text:
\section{Introduction}\label{sec1}

Contempoary representation theory uses a wide range of homological
methods. This makes understanding homological properties of
algebraic structures an interesting and important problem.

In Lie theory, one of the most important representation theoretic
objects of study is the Bernstein-Gelfand-Gelfand (BGG)
category $\mathcal{O}$ associated to a fixed triangular
decomposition of a semisimple finite dimensional complex Lie
algebra, introduced in \cite{BGG0,BGG}. This category has a number of
remarkable properties and applications, see \cite{Hu08} and
references therein.
Study of various homological invariants of category $\mathcal{O}$
goes back to the original paper \cite{BGG} which determines its
global dimension. Some time ago, the second author started a
series of papers \cite{Ma1,Ma2} aimed at a systematic
study of homological invariants of various classes of structural
modules in category $\mathcal{O}$. This series was
subsequently continued with \cite{CM1,CoM2,KMM1} covering a wider
class of questions and objects and also extending the results
to Rocha-Caridi's parabolic generalization of $\mathcal{O}$
from \cite{RC}.

Both category $\mathcal{O}$ and its parabolic generalization,
can be defined, in a natural way, in the setup of Lie superalgebras,
see \cite{Ka,Mu12,ChWa12,Ma,AM}. The objects obtained in this way
are very interesting and highly non-trivial, see \cite{Mu12,ChWa12}
and references therein. Therefore it is natural to study homological
invariants for category $\mathcal{O}$ for Lie superalgebras.
The first step here would be to find out which of the classical
results (i.e. results for Lie algebras) can be generalized to the
super-setting and how. This is one of the motivations behind
some parts of the present paper.

The recent paper \cite{KKM} discovered an interesting connection
between bigrassmannian permutations and cokernels of inclusions
of Verma modules for the Lie algebra $\mathfrak{sl}_n$.
The second motivation of this paper is to investigate to which
extent this result generalizes to the super-setting. This is
a non-trivial challenge as it is very well-known that, unlike
the classical case, homomorphisms between Verma supermodules over
Lie superalgebras are often not injective.

To our surpize, it turns out that, both for the homological
properties of structural supermodules and for the description of
cokernels of inclusions of Verma supermodules, it is possible to
obtain fairly complete results for the periplectic
Lie superalgebra $\pn$. This is the main aim of the
present paper.

Let us now briefly describe the structure of the
paper and the main results. In Section~\ref{Sect::pre}
we collect preliminaries on classical Lie superalgebras
which are relevant for the rest of the paper.

In Section~\ref{Sect::HomVerSuper} we describe Verma supermodules
for classical Lie superalgebras of type I having the property that
any homomorphism from any other Verma supermodule to
them is injective. As an application, we prove that every
non-zero homomorphism between Verma supermodules for $\pn$ is an
embedding.

In Section~\ref{Sect::soccokerpn}, we describe the socle of the
cokernel of a nonzero homomorphism between Verma supermodules
for  $\pn$. We manage to reduce the problem for $\pn$ to the
corresponding problem for $\gl(n)$, where the main result of
\cite{KKM} applies.

Section~\ref{Sect::homodim} is devoted to the study of possible connections
between homological invariants for a Lie superalgebra
and the corresponding homological invariants for its even
part, which is a Lie algebra. In particular, we relate
the finitistic dimension of the category of
supermodules for a classical Lie superalgebra to the
global dimension of the category of modules for its
even part.  In Sections~\ref{sect::52} and \ref{sect::54} we
focus on projective and injective dimensions of structural
modules in an arbitrary parabolic cateogry $\mc O^\fp$.
In Section~\ref{sect::55} we introduce the notion of
the associated variety and discuss its relation to
projective dimension.

Finally, in  Section~\ref{Sect::6}  we investigate the
projective dimensions of structural modules in the
parabolic category $\mc O^\fp$ for $\pn$  and the ortho-symplectic Lie
	superalgebra $\mf{osp}(2|2n)$.
\vspace{3mm}

\noindent
{\bf Acknowledgments.} The first author is partially supported by MoST grant
108-2115-M-008-018-MY2. For the second author the research was
partially supported by the Swedish Research Council
and G{\"o}ran Gustafssons Stiftelse.  We thank Kevin Coulembier for helpful comments.

\section{Preliminaries} \label{Sect::pre}

Throughout, we let   $\mf g = \mf g_\oo\oplus \mf g_{\ob}$
be a finite-dimensional classical Lie superalgebra.
This means that $\mf g_\oa$ is a reductive Lie algebra
and  $\mf g_\ob$ is a completely reducible $\mf g_\oa$-mo\-du\-le
under the adjoint action.

For a given classical Lie superalgebra, we fix triangular decomposition
\begin{align} \label{eq::tridec}
& \mf g_\oa =\mf n_\oa \oplus \mf h_\oa \oplus \mf n^-_\oa
\quad\text{ and }\quad
\mf g =\mf n \oplus \mf h \oplus \mf n^-
\end{align}
of $\mf g_\oa$ and of $\g$, respectively, in the sense of
\cite[Section~2.4]{Ma}. Here
$\mf b_\oa:=\mf n_\oa \oplus \mf h_\oa$ and
$\mf b:=\mf n \oplus \mf h$ are  Borel subalgebras in
$\mf g_\oa$ and $\mf g$, respectively. In the
present paper we assume that the Cartan subalgebra
$\mf h$ is purely  even, i.e., $\mf h=\mf h_\oa$.
We denote by $\displaystyle \mf g=\bigoplus_{\alpha\in \h^\ast}\mf g^\alpha$
the root space decomposition of $\mf g$ with respect to $\mf h^\ast$.
We denote by $W$ the Weyl group of
$\mf g_\oa$.  The usual dot-action of $W$ on
$\mf h^\ast$ is defined as
$$w\cdot \la := w(\la+\rho_\oa)-\rho_\oa,~\text{for any }w\in W,~\la \in \h^\ast,$$ where $\rho_\oa$ is the half of the sum of all
positive roots of $\mf g_\oa$.
For a given weight $\mu \in \h^\ast$, we let $W_\mu$ denote
the stabilizer of $\mu$ under the dot-action of $W$.
Let $\ell: W\rightarrow \mathbb{Z}_{\geq 0}$
denote the usual length function.

Finally, a weight is called {\em integral},
{\em dominant} or {\em anti-dominant} if it is
integral, dominant or anti-dominant as a $\mf g_\oa$-weight,
respectively.

Finally, we denote by $\Phi^+, \Phi^+_\oa$ and $\Phi^+_\ob$ the sets of positive roots, even positive roots and odd positive roots with respect to the triangular decomposition \eqref{eq::tridec}, respectively. In other words, these are exactly the non-zero
weights of $\mf b$, $\mf b_\oa$ and $\mf b_\ob$, respectively.

\subsection{BGG category $\mc O$}

Associated to the triangular decomposition \eqref{eq::tridec}
above, the BGG category $\mc O  $ (resp. $\mc O_\oa$)
is defined as the full subcategory of the category of
finitely generated  $\g$-modules (resp. $\mf g_\oa$-modules)
which are semisimple over $\mf h$ and locally $U(\fn^+)$-finite
(resp. $U(\fn^+_\oa)$-finite), see \cite{BGG, Hu08}.

For $\la \in \h^\ast$, the corresponding Verma modules
over $\mf g_\oa$ and $\mf g$ are defined as
$$\Delta_{\oa}(\lambda)=U(\fg_{\oa})\otimes_{U(\fb_{\oa})}\mC_{\la}\quad\text{ and }\quad\Delta(\la) := U(\mf g)\otimes_{U(\mf b)}   \mC_{\la},$$
respectively. Here $\mC_{\la}$ is the one-dimensional module over
the Borel subalgebra corresponding to $\lambda$.
The unique simple quotients of $\Delta_\oa(\la)$ and
$\Delta(\la)$ are denoted by $L_\oa(\la)$ and $L(\la)$, respectively.
The corresponding dual Verma modules $\nabla_\oa(\la)$ and
$\nabla(\la)$ are similarly defined as coinduced modules,
see \cite[Subsection~3.2]{Hu08} and
\cite[Definition~3.2]{CCC} for more details.
We note that $L(\la) \hookrightarrow \nabla(\la).$ Also, we denote by $T_\oa(\la)$ and $T(\la)$ the tilting modules of highest weight $\la$ in $\mc O_\oa$ and $\mc O$, respectively, see \cite[Section 4.3]{Ma} and \cite[Theorem 3.5]{CCC} for more details.

Let us denote by $\Res:=\Res_{\mf g_\oa}^\g$,
$\Ind:=\Ind_{\mf g_\oa}^\g$ and $\Coind:=\Coind_{\mf g_\oa}^\g$ the
corresponding restriction, induction and coinduction functors,
respectively. It is well-known, see \cite[Theorem 2.2]{BF}
and \cite{Gor1}, that there is an isomorphism of functors
\[\Ind \cong \Coind(  \Lambda^{\text{top}}\mf g_\ob\otimes-).\]
\vskip 0.2cm

\subsection{Lie superalgebras of type I} \label{subsect::21}

A classical Lie superalgebra   $\mf g = \mf g_\oo\oplus \mf g_{\ob}$
is said to be of {\em type I} if $\mf g$ has a compatible
$\mathbb{Z}$-grading $\mf g= \mf g_{-1}\oplus \mf g_0 \oplus \mf g_{1}$
such that $\mf g_0 =\mf g_\oo$ and
$\mf g_\ob = \mf g_1 \oplus \mf g_{-1}$, where $\mf g_{\pm 1}$
are $\mf g_\oo$-submodules of $\mf g_\ob$ with
$[\mf g_{1},\mf g_{ 1}] =[\mf g_{-1},\mf g_{-1}]  =0$.
We set $\mf g_{\geq 0}: = \mf g_0 \oplus \mf g_1$ and
$\mf g_{\leq 0}: = \mf g_0 \oplus \mf g_{-1}$.

