\Bbbk)\cong \pi^{l}_{\boldsymbol{j}}(R\otimes_S K$.
\end{remark}

Next we will make use of the LCM lattice $L_I$, \cite{GaPeWe99}
and the GCD graph $G_I$,\cite{Av02}. We recall that $L_I$ and
$G_I$ have vertices the least common multiples of the monomial
generators of $I$ while the edges of $G_I$ join least common
multiples that are relatively prime. In \cite{Av02} it is shown
that if $I$ and $I'$ are two monomial ideals of $S$ and $S'$
respectively with an isomorphism of lattices $\lambda:
L_I\mapright{} L_{I'}$ which induces an isomorphism of the GCD
graphs $\lambda: G_I\mapright{} G_{I'}$ then the homotopy Lie
algebras of $R=S/I$ and $R'=S/I'$ are isomorphic and the
Poincar\'e series of $R$ and $R'$ have the same denominator. We
will show that the multigraded version of this result is actually
true. We consider the multigraded acyclic closures $R<X>$ and
$R'<X'>$ of $\Bbbk$ over $R$ and $R'$.

\begin{remark}
Let $I\subset S=\Bbbk[x]$ and $I'\subset S'=\Bbbk[x']$ be two
monomial ideals, $K, K'$ the Koszul complexes on the $x_i$ and
$x_i'$ respectively and $\lambda: L_I\mapright{} L_{I'}$ an
isomorphism of lattices. If $\lambda$  induces an isomorphism of
the GCD  graphs $\lambda: G_I\mapright{} G_{I'}$ then $H(K\otimes
R)_{\bf j}\cong H(K'\otimes R')_{\lambda(\bf j)}$ through the
isomorphism that sends the algebra generators $X_A$ of $H(K\otimes
R)=\Bbbk[X_A]/L$ to the generators $X'_{\lambda(A)}$ of
$H(K'\otimes R')=\Bbbk[X'_{\lambda(A)}]/L'$, (here our notation is
as in the remarks preceeding Proposition 2.2). Indeed, let
$\lambda$ be as above. Since $\lambda$ is an isomorphism of the
LCM lattices the minimal resolution of $I$ gives rise to a minimal
resolution of $I'$ and the generators $X_A$ of  $H(K\otimes R)$
correspond to generators $X'_{\lambda(A)}$ of $H(K'\otimes R')$.
Since $\lambda$ is an isomorphism of the GCD graphs the relations
among  the $X_A$ correspond to relations among the
$X'_{\lambda(A)}$. Conversely if the Koszul homology algebras are
isomorphic as above then there is a an isomorphism of multigraded
vector spaces between the minimal resolutions of the two ideals
and a map $\lambda$ which is an isomorphism of lattices and GCD
graphs for the multidegrees that appear in the resolutions.
\end{remark}

\begin{proposition}
Let $I'$ be a monomial ideals of $S'=\Bbbk[x']$ such that
$\lambda: L_I\mapright{} L_{I'}$ is an isomorphism of lattices
that induces an isomorphism of the GCD  graphs $\lambda:
G_I\mapright{} G_{I'}$.  Then there is a map $\hat\lambda$ from
the set of the multidegrees of the variables $X_n$ to the set of
the multidegrees of the variables $X'_n$ such that $\card
X_{n,\boldsymbol{j}}=\card X'_{n,\hat\lambda(\boldsymbol{j})}$ and $\epsilon_{n,\boldsymbol{j}}=\epsilon'_{n,
\hat\lambda(\boldsymbol{j})}$.
\end{proposition}
\begin{proof} We define $\hat \lambda$ by induction on $n$.
Let $T'$ be the Taylor complex of $I'$. Then there is an
isomorphism of vector spaces $(T\otimes_S \Bbbk)_{\boldsymbol{j}}\cong
(T'\otimes_{S'} \Bbbk)_{\lambda(\boldsymbol{j})}$, for  $\boldsymbol{j} \in L_I$.
Suppose that $\hat \lambda$ is an extension of $\lambda$ and
$R[X_{\le n,\boldsymbol{j}}]\otimes \Bbbk \cong R'[X'_{\le
n,\hat\lambda(\boldsymbol{j})}]\otimes \Bbbk$. Extend $\hat\lambda$ to the
multidegrees of the multigraded generators of $H_n(R[X_{\le n}])$
and $H_n(R'[X'_{\le n}])$. It follows that there is an isomorphism
of homotopy Lie algebras $\pi^{\bullet}_{-\boldsymbol{j}} (T\otimes_S
\Bbbk)\cong \pi^{\bullet}_{-{\hat \lambda}(\boldsymbol{j})} (T'\otimes_{S'}
\Bbbk)$, and the proposition now follows.
\end{proof}

\begin{remark} Let $I$ be a monomial ideal and $I_{pol}$ be the square
free monomial ideal in $A=\Bbbk[z]$ that corresponds to $I$,
\cite{Fr82}. The map $\lambda $ that sends a monomial of $I$ to a
squarefree monomial in $I_{pol}$ is an isomorphism of lattices and
of GCD graphs. Moreover $Q_R(y,t)$ can be obtained from $Q_A(z,t)$
by applying $\lambda^{-1}$ to the monomial coefficients of $t$,
\cite{Ba82}.
\end{remark}

Theorem \ref{denomisom} now completes the multigraded version of
Theorem 1, \cite{Av02}.

\begin{theorem}\label{denomisom} Let $S=\Bbbk[x]$ and $S'=\Bbbk[x']$ be
polynomial rings over a field $\Bbbk$,  and $I\subset S$,
$I'\subset S'$ be ideals generated by monomials of degree at least
2. Let $\lambda: L_I\mapright{} L_{I'}$ and the induced $\lambda:
G_I\mapright{} G_{I'}$ be isomorphisms.  Suppose that  $Q_R(y,
t)=1+\sum (\sum c_{\boldsymbol{j}} y^{\boldsymbol{j}}) t^i$ then $Q_{R'}(y',t)= 1+\sum
(\sum c_{\boldsymbol{j}} y^{\lambda(\boldsymbol{j})}) t^i$.
\end{theorem}

\begin{proof}
By Proposition \ref{lcmcoefficients} the multidegrees $\boldsymbol{j}$ in
$Q_R(y,t)$ are in $L_I$, so $\lambda(\boldsymbol{j})$ makes sense. By the
above remark it is enough to examine the case where $I$ and $I'$
are both squarefree.  The key point is that since $I$ and $I'$ are
squarefree, $\lambda$ is  additive: $\lambda(\bf j+\bf
i)=\lambda(\bf j)+\lambda(\bf i)$. It follows that its extension
$\hat\lambda$ is also additive. Since
$$Q_R(y,t)={ \prod_{i=1}^n(1+ty_i)\over P_R(y,t)}=
 {{\prod_{i=1}^\infty \prod_{\boldsymbol{j}}(1-y^{\boldsymbol{j}} t^{2i})^{\rank\pi^{2i}_{- \boldsymbol{j}}}  }
\over{\prod_{i=2}^\infty \prod_{\boldsymbol{j}}(1+y^{\boldsymbol{j}} t^{2i-1})^{\rank
\pi^{2i-1}_{-\boldsymbol{j}}} }}
$$
the linear coefficient $c_{\boldsymbol{j}}$ of the monomial $ y^{\boldsymbol{j}}t^n$ of
$Q_R(y,t)$, depends on the deviations $\epsilon_{i,\boldsymbol{f}}$ when $i\le n$.
The corresponding combination of the deviations
$\epsilon'_{i,\hat\lambda(\boldsymbol{f})}$ gives a term of $Q_{R'}(y',t)$ of
$t$-degree $n$. The theorem now follows.
\end{proof}

The condition that $\lambda$ is an isomorphism of GCD graphs is a
necessary condition as the example from \cite{Ga00} shows. The
isomorphism of LCM lattices for $I= (x_1^2, x_2^2x_3)$ and
$I'=(x_1x_2^2, x_1x_3^2)$ is not an isomorphism of GCD graphs and
$$Q_{S/I}(y,t)=1-t^2(y_1^2-y_2^2y_3)+t^4(y_1^2y_2^2y_3)$$
while
$$Q_{S/I'}(y,t)=1-t^2(y_1y_2^2+y_1y_3^2)-t^3(y_1y_2^2y_3^2).$$


\section{Introductory remarks about hyperbolicity}

In this section we define Brody and Kobayashi hyperbolicity and state
their basic properties. We also give examples of hyperbolic and
non-hyperbolic manifolds. The reader can refer to
\cite{lang:hyperbolic} and \cite{demailly:hyperbolic} for more
details.\footnote{After these notes were written, I discovered that O.
  Debarre also has some very nice notes on hyperbolicity
  \cite{debarre:hyperbolicity}. The reader is encouraged to consult
  these notes for additional information on hyperbolic varieties,
  especially about those varieties with ample cotangent bundle.}  Let
$\mathbb{C}$ denote the complex plane. We use $X$ and $Y$ to denote complex
manifolds. We reserve $C$ for curves.  \smallskip

\noindent {\bf Observation:} If $C$ is a curve of genus at least two,
then any holomorphic map $f: \mathbb{C} \rightarrow C$ is necessarily
constant.  \smallskip

\noindent {\bf Proof:}  Any map $f:\mathbb{C} \rightarrow C$ factors through
the universal cover of $C$, which is the unit disc. Such a map is
constant since by Liouville's Theorem every bounded entire function is
constant.  $\Box$ \smallskip

In contrast, curves of genus zero and one do admit non-constant
holomorphic maps from $\mathbb{C}$. This property of higher genus curves can
be generalized to higher dimensional manifolds and leads to the
concept of Brody hyperbolicity.

\begin{definition}
  A complex manifold $X$ is {\bf Brody hyperbolic} if there are no
  non-constant holomorphic maps from $\mathbb{C}$ to $X$.
\end{definition}
{\bf Examples:} $\bullet$ Any variety of the form $\prod_{i=1}^n C_i$,
where $C_i$ are curves of genus at least two, is Brody
hyperbolic. More generally, a finite product of Brody hyperbolic
manifolds is Brody hyperbolic.
 \smallskip

 $\bullet$ If a complex manifold $X$ contains a rational curve or a
 complex torus, then $X$ is not Brody hyperbolic. If $X$ contains
 a rational curve, then the inclusion of $\mathbb{C}$ in the rational curve
 followed by the inclusion of the rational curve in $X$ gives a
 non-constant holomorphic map from $\mathbb{C}$ to $X$.