\subsubsection{Triangular decompositions for $\mf g$ of type I}

For classical Lie superalgebras $\mf g$ of type I considered
in the present paper, we always assume that the triangular
decomposition in \eqref{eq::tridec} has the property
\begin{align}
&\mf b = \mf b_\oa \oplus \mf g_1, \label{eq::typeItridec}
\end{align}
that is, $\mf b_\ob=\g_1$.

In particular, we are interested in the following
superalgebras from Kac's list (cf. \cite{Ka}):
\begin{align}
&(\text{Type \bf A}): ~\mf{gl}(m|n), \mf{sl}(m|n)
\text{ and } \mf{sl}(n|n)/\mathbb{C}I_{n|n};\label{cLIa} \\
&(\text{Type \bf C}): ~\mf{osp}(2|2n); \label{cLIc} \\
&(\text{Type \bf P}): ~\pn\text{ and } [\pn, \pn]; \label{cLIp}
\end{align}
where $m>n \geq 1$ are integers (see, e.g.,
\cite[Chapter 2, 3]{Mu12} and \cite[Section 1.1]{ChWa12}).
See, Section~\ref{sect::32} and Section~\ref{eq::osp} for more details
on $\pn$ and $\mf{osp}(2|2n)$.

\subsubsection{Induced and coinduced modules}

For a given object $V \in \mc O_\oa$, we extend $V$
trivially to a $\mf g_{\geq 0}$-module  and define
the {\em Kac module} $$K(V):=\Ind_{\mf g_{\geq 0}}^{\mf g}V.$$
This defines an exact functor
$K({}_-): \mc O_\oa \rightarrow \mc O$ which is called the
Kac functor. For $\la \in \h^\ast$, we define
$K(\la) := K(L_\oa(\la))$.  Also, we note that
$\Delta(\la)  \cong K(\Delta_\oa(\la))$ and
$\nabla(\la):= \text{Coind}_{\mf g_{\leq 0}}^{\mf g} (\nabla_\oa(\la)).$

\section{Homomorphisms between Verma supermodules} \label{Sect::HomVerSuper}

In  this section  we assume that
\begin{enumerate}
\item $\mf g$ is a classical  Lie superalgebra of type I
such that the triangular decomposition in \eqref{eq::tridec}
satisfies the condition in \eqref{eq::typeItridec}. \label{eq::cond1ch3}
\item There is an   element $d\in \mc Z(\mf g_\oa)$ and a  non-zero scalar $D\in \C$ such that $$[d,x]=Dx,~ \text{for any} ~ x\in \mf g_{-1}.$$  \label{eq::cond2ch3}
\end{enumerate}
On can check that the superalgebras $\mf{gl}(m|n)$, $\mf{osp}(2|2n)$ and $\mf{pe}(n)$
from the list \eqref{cLIa}-\eqref{cLIp} satisfy conditions \eqref{eq::cond1ch3} and \eqref{eq::cond2ch3}. We refer to \cite[Section 4.2]{CM} for a description of the
elements $d$.

One of the goals in this subsection is to give a necessary and sufficient condition for the homomorphism
$$K({}_-): \text{Hom}_{\mc O_\oa}(\Delta_\oa(\mu),\Delta_\oa(\la)) \longrightarrow \text{Hom}_{\mc O}(\Delta(\mu),\Delta(\la)),$$
where $K({}_-)$ is the Kac functor defined in
Subsection~\ref{subsect::21}, to be an isomorphism.

\subsection{Kac functor and homomorphisms between Verma supermodules}
Using the same argument as in the proof of  \cite[Lemma 4.3]{CP},
we have the following useful lemma.

\begin{lem}\label{lem::CP43}
Let $\mu, \nu \in \h^\ast$ be integral weights such that  $[K(\mu):L(\nu)]>0$  and $K(\nu )=L(\nu)$. Then $\mu=\nu$.
\end{lem}

Our  first main result is the following.

\begin{thm} \label{mainthm}
Let $\la\in \h^\ast$ be integral. Then the following
conditions are equivalent:
\begin{itemize}
\item[(1)] The Kac module $K(\uv \la)$ is simple, where $\uv \la \in W\cdot \la$ is anti-dominant.
\item[(2)] For any $\mu \in \h^\ast$, every non-zero homomorphism
$f:\Delta(\mu)\rightarrow \Delta(\la)$ is an embedding.
\item[(3)] For any $\mu \in \h^\ast$, the Kac functor $K({}_-)$
gives rise to an isomorphism,
\begin{align*}
&K({}_-): \emph{Hom}_{\mc O_\oa}(\Delta_\oa(\mu),\Delta_\oa(\la)) \xrightarrow{\cong} \emph{Hom}_{\mc O}(\Delta(\mu),\Delta(\la)).\end{align*}
\end{itemize}
\end{thm}

\begin{proof}
In the proof we regard $\Delta_\oa(\la)$ and $\Delta_\oa(\mu)$
as direct summands of $\Res \Delta(\la)$ and
$\Res \Delta(\mu)$, respectively.

We first prove $(1)\Rightarrow (3)$. Using the exactness of
Kac functor, the characters of $\Delta(\la)$ and $\Delta(\mu)$
are given as follows:
\begin{align*}
&\text{ch}\Delta(\la) = \sum_{\zeta} [\Delta_\oa(\la): L_\oa(\zeta)] \text{ch}K(\zeta),\qquad
\text{ch}\Delta(\mu) = \sum_{\zeta} [\Delta_\oa(\mu): L_\oa(\zeta)] \text{ch}K(\zeta).
\end{align*}
This implies that there exists $\zeta \in W\cdot \mu$ such that
$[K(\zeta):L(\uv \la)]>0$ since $f$ is non-zero and
$\text{soc}(\Delta(\la)) =L(\uv \la)$  by \cite[Theorem~51]{CM}.

Since $d\in \mc Z(\mf g_\oa)$, the element $d$ acts on
$\Delta_\oa(\la)$ and $\Delta_\oa(\mu)$ via some scalars,
say $c_\la$ and $c_\mu$, respectively. From Lemma~\ref{lem::CP43},
we conclude that $\zeta =\uv \la$ since
$[K(\zeta):L(\uv \la)]>0$  and $K(\uv \la )=L(\uv \la)$.
Therefore $\mu \in W\cdot \la$ and hence $c_\la =c_\mu$.

Applying the restriction functor to $f$, we obtain   a
$\g_\oa$-homomorphism
$$\Res(f): \Lambda^{\bullet} \mf g_{-1} \otimes \Delta_\oa(\mu) \rightarrow \Lambda^{\bullet} \mf g_{-1}\otimes \Delta_\oa(\la).$$
We now consider the eigenspaces with respect to the operator
$d$.  Observe that, for any $k\geq 0$, the subspace
$\Lambda^{k} \mf g_{-1}\otimes \Delta_\oa(\mu)$
of $\Res \Delta(\mu)$ is the eigenspace for $d$
corresponding to the eigenvalue $c_\mu+kD$. Similarly, the subspace
$\Lambda^{k} \mf g_{-1}\otimes \Delta_\oa(\la)$
of $\Res \Delta(\lambda)$ is the eigenspace for $d$
corresponding to the with eigenvalue $c_\la+kD$. Therefore, we have
\begin{align} \label{eq::11}
0\neq \Res(f)(\Delta_\oa(\mu)) \subseteq \Delta_\oa(\la)
\end{align}
since $f\neq 0$ and $c_\la =c_\mu.$ Consequently, we may conclude that
\begin{align} \label{eq::12}
\Res(f)\bigg{|}_{\Delta_\oa(\mu)}: \Delta_\oa(\mu) \rightarrow  \Delta_\oa(\la)
\end{align}
is an embedding by \cite[Theorem~7.6.6]{Di}.

Let $K$ be the kernel of $f$. If we assume that $f$ is not an
embedding, then we have  $K \supseteq \text{soc}(\Delta(\mu))$ by
\cite[Theorem~51]{CCM}. This gives
\begin{align}
&K \supseteq \text{soc}(\Delta(\mu)) =
U(\mf g)\cdot (\Lambda^{\text{top}}\mf
g_{-1}\otimes \text{soc}(\Delta_\oa(\mu))),
\end{align}
which means that
\[\Lambda^{\text{top}}\mf g_{-1}\otimes \Res(f)(\text{soc}
(\Delta_\oa(\mu)))=f(\Lambda^{\text{top}}\mf g_{-1}\otimes
\text{soc}(\Delta_\oa(\mu))) =0,\]
a contradiction to \eqref{eq::11}. Note that this proves the
direction  $(1)\Rightarrow (2)$. We proceed, however, with
the proof of the direction  $(1)\Rightarrow (3)$.

Observe that the Kac functor gives rise to a
monomorphism
$$K({}_-): \text{Hom}_{\mc O_\oa}(\Delta_\oa(\mu),\Delta_\oa(\la)) \rightarrow \text{Hom}_{\mc O}(\Delta(\mu),\Delta(\la)).$$
It suffices to show that, for given non-zero
$f,g\in \text{Hom}_{\mc O}(\Delta(\mu),\Delta(\la))$,
there exists $\alpha \in \C$ such that $f=\alpha g.$
By Equation~\eqref{eq::12}, both $f$ and $g$ restrict
to non-zero $\mf g_\oa$-homomorphisms
$\Res(f)\bigg{|}_{\Delta_\oa(\mu)}$ and
$\Res(g)\bigg{|}_{\Delta_\oa(\mu)}$ from  $\Delta_\oa(\mu)$
to  $\Delta_\oa(\la)$, respectively. From \cite[Theorem~7.6.6]{Di}
we have that there
exists $\alpha \in \C$ such that
$$\Res(f)\bigg{|}_{\Delta_\oa(\mu)} =\alpha \Res(g)\bigg{|}_{\Delta_\oa(\mu)}.$$
This proves the part $(1)\Rightarrow(3)$   since
$\Delta(\mu)$ is generated by $\Delta_\oa(\mu)$ as a $\mf g$-module.

Note that the implication $(3)\Rightarrow (2)$ is clear.
It remains to prove the implication $(2)\Rightarrow (1)$,
so we now assume  $(2)$.
We set $L(\nu):=\text{soc}(K(\uv \la))$. Then we have the map
$\Delta(\nu) \onto L(\nu) \hookrightarrow K(\uv \la)= \Delta(\uv \la)$.
Since this map is an embedding by $(2)$, it follows that
$\Delta(\nu) = L(\nu)  \hookrightarrow K(\uv \la)= \Delta(\uv \la)$.
By Lemma \ref{lem::CP43}, from $[K(\uv \la):L(\nu)]\neq 0$ and
$K(\nu) =L(\nu)$, we deduce that $\uv \la = \nu$.
This gives $K(\uv \la) =L(\uv \la)$ and  completes the proof.
\end{proof}

\subsection{Example: The periplectic Lie superalgebras $\pn$} \label{sect::32}
\subsubsection{} \label{sect::321}
For positive integers $m,n$, the general linear Lie superalgebra $\mathfrak{gl}(m|n)$
can be realized as the space of $(m+n) \times (m+n)$ complex matrices
\begin{align*}
\left( \begin{array}{cc} A & B\\
C & D\\
\end{array} \right),
\end{align*}
where $A,B,C$ and $D$ are $m\times m, m\times n, n\times m, n\times n$ matrices,
respectively. The Lie bracket of $\gl(m|n)$ is given by the super commutator.
Let $E_{ab}$, for $1\leq a,b \leq m+n$ be the elementary matrix in $\mathfrak{gl}(m|n)$.
Its $(a,b)$-entry  is equal to $1$ and all other entries are $0$.

\subsubsection{}

The standard matrix realization of the periplectic Lie superalgebra
$\pn$ inside the general linear Lie superalgebra
$\mathfrak{gl}(n|n)$ is given by
\begin{align}\label{plrealization}
\mf g= \mf{pe}(n):=
\left\{ \left( \begin{array}{cc} A & B\\
C & -A^t\\
\end{array} \right)\| ~ A,B,C\in \C^{n\times n},~\text{$B^t=B$ and $C^t=-C$} \right\}.
\end{align}
Throughout the present paper, we fix the Cartan subalgebra $\mf h= \mf h_\oa \subset \mf g_\oo$ consisting of diagonal matrices. We denote the dual basis of $\mf h^*$ by $\{\vare_1, \vare_2, \ldots, \vare_n\}$ with respect to the standard basis of $\mf h$ defined as
\begin{align}
\{H_i:=E_{i,i}-E_{n+i,n+i}|~1\leq i \leq n \}\subset \pn, \label{eq::cartan}
\end{align}
where $E_{a,b}$ denotes the $(a,b)$-matrix unit,
for $1\leq a,b \leq 2n$.
The sets $\Phi_\oa$, $\Phi_\ob^\pm$ and $\Phi_\ob$ of even,
odd positive, odd negative and odd roots are, respectively,  given by
\begin{align}\label{eqroots}
&\Phi_\oa=\{\epsilon_i-\epsilon_j\,|\, 1\le i\not=j\le n\}, \\
&\Phi^+_\ob=\{\epsilon_i+\epsilon_j\,|\, 1\le i\le j\le n\} ,\\
&\Phi^-_\ob=  \{- \epsilon_i-\epsilon_j\,|\, 1\le i< j\le n\},\\
&\Phi_\ob:=\Phi_\ob^+\cup \Phi_\ob^-.
\end{align}

The space $\mf h^\ast = \bigoplus_{i=1}^n\C \vare_i$ is equipped with
a natural bilinear form
$\langle \vare_i , \vare_j \rangle =\delta_{i,j}$, for any
$1\leq i,j \leq n$. The Weyl group $W=\mf S_n$   is the symmetric
group    acting on $\h^\ast$.  For any $\alpha\in\Phi_\oa$, we let
$s_\alpha$ denote the corresponding reflection in $W$.

\subsubsection{}

We  fix the Borel subalgebra $\fb_{\oa}$ of $\g_{\oa}=\mathfrak{gl}(n)$
consisting of matrices in \eqref{plrealization} with $B=C=0$ and
$A$ upper triangular.

The subalgebras $\mf g_1$ and $\mf g_{-1}$ are given by
\begin{align*}
\mf g_1:=
\{\begin{pmatrix}
0 & B \\
0 & 0
\end{pmatrix}|B^t=B\}\quad\mbox{and}\quad \mf g_{-1}:=
\{\begin{pmatrix}
0 & 0 \\
C & 0
\end{pmatrix}|C^t=-C\}.
\end{align*}
We note that $\Phi_\ob^+$ and $\Phi_\ob^-$ are the sets of
weights of the $\mf g_0$-modules $\mf g_1$ and $\mf g_{-1}$, respectively.

The Borel subalgebra is defined as $\mf b:=\mf b_\oa \oplus \g_{1}$, while the reverse Borel subalgebra $\mf b^r$ is defined as
$\mf b^r:=\mf b_\oa \oplus \mf g_{-1}.$  It can be shown
that $\mf b^r$ is still a Borel subalgebra in thee sense
of \cite[Section~2.4]{Ma} (see, e.g., \cite[Section~5]{CCC}).

Furthermore, we let $L^{r}(\la)$ be the irreducible module
in the corresponding ``reversed'' category $\mc O$ of $\mf b^r$-highest weight $\la$.
We let  $\la^+\in \h^\ast$ be uniquely determined by
\begin{align}
&L(\la^+) =L^r(\la). \label{eq::oddref}
\end{align}
It is known that the  weight $\la^+$  can be computed using
odd reflection, and we refer to \cite[Section~2.2]{PS89} and \cite[Theorem 3.6.10]{Mu12}
for a treatment of odd reflections for the periplectic Lie superalgebras $\pn$.
In particular,  the effect on the highest weight of a
simple $\pn$-module under odd reflection and inclusion
was computed in \cite[Lemma~1]{PS89}.

Finally, we set  $\displaystyle X:=\bigoplus_{i=1}^n\mZ\ep_i$;
$\omega_k:=\vare_1+\cdots +\vare_k$, for any $1\leq k\leq n$;
and also $\eta:=(1-n)(\vare_{1}+\cdots +\vare_{n})$.
Then we have $\Lambda^{\text{top}}\mf g_{-1}\cong \C_{\eta}$
as a $\mf g_\oa$-module.

\subsubsection{}

Let $\la, \mu\in \h^\ast$ be two integer weights.
If there are  positive roots
$$\alpha_1,\alpha_2,\ldots, \alpha_k\in \Phi_\oa^+,$$
such that
$\la =s_{\alpha_k} s_{\alpha_{k-1}}\cdots s_{\alpha_1}\cdot \mu $
and, for each $1\leq q \leq k-1$, we have
\[\langle s_{\alpha_q}s_{\alpha_{q-1}}\cdots s_{\alpha_1}\cdot \mu,~ \alpha_{q+1} \rangle \geq 0, \]
then  we write $\mu \uparrow \la$, cf. \cite[Subsection~7.6]{Di}.

\begin{prop} \label{matincoro0}
Let $\mf g=\pn$ and $\la \in \h^\ast$ be an integral weight.
\begin{itemize}
\item[(1)] For any integral weight $\mu\in \h^\ast$, we have that
$$\emph{Hom}_{\mc O}(\Delta(\mu),\Delta(\la)) \cong \begin{cases}
\C, \emph{ if } \mu \uparrow \la,\\
0, \emph{ otherwise. }
\end{cases}$$
\item[(2)] A weight vector $v\in \Delta(\la)$ satisfies $\mf nv=0$
if and only if $$v=1\otimes v_0\in  U(\mf g) \otimes_{\mf g_{\geq 0}}
\Delta_\oa(\la) \equiv  \Res \Delta(\la),$$
for some vector $v_0$ satisfying $\mf n_\oa v_0=0$.
\end{itemize}
\end{prop}

\begin{proof}
We recall that $\pn$ admits a grading operator
$d:=\text{diag$(1,1,\ldots,1)$}\in \g_\oa$ which satisfies
Condition~\eqref{eq::cond1ch3} at the beginning of
Section~\ref{Sect::HomVerSuper}. Also, it follows from
\cite[Lemma~5.11]{CC} that every weight satisfies
Assumption $(1)$ in Theorem~\ref{mainthm}. Therefore the proof follows
from Theorem~\ref{mainthm} and \cite[Theorem~4.2, Theorem~5.1]{Hu08}.
\end{proof}