 The universal cover of a $g$-dimensional complex torus is $\mathbb{C}^g$.
 Taking the image of a general complex line in $\mathbb{C}^g$ under the
 quotient map gives a non-constant holomorphic map from $\mathbb{C}$ into the
 complex torus. If $X$ contains a complex torus, composing the
 previous map with the inclusion of the complex torus in $X$ gives a
 non-constant holomorphic map from $\mathbb{C}$ into $X$. \smallskip

 $\bullet$ The blow-up of a variety is not Brody hyperbolic.
 Consequently, Brody hyperbolicity is not a birational invariant.
 \smallskip

 $\bullet$ Consider $X = \P^2 - \cup_{i=1}^4 l_i$ where $l_i$ are
 general lines.  Then $X$ is not Brody hyperbolic since there are
 $\mathbb{C}^*$'s in $X$---take the intersection of $X$ with the line passing
 through $l_1 \cap l_2$ and $l_3 \cap l_4$.  \smallskip

\begin{figure}[htbp]
\begin{center}
\epsfig{figure=lines.eps}
\end{center}
\caption{A Brody hyperbolic manifold.}
\label{lines}
\end{figure}

$\bullet$ Consider $Y = \P^2 - \cup_{i=1}^5 l_i$ where $l_i$ are the
lines pictured in Figure \ref{lines}.  $Y$ is Brody hyperbolic.
Consider the pencil of lines in $\P^2$ based at one of the points
where three of the lines $l_i$ intersect. Restricting this pencil to
$Y$, we see that $Y$ is fibered over $\P^1$ punctured at three points
with fibers isomorphic to $\P^1$ punctured at three points. Composing
any  holomorphic map from $\mathbb{C}$ with the map to the base of the
fibration, we conclude that the image of the map must lie in a fiber.
Since the fibers are Brody hyperbolic, the map must be constant.
Using the following easy lemma, which generalizes the idea of this
example, one can generate many examples of Brody hyperbolic manifolds.

\begin{lemma}\label{fibration}
  Let $f: X \rightarrow Y$ be a smooth morphism of complex manifolds,
  where $Y$ is Brody hyperbolic and $f^{-1} (y)$ is Brody hyperbolic
  for every $y \in Y$. Then $X$ is Brody hyperbolic.
\end{lemma}

The Schwarz-Pick Lemma  states that a holomorphic
map between two hyperbolic Riemann surfaces is either a local isometry
or distance decreasing. The hyperbolic distance between any two points
$p$ and $q$ on a hyperbolic Riemann surface $C$ is equal to the
shortest hyperbolic distance between lifts of the points $p$ and $q$
to the universal cover, the unit disc $\Delta$. Any holomorphic map
$f: \Delta \rightarrow C$ factors through the universal cover.
Consequently, the Schwarz-Pick Lemma implies that the infimum of the
hyperbolic distances between $p'$ and $q'$ in $\Delta$ for which there
exists a holomorphic map $f: \Delta \rightarrow C$ such that $f(p')=p$
and $f(q') = q$ is achieved when $f$ is the universal covering
map. Moreover, this infimum
is equal to the hyperbolic distance between $p$ and $q$. In this form
we can generalize the hyperbolic distance to higher dimensional
manifolds. \smallskip

Let $\Delta_r$ denote the disc of radius $r$ in the complex plane.  We
will denote the unit disc by $\Delta$.  If $X$ is a complex manifold,
then $X$ can be endowed with a pseudo-distance due to Kobayashi.

\begin{definition}
  Given a tangent vector $\xi \in T_{X,x}$ at $x \in X$ we define its
  {\it Kobayashi pseudo-norm} to be
  $$k(\xi) := \inf_{\lambda} \{ \lambda : \exists \ f : \Delta
  \rightarrow X, \ f(0) = x, \ \lambda f'(0) = \xi \}.$$
  The {\it
    Kobayashi pseudo-distance} $d_{X}$ is the geodesic pseudo-distance
  obtained by integrating this pseudo-norm.
\end{definition}

\begin{figure}[htbp]
\begin{center}
\epsfig{figure=kob.eps}
\end{center}
\caption{Measuring the Kobayashi distance.}
\label{kob}
\end{figure}

More visually, we can describe how to compute the distance between any
two points $p,q \in X$ as follows. We find chains of maps $f_i :
\Delta \rightarrow X$ with two distinguished points $p_i, q_i$, where
$f_1(p_1) = p$, $f_n(q_n)= q$ and $f_i(q_i) = f_{i+1}(p_{i+1})$. The
Kobayashi pseudo-distance is the infimum of the sum of the hyperbolic
distances between $p_i$ and $q_i$ over all such chains. \smallskip

The Schwarz-Pick Lemma generalizes to holomorphic maps between higher
dimensional manifolds endowed with the Kobayashi pseudo-distance.

\begin{lemma}\label{Pick}
  If $f: X \rightarrow Y$ is a holomorphic map between two complex
  manifolds, then $d_X(p,q) \geq d_Y (f(p), f(q))$.
\end{lemma}
Unfortunately, $d_X$ does not have to be non-degenerate. \smallskip

\noindent {\bf Main Counterexample:} The Kobayashi pseudo-distance on
$\mathbb{C}$ is identically zero. To compute the distance between two points
$x$ and $y$ in $\mathbb{C}$,  consider for each integer $n>1$ the functions
$f_n(z)= n(y-x)z + x$ from
the unit disc to $\mathbb{C}$. The function $f_n(z)$ maps 0
to $x$ and $1/n$ to $y$. Since the hyperbolic distance between $0$ and
$1/n$ in the unit disc tends to zero as $n$ tends to infinity, we
conclude that the Kobayashi pseudo-distance between $x$ and $y$ in
$\mathbb{C}$ is zero. \smallskip

\begin{definition}
  A complex manifold $X$ is called {\it Kobayashi hyperbolic} if its
  Kobayashi pseudo-distance is non-degenerate.
\end{definition}

\noindent {\bf Kobayashi hyperbolicity implies Brody hyperbolicity. }
  As a consequence of Lemma \ref{Pick} and the above example we see that
if a complex manifold $X$ admits a non-constant holomorphic map from
$\mathbb{C}$, then $X$ cannot be Kobayashi hyperbolic. In other words,
Kobayashi hyperbolicity implies Brody hyperbolicity.  For compact
complex manifolds the converse of this result is true
(\cite{brody:hyperbolic}).

\begin{theorem}[Brody]\label{Br=Ko}
  A compact complex manifold $X$ is Kobayashi hyperbolic if and only
  if it is Brody hyperbolic.
\end{theorem}

\noindent{\bf Sketch of proof:} If $X$ is not Kobayashi hyperbolic, then there
exists a sequence of maps $f_n : \Delta \rightarrow X$ such that
$|df_n (0 )|$ tends to $\infty$, where $| \cdot |$ denotes a fixed
Hermitian metric on $X$. By rescaling the map we can assume that we
have a sequence of maps $f_n : \Delta_{r_n} \rightarrow X$ where
$|df_n(0)| = 1$ and the radii $r_n$ tend to $\infty$. \smallskip

Let $f:\Delta_r \rightarrow X$ be a holomorphic map with $|df(0)| =
c$. Let $f_t(z) = f(tz)$. Consider the function $$g(t)= \sup_{z \in
  \Delta_r} |df_t (z)|.$$
The function $g(t)$ is  increasing
for $0 \leq t \leq 1$ and continuous for $0 \leq t <1$. Moreover, as
$t$ tends to 1 from below $g(t)$ tends to $g(1)$.  Hence there exists
$t$ in the interval $[0,1]$ and an automorphism $h$ of
$\Delta_r$ such that $$\sup_{z \in \Delta_r} |d(f_t \circ h)
(z)| = |d(f_t \circ h) (0)| = c .$$
This is known as the Brody
reparametrization lemma. \smallskip

Applying the Brody reparametrization lemma to our family of maps $f_n$, we
 obtain
 a new family of
maps $\tilde{f}_n :\Delta_{r_n} \rightarrow X$ whose derivatives are bounded
in norm by 1 in $\Delta_{r_n}$. Moreover, $|d \tilde{f}_n (0)| = 1$
 for every $n$. We would like to show that we can select a subsequence
 from this family that is uniformly convergent on every compact subset
 of $\mathbb{C}$. Since the derivatives at $0$ all have norm 1, then the
 family converges to a non-constant holomorphic map from $\mathbb{C}$ to
 $X$. This violates Brody hyperbolicity. \smallskip

Using the compactness of $X$ we can check uniform convergence in a
neighborhood of every point $z_0$ in $\mathbb{C}$. Again using the compactness
of $X$ (and passing to a subsequence if necessary) we can assume the
family converges at $z_0$. In this situation the derivative bounds
imply that the  family forms an equicontinuous family of
holomorphic maps around $z_0$.  We can conclude that a subsequence
converges uniformly on compact sets by Ascoli's theorem. We thus
obtain a holomorphic map from $\mathbb{C}$ to $X$. $\Box$

\smallskip

Note that Brody's theorem does not have to hold for non-compact
manifolds (see \cite{brody:thesis}).  Consider the domain in $\mathbb{C}^2$
given by
$$
D : = \{ (z,w) \ | \ |z|<1, \ |zw|<1 \ \mbox{and} \ |w| < 1 \
\mbox{if} \ z = 0 \}. $$
The projection of the domain $D$ to the first
coordinate gives rise to a family of discs parameterized by the unit
disc in the $z$-plane.  Consequently, Lemma \ref{fibration} implies
that $D$ is Brody hyperbolic. On the other hand, when $z$ approaches
zero, the radius of the disc lying over $z$ tends to infinity. Using
this we can show that the Kobayashi pseudo-distance between $p =
(0,0)$ and $q =(0,w_0)$ in $D$ vanishes.  We can find two points
$p'$ and $q'$ having the same $w$ coordinates as $p$ and $q$,
respectively, in a fiber close to zero. Since $p'$ and $q'$ are in a
very large disc, the hyperbolic distance between them is very small.
Since the hyperbolic distances between $p$ and $p'$ and $q$ and $q'$
are very small, the sum of the hyperbolic distances can be made
arbitrarily small. \smallskip

More explicitly, consider  the three maps
$$f_1 (z) = (z,0), \ f_2(z) = \left( \frac{1}{n}, nz \right) , \ f_3
(z) = \left( \frac{1}{n} + \frac{1}{2} z, w_0 \right)$$
from the unit
disc into $D$. Note that $f_1(0)= (0,0)$, $f_1(1/n)=(1/n,0)$, $f_2(0)=
(1/n,0)$, $f_2(w_0/n) = (1/n, w_0)$ and $f_3(0)= (1/n,w_0)$,
$f_3(-2/n)= (0,w_0)$.  Hence these maps form a chain of maps from the
unit disc to $D$ connecting $(0,0)$ to $(0,w_0)$. As $n$ tends to
infinity, the sum of the hyperbolic distances between the chosen
points in the unit disc tends to zero. We conclude that the Kobayashi
pseudo-distance between $(0,0)$ and $(0,w_0)$ is zero. Observe that
this example also shows that the analogue of Lemma \ref{fibration}
does not hold for Kobayashi hyperbolicity.  \smallskip

Under suitable hypotheses on a subvariety $Y$ of $X$, we can assert
that $X-Y$, the complement of $Y$ in $X$, is Kobayashi hyperbolic. For
example, the following theorem is often useful in proving that the
complement of a subvariety is Kobayashi hyperbolic.

\begin{theorem}
  Let $X$ be a compact variety and let $Y$ be a proper algebraic subset. If
  $Y$ and $X - Y$ are Brody hyperbolic, then $X-Y$ is Kobayashi
  hyperbolic.
\end{theorem}

\noindent {\bf Sketch of proof:} If $X-Y$ is not Kobayashi hyperbolic,
we can get a sequence of maps $f_n: \Delta_{r_n} \rightarrow X-Y$ that
converges to a holomorphic map $g$ from $\mathbb{C}$ to $X$.  The question is
whether the image of $g$ intersects $Y$.  The image cannot be
contained in $Y$ since $Y$ is Brody hyperbolic.