\begin{rem}
	In \cite{LLW}, Liu, Li and Wang  established a closed formula for a singular vector of weight $\la -\beta$ in the Verma
	module of highest weight $\la$ for Lie superalgebra $\gl(m|n)$ when $\la$ is atypical with respect to an odd positive root $\beta$. In particular, the authors proved the following identity
	\begin{align}
	&\text{Hom}(\Delta(\la -\beta),\Delta(\la))=\C, \label{eq::LLWid}
	\end{align}  where $\Delta$'s denote Verma supermodules for $\g =\gl(m|n)$. This identity was also known in \cite{ChWa18} for the exceptional Lie superalgebra $D(2|1;\zeta)$. By Proposition \ref{matincoro0}, it follows that the identity \eqref{eq::LLWid} fails to hold for $\pn$.
\end{rem}

\begin{rem}  The Borel subalgebras have been classified  in \cite[Section 5]{CCC}. It would be interesting to determine for which other Borel  subalgebras one has an analogue of
	Proposition \ref{matincoro0}. We note that Proposition \ref{matincoro0} fails in 	the case of the reverse Borel subalgebra $\mf b^r$. To see this, we define Verma supermodules with respect to $\mf b^r$ as follows
	$$\Delta^r(\la):=U(\mf g)\otimes_{\mf b^r}\C_\la \cong \text{Ind}^\g_{\mf g_{\leq 0}}\Delta_\oa(\la),$$
	which admits a filtration with subquotients   $ \text{Ind}^\g_{\mf g_{\leq 0}}L_\oa(\zeta)$, for $\zeta \in \h^\ast$.	By \cite[Proposition 4.15]{CM}, the modules $\text{Ind}^\g_{\mf g_{\leq 0}}L_\oa(\zeta)$ are never simple, for any $\zeta\in \h^\ast$. By similar argument as used in the direction $(3)\Rightarrow (2)$ and $(2)\Rightarrow (1)$ in the proof of Proposition \ref{mathcoro}, it follows that  Proposition \ref{mathcoro} fails for these modules $\Delta^r(\la)$.
\end{rem}

\begin{rem}
Consider a basic Lie superalgebras $\mf g$ of type I, for instance,
$\mf g=\gl(m|n)$, or $\mf{osp}(2|2n)$. It has been shown in  by Kac \cite{Ka2}
that a weight $\la$ satisfies Assumption (1) in Theorem~\ref{mainthm}
if and only if $\la$ is typical. Since the notions of strongly
typical and typical are identical in these cases
 (see, e.g., \cite[Section 2.4]{Gor2} or \cite{Gor3}), the statement
of Theorem~\ref{mainthm} follows from Gorelik's  theorem,
see \cite[Theorem 1.3.1, Theorem 1.4.1]{Gor2}.
\end{rem}

\section{The socle of the cokernel of inclusions of Verma supermodules
for  $\pn$} \label{Sect::soccokerpn}

In this section, we assume that $\mf g=\pn$.  The goal of this
section is to describe the socle of the cokernel of an arbitrary
non-zero homomorphism between Verma $\pn$-supermodules.

Let $\la,\gamma\in \h^\ast$ and
$\Delta(\gamma) \xrightarrow{f} \Delta(\la)$ be a
non-zero homomorphism between the corresponding
Verma supermodules  over $\mf g$.
By Theorem \ref{mainthm}, there exists a homomorphism
$\tilde{f}\in \text{Hom}_{\mc O}(\Delta_\oa(\gamma), \Delta_\oa(\la))$
such that $K(\tilde{f}) =f$. Applying the Kac functor to the
short exact sequence
$$0\rightarrow \Delta_\oa(\gamma) \xrightarrow{\tilde{f}} \Delta_\oa(\la) \rightarrow \Delta_\oa(\la)/\Delta_\oa(\gamma)\rightarrow 0$$
in $\mc O_\oa$,
we obtain the  short exact sequence
$$0\rightarrow \Delta(\gamma) \xrightarrow{f} \Delta(\la) \rightarrow K(\Delta_\oa(\la)/\Delta_\oa(\gamma))\rightarrow 0$$
in $\mc O$. This  gives rise to the isomorphism
$$\text{soc}(\Delta(\la)/\Delta(\gamma)) \cong
\text{soc}(K(\Delta_\oa(\la)/\Delta_\oa(\gamma))) ,$$
where we write $\Delta(\la)/\Delta(\gamma)$ for
$\Delta(\la)/f(\Delta(\gamma))$.

Also, by Theorem~\ref{mainthm}, there exists a dominant integral
$\mu$ such that  $\la = x\cdot \mu$ and $\gamma =y\cdot \mu$,
for some $x,y\in W$ with $x<y$ with respect to the Bruhat order.
Let
$$\theta_\mu^{\text{on}}:(\mc O_\oa)_0 \rightarrow (\mc O_\oa)_\mu$$
denote the translation functor to the $\mu$-wall, where
$(\mc O_\oa)_0$ and $(\mc O_\oa)_\mu$ denote the principal
block and the block containing $L_\oa(\mu)$, respectively.
For any $ s\in W$, we set  $n_{x, y, z}$ to be the integer
determined by
$$\theta_\mu^{\text{on}}\left(\text{soc}(\Delta_\oa(x'\cdot 0)/\Delta_\oa(y'\cdot 0))\right)=\text{soc}(\Delta_\oa(x\cdot \mu)/\Delta_\oa(y\cdot \mu))=\bigoplus_{z\in W}L_\oa(z\cdot \mu)^{\oplus n_{x, y, z}},$$
where $x',y'$ are the shortest representatives in the cosets $xW_\mu$ and
$yW_\mu$, respectively, see \cite[Proposition~15]{KKM}.

For $\nu \in \h^\ast$ we recall the weight $\nu^+$ defined in
\eqref{eq::oddref}. We are now in a position to state the main
result of this section which describes the socle of the cokernel
of homomorphism between Verma supermodules  over $\mf g$.

\begin{thm} \label{2ndmainthm}
We have the following description of
the socle of the cokernel of a non-zero homomorphism
$\Delta(\gamma) \rightarrow \Delta(\la)$:
\[\emph{soc}(\Delta(\la)/\Delta(\gamma))\cong
\bigoplus_{z\in W}L((z\cdot \mu)^++\eta)^{\oplus n_{x,y,z}}. \]
\end{thm}

\begin{proof}
By \cite[Lemma~3.2]{CM} (see also \cite[Theorem 51]{CCM}), we have  \begin{displaymath}
\begin{array}{rcl}
\text{soc}(\Delta(\la )/\Delta(\gamma))
&\cong &\text{soc}(K(\Delta_\oa(\la)/\Delta_\oa(\gamma))) \\
&= &U(\mf g)\cdot  \left(\Lambda^{\text{top}} \mf g_{-1}\otimes
\text{soc}(\Delta_\oa(\la)/\Delta_\oa(\gamma))\right) \\
&\cong& \text{head}\,\Ind_{\g_{\leq 0}}^\g
\left(\bigoplus\limits_{z\in W}
L_\oa(z\cdot 0 +\eta)^{\oplus n_{x,y,z}}\right).
\end{array}
\end{displaymath}
It follows that
$$\text{soc}(\Delta(\la)/\Delta(\gamma))\cong
\bigoplus\limits_{z\in W}L^r(z\cdot 0 +\eta)^{\oplus n_{x,y,z}}=
\bigoplus_{z\in W}L((z\cdot \mu)^++\eta)^{\oplus n_{x,y,z}}.$$
This completes the proof.
\end{proof}

\begin{rem}
Let $\la$ be a dominant and regular weight.
By Theorem~\ref{2ndmainthm}, for any $y\in W$, the quotient
$\Delta(\la)/\Delta(y\cdot \la)$ has simple socle if and only if
$y$ is a bigrassmannian permutation, see
\cite[Section~1.1, Theorem~1.3]{KKM}.
\end{rem}

For an integral weight $\la \in \h^\ast$, we let
$\ov \la\in W\cdot \la$ and
$\uv \la\in W\cdot \la$ be the unique dominant and
the unique anti-dominant element, respectively.
Theorem~\ref{2ndmainthm} has the following consequence.

\begin{cor} \label{mathcoro}
Suppose that $\mu \in \h^\ast$ is not anti-dominant. Then we have
$$\emph{dim Ext}_{\mc O}^1(L(\mu),\Delta(\la)) =
[\emph{soc}\left(\Delta(\ov \la)/\Delta(\la)\right): L(\mu)].$$
\end{cor}

\begin{proof}
Let
\begin{align} \label{eq::SES1}
&0\rightarrow \Delta(\la) \rightarrow M \rightarrow L(\mu) \rightarrow 0,
\end{align}
be a non-split short exact sequence. Since it is non-split,
the socle of $M$
coincides with the socle   $L(\uv \la)$ of $ \Delta(\la).$
Therefore $M$ is a submodule of the injective envelope
$I(\uv \la) =T(\ov \la)$. Now, we consider a two step
projective-injective copresentation of $\Delta(\ov \la)$
(whose existence follows, for example, from \cite[Theorem~7.3]{AM}):
\[0\rightarrow \Delta(\ov \la) \rightarrow T(\ov \la) \xrightarrow{f} U. \]
By our assumption, $[M/\text{soc}(M): L(\zeta)] >0$ implies that $\zeta$
is not antidominant. As a consequence, $f(M)=0$ since the socle
of the projective-injective module $U$ is a direct sum of simple
modules of anti-dominant   highest weights. This means that $M$
is a submodule of $\Delta(\ov \la)$. Consequently,
$L(\mu)\cong M/\Delta(\la)$ corresponds to a socle constituent of
$\Delta(\ov \la)/\Delta(\la)$. This completes the proof.
\end{proof}

\begin{ex}
We consider $\mf g= \mf{pe}(2).$ Let $\la, \mu\in \h^\ast$ be integral. Suppose that  $\mu$ is not antidominant.  Set $\ov \la =a \vare_2+b\vare_2$. Then, by \cite[Lemma 6.1]{CC}, we have
\begin{align}
\text{soc}(\Delta(\ov \la) /\Delta(\la))=
\begin{cases}
L(\ov \la-\omega_2), \text{ if $a =b$.} \\
L(\ov \la), \text{ if $a >b$.}\\
0,\text{ if $\la =\ov \la.$}
\end{cases}
\end{align}
It follows that
\begin{align*}
&\text{dim}\,\text{Ext}^1_{\mc O}(L(\mu), \Delta(\la)) = \begin{cases}
1, \text{ if either $\mu =\ov \la-\omega_2$, for $a=b$, or $\mu=\ov \la$, for $a>b$};\\
0, \text{ otherwise.}
\end{cases}
\end{align*}
\end{ex}

\begin{rem}
Proposition~\ref{2ndmainthm} and Corollary~\ref{mathcoro} can be generalized to other classical Lie superalgebras of type I which
have type A even part, under the condition that
$\text{dim}\,\text{Hom}_{\mc O}(\Delta(\gamma), \Delta(\la))\leq 1$.
\end{rem}