If we rule out that the image of $g$ intersects $Y$, then we obtain a
contradiction since $X-Y$ is Brody hyperbolic. One can prove that the
intersection points of the image of $g$ with $Y$ have to be isolated.
Suppose $g(z_0) \in Y$. Take a circle $S$ around $z_0$ such that $g(S)
\subset X-Y$. The winding number is zero because $f_n(\Delta_{r_n})$
does not meet $Y$. Hence $Y$ cannot intersect the image of the
interior of $S$, contradicting that $g(z_0) \in Y$. $\Box$ \smallskip

We now discuss some examples of Kobayashi hyperbolic
manifolds. From now on whenever we say hyperbolic without further
qualification, we will always mean Kobayashi hyperbolic.  The
following example due to Green (\cite{green:hyperplane}) gives the
first non-trivial examples.

\begin{theorem}[Green]
  The complement of $2n+1$ general hyperplanes in $\P^n$ is
  hyperbolic.
\end{theorem}

Generalizing further one can ask whether the complement of a very
general irreducible hypersurface in $\P^n$ of large enough degree is
hyperbolic.  Siu and Yeung answer this question positively in $\P^2$
(\cite{siuyeung:complement}). Here and in the following by `very
general' we will refer to the complement of the union of countably
many proper subvarieties.

\begin{theorem}[Siu-Yeung]\label{curve}
  Let $C$ be a very general curve of sufficiently large degree
  in $\P^2$. Then $\P^2 - C$ is Kobayashi hyperbolic.
\end{theorem}

Note that this theorem can be interpreted as a generalization of
Picard's theorem to higher dimensions. Recall that Picard's theorem
says that any entire map which omits two values is constant. This
theorem implies that any holomorphic map from $\mathbb{C}$ into $\P^n$ which
omits a (very general) hypersurface of a large enough degree is constant.
\smallskip

Siu and Yeung also prove the analogue of this result for the
complement of very general ample divisors in abelian varieties
(\cite{siuyeung:4} \cite{siuyeung:one}, \cite{siuyeung:2}
\cite{siuyeung:3}).

\begin{theorem}[Siu-Yeung]
  Let $A$ be an abelian variety. Let $D$ be a very general ample
  divisor in $A$. Then $A - D$ is Kobayashi hyperbolic.
\end{theorem}

\noindent  There is a close relation between the hyperbolicity of
complements of divisors in projective space and the hyperbolicity of
general hypersurfaces in projective space. We have the following
theorem due to Siu (\cite{siu:hyperbolic}, \cite{siu:hypersurface}).

\begin{theorem}[Siu]\label{siuhyp}
  A very general surface of sufficiently high degree in $\P^3$ is
  Kobayashi hyperbolic.
\end{theorem}

\noindent In fact, Siu recently has generalized Theorems \ref{curve} and
\ref{siuhyp} to hypersurfaces in $\P^n$. Not all the proofs
have appeared in print.

\begin{theorem}[Siu]
  Let $X$ be a very general hypersurface of sufficiently large degree
  in $\P^n$. Then $\P^n - X$ is Kobayashi hyperbolic.
\end{theorem}

\begin{theorem}[Siu]\label{higherdim}
  A very general hypersurface of sufficiently high degree in $\P^n$ is
  Kobayashi hyperbolic.
\end{theorem}

\noindent For surfaces in $\P^3$ the term `sufficiently high degree'
can be taken to mean at least 11. Conjecturally a very general
hypersurface of degree larger than $2n$ in $\P^n$ is expected to be
hyperbolic.  However, the current known bounds are much larger than the
expected bounds.  \smallskip

Finally, we observe that some properties of the tangent or cotangent
bundle of a complex manifold forces the manifold to be hyperbolic. For
example, if $X$ admits a metric with negative sectional curvature; or
if $X$ has ample cotangent bundle; or if $X$ is the quotient of a
bounded domain in $\mathbb{C}^n$ by a free group action, then $X$ is Kobayashi
hyperbolic. \smallskip

As an amusing corollary we note that $M_g^0$, the moduli space of
automorphism-free, smooth curves of genus $g > 2$, is Kobayashi
hyperbolic since the Weil-Petersson metric is a metric on $M_g^0$ with
negative sectional curvature.  Consequently, $M_g^0$ does not contain
any complete rational or elliptic curves. It would be interesting to
give lower bounds on the genus of complete curves contained in $M_g^0$.

\section{The geometry of Kobayashi hyperbolicity}

In this section we discuss some proven and conjectural geometric
characterizations of hyperbolicity. Hyperbolicity imposes strong
restrictions on the geometry of a variety. In particular it constrains
the type of subvarieties the variety can have.

\begin{proposition}
  Let $X$ be a compact complex hyperbolic manifold with Hermitian
  metric $\omega$. Then $\exists  \ \epsilon > 0$ such that for any
  reduced, irreducible curve $C \subset X$, the genus $g$ of the
  normalization satisfies $$g \geq \epsilon \deg_{\omega} (C)$$
\end{proposition}

\noindent {\bf Sketch of proof:} If $X$ is hyperbolic, then there
exists a constant such that $d_X(\xi) \geq \epsilon_0 || \xi
||_{\omega}$ for every tangent vector $\xi$. Let $\nu: C^{\nu}
\rightarrow C$ be the normalization of a curve $C \subset X$. If we
denote the hyperbolic metric on $C^{\nu}$ by $k_{C^{\nu}}$, then the
Gauss-Bonnet formula implies that $$-\frac{1}{4} \int_{C^{\nu}}
\mbox{curv}(k_{C^{\nu}}) = -\frac{\pi}{2} \chi(C^{\nu}).$$
There is a
natural holomorphic map $i \circ \nu: C^{\nu} \rightarrow X$ obtained
by composing the normalization map $\nu$ from $C^{\nu}$ to $C$ by the
inclusion $i$ of $C$ in $X$. Since the Kobayashi distance can only
decrease under compositions of holomorphic maps, we conclude that
$$k_{C^{\nu}}(\xi) \geq \epsilon_0 ||(i \circ \nu)_*
(\xi)||_{\omega}$$
for any tangent vector $\xi \in T_{C^{\nu}}$.
Integrating both sides of the inequality yields the proposition. $\Box$
\smallskip

Lang has conjectured that the converse also holds.  Let $X$ be a
compact, complex manifold endowed with a Hermitian metric.  Lang
conjectures that if there exists a positive constant $\epsilon$ such
that for every reduced, irreducible curve $C$ in $X$ the ratio of the
genus of the normalization of $C$ to the degree of $C$ (with respect
to the Hermitian metric) is bounded below by $\epsilon$, then $X$ is
hyperbolic. \smallskip

If the geometric genus of every curve in a variety $X$ is bounded
below by some fixed positive multiple of their degree, then $X$ does not
admit any non-constant holomorphic maps from any abelian variety. More
generally, Lang has conjectured that a projective variety $X$ is
hyperbolic if and only if it does not admit any holomorphic maps from
an abelian variety. The latter conjecture, of course, implies the
former one for projective varieties. \smallskip

One can ask for the relation between varieties of general type and
hyperbolic varieties. Since varieties of general type can
contain rationally connected subvarieties or abelian subvarieties, an
arbitrary variety of general type cannot be hyperbolic. However, Lang
has conjectured that the existence of subvarieties which are not of
general type accounts for the failure of hyperbolicity.

\begin{conjecture}\label{generaltype}
  A projective algebraic variety $X$ is hyperbolic if and only if
  every subvariety of $X$ is of general type.
\end{conjecture}
Ein (\cite{ein:generaltype}) and later Voisin (\cite{voisin:clemens})
have shown that any subvariety of a very general hypersurface of
degree at least $2n+1$ in $\P^n$ is of general type. Combining their
theorem with this conjecture one obtains a conjectural sharp form of
Siu's Theorem \ref{higherdim}. \smallskip

Proving that a projective variety $X$ is hyperbolic usually has two
components.  One has to show that there are no non-algebraic maps of
$\mathbb{C}$ into $X$ and that there are no rational or elliptic curves in
$X$. When the problem is broken into these two components, then some
progress can be made on each component of the problem under some
geometric restrictions. We now survey some of the results on these
questions. \smallskip

One of the first people to make important progress on hyperbolicity
questions was the French mathematician Andr\'e Bloch, even though at
the time hyperbolicity was not defined
(see \cite{bloch:hyperbolicity}). Bloch begins by determining the
Zariski closure of an entire map into a complex torus.

\begin{theorem}\label{abelian}
  The Zariski closure of a holomorphic map $\mathbb{C} \rightarrow T$ to a
  complex torus $T$ is the translate of a subtorus of $T$.
\end{theorem}
This theorem leads to a fairly complete understanding of the
hyperbolic subvarieties of complex tori.

\begin{corollary}
  A subvariety $X$ of an abelian variety $A$ which does not contain
  any translates of subtori of $A$ is hyperbolic.
\end{corollary}
Originally Bloch used these ideas to prove the following theorem often
referred to as Bloch's Theorem (\cite{bloch:hyperbolicity}).