\section{Homological dimensions} \label{Sect::homodim}

Let $\mc A$ be an abelian category. Suppose that $\mc A$ has
enough projective and injective objects. For any $M\in \mc A$,
we let $\pd_{\mc A}M$   denote the projective dimension of  $M$.
We define $\gld\,\mc A$ to be the global dimension of  $\mc A$, namely,
\[\gld\,\mc A:=\mathop{\text{sup}}\limits_{M \in\mc A} \{\pd_{\mc A} M\}.\]
The finitistic dimension of $\mc A$ is defined as
\[ \text{fin.dim}\,\mc A : =\mathop{\text{sup}}\limits_{M \in \mc A} \{\pd_{\mc A}M|~\pd_{\mc A}M<\infty \}. \]
We will simply use $\pd({}_-)$ and $\gld({}_-)$ when the context is clear. Similarly, we have the injective dimension $\id\,M= \id_{\mc A}M$
of an object $M\in\mc A$.

In this section, we assume that $\mf g$ is an arbitrary classical Lie superalgebra.

\subsection{Finitistic dimension of various categories of supermodules}  \label{sect::51}

We denote by $\mf g$-Mod and $\mf g_\oa$-Mod the category
of all $\mf g$-modules and the category of all
$\mf g_\oa$-mo\-du\-les, respectively.

Let $\tilde{\mc C}$ and  ${\mc  C}$ be abelian subcategories of
$\mf g$-Mod and $\mf g_\oa$-Mod, respectively. Assume that
\begin{enumerate}[(i)]
\item $\Ind, \Coind$ and $\Res$ restricts to well-defined functors between ${\mc C}
$ and $\tilde{\mc C}$.  \label{eq::111}
\item $\mc C$ has enough injective and projective modules.  \label{eq::112}
\item $\text{gl.dim}\,\mc C <\infty$.\label{eq::113}
\end{enumerate}

 The following theorem generalizes \cite[Theorem~4.1]{CS} (also see \cite[Lemma 4.3]{CS}) where the case
of the superalgebra $\gl(m|n)$ was considered.
\begin{thm} \label{thm::findimthm}
We have $\emph{fin.dim} \,\tilde{\mc C} = \emph{gl.dim}\,\mc C$.
\end{thm}
\begin{proof}
Since $\Ind$ is isomorphic to $\Coind$, up to parity
change, it follows that the functors $\Ind$,
$\Coind$ and $\Res$ send injective
(resp. projective) objects to injective (resp. projective) object.

The category $\tilde{\mc C}$ is a Frobenius extension of $\mc C$ in
the sense of \cite[Definition~2.2.1]{Co}. Therefore   $\tilde{\mc C}$
has enough projective and injective objects by
\cite[Proposition~2.2.1]{Co}. Moreover, every injective
(resp. projective) object in
$\tilde{\mc C}$ is a direct summand of an object induced
from an injective (resp. projective) object of $\mc C$.

Consequently, all injective objects in $\tilde{\mc C}$
have finite projective dimension.
By an argument used in the proof   \cite[Theorem~3]{Ma}
(see also  \cite[Corollary~3.6.10]{Co}), we have that
$\text{fin.dim}\, \tilde{\mc C}$ is the maximum of projective
dimensions of injective objects in $\tilde{\mc C}$. Note also that the
global dimension of $\mc C$ is the maximum of projective
dimensions of injective objects in $\mc C$, as well.

We observe that, for any injective
$I$ in $\mc C$ and $Y\in \tilde{C}$, it follows from the
Frobenius reciprocity  that
\[\Ext^d_{\tilde{\mc C}}(\Ind I, Y) =\Ext^d_{\mc C}(I,\Res Y),  \]
which implies that
$\text{fin.dim} \,\tilde{\mc C}  \leq  \text{gl.dim}\,\mc C$.

Finally, let $I\in \mc C$ be injective. Then $I$ is a direct
summand of $\Res \Ind I\cong \Lambda^{\bullet} \mf g_\ob \otimes I$,
which implies that
\[\pd\, I \geq \pd\, \Ind I \geq \pd \,\Res \Ind I \geq \pd\, I. \]
This shows that
$\text{fin.dim} \,\tilde{\mc C}  \geq  \text{gl.dim}\,\mc C$
and the claim follows.
\end{proof}

Below we apply Theorem~\ref{thm::findimthm} to various examples.

\begin{ex}
Let  $\tilde{\mc  C}= \mf g$-Mod and ${\mc  C}= g_\oa$-Mod.
Condition~\eqref{eq::111} is obvious and Conditions~\eqref{eq::112}
and \eqref{eq::113} are clear since $U(g_\oa)$ is noetherian.
We have $\text{gl.dim}\,\mc C =\text{dim}\,\mf g_\oa$.   Therefore,
by Theorem~\ref{thm::findimthm}, we have
$$\text{fin.dim}~\mf g\text{-Mod} = \dim \g_\oa.$$
\end{ex}

\begin{ex}
Consider the super analog $\mc C(\mf g,\h)$ of the category
$\mc C(\mf g_\oa,\h)$ of weight $\mf g$-modules from \cite{Zu},
that is, $\mc C(\mf g,\h)$ the full subcategory of $\mf g$-Mod
consisting of $\mf g$-modules which are semisimple over $\mf h$.

For $\la \in \h^\ast$, let $I_\la$ be the left ideal of
$U(\mf g)$ generated by $h-\la(h)$, for $h\in\mf h$.
Then the quotient of $U(\mf g)$ by $I_\la$
is projective in $\mc C(\g,\h)$. Also, every weight $\mf g$-module
is a quotient of a direct sum of the $I_\la$'s. Therefore
$\mc C(\mf g_\oa, \h)$ has enough projective objects. By
\cite[Lemma 2.2]{Zu}  $\mc C(\mf g_\oa,\mf h)$ has
enough injective objects as well.

Let $M\in \mc C(\mf g_\oa,\h)$. By \cite[Corollary 3.1.8]{Ku},
the trivial module $\C$ in $\mc C(\mf g_\oa,\h)$ has the following
finite projective resolution  in $\mc C(\mf g_\oa,\h)$:
\[\cdots  \rightarrow D_2\rightarrow D_1 \rightarrow D_0 \rightarrow \C \rightarrow 0, \]
where $D_p$ is the induced module  $U(\mf g_\oa)\otimes_{U(\mf h)}\Lambda^p(\mf g_\oa/\mf h)$, for $p\geq  0$. Tensoring with this
resolution (over $\mathbb{C}$), we conclude that every
$M\in \mc C(\mf g_\oa,\mf h)$
has finite projective dimension.

In particular, we have
\begin{align*}
&\text{fin.dim}\, \mc C(\mf \g,\h)=
\text{fin.dim} \,\mc C(\mf \g_\oa,\h) =
\text{dim}\g_\oa -\text{dim}\h.
\end{align*}
\end{ex}

\begin{ex}
Let $\mc O^\fp$ be the parabolic BGG category of $\mf g$-modules
in the sense of \cite[Section 3]{Ma}. By \cite{CoM2}, we have
\begin{align} \label{eq::egfindim}
&\text{fin.dim}\, \mc O^\fp  =
\gld \,\mc O_\oa^\fp=2\ell(w_0) -2\ell(w_0^\fp),
\end{align}
where $w_0$ is the longest element in $W$ and
$w_0^\fp$ is the longest element in the Weyl group of
the Levi subalgebra of $\fp$.
\end{ex}

\begin{ex}
Consider $\mf g=\pn$ with a reduced parabolic subalgebra $\fp$.
For $\la \in \h^\ast$, let $\text{Irr}\,\mc O_{\la}^\fp$ denote
the set of all highest weights of irreducible modules in the
parabolic block $\mc O_\la^\fp$.   Recall the equivalence
relation $\sim$ on $\h^*$ defined in~\cite[Subsection~5.2]{CC}
which is transitively generated by
$$\begin{cases}\la\sim \la \pm2\vare_k, &\mbox{ for~$1\le k\le n$;}\\
\la\sim w \cdot \la,&\mbox{ for~$w\in W_{[\lambda]}$.}
\end{cases}$$
Here   $W_{[\la]}$ denotes the integral Weyl group associated
to $\la$. It is generated by all  reflection $s_\alpha$ with
$\langle\la, \alpha \rangle \in \Z.$ It is well-known that
$W_{[\la]}$ is the Weyl group of a certain semisimple Lie
algebra (see, e.g., \cite[Theorem 3.4]{Hu08}). We define
$w_0^{\la}$ to be the longest element of $W_{[\la]}$.
For $\la ,\nu  \in \Sigma^+_\fp$, it follows from
\cite[Theorem~B]{CP} (see also \cite[Theorem~3.2]{CP})
that  $\mc O_\la^\fp =\mc O_\nu^\fp$ if and only if $\la \sim \nu$.
As a consequence, there exists $\nu \in \text{Irr}\,\mc O_{\la}^\fp$
which is dominant and regular.

Let $\la\in \h^\ast$ be a dominant weight such that the
parabolic block $\mc O^\fp_\la$ (see Section~\ref{Sect::projdimpn}
for the precise definition) is non-zero. Let
$  (\mc O^\fp_{\oa})_{\la +X}$ denote the full subcategory
of $\mc O^\fp_{\oa}$ consisting of all modules modules whose weights
belong to the set $\la +X$. Applying, if necessary, the equivalence established in \cite[Section~2.3]{CMW} (see also
\cite[Section 4.3]{CC}), we can assume that
$W_{[\la]}$ is a parabolic subgroup of $W$. Then the problem
of global dimension of $(\mc O^\fp_{\oa})_{\la +X}$ reduces
to that of an integral block for a semisimple Lie algebra of
type $\bf A$ by \cite[Proposition~2.3]{CMW}. By
\cite[Theorem~5.2]{CoM2} we have
\begin{align}
&\text{gl.dim}   (\mc O^\fp_{\oa})_{\la +X} =2\ell(w_0^{\la})
-2\ell(w_0^\fp). \label{lem::34}
\end{align}

Take $\nu \in \text{Irr}\,\mc O_{\la}^\fp$ dominant
and regular. In this case
$\pd\, I_\oa^\fp(\nu) =2 \ell(w_0^{\la}) -2\ell(w_0^\fp)$ by
\cite[Theorem~4.1, Proposition~6.9]{CoM2}.   By
Theorem \ref{thm::findimthm}, we thus have
\begin{align*}
&\text{fin.dim}\, \mc O_\la^\fp =2\ell(w_0^{\la}) - 2\ell(w_0^{\fp}).
\end{align*}
\end{ex}

\subsection{The parabolic BGG category $\mc O^\fp$}
\label{Sbusect::parabolicCate}

In this section, we let $\mf g$ be an arbitrary  classical
Lie superalgebra. We let $\langle {}_-,{}_-\rangle$ denote the usual
bilinear form on $\mf h^\ast$.

In the remaining parts of the present paper, we use the notion of
parabolic decomposition as defined in \cite[Section~2.4]{Ma} and
consider a {\em reduced} parabolic subalgebra $\fp$ from
\cite[Subsection~1.4]{CCC}. Namely, we assume that
$\mf b\subseteq \fp\subseteq \mf g$ with a purely even
Levi subalgebra $\mf l\subseteq \fp$.   The even part
$\mf p_\oa$ of $\mf p$ is a parabolic subalgebra of $\mf g_\oa$
arising from a parabolic decomposition of $\mf g_\oa$.
{\em In what follows, we always assume that $\fp_\ob =\g_1$
in the case when $\mf g$ is of type I.}

The {\em parabolic category} $\mc O^\fp$ (resp. $\mc O^{\fp_\oa}_\oa$)
is the full subcategory of $\mc O$ (resp. $\mc O_\oa$) consisting
of all finitely generated $\mf g$-modules (resp. $\mf g_\oa$-modules)
on which $\fp$ (resp. $\fp_\oa$) acts locally finitely.
For simplicity of notation, we define
$\mc O^\fp_\oa:= \mc O^{\fp_\oa}_\oa$, and, in what follows,
we omit $\oa$ from   $(-)^{\fp_\oa}$ or $(-)_{\fp_\oa}$ when
defining notation for $\g_\oa$-modules.

For any $\la \in \h^\ast$, let $L^{\mf l}(\la)$ denote the
irreducible $\mf l$-module with highest weight $\la$.
We define the set of {\em $\mf l$-dominant weights}
(called $\fp$-dominant weights in \cite[Section~3]{CCC}) as follows:
\begin{align}
&\Sigma^+_\fp:=\{\la\in \mf h^\ast|~ \langle\la, \alpha\rangle
\in \mathbb Z_{\geq 0},\text{ for all } \alpha \in \Phi^+(\mf l) \}. \label{eq::ldomi}
\end{align}
The corresponding  parabolic category $\mc O^\fp$ is the Serre
subcategory of $\mc O$ generated by
$\{L(\la) \, | \,  \la \in \Sigma_\fp^+\}$.

Let $\mc O^\fp_\la$ denote the (indecomposable) block in
$\mc O^\fp$ containing $L(\la)$.
In particular, $\mc O_\la:=\mc O_\la^{\mf  b}$ is the
(indecomposable) block  containing $L(\la)$
in  category $\mc O$.

For $\la \in \Sigma_\fp^+$, we define the parabolic Verma
$\mf g_\oa$-module and  the parabolic Verma $\mf g$-module
respectively as
\begin{align}
&\Delta^\fp_\oa(\la) : =
\Ind_{\mf p_\oa}^{\mf g_\oa}L^{\mf l}(\la)\quad\text{ and }\quad
\Delta^\fp(\la):=
\Ind_{\mf p}^{\mf g}L^{\mf l}(\la) . \label{Eq::paraV}
\end{align}
Similarly, we can define the  dual parabolic Verma modules
$\nabla^\fp_\oa(\la)$ and $\nabla^\fp(\la)$  in
$\mc O^\fp_\oo$ and $\mc O^\fp$, respectively
(see \cite[Definition~3.2]{CCC}). We have
$$\Delta^\fp(\la) \twoheadrightarrow L(\la)\quad\text{ and }\quad L(\la)
\hookrightarrow \nabla^\fp(\la).$$

Let $P^\fp(\la)$ be the projective cover of $L(\la)$
in $\mc O^\fp$ and $I^\fp(\la)$ be the injective envelope of $L(\la)$
in $\mc O^\fp$. For $\mc O^\fp =\mc O$, we
denote the corresponding objects by $P(\la)$ and $I(\la)$.
Similarly, we define the indecomposable
projectives $P_\oa(\la)$, $P_\oa^\fp(\la)$ and
injectives $I_\oa(\la)$, $I^\fp_\oa(\la)$ in
$\mc O_\oa$ and $\mc O_\oa^\fp$, respectively.  Also, we denote by $T^\fp(\la)$ and $T_\oa^\fp(\la)$ the tilting modules of highest weight $\la$ in $\mc O^\fp$ and $\mc O_\oa^\fp$, respectively. Finally,
we note that $\Ind: \mc O_\oa^\fp\rightarrow \mc O^\fp$
and $\Res : \mc O^\fp\rightarrow \mc O_\oa^\fp$ are well-defined.

Also, we note that
$$\Delta^\fp_\la= K(\Delta^\fp_\oa(\la)),~\nabla^\fp(\la):=
\Ind_{\mf g_\oa+\mf g _{-1}}^{\mf g}(\nabla^\fp_\oa(\la)
\otimes \Lambda^{\text{top}}\mf g_1^\ast)\cong
\Coind_{\mf p_\oa+\mf g_{-1}}^{\mf g}L^{\mf l}(\la),$$
in the case when $\mf g$ is of type I.

We define $w_0^\fp$ to be the longest element
in the Weyl group of $\mf l$.  Finally, we set $w_0:=w_{0}^{\mf b}$.

\subsection{Projective dimension of modules in $\mc O^\fp$}
\label{sect::52}

\subsubsection{Preliminary results}

The following lemma is the parabolic version  of \cite[Lemma~4.3]{CS}.

\begin{lem}\label{lem::31}
{\hspace{2mm}}

\begin{itemize}
\item[(1).] For $M\in \mc O^\fp$, we have $\epd\, M \geq \epd\, \Res M$.
\item[(2).] For $N\in \mc O^\fp_\oa$, we have
$\epd\, N = \epd\,  \Ind N =\epd\, \Coind N$.
\end{itemize}
\end{lem}

\begin{proof}
As already mentioned, $\Res$, $\Ind$ and $\Coind$ send
projective resolutions to projective resolutions.
Also, $N$ is a direct summand of $\Res\Ind N$.
This implies all statements.
\end{proof}

\begin{cor} \label{coro::53}
For any weight $\zeta\in \Sigma^+_\fp,$ we have
\[\epd\, \Delta^\fp(\zeta) \geq  \epd \, \Delta_\oa^\fp(\zeta)
\quad\text{ and }\quad
\epd\, \nabla^\fp(\zeta) \geq  \epd \, \nabla_\oa^\fp(\zeta).\]
\end{cor}

\begin{proof}
Noting that the modules
$\Delta_\oa^\fp(\zeta)$ and  $\nabla_\oa^\fp(\zeta)$
are  direct summands of $\Res \Delta^\fp(\zeta)$ and
$\Res \nabla^\fp(\zeta)$, respectively, the statement follows
from Lemma~\ref{lem::31}.
\end{proof}

The following result is a consequence of the combination of  \cite[Corollary~3.2.5]{Co} and \cite[Theorem~8.2.1]{Co}.
We provide more details in the proof.