\begin{theorem}[Bloch's Theorem]
  Any holomorphic map of $\mathbb{C}$ into a smooth, compact K\"ahler variety $X$
  whose irregularity ($h^0(X, \Omega_X^1)$) is bigger than its
  dimension is analytically degenerate, i.e. its image lies in a
  proper analytic subvariety.
\end{theorem}
{\bf Proof:} Recall that given a smooth, compact K\"ahler variety $X$, we
can associate to it a complex torus $\mbox{Alb}(X)$ of dimension
$h^0(X, \Omega_X^1)$ and a map $a: X \rightarrow \mbox{Alb}(X)$ with
the following universal property: for any complex torus $T$ and any
morphism $f: X \rightarrow T$, there exists a unique morphism $g:
\mbox{Alb}(X) \rightarrow T$ such that $f= g \circ a$. The complex
torus $\mbox{Alb}(X)$ is referred as the {\it Albanese}  variety of
$X$ and the
map $a$ is called the {\it Albanese map}.  $\mbox{Alb}(X)$ is
unique up to isomorphism. \smallskip

Bloch's theorem follows by considering the Albanese map. Let $f: \mathbb{C}
\rightarrow X$ be a holomorphic map. Consider $a \circ f: \mathbb{C}
\rightarrow \mbox{Alb}(X)$.  Since the irregularity of $X$ is larger
than the dimension of $X$, the image of $X$ under the Albanese map
lies in a proper subvariety of $\mbox{Alb}(X)$.  By the universality
of the Albanese variety and its uniqueness, the Albanese image of $X$
cannot be the translate of a subtorus. Hence the image of $\mathbb{C}$ under
the map $a \circ f$ in the Albanese variety has to be analytically
degenerate in the image of $X$.  It follows that the image of the
original map $f: \mathbb{C} \rightarrow X$ has to be analytically degenerate.
$\Box$ \smallskip

Using these ideas one also obtains information about the hyperbolicity
of complements of ample divisors in abelian varieties.
\begin{theorem}\label{abeliancomplement}
  Let $A$ be an abelian variety. Let $D$ be an ample divisor that
  does not contain any translates of  abelian subvarieties.  Then
  $A-D$ is hyperbolic.
\end{theorem}

Recall that a holomorphic map from $\mathbb{C}$ into an algebraic variety is
called algebraically degenerate if the Zariski closure of the image
lies in a proper algebraic subvariety.  More recently McQuillan
(\cite{mcquillan:foliation}) has proved the algebraic degeneracy of
entire maps into surfaces of general type with $c_1^2 > c_2$.  More
precisely,

\begin{theorem}[McQuillan]
  If $X$ is a surface of general type satisfying $c_1^2 > c_2$, then
  all entire curves on $X$ are algebraically degenerate. In
  particular, if $X$ does not contain any rational or elliptic curves,
  then $X$ is hyperbolic.
\end{theorem}
McQuillan obtains this result by first proving a result about the
algebraic degeneracy of leaves of certain foliations on surfaces of
general type, then showing that under the assumptions on $X$ there
always is a foliation which contains the image of any entire map in
one of its leaves. \smallskip

Most ways of proving the algebraic degeneracy of entire maps into
complex projective manifolds depend on producing enough differential
relations on the manifold that every entire map has to satisfy. One
then shows that these relations have a small base locus. This forces
the holomorphic map to lie in this base locus. One often formalizes
these ideas in terms of jet bundles (see \cite{demailly:hyphyp},
\cite{demailly:hyperbolic} or any of Siu's papers cited above).
\smallskip

In the other direction, there has been extensive work on showing that
certain varieties do not contain any rational or elliptic curves. Here
we mention the work of G. Xu bounding below the geometric genus of any curve
on a very general surface of degree at least 5 in $\P^3$ (\cite{xu:bound},
see also Clemens' paper \cite{clemens:bound}).

\begin{theorem}[Xu]
On a very general surface of degree $d$ in $\P^3$, the geometric genus of
any curve is greater than or equal to $d(d-3)/2 - 2$. Tritangent plane
sections achieve this bound. For $d \geq 6$ the tritangent plane
sections are the only curves that achieve the bound.
\end{theorem}
A very simple consequence of the theorem is that a general
hypersurface of degree at least 5 in $\P^3$ contains no rational or
elliptic curves. Thus the problem of showing the hyperbolicity of a
very general hypersurface in $\P^3$ of degree at least 5 reduces to
showing that any entire map into such a surface is algebraically
degenerate. \smallskip

Motivated by our discussion one can ask the following question:
\smallskip

\noindent {\bf Question:} Can the closure of the image of a
holomorphic map $\mathbb{C} \rightarrow \P^n$ be a variety of general type?
\smallskip

\noindent A negative answer to this question would imply one direction
of Conjecture \ref{generaltype}. If all the subvarieties of a compact
variety are of general type, then the variety would have to be
hyperbolic. At present this question seems very hard to answer.
\smallskip

We close this section with a discussion of how hyperbolicity varies in
families. Let $X$ be a compact $C^{\infty}$ manifold with Hermitian
metric $| \ |$. Brody proved that if one considers the various complex
structures that one can put on $X$, the set of those that are
hyperbolic is open in the analytic topology. More precisely, one can
consider the function
$$D(s) = \sup_{f \in \mbox{Hol} (\Delta, X_s)} |f'(0)|$$ on the moduli
spaces of complex structures parameterized by $S$. Brody proves

\begin{theorem}
  $D(s)$ is a continuous function. Since the hyperbolic complex
  structures correspond to those for which $D(s)$ is finite, the
  hyperbolic complex structures form an open set in the analytic
  topology.
\end{theorem}
Note that if one considers an algebraic family of quasiprojective
smooth varieties, then Brody's theorem may fail. In fact in
such a family (at least if we do not fix the $C^{\infty}$ type) the
set of fibers
that are Brody  hyperbolic can be Zariski locally closed. Varying two of the
lines that meet a third at the same point in Figure \ref{lines}
provides such an example. \smallskip

Brody's theorem naturally raises the following question. \smallskip

\noindent {\bf Question:} Is hyperbolicity Zariski open in algebraic
families of projective varieties? \smallskip

\noindent  This question, to the best of my knowledge, is still
open. A positive answer would have interesting geometric implications.
For example, it is not known whether the space of high degree surfaces
in $\P^3$ that contain rational and elliptic curves form an algebraic
variety or a countable union of subvarieties of the space of surfaces.
A positive answer to the Zariski openness of hyperbolicity would
settle this and similar questions.

\section{The arithmetic of Kobayashi hyperbolicity}

In this section we summarize some conjectures, mainly due to Lang,
about rational points on hyperbolic varieties. We also give some
surprising consequences of the conjectures about the distribution of
rational points on curves. The main references for this section are
\cite{lang:hyperbolic}, \cite{caporaso:lang},
\cite{capjoemazur:points} and \cite{capjoemazur:lang}.

The arithmetic of curves of genus two or more and those of genus one
and zero exhibit drastically different properties. If $C$ is a curve
of genus zero or one defined over a number field $K$, then after
passing to a finite field extension $L$ there are infinitely many
$L$-rational points on $C$. In fact, $L$ can be chosen so that these
points are dense in the analytic topology. The Mordell-Faltings
Theorem stands in sharp contrast to this result.

\begin{theorem}[Mordell-Faltings]
  Let $C$ be a smooth curve of genus $g \geq 2$ defined over a number
  field $K$. Then $C$ has only finitely many rational points over any
  finite field extension $L$ of $K$.
\end{theorem}

\begin{definition} We say a variety $V$ defined over a number field
  $K$ is of {\it Mordell type} if $V$ contains only finitely many
  rational points over any finite field extension of $K$.
\end{definition}

It is natural to ask which, if any, of the geometric properties we
described so far imply that a variety is of Mordell type.  Clearly a
variety of Mordell type does not contain any rational or elliptic
curves. Similarly any map from an abelian variety to a variety of
Mordell type has to be constant. In view of these observations it is
not unreasonable to formulate the following conjecture due to Lang.

\begin{conjecture}\label{mordellic}
  A complex projective variety $V$ defined over a number field is of
  Mordell type if and only if it is hyperbolic.
\end{conjecture}

In fact Lang has conjectured much more precise statements.

\begin{conjecture}[weak form]\label{weak}
  If $X$ is a variety of general type defined over a number field $K$,
  then the set of $K$-rational points of $X$ is not Zariski dense.
\end{conjecture}
The conjecture that a hyperbolic manifold is of Mordell type follows
from Conjecture \ref{weak} and the geometric Conjecture
\ref{generaltype} by the following argument.  If a hyperbolic manifold
$X$ has infinitely many rational points, then the Zariski closure of
these points is a subvariety not of general type by Conjecture
\ref{weak}. By Conjecture \ref{generaltype} every subvariety of $X$ is
of general type. Lang has conjectured an even stronger statement.

\begin{conjecture}[strong form]\label{strong}
  If $X$ is of general type, then there exists a proper algebraic
  subset $Z$ of $X$, such that over any finite extension $L$ of $K$,
  the number of $L$-rational points of $X-Z$ is finite.
\end{conjecture}

The Lang conjectures are currently open except for the case of curves
and more generally for subvarieties of abelian varieties. If true,
they would provide a fundamental understanding of the arithmetic of
varieties of general type.  \smallskip

The Lang Conjectures \ref{weak} and \ref{strong} have some surprising
consequences for rational points on curves of genus at least two.
These consequences were investigated by Caporaso, Harris and Mazur in
the papers cited above.

\begin{theorem}
  The weak form of Lang's conjecture implies that for every number
  field $K$ and genus $g \geq 2$, there exists a constant $B(K,g)$
  such that any curve of genus $g$ defined over $K$ has at most
  $B(K,g)$ $K$-rational points.
\end{theorem}

The strong form of Lang's conjecture implies an even more surprising
bound.
\begin{theorem}
  The strong Lang conjecture implies that for any $g \geq 2$ there
  exists an integer $N(g)$ such that there are only finitely many
  curves defined over a number field $K$ that have more than $N(g)$
  $K$-rational points.
\end{theorem}

\noindent {\bf Ideas behind the proof:} These theorems depend on the
  following geometric theorem Caporaso, Harris and Mazur prove.

\begin{theorem}(Correlation) \label{correlation}
  Let $f:X \rightarrow B$ be a proper morphism of irreducible and
  reduced schemes whose general fiber is a smooth curve of genus at
  least two. Then for $n$ sufficiently large the fiber product $h:
  X_B^n \dashrightarrow W$ admits a dominant rational map to a variety
  of general type. Moreover, if $f:X \rightarrow B$ is defined over a
  number field $K$, $h$ and $W$ can also be defined over $K$.
\end{theorem}
D. Abramovich proved a generalization of the Correlation Theorem for
morphisms with higher dimensional fibers of general type under
suitable hypotheses (see \cite{abramovich:correlation}). However, in
order to deduce the bounds on the rational points on curves, we only
need the case of curves. \smallskip

Caporaso, Harris and Mazur deduce the bounds on rational points on
curves by applying the Lang conjectures to a global family. More
precisely, they start with a family of curves $f : X \rightarrow B$
such that for every curve $C$ defined over $K$, there exists a
$K$-rational point on $B$ such that the fiber over it is isomorphic to
$C$ over $K$. \smallskip

Using Theorem \ref{correlation}, $X_B^n$, the $n-$th fiber product of
$X$ over $B$, admits a rational map to a variety $W$ of general type.
By Lang's conjecture there exists a smallest closed, proper subvariety
$V$ of $X_B^n$ that contains all the $K$-rational points. \smallskip

By studying the successive images of the complement of $V$ under the
projections to various factors of $X_B^n$, they produce a non-empty
open subset $U$ of $B$ and an integer $N$ such that for every rational
point $b \in U$ the fiber over $b$ has at most $N$ points. Then by
Noetherian induction on the complement of $U$, they conclude the
uniform bound. $\Box$

\bibliographystyle{math}


\section{Introduction}

Let $p$ be a prime, and let $RG(p)$ denote the set of equivalence
classes of radically graded finite dimensional quasi-Hopf algebras
over $\mathbb C$, whose radical has codimension $p$. In
\cite{eg1},\cite{eg2} we completely describe the set $RG(p)$.
Namely, we show that for $p>2$, $RG(p)$ consists of the quasi-Hopf
algebras $A(q)$ constructed in \cite{g} for each primitive root of
unity $q$ of order $p^2$, the Andruskiewitsch-Schneider Hopf
algebras \cite{as1}, and semisimple quasi-Hopf algebras $H_+(p)$
and $H_-(p)$ of dimension $p$; on the other hand, $RG(2)$ consists
of the Nichols Hopf algebras of dimension $2^n$, $n\ge 1$, and the
special quasi-Hopf algebras $H(2)$, $H_\pm(8)=A(\pm i)$, and
$H(32)$ (of dimensions 2,8,32) constructed in \cite{eg1}.

The purpose of this paper is to classify finite dimensional quasi-Hopf
algebras $A$ whose radical is a quasi-Hopf ideal and
has codimension $p$; that is, $A$ with $\text{gr} A\in RG(p)$,
where $\text{gr} A$ is the associated graded algebra
taken with respect to the radical filtration on $A$.
The main result of this paper is the following theorem.

\begin{theorem}\label{main}
Let $A$ be a finite dimensional quasi-Hopf algebra whose radical
is a quasi-Hopf ideal of
prime codimension $p$. Then either $A$ is twist equivalent to a
Hopf algebra, or it is twist equivalent to $H(2)$, $H_\pm(p)$,
$A(q)$, or $H(32)$.
\end{theorem}

Note that if $A$ is a Hopf algebra (up to twist) then for $p=2$
$A$ is a Nichols Hopf algebra of dimension $2^n$, while for $p>2$
it is the dual to a finite dimensional pointed Hopf algebra with
group of grouplike elements $\Bbb Z_p$. Such Hopf algebras are
completely classified for $p\ne 5,7$ in \cite{as2} (see Remark
1.10(v), \cite{mu}).

Note also that any finite tensor category whose simple objects are
invertible and form a group of order $p$ under tensor is the
representation category of a quasi-Hopf algebra $A$ as above. Thus
this paper provides a classification of such categories.

The organization of the paper is as follows. Section 2 is devoted
to preliminaries. In Section 3 we introduce a construction
of semidirect products for quasi-Hopf algebras, which is used in
the sequel. In Section 4 we consider
liftings of the quasi-Hopf algebras $A(q)$ from [G],
and of $H(32)$. We prove:

\begin{theorem}\label{main1}
Let ${\rm gr}(A)$ be $A(q)$ or $H(32)$.
Then $A$ is equivalent to ${\rm gr}(A)$ as a quasi-Hopf algebra.
\end{theorem}

In Section 5 we consider liftings of Andruskiewitsch-Schneider
Hopf algebras \cite{as1}. We prove:

\begin{theorem}\label{main2}
Let ${\rm gr}(A)$ be an Andruskiewitsch-Schneider Hopf algebra.
Then $A$ is twist equivalent to a Hopf algebra.
\end{theorem}

By \cite{eg1}, \cite{eg2}, Theorem \ref{main} follows from
Theorems \ref{main1} -- \ref{main2}.

{\bf Acknowledgments.} The research of the first author was
partially supported by the NSF grant DMS-9988796.
The second author was supported by Technion V.P.R.
Fund - Dent Charitable Trust - Non Military Research Fund, and by The Israel
Science Foundation (grant No. 70/02-1). He also thanks MIT for its
warm hospitality.
Both authors were supported by BSF grant No. 2002040.

\section{Preliminaries}

All constructions in this paper are done over the field of complex
numbers $\mathbb{C}$.

We refer the reader to \cite{d} for the definition of a quasi-Hopf
algebra and a twist of a quasi-Hopf algebra.

\subsection{The quasi-Hopf algebras $A(q)$}
\begin{theorem} \cite{g}\label{propp3}
Let $n\ge 2$ be an integer.

\noindent
 There exist $n^3-$dimensional quasi-Hopf algebras $A(q)$, parametrized
by primitive roots of unity $q$ of order $n^2$,
which
have the following structure. As algebras $A(q)$ are generated
by $a,x$ with the relations $ax=q^n xa$, $a^n=1$, $x^{n^2}=0$.
The element $a$
is grouplike, while the coproduct of $x$ is given by the formula
\begin{equation*}
\Delta(x)=x\otimes \sum_{y=0}^{n-1}q ^y {\bf 1}_y + 1\otimes (1-{\bf
1}_{0})x + a^{-1}\otimes {\bf 1}_{0}x,
\end{equation*}
where $\{\mathbf{1}_i|0\le
i\le n-1\}$ is the set of primitive idempotents of ${\mathbb C}[a]$
defined by the condition $a{\bf 1}_i=q ^{ni}{\bf 1}_i$. The associator
is
$$\Phi:=\sum_{i,j,k=0}^{n-1}q^{\frac{-i(j+k-(j+k)')}{n}}{\bf 1}_i\otimes
{\bf 1}_j\otimes {\bf 1}_k,$$ the distinguished
elements are $\alpha=a$,
$\beta=1$, and the antipode is $S(a)=a^{-1}$,
$S(x)=-x\sum_{z=0}^{n-1}q ^{n-z}{\bf 1}_z$.
\end{theorem}

Note that $S^2(x)=q^{n+1}x$, $S^2(a)=a$, so $S^{2n}=\text{Ad}(a)$.

\subsection{The quasi-Hopf algebra $H(32)$}
\begin{theorem} \cite{eg1} \label{32}
There exists a $32-$ dimensional quasi-Hopf algebra $H(32)$, which
has the following structure. As an algebra $H(32)$ is generated by
$a,x,y$ with the relations $ax=-xa$, $ay=-ya$, $a^2=1$, $x^4=0$,
$y^4=0$, $xy+iyx=0$. The element $a$ is grouplike, while the
coproducts of $x,y$ are given by the formulas $$\Delta(x)=x\otimes
(p_+ + ip_-) +1\otimes p_+x + a\otimes p_-x,$$
$$\Delta(y)=y\otimes (p_+ - ip_-) +1\otimes p_+y + a\otimes
p_-y,$$ where $p_+:=(1+ a)/2, p_-:=(1-a)/2$. The associator is
$\Phi=1-2p_-\otimes p_-\otimes p_-$, the distinguished elements
are $\alpha=a$, $\beta=1$, and the antipode is $S(a)=a$,
$S(x)=-x(p_+ + ip_-)$, $S(y)=-y(p_+-ip_-)$. Thus, $H(32)$ is
generated by its quasi-Hopf subalgebras $A(i)$ and $A(-i)$
generated by $a,x$ and $a,y$, respectively.
\end{theorem}

Note that $S^2(x)=ix$ and $S^2(y)=-iy$, so $S^4=\text{Ad}(a)$.

\subsection{Andruskiewitsch-Schneider Hopf algebras}

In \cite{as1}, Theorem 1.3, Andruskiewitsch and Schneider
completely classified finite dimensional coradically graded Hopf
algebras with coradical of prime dimension $p\ge 3$. Since this
collection of Hopf algebras is obviously self-dual, the list of
Theorem 1.3 of \cite{as1} also classifies radically graded Hopf
algebras with radical of prime codimension $p\ge 3$.

Let us recall this list.

1. Quantum lines (Taft Hopf algebras) of dimension $p^2$ (type
$A_1$). They are generated by $a,x$ where
$$
a^p=1,\ x^p=0,\ ax=qxa,
$$
and
$$
\Delta(a)=a\otimes a, \Delta(x)=x\otimes a+1\otimes x.
$$
Here $q$ is a primitive root of unity of degree $p$.

2. Quantum planes (book Hopf algebras) of dimension $p^3$
(type $A_1\times A_1$).
They are generated by $a,x,y$ where
$$
a^p=1,\ x^p=0,\ y^p=0,\ xy-yx=0,\ ax=qxa,\ ay=q^mya
$$
and
$$
\Delta(a)=a\otimes a, \Delta(x)=x\otimes a+1\otimes x,
\Delta(y)=y\otimes 1+a^m\otimes y.
$$
Here $q$ is primitive a root of unity of degree $p$,
and $m\in \Bbb Z_p$ is a nonzero element.

3. Algebras of type $A_2,B_2,G_2$. They are generated by $a,x,y$
satisfying the relations $axa^{-1}=qx$ and $aya^{-1}=q^by$, where
$b\in {\mathbb Z}_p$ is a nonzero element, and additional Serre relations
coming from the Nichols algebra ${\mathfrak u}_q^+$ (of type
$A_2,B_2$, or $G_2$, respectively) generated by $x,y$. The element
$a$ is grouplike, while the coproducts of $x,y$ are given by the
formulas
$$\Delta(x)=x\otimes a+1\otimes x,\;\Delta(y)=y\otimes a^d+1\otimes y,$$ where
$0\ne d\in {\mathbb Z}_p$, $b+d=-1$, and the following conditions hold:

1) The case $A_2$: $p=3$, or $p=1$ modulo 3, and $d^2+d+1=0$.

2) The case $B_2$: $p=1$ modulo 4, and $2d^2+2d+1=0$.

3) The case $G_2$: $p=1$ modulo 3, and $3d^2+3d+1=0$.

Here, as before, $q$ is a primitive root of unity of degree $p$.

4. Algebras of type $A_2\times A_1$ ($p=3$). They are generated by
$a,x,y$ as above (with $b=d=1$), and a new element $z$ such that
$aza^{-1}=q^fz$, $$\Delta(z)=z\otimes a^{-f}+1\otimes z,$$ where
$f=1$ or $2$ (subject to appropriate Serre relations).