\begin{prop}\label{prop::53}
Suppose that $\mf g$ is of type I.
Then, for each $\la \in \Sigma^+_\fp$, we have
$$\epd_{\mc O^\fp}\, I^\fp(\la) =
\epd_{\mc O^\fp_\oa}\, I^\fp_\oa(\la).$$
\end{prop}

\begin{proof}
The proof of \cite[Corollary 4.7]{CCC} shows that
$L_\oa(\la)$ is a quotient of $\Res L(\la)$.
Therefore we have
\begin{align*}
&\text{Hom}_{\mc O^\fp}(L(\la), \Coind\, I^\fp_\oa (\la)) =
\text{Hom}_{\mc O_\oa^\fp}(\Res L(\la), I^\fp_\oa (\la))  =
[\Res L(\la): L_\oa(\la)]\neq 0.
\end{align*}
This means that $L(\la) \hookrightarrow \Coind I_\oa^\fp(\la)$ and
so $I^\fp(\la)$ must be a direct summand of $\Coind I_\oa^\fp(\la)$.
Consequently $\pd\, I^\fp(\la) \leq \pd\, I_\oa^\fp(\la).$

Next, we claim that $L_\oa(\la)$ is a direct summand of the socle
of $\Res I^\fp(\la)$. We start by observing that
\begin{align}
&\Ind L_\oa(\la) =\Ind^\g_{\mf g_{\geq 0}} \Ind^{\g_{\geq 0}}_{\mf g_\oa} L_\oa(\la) \cong
\Ind^\g_{\mf g_{\geq 0}} (\Lambda^{\bullet} \mf g_{1}\otimes L_\oa(\la)).
\end{align}
This implies that
\begin{align*}
&\text{Hom}_{\mc O^\fp_\oa}(L_\oa(\la), \Res I^\fp (\la)) =
\text{Hom}_{\mc O^\fp}(\Ind L_\oa(\la), I^\fp (\la))\\
& = [\Ind L_\oa(\la): L(\la)]\geq[\Ind^{\mf g}_{\mf g_{\geq 0}} L_\oa(\la): L(\la)] \neq 0.
\end{align*}
Taking Lemma \ref{lem::31} into account, we thus get
$\pd\, I_\oa^\fp(\la)\leq \pd\, \Res I^\fp(\la) \leq \pd \,I^\fp(\la)$.
This completes the proof.
\end{proof}

The following result is also a consequence of the combination
of \cite[Corollary~3.2.5]{Co} and  \cite[Theorem~8.2.1]{Co}.
We provide more details in the proof.

\begin{prop} \label{thm::32}
Suppose that $\mf g$ is of type I.
Then, for each $\la \in \Sigma^+_\fp$, we have
$$\epd_{\mc O^\fp}\, T^\fp(\la) = \epd_{\mc O^\fp_\oa}\,  T^\fp_\oa(\la).$$
\end{prop}

\begin{proof}
Define $\xi_{\pm}\in \h^\ast$ such that
$\C_{\xi_\pm }\cong \Lambda^{\text{top}} \mf g_{\pm 1}$, as
$\mf g_\oa$-modules. We note that
\begin{align*}
&\Ind \Delta_\oo^{\fp}(\lambda) \cong \Ind_{\mf g_\oo}^{\mf g}
\Ind_{\mf p_\oo}^{\mf g_\oo}L_{\mf l}(\lambda) \cong
\Ind_{\mf p}^{\mf g} \Ind_{\mf p_\oo}^{\mf p}L_{\mf l}(\lambda).
\end{align*}
This means that $\Ind \Delta_\oo^{\fp}(\lambda) $ has a
$\Delta^{\fp}$-flag starting at $\Delta^{\fp}(\la+\xi_+)$.
Therefore, the module $T^\fp(\la  +\xi_+)$
is a direct summand of the module $\Ind T_\oa^\fp(\la)$.
Using Lemma~\ref{lem::31}, we conclude that
$$\pd\, T^\fp(\la) \leq \pd\, \Ind T^\fp_\oa (\la -\xi_+)
=\pd \,T^\fp_\oa(\la-\xi_+)=\pd \,T^\fp_\oa(\la).$$
Here the last equality follows by tensoring with
$\Lambda^{\text{top}} \mf g_{\pm 1}$.

Next, we claim that
$ T_\oa^\fp(\la+\xi_-) \hookrightarrow \Res T^\fp(\la)$.
To see this, we note that
\begin{align}
&\Delta^\fp_\oa(\la +\xi_-) \hookrightarrow \Lambda^{\bullet} \mf g_{-1}\otimes \Delta^\fp_\oa(\la) \cong  \Res \Delta^\fp(\la)  \hookrightarrow  \Res T^\fp(\la).
\end{align}
This implies that there is an indecomposable  direct summand of
$\Res T^\fp(\la)$ which has a $\Delta^\fp_\oa$-flag starting at
$\Delta^\fp_\oa(\la+\xi_-)$. In particular,
$T_\oa^\fp(\la+\xi_-)$ is a direct summand
of $\Res T^\fp(\la)$.
Consequently, we have
$$\pd \, T_\oa^\fp(\la)= \pd\, T_\oa^\fp(\la+\xi_-) \leq \pd\, \Res T^\fp(\la)\leq \pd\, T^\fp(\la)$$
by Lemma \ref{lem::31}.
This claim of the proposition follows.
\end{proof}

\subsubsection{Parabolic dimension shift}

This subsection is devoted to the following formula
for projective dimension.

\begin{thm} \label{thm::510}
Suppose that $M\in \mc O^\fp$ with $\epd_{\mc O^\fp}M<\infty$. Then we have
\begin{align*}
&\epd_{\mc O^\fp}M = \epd_{\mc O}M   -2\ell(w_0^\fp).
\end{align*}
\end{thm}

\begin{proof}
We first claim that
\[\pd_{\mc O^\fp}M < \infty \Rightarrow \pd_{\mc O}M < \infty. \]
To see this, we let $Q$ be a projective module in $\mc O^\fp$.
The adjoint pair  $\Ind$ and $\Res$ of functors between
$\mc O_\oa^\fp$ and $\mc O^\fp$ gives an epimorphism
$\Ind\Res Q \rightarrow Q$, and so $Q$ is a direct summand of
$\Ind \Res Q$. Using Frobenius reciprocity, it follows that
\[\text{Ext}_{\mc O}^d(\Ind \Res Q, N)\cong
\text{Ext}_{\mc O_\oa}^d( \Res Q, \Res N)=0,\]
for any $d>\pd_{\mc O_\oa}\Res Q$ and $N\in \mc O$.
As a consequence, $\pd_{\mc O}Q<\infty.$

Now suppose that $\pd_{\mc O^\fp}M < \infty$. This means that
$M$ has a finite projective resolution in $\mc O^\fp$:
\[   0 \rightarrow Q^d\xrightarrow{f^d} Q^{d-1}\xrightarrow{f^{d-1}}
\cdots \xrightarrow{f^2} Q^1\xrightarrow{f^1} Q^0 \xrightarrow{f^0} M \rightarrow 0.\]
We define $K^s$ to be the kernel of the map $f^s$, for each $s$. Then
the inequality $\pd_{\mc O}K^{s}<\infty$, together with the exactness
of $0\rightarrow K^s \rightarrow Q^{s-1} \rightarrow K^{s-1}\rightarrow 0$,
imply that $\pd_{\mc O}K^{s-1}<\infty$. Consequently,
we have $\pd_{\mc O}M < \infty$.

Now suppose that both $\pd_{\mc O}M$ and $\pd_{\mc O^\fp}M$ are finite. Then, by \cite[Corollary~3.2.5]{Co} and \cite[Theorem~7.2.1(iii)]{Co},
we have
$$\pd_{\mc O}M=\pd_{\mc O_\oa}\Res M
\quad\text{ and }\quad\pd_{\mc O^p}M=\pd_{\mc O^\fp_0}\Res M.$$
The proof now follows from \cite[Theorem~4.1]{CoM2}.
\end{proof}

\begin{cor} Let $M \in \mc O^\fp$ with $\epd_{\mc O^\fp} M<\infty.$
\begin{enumerate}
\item[(1)] 	$M$ is projective in $\mc O^\fp$ if
and only if  $\epd_{\mc O}M =2\ell(w_0^\fp)$.
\item[(2)] If $\fp=\fg$, then $\epd_{\mc O}M =2\ell(w_0)$.
\end{enumerate}
\end{cor}
\begin{proof}
Claim~$(1)$ follows directly from Theorem~\ref{thm::510}.

If $\fp=\fg$, then $\mc O^\fp$ is the category of finite dimensional modules. In this case, we have
$\text{fin.dim}\,\mc O^\fp =  2\ell(w_0) - 2\ell(w_0^\fp) =0$
by \eqref{eq::egfindim}, which implies that $\pd_{\mc O^\fp}M=0$.
Claim~$(2)$ now follows from Theorem~\ref{thm::510}.
\end{proof}

\subsection{Injective   dimension of modules in $\mc O^\fp$} \label{sect::54}

Let $\mu$ be a dominant integral weight whose singularity determines the parabolic
subalgebra $\fp_\oa$.  Following \cite[Section~1.4]{CCC}, there is a parabolic
subalgebra $\hat \fp$ (in the sense of \cite[Section~2.4]{Ma}) of $\mf g$ such
that $\hat \fp_\oa$ is determined by $\hat \mu:= -w_0\mu$. For instance, if $\mf g$
is of type I, then $\hat \fp= \hat \fp_\oa \oplus \mf g_{-1}$.
Also, we have $\hat{\mf b} =\mf b^r$ for $\pn$ from Section \ref{sect::32}.