5. Algebras of type $A_2\times A_2$ ($p=3$). They are generated by
$a,x,y$ as above (with $b=d=1$), and new elements $x',y'$, such
that $ax'a^{-1}=q^fx'$, $ay'a^{-1}=q^{f}y'$,
$$\Delta(x')=x'\otimes a^{-f}+1\otimes x',\;\Delta(y')=y'\otimes
a^{-f}+1\otimes y',$$ where $f=1$ or $2$ (subject to appropriate
Serre relations).

In all cases the Hopf algebra is isomorphic
${\mathbb C}[{\mathbb Z}_p]{\triangleright\!\!\!<} {\mathfrak u}_q^+$ as an {\em algebra}
(where ${\mathfrak u}_q^+$ is the corresponding Nichols algebra).

\subsection{Deformations}

Let $A$ be a finite dimensional quasi-Hopf algebra, in which the
radical is a quasi-Hopf ideal. Let $A_0$ be the associated graded
quasi-Hopf algebra of $A$ under the radical filtration. The
algebra $A_0$ has a grading $A_0=\bigoplus_{i=0}^{N} A_0[i]$ by
nonnegative integers. Thus we have gradings on the spaces of
linear maps between tensor powers of $A_0$. If $f_0$ is such a map
of some degree $d$ and $f$ is obtained from $f_0$ by adding terms
of degree higher than $d$, we will write $f=f_0+{\rm hdt}$ (where
hdt stands for higher degree terms) and say that $f$ is a
deformation (or lifting) of $f_0$. \footnote{More generally, if
$f_1,...,f_n$ have degrees $d_1,...,d_n$ then $f_1+...+f_n$ + hdt
will mean the sum of $f_i$ plus a sum of corrections of degrees
higher than all $d_i$.} Thus we can talk about deformations of
algebra, coalgebra structures, associator, antipode, etc. In
particular, we may say that $A$ is a deformation of $A_0$, in the
sense that $A$ is identified with $A_0$ as a graded vector space,
so that the structure maps of $A$ are given by some deformations
of the structure maps of $A_0$. This is the way we will think
about $A$ in the considerations below.

\subsection{Quasi-Hopf liftings of Hopf algebras}

\begin{proposition} \label{qhl}
Let $A$ be a finite dimensional quasi-Hopf algebra whose
radical is a quasi-Hopf ideal, such that $A_0={\rm gr}(A)$ is a
Hopf algebra (i.e., the associator of $A_0$ is equal to $1$). If
$H^3(A_0^*,\Bbb C)=0$ then $A$ is twist equivalent to a Hopf
algebra.
\end{proposition}

\begin{proof}
Let $\Phi$ be the associator of $A$.
We have $\Phi=1+$ hdt. Write $\Phi=1+\gamma +$ hdt, where $\gamma$
is of degree $d$. Then $\gamma$ is a $3-$cocycle of $A_0^*$ with
coefficients in the trivial $A_0^*-$bimodule ${\mathbb C}$. Since
$H^3(A_0^*,\Bbb C)=0$, we have $\gamma=dj$, where $j\in
A_0^{\otimes 2}[d]$, and the twisted associator $\Phi^{1+j}$ is
equal to $1+$ terms of degree $\ge d+1$. By continuing this
procedure, we will come to a situation where $\Phi=1$, as desired.
\end{proof}

\section{Semidirect products}

Let $H$ be a quasi-Hopf algebra with associator $\Phi$,
 $g:H\to H$ an algebra automorphism.
Define $\Delta_{g}(h):=(g\otimes g)(\Delta(g^{-1}(h)))$.
Assume that
\begin{equation}\label{autg}
\Delta_{g}(h)=
K\Delta(h)K^{-1},\;h\in H,
\end{equation}
for some invertible element $K\in H\otimes H$ such that
$(\varepsilon\otimes 1)(K)=(1\otimes \varepsilon)(K)=1$, and
$$
g^{\otimes 3}
(\Phi)=\Phi^K:=(1\otimes K)(id\otimes \Delta)(K)\Phi(\Delta\otimes
id)(K^{-1})(K^{-1}\otimes 1).
$$

\begin{lemma} The semidirect product algebra ${\mathbb C}[g,g^{-1}]
{\triangleright\!\!\!<} H$ is a quasi-Hopf algebra with associator $\Phi$,
$\Delta(g)=K^{-1}(g\otimes g)$ and $\varepsilon(g)=1$.
\end{lemma}

\begin{proof}
Using equation (\ref{autg}), we see that
there exists a well defined coproduct on the algebra
${\mathbb C}[g,g^{-1}]{\triangleright\!\!\!<} H$ which extends the coproduct on $H$
and such that $\Delta(g)=K^{-1}(g\otimes g)$.
It is straightforward to verify that $(\Delta\otimes id)\Delta(g)=
\Phi^{-1}(id\otimes \Delta)\Delta(g)\Phi$.
It is also easy to show that the antipode extends to the semidirect
product. The lemma is proved.
\end{proof}

Suppose further that $g^n(h)=aha^{-1}$ for some $a\in H$,
and that
\begin{equation}\label{jg}
(g^{n-1})^{\otimes 2}(K)\dots g^{\otimes 2}(K)K =(a\otimes a)\Delta(a)^{-1}.
\end{equation}

\begin{lemma}\label{sd} The ideal generated by $g^n-a$ is a quasi-Hopf
ideal, so the quotient $\tilde H:=
({\mathbb C}[g,g^{-1}]{\triangleright\!\!\!<} H)/<g^n-a>$ is a quasi-Hopf algebra.
\end{lemma}

\begin{proof}
Using condition (\ref{jg}) it is straightforward to see that
\begin{eqnarray*}
\lefteqn{\Delta(g^n-a)}\\
& = &
\Delta(a)(a^{-1}g^n\otimes a^{-1}g^n -1)
\\
& = & \Delta(a)(a^{-1}(g^n-a)\otimes a^{-1}g^n +
1\otimes a^{-1}(g^n-a)).
\end{eqnarray*}
It is also straightforward to verify that $<g^n-a>$ is invariant under the
antipode. Thus $<g^n-a>$ is a quasi-Hopf ideal.
\end{proof}

\section{Proof of Theorem \ref{main1}}

Let $A_0=A(q)$, where $q$ is a
primitive
root of unity of order $n^2$,  or $A_0=H(32)$ (in which case we
set $n=2$). Let $A$ be a lifting of $A_0$.
We apply the semidirect product construction to the
automorphism $g:=S^2$ of the algebra $A$.

Proposition 1.2 of \cite{d} provides a twist $K$ such that $\Delta_{g}(y)=
K\Delta(y)K^{-1}$, for all $y\in A$. Let $K_0$ be the degree zero
part of $K$. Then $K_0$ commutes with $\Delta_0(y)$ for any $y\in
A_0$, where $\Delta_0$ is the comultiplication of $A_0$.
This implies, using straightforward computations, that $K_0=1$,
and hence $K=1+$ hdt.

\begin{lemma}
The automorphism $S^{2n}$ of $A$ is inner.
\end{lemma}

\begin{proof}
Let $G:={\rm Aut}(A)/{\rm Inn}(A)$ be the group of outer algebra
automorphisms of $A$; it is a linear algebraic group. Let $S_0$ be
the antipode of $A_0$. Then $S_0^{2n}={\rm Ad}(a)$. Hence the
class $[S^{2n}]$ of $S^{2n}$ is unipotent in $G$. On the other
hand, from the main result of \cite{eno} (see also \cite{hn}) we
know that on the category of representations of $A$ there exists
an isomorphism of tensor functors $\theta_V:V\to \chi\otimes V^{****}\otimes
\chi^{-1}$, where $\chi$ is a $1-$dimensional representation of
$A$. In our case, $\chi$ has order $n$. Therefore there exists a
tensor isomorphism $c_V:V\to V^{**^{2n}}$, and thus $S^{4n}$ is an
inner automorphism of $A$: $S^{4n}(z)=czc^{-1}$ where
$c\in A$ is the element realizing the automorphism $c_V$ (i.e., $c_V=c|_V$).
Hence, $[S^{2n}]=1$ in $G$ and $S^{2n}$
is inner.
\end{proof}

Thus, $S^{2n}(z)=b zb^{-1}$ for some element $b\in
A$, where $b=a+$ hdt is a deformation of the grouplike
element $a\in A_0$.

\begin{lemma}\label{choi} The element $b$ can be chosen so that
\begin{equation}\label{jg1}
K^{-1}g^{\otimes 2}(K^{-1})\dots (g^{n-1})^{\otimes
2}(K^{-1})=\Delta(b)(b^{-1}\otimes b^{-1}).
\end{equation}
\end{lemma}

\begin{proof} Denote the left hand side of (\ref{jg1}) by $T$.

Let $c$ be the element defined above such that $S^{4n}(z)=czc^{-1}$.
It is easy to check that $c=a^2+$ hdt.

We claim that
$$
T(b\otimes b)T(b^{-1}\otimes b^{-1})=\Delta(c)(c^{-1}\otimes c^{-1}).
$$
Indeed, since $\{c_V\}$ is an isomorphism of tensor functors, $L(c_V\otimes c_W)=
c_{V\otimes W}$, where $L$ is the tensor structure on the functor
$**^{2n}$. This defines an element $L\in A\otimes A$ such that $L=\Delta(c)
(c^{-1}\otimes c^{-1})$. However, we know from \cite{d} that the tensor structure
on $**$ is given by $K^{-1}$, and hence the tensor structure on $**^{2n}$
is given by $K^{-1}(S^2\otimes S^2)(K^{-1})\cdots (S^{4n-2}\otimes S^{4n-2})(K^{-1})$,
which is equal to $T(S^{2n}\otimes S^{2n})(T)=T\text{Ad}(b\otimes b)(T)$, as desired.

Thus we have that $z:=b^2c^{-1}=1+$ hdt is a central element.
Replacing $b$ with $bz^{-1/2}$, we may assume without
loss of generality that $b$ is chosen to satisfy the condition
$z=1$, i.e. $b^2=c$.

We claim that for this choice of $b$,
equation (\ref{jg1}) holds.
Indeed, assume the contrary,
and let the lowest degree of nontrivial terms in the difference
between the two sides of
(\ref{jg1}) be $d$. Let $\gamma\ne 0$ be the part of the difference that is
homogeneous of degree $d$; so $T=\Delta(b)(b^{-1}\otimes
b^{-1})-\gamma$ modulo elements of degree $\ge d+1$.
Then modulo elements of degree $\ge
d+1$, we have for any $y\in A$,
\begin{eqnarray*}
\lefteqn{\Delta_{{\rm Ad}(b)}(y)}\\ & = &
(b\otimes b)\Delta(b^{-1}yb)(b^{-1}\otimes b^{-1})=T^{-1}\Delta(y)T\\
& = & ((b\otimes b)\Delta(b)^{-1}+\gamma)
\Delta(y)(\Delta(b)(b^{-1}\otimes b^{-1})-\gamma).
\end{eqnarray*}
This means that for any $y\in A_0$ one has
$[\gamma,\Delta_0(y)]=0$, and in particular
$[\gamma,a\otimes a]=0$.

Also, modulo terms of degree $\ge d+1$,
$$
(\Delta(b)(b^{-1}\otimes b^{-1})-\gamma)(b\otimes
b)(\Delta(b)(b^{-1}\otimes b^{-1})-\gamma)
(b^{-1}\otimes b^{-1})=\Delta(c)(c^{-1}\otimes c^{-1}).
$$
Since $b^2=c$, this implies that
$$
\gamma+(a\otimes a)\gamma(a^{-1}\otimes a^{-1})=0.
$$
Since $[\gamma,a\otimes a]=0$, this is equivalent to
$2\gamma=0$. Contradiction.
\end{proof}

Lemma \ref{sd} and Lemma \ref{choi} imply that we can define the
semidirect product quasi-Hopf algebra $$\tilde
H:=({\mathbb C}[g,g^{-1}]{\triangleright\!\!\!<} A)/<g^n -b>;$$ it is of dimension $n^4$
for $A=A(q)$, and 64 for $A=H(32)$.

Let $J\in \tilde H\otimes \tilde H$ be a deformation of the twist
$J_0$ from \cite{g}. Then it follows from \cite{g} that
$H:={\tilde H}^ {J^{-1}}$ is a quasi-Hopf lifting (=deformation)
of a Hopf algebra $H_0$. If $A_0=A(q)$ then $H_0$ is the Taft Hopf
algebra of dimension $n^4$. On the other hand, if $A_0=H(32)$ then
$H_0$ is a book Hopf algebra of dimension $64$, generated by
$g,x_+,x_-$ satisfying $gx_+g^{-1}=ix_+$, $gx_-g^{-1}=-ix_-$,
$x_+x_-=x_-x_+$.  The element $g$ is grouplike and
$$\Delta(x_+)=x_+\otimes g+1\otimes x_+,\;\Delta(x_-)=x_-\otimes 1+g^{-1}\otimes x_-.$$

\begin{theorem}\label{taft}
If $H$ is a basic quasi-Hopf algebra such that $H_0=\text{gr} H$
then there exists a quasi-Hopf twist $F$
such that $H\cong H_0^F$ and $F=1$ modulo higher terms.
\end{theorem}

\begin{proof}
Let us first show that $H$ is twist equivalent to a Hopf algebra.
By Proposition \ref{qhl}, for this it is sufficient to show that
$H^3(H_0^*,\Bbb C)=0$. If $H_0$ is the Taft algebra then
$H_0^*\cong H_0$ and one finds (see  \cite{gk}), that
$H^\bullet(H_0^*,{\mathbb C})$ is the polynomial algebra on an element of
degree 2. In particular, $H^3(H_0,{\mathbb C})=0$, as required. On the
other hand, if $H_0$ is the book algebra of dimension $64$ then
$H_0^*$ is generated by $g,x,y$ with $g^4=1$, $x^4=y^4=xy-yx=0$,
$gxg^{-1}=ix$, $gyg^{-1}=iy$. From this it is easy to obtain using
K\"unneth formula that $H^\bullet(H_0^*,\Bbb C)$ is the polynomial
algebra in two elements of degree 2 (see \cite{gk}). In
particular, again $H^3(H_0^*,\Bbb C)=0$.

Now, it is easy to see, using the
methods of \cite{as2}, that in both cases
(the Taft algebra and the book algebra),
$H_0$ has no non-trivial Hopf liftings,
so there exists a twist $F=1+$ hdt and an isomorphism
of quasi-Hopf algebras $\eta:H_0^{F}\to H$, as desired.
\footnote{We note that if $H_0$ is the book algebra of
dimension 64, then $H_0^*$, unlike $H_0$,
has a nontrivial Hopf lifting, which
is the dual of the quantum $sl_2$ at the 4-th root of unity.}
\end{proof}

Replacing $J$ with $JF$, we may assume that $F=1$. Thus we find
that $\tilde H=H_0^{J}$. So we have a homomorphism of quasi-Hopf
algebras
\begin{equation}\label{eta}
\eta: A\to H_0^{J}.
\end{equation}

The map $\eta$ may be regarded as an injective linear map $A_0\to
H_0^J$, whose image is a quasi-Hopf subalgebra of $H_0^J$, and
whose degree $0$ part is the ``tautological'' homomorphism
$\eta_0: A_0\to H_0^{J_0}$\footnote{Note the difference between
the map $\eta_0: A_0\to H_0$ and the map $A_0\to H_0^*$ (not
$H_0$!) constructed in \cite{eg1}.}. Let $E$ be the set of pairs
$(\eta,J)$ satisfying this condition. On $E$ there is an action of
the ``gauge group'' ${\mathcal G}$, generated by the following
three types of transformations.

1. $(\eta,J)\to (\eta\circ \xi,J)$, where $\xi=1$+hdt is a linear
automorphism of $A_0$.

2. $(\eta,J)\to ({\rm Ad}(h)\circ \eta,(h\otimes
h)J\Delta(h)^{-1})$,
where $h=1+$hdt is an element of $H_0$, and $\Delta$ is the
coproduct of $H_0$.

3. $(\eta,J)\to (\eta,TJ)$, where $T=1$+hdt belongs to
$\eta(A_0)^{\otimes 2}$.

It is clear that the transformations from ${\mathcal G}$ do not
change the equivalence class of the quasi-Hopf algebra
$\eta(A_0)$. Thus, Theorem \ref{main1}
follows from the following proposition.

\begin{proposition}
Any pair $(\eta,J)\in E$ is equivalent under
${\mathcal G}$ to $(\eta_0,J_0)$.
\end{proposition}

\begin{proof} It suffices to prove that for any $d>0$,
any pair $(\eta,J)\in E$ can be brought by the action of
${\mathcal G}$ to a pair equal to $(\eta_0,J_0)$ modulo terms of
degree $\ge d$. We will prove it by induction in $d$. The base of
induction is obvious. To prove the induction step, we need to take
a pair $(\eta,J)$ which is equal to $(\eta_0,J_0)$ modulo terms of
degree $d$, and show it is equivalent to a pair equal to
$(\eta_0,J_0)$ modulo terms of degree $d+1$.

Thus, the multiplication in $A_0$ obtained by pullback of the
multiplication of $H_0$ by $\eta$
is given by $z*w=zw+c_d(z,w)$+hdt, where $zw$ is the
multiplication in $A_0$, and $c_d$ is a bilinear operator
$A_0\times A_0\to A_0$ of degree $d$, which is a Hochschild 2-cocycle.
Similarly, $\eta(z)=\eta_0(z)+\eta_d(z)+$ hdt, where $\eta_d$ is a degree $d$
operator $A_0\to H_0$, and $J=TJ_0$, $T=1+t_d+$ hdt, where
$t_d\in H_0\otimes H_0$ has degree $d$.

We have $\eta(z*w)=\eta(z)\eta(w)$ which is equivalent to
$$(\eta_0+\eta_d+\cdots)(zw+c_d(z,w)+\cdots)=(\eta_0+\eta_d+\cdots)
(z)(\eta_0+\eta_d+\cdots)(w).$$ Taking $z$ of degree $d_z$ and $w$
of degree $d_w$, and looking at terms of degree $d+d_z+d_w$ we get
$\eta_d(zw)+\eta_0(c_d(z,w))=\eta_d(z)\eta_0(w)+\eta_0(z)\eta_d(w)$,
which implies $\eta_0(c_d(z,w))=\eta_d(z)\eta_0(w)+\eta_0(z)\eta_d(w)-\eta_d(zw)$.
Therefore, $\eta_0(c_d)$ is the differential of a 1-cochain $\eta_d\in
C^1(A_0,H_0)$. So $\eta_0$ sends the
cohomology class of $c_d\in H^2(A_0,A_0)$ into 0 in
$H^2(A_0,H_0)$. Since $A_0$ is a direct summand in $H_0$ as an
$A_0$-bimodule, this implies that the class $[c_d]\in H^2(A_0,A_0)$ is
zero. Thus, $c_d$ is the differential of some element $\tau$ of degree
$d$ in $C^1(A_0,A_0)$. So, using an element $g_1\in {\mathcal G}$ of
type $1$, we can replace $(\eta,J)$ with an equivalent pair for
which $c_d=0$. Thus in the sequel we may assume that $c_d=0$.

Now, $\eta_d$ is a Hochschild 1-cocycle of $A_0$ with coefficients
in $H_0$. Consider the cohomology class $[\eta_d]$ in
$H^1(A_0,H_0)$. We claim that in fact, $[\eta_d]$ belongs
to the subspace $H^1(A_0,A_0)\subset H^1(A_0,H_0)$. To prove
this, note that since $\eta(A_0)$ is closed under coproduct,
for any $z\in A_0$ we have
$$
\Delta_0(\eta_d(z))-(\eta_d\otimes \eta_0+\eta_0\otimes
\eta_d)(\Delta_0(z))=[\Delta_0(z),t]\; {\rm mod }\;A_0\otimes A_0,
$$
where $\Delta_0$ denotes the coproduct of $H_0^{J_0}$ and
$A_0$, and $t:=t_d$.

Now recall that $H_0=\oplus_{k=0}^{n-1}A_0g^k$.
Thus, $\eta_d=\sum_{k=0}^{n-1} \omega_k$, where
$\omega_k\in Z^1(A_0,A_0g^k)$ (a 1-cocycle). Similarly,
$t=\sum_{k,l=0}^{n-1} t_{kl}$, $t_{kl}\in A_0g^k\otimes A_0g^l$.
Thus for $k\ne 0$ we have
$$
-(\omega_k\otimes \eta_0)(\Delta_0(z))=[\Delta_0(z),t_{k0}]
$$
Applying the counit in the second component, we get that
$\omega_k(z)=[z,y_k]$
for some $y_k\in A_0g^k$, which implies
that $\omega_k$ is an inner derivation.

Thus $[\eta_d]\in
H^1(A_0,A_0)$, and hence we could have chosen $g_1$ so that
$[\eta_d]=0$. So we may assume $[\eta_d]=0$ and
thus $\eta_d$ is the bracketing operator with an element $y\in H_0$.
Thus using an appropriate element $g_2\in {\mathcal G}$ of type
2, we can assume in the sequel that $\eta_d=0$.

Now we will show that in the situation when $c_d=0$ and
$\eta_d=0$, we must have $t\in A_0\otimes A_0$. Then by using an
element $g_3\in {\mathcal G}$ of type 3, we can come to a
situation with $t=0$ which completes the induction step.

Since the associator $\Phi:=\Phi_0^T$ lies in
$\eta(A_0)^{\otimes 3}$, we find that
\begin{equation}\label{dt}
t\otimes 1 + (\Delta_0\otimes id)(t)-\Phi_0 (id\otimes \Delta_0)(t)
\Phi_0^{-1}+\Phi_0 (1\otimes t) \Phi_0^{-1}\in A_0^{\otimes 3}.
\end{equation}
Let $H_0=A_0\oplus H'$, where $H'$ is the direct sum of $A_0g^k$ for
$k=1,...,n-1$.
Let $t'$ be the projection
of $t$ to $H'\otimes H_0$. Since the first component of
$t-t'$ is in $A_0$, we get from (\ref{dt}):
\begin{equation}\label{dt'}
t'\otimes 1 + (\Delta_0\otimes id)(t')-\Phi_0 (id\otimes \Delta_0)(t')
\Phi_0^{-1}+\Phi_0 (1\otimes t') \Phi_0^{-1}\in A_0\otimes H_0\otimes H_0.
\end{equation}
But out of the four summands in this expression,
all but the last one belong to $H'\otimes H_0\otimes H_0$, while the
last summand is in $A_0\otimes H_0\otimes H_0$.
Hence the last summand is zero and $t'=0$.
Similarly, one shows that the projection $t''$ of $t$ to
$H_0\otimes H'$ is zero. Hence, $t\in A_0\otimes A_0$, as
desired.
\end{proof}

The theorem is proved.

\section{Proof of theorem \ref{main2}}

Let $H$ be a Hopf algebra
from the list of Subsection 2.3.
Since this list is self-dual, Theorem \ref{main2}
follows from Proposition \ref{qhl} and the following proposition.

\begin{proposition} We have $H^3(H,{\mathbb C})=0$.
\end{proposition}

\begin{proof}
By a standard argument, $H^3(H,{\mathbb C})=H^3({\mathbb C}[{\mathbb Z}_p]{\triangleright\!\!\!<} {\mathfrak u}_q^+,{\mathbb C})=
H^3({\mathfrak u}_q^+,{\mathbb C})^{{\mathbb Z}_p}$.

Let us first assume that $p>3$. In this case, according to Section
2.3, the rank of the corresponding simple Lie algebra is $\le 2$,
and $p$ is greater than its Coxeter number. Thus we can apply
\cite{gk}, Theorem 2.5. In this theorem, it is shown that
$H^\bullet({\mathfrak u}_q^+,{\mathbb C}) =\bigoplus_{w\in W}{\mathbb C} {\eta_w}\otimes
S({\frak n}_+)$, where $S({\frak n}_+)$ is the symmetric algebra
of the positive part of the associated Lie algebra, sitting in
degree 2, $W$ is the Weyl group, and $\eta_{w}$ has degree $l(w)$
(=length of $w$) .

The action of the generator
$a$ of ${\mathbb Z}_p$ on $H^\bullet({\mathfrak u}_q^+,{\mathbb C})$ is given as follows: $a$ acts
trivially on $S({\frak n}_+)$ and $a\circ \eta_{w}=\lambda_w\eta_w$, where
in the $A_1$-case $\lambda_w=q^{-1}$ for the nontrivial element
$w\in W$, and in the other (rank 2) cases
$\lambda_w=q^{-(m+nd)}$, where $m,n$ are determined by writing
the weight $\gamma_w:=
\sum_{\alpha\in R^+,\;w(\alpha)<0}\alpha=\rho-w(\rho)$ as an
integral combination $m\alpha_1+n\alpha_2$ of the simple roots (here
$\rho$ is half the sum of positive roots, and $R_+$ is the set of
positive roots; in the non-simply laced case,
$\alpha_2$ always denotes the short root).

We claim that $H^3({\mathfrak u}_q^+,{\mathbb C})^{{\mathbb Z}_p}=0$. Indeed, since ${\mathbb Z}_p$ acts
trivially on $S({\frak n}_+)$, and since $\lambda_w\ne 1$ if $w$
is a simple reflection, we have $H^3({\mathfrak u}_q^+,{\mathbb C})^{{\mathbb Z}_p}=
(\bigoplus_{w\in W,\;l(w)=3}{\mathbb C}\eta_w)^{{\mathbb Z}_p}$. We will now show that
in each of the four cases $A_1\times A_1$, $A_2$, $B_2$ and $G_2$,
$\lambda_w\ne 1$, and hence $H^3({\mathfrak u}_q^+,{\mathbb C})^{{\mathbb Z}_p}=0$.

In the case $A_1\times A_1$ there are no elements $w$ of length 3, so the result
is clear.

In the case $A_2$, there is one element $w$ of length 3 and one computes
that $\gamma_w=2\alpha_1+2\alpha_2$. Therefore $\lambda_w=(q^{-1})^{2+2d}$.
Since $d^2+d+1=0$, $d+1\ne 0$, so $\lambda_w\ne 1$.

In the case $B_2$, there are two elements $w_1,w_2$ of length 3 and one computes
that $\gamma_{w_1}=3\alpha_1+3\alpha_2$, $\gamma_{w_2}=2\alpha_1+4\alpha_2$.
Therefore $\lambda_{w_1}=(q^{-1})^{3+3d}$ and $\lambda_{w_2}=(q^{-1})^{2+4d}$.
Since $2d^2+2d+1=0$, we have $d+1,2d+1\ne 0$, so $\lambda_{w_1},\lambda_{w_2}\ne 1$.

In the case $G_2$, there are two elements $w_1,w_2$ of length 3 and one computes
that $\gamma_{w_1}=4\alpha_1+4\alpha_2$, $\gamma_{w_2}=2\alpha_1+6\alpha_2$.
Therefore $\lambda_{w_1}=(q^{-1})^{4+4d}$ and $\lambda_{w_2}=(q^{-1})^{2+6d}$.
Since $3d^2+3d+1=0$, we have $d+1,3d+1\ne 0$, so $\lambda_{w_1},\lambda_{w_2}\ne 1$.
Thus the result for $p>3$ follows.

Finally, let us consider the case $p=3$. In the rank 1 case the
above argument applies, so we consider rank $\ge 2$. Then the
argument above does not quite apply, since the rank of the Lie
algebra can now be bigger than 2, and $p$ may be equal to the
Coxeter number (a case not covered by Theorem 2.5 in \cite{gk}).
Thus we will use a slightly different argument. By \cite{gk}, the
Nichols algebra ${\frak u}_q^+$ has a filtration under which
$H^\bullet (\text{gr} {\mathfrak u}_q^+,{\mathbb C})=\Lambda_q\otimes S({\frak
n}_+)$, where $\Lambda_q$ is the $q-$analogue of the exterior
algebra generated by $\varepsilon_{\alpha}$, $\alpha$ a positive
root, such that
$\varepsilon_{\alpha}\varepsilon_{\beta}+q^{<\alpha,\beta>}
\varepsilon_{\beta}\varepsilon_{\alpha}=0$ for $\beta<\alpha$ and
$\varepsilon_{\alpha}^2=0$ for any $\alpha$.

The generator $a$ of ${\mathbb Z}_p$ acts trivially on $S({\frak n}_+)$ and
by $a\varepsilon_{\alpha} a^{-1}=q^{-(m+nd)}\varepsilon_{\alpha}$,
where $\alpha=m\alpha_1+n\alpha_2$. The group $G$ of grouplikes in
the Frobenius-Lusztig quantum group acts on both $H^\bullet(\text{gr}
{\mathfrak u}_q^+,{\mathbb C})$ and $H^\bullet({\mathfrak u}_q^+,{\mathbb C})$, and
the second one is isomorphic to a subrepresentation of the first
one by a spectral sequence argument \cite{gk}.

In the case $A_2$ ($d=1$) we claim that $H^3(\text{gr} {\mathfrak u}_q^+,{\mathbb C})^{{\mathbb Z}_3}=0$.
Indeed, we have
$H^3(\text{gr} {\mathfrak u}_q^+,{\mathbb C})=<\varepsilon_{\alpha_1}\varepsilon_{\alpha_2}
\varepsilon_{\alpha_1+\alpha_2}>$, and the generator $a$ of ${\mathbb Z}_p$ acts on this $1-$
dimensional space by $q^{-(1+d+1+d)}=q^{-4}\ne 1$.

In the case $A_2\times A_1$ ($d=1$), we have that the Nichols
algebra
${\frak u}_q^+$ is generated
by $x,y,z$ with $aza^{-1}=q^fz$. Thus, if $\beta$ is the simple
root corresponding to $z$, we have,
$H^3(\text{gr} {\mathfrak u}_q^+,{\mathbb C})^{{\mathbb Z}_3}
=<\varepsilon_{\beta}\varepsilon_{\alpha_1}\varepsilon_{\alpha_2}>$ if $f=2$, and is
zero if $f=1$.

In the case $A_2\times A_2$ ($d=1$), we have that the Nichols
algebra ${\frak u}_q^+$ is generated by $x,y,x',y'$ with
$axa^{-1}=qx$, $aya^{-1}=qy$, $ax'a^{-1}=q^fx'$ and
$ay'a^{-1}=q^fy'$. Hence, if $\beta_1,\beta_2$ are the roots
attached to $x',y'$, we find that $$H^3(\text{gr} {\mathfrak
u}_q^+,{\mathbb C})^{{\mathbb Z}_3}
=<\varepsilon_{\beta_i}\varepsilon_{\alpha_1}\varepsilon_{\alpha_2},
\varepsilon_{\alpha_i}\varepsilon_{\beta_1}\varepsilon_{\beta_2}|i=1,2>$$
if $f=1$, and is zero if $f=2$.

But even in cases where $H^3(\text{gr} {\mathfrak u}_q^+,{\mathbb C})^{{\mathbb Z}_3}\ne
0$, this cohomology is killed in the spectral sequence $H^\bullet(\text{gr}
{\mathfrak u}_q^+,{\mathbb C})^{{\mathbb Z}_3}\to H^\bullet({\mathfrak u}_q^+,{\mathbb C})^{{\mathbb Z}_3}$
since by \cite{gk}, the only weights of terms under $G$ which can
survive in this sequence must be of the form $w(\rho)-\rho$ while
$-(\alpha_1+\alpha_2)\ne w(\rho)-\rho$ modulo 3. So in all cases
$H^3({\mathfrak u}_q^+,{\mathbb C})^{{\mathbb Z}_3}=0$, and the result follows.
\end{proof}