We recall, see \cite[Section~1.3]{CCC}, that there is a duality
(i.e. an anti-equivalence) $\bf D: \mc O^\fp\rightarrow \mc O^{\hat \fp}$ such that
\begin{align} \label{eq::D}
&{\bf D} L^\fp(\la) =L^{\hat \fp}(-w_0\la),  ~{\bf D}\Delta^\fp(\la) = \nabla^{\hat \fp}(-w_0\la), \text{ and }{\bf D}T^\fp(\la) = T^{\hat \fp}(-w_0\la).
\end{align}
We refer   to \cite[Proposition~3.4, Lemma~3.6]{CCC} for more details.
Combined with the results in Subsection~\ref{sect::52}, we have the following statement.

\begin{prop}\label{prop::injdim}
For any $M\in \mc O^\fp$, we have $ \eid_{\mc O^\fp}M= \epd_{\mc O^{\hat \fp}}{\bf D}M.$

	Furthermore, we have the following properties:
	\begin{itemize}
		\item[(1)]  Suppose that $\epd_{\mc O^{ \fp}} M<\infty$, then
		\begin{align}
		&\eid_{\mc O^\fp} M = \eid_{\mc O_\oa^{\fp}} \Res M. \label{eq::551}
		\end{align}

		\item[(2)] Suppose that $\eid_{\mc O^{ \fp}} M<\infty$, then
	\begin{align}
	&\eid_{\mc O^\fp}M = \eid_{\mc O}M   -2\ell(w_0^\fp). \label{eq::552}
	\end{align}

		\item[(3)] Suppose that $\g$ is of type I. Then, for any $\la \in \Sigma^+_\fp$, we have
	\begin{align}
	&\eid_{\mc O^\fp}\, P^\fp(\la) =\eid_{\mc O^\fp_\oa}\, P_\oa^{ \fp}(\la), \label{eq::553}\\
	&\eid_{\mc O^\fp}\, T^\fp(\la) = \eid_{\mc O^\fp_\oa} \,T_\oa^{  \fp}(\la).  \label{eq::554}
	\end{align}
		\end{itemize}
\end{prop}

\begin{proof}
Applying $\bf D$, we have
	\begin{align}
	&\id_{\mc O^\fp} M =\pd_{\mc O^{\hat \fp}}{\bf D}M.
	\end{align}
Suppose that $\pd_{\mc O^{\hat \fp}}{\bf D}M <\infty$. We note that
$\Res$ intertwines $\bf D$ and the usual  duality on $\mc O^\fp_\oa$
which we also denote by $\bf D$, abusing notation
(cf. the proof of \cite[Theorem~3.7]{CCC}).
Therefore, from \cite[Corollary~3.2.5]{Co} and \cite[Theorem~7.2.1]{Co},  it follows that
	$$\pd_{\mc O^{\hat \fp}}{\bf D}M= \pd_{\mc O_\oa^{\hat \fp}}\Res{\bf D}M=\pd_{\mc O_\oa^{\hat \fp}}{\bf D}\Res M=\id_{\mc O_\oa^{\fp}} \Res M.$$
	This proves Part $(1)$ and Equation \eqref{eq::551}.

 To prove Part $(2)$, we note that, from Theorem \ref{thm::510} and  the proof of \cite[Theorem~3]{Ma}, we obtain $\id_{\mc O^\fp} M<\infty \Rightarrow \id_{\mc O}M <\infty$. Therefore we may assume that both $\id_{\mc O} M$ and $\id_{\mc O^\fp}M$ are finite. From Equation \eqref{eq::551} and \cite[Theorem 4.1]{CoM2}, we have
	\begin{align}
	&\id_{\mc O^\fp} M = \id_{\mc O_\oa^{\fp}} \Res M =
	\id_{\mc O_\oa} \Res M-2\ell(w_0^\fp) =\id_{\mc O} M -2\ell(w_0^\fp).
	\end{align} This proves Equation \eqref{eq::552} and Part $(2)$.
	An alternative proof of Part $(2)$ is due the following observation:
	\begin{align*}
	&\pd_{\mc O^\fp}M = \pd_{\mc O^{\hat \fp}}{\bf D}M = \pd_{\mc O}{\bf D}M   -2\ell(w_0^{\hat \fp}) \\
	&=  \pd_{\mc O}{\bf D}M   -2\ell(w_0^\fp) = \pd_{\mc O^\fp}{\bf D}M=\id_{\mc O^\fp}M.
	\end{align*}

	 It remains to prove Part $(3)$.
By \eqref{eq::D}, we have $$
\id_{\mc O^\fp} P^\fp(\la) = \pd_{\mc O^{\hat \fp}} I^{\hat \fp}(-w_0\la),~
\id_{\mc O^\fp} T^\fp(\la) = \pd_{\mc O^{\hat \fp}} T^{\hat \fp}(-w_0\la).$$
 It follows that
 \[\id_{\mc O^\fp} P^\fp(\la) = \pd_{\mc O_\oa^{\hat \fp}} I_\oa^{\hat \fp}(-w_0\la)
 =\pd_{\mc O_\oa^{\fp}} I_\oa^{ \fp}(\la)=\id_{\mc O_\oa^{\fp}} P_\oa^{ \fp}(\la), \]
  \[ \id_{\mc O^\fp} T^\fp(\la) = \pd_{\mc O_\oa^{\hat \fp}} T^{\hat \fp}_\oa(-w_0\la)=
  \pd_{\mc O_\oa^{\fp}} T_\oa^{ \fp}(\la), \]
 by  Proposition \ref{prop::53}, Theorem \ref{thm::32} and \cite[Theorem 11]{So90}.
\end{proof}

\subsection{A sufficient condition for infiniteness of projective dimension}  \label{sect::55}
We recall the  definitions of the {\em associated variety} from \cite{DS}.
Let $M\in \g$-Mod and $x\in \mf g_{\ov 1}$ with $[x,x]=0$.
Define $$M_x:= \text{ker}_x/xM,$$ where $\text{ker}_x$ is the kernel
of the map $x: M\rightarrow M$ sending $m$ to $xm$, where $m\in M.$
Then the associated variety $X_M$ of $M$ is defined as
\[X_M:=\{x\in \g_\ob|~[x,x]=0 \text{ and } M_x\neq 0\}.\]

The following statement is \cite[Lemma~2.2(i)]{DS}:

\begin{lem}\label{lem::DS}
Let $M\in \g\emph{-Mod}$ be a summand of $\Ind N$, for some $N\in \mf g_\oa\emph{-Mod}$. Then $X_M=0$.
\end{lem}

\begin{lem} \label{lem::36}
Let $0\rightarrow A \rightarrow B  \xrightarrow{f}  C  \rightarrow 0$ be a short exact
sequence in $\g\emph{-Mod}$. Let $x\in \mf g_\ob$ be nonzero with $[x,x]=0$.
Suppose that $A_x =B_x =0$. Then we have  $C_x =0.$

In particular, if $X_A=X_B=0$, then $X_C=0$.
\end{lem}

\begin{proof}
Let $c\in C$ be such that $xc=0$. We want to show that there exists
$b\in B$ such that $c=xf(b)$, which implies that $C_x =0$.
To see this, we first pick $d\in B$ such that $f(d)=c$.
Then $f(xd)=0$ implies that $xd\in A$.

Now $x(xd) =x^2d =0$ (due to $x^2=0$) implies $xd\in  xA$ and so
there is $a\in A$ such that $xd =xa$ since $A_x =0$.
Finally, we note that $x(d-a)=0$ implies that $d-a \in xB$ since $B_x=0$.
Therefore there is $b\in B$ such that $d-a =xb$. We now apply $f$ and obtain
$$c=f(d)=f(d-a)=xf(b),$$ as desired.
\end{proof}

The following observation is known (see, e.g., \cite[Section 4, proof of Theorem 4.1]{CS} and \cite[Lemma 2.2]{DS}).

\begin{cor} \label{cor::59}
Suppose $\tilde{\mc C}$ and $\mc C$ are respective  subcategories of $\g$\emph{-Mod} and
$\g_\oa$\emph{-Mod} satisfying assumptions \eqref{eq::111}-\eqref{eq::113}.
If $M \in\tilde{\mc C}$ is such that $X_M\neq 0$, then $\epd_{\tilde{\mc C}} M=\infty$.

In particular,	if $M \in \mc O^\fp$ is such that $X_M\neq 0$,
then $\epd_{\mc O^\fp} M=\infty$.
\end{cor}

\begin{proof}
Suppose that there exists a finite projective resolution of $M$ in $\tilde{\mc C}$:
\[0 \xrightarrow{f^{d+1}} P^d \xrightarrow{f^d} \cdots \xrightarrow{} P^2\xrightarrow{f^2}  P^1\xrightarrow{f^1}  P^0 \rightarrow M \xrightarrow{f^0}  0.\]

Let $0\leq s\leq d$. By \cite[Proposition~2.2.1]{Co}, every projective module $P^s$ is a direct summand of a module induced from $\mc C$ to $\tilde{\mc C}$. By Lemma \ref{lem::DS} we
thus have $X_{P^s} = 0$.

Define $K^s$ to be the kernel of $f^s$. For any $s$, we have an exact sequences
$$0\rightarrow K^s \rightarrow P^{s-1} \rightarrow K^{s-1}\rightarrow 0.$$
In particular, the fact that $K^d=P^d$ implies that $K^{d-1}_x=0$ by Lemma \ref{lem::36}.  By induction on $s$, we may conclude that $M_x=K^0_x =0$.  The claim of the lemma follows.
\end{proof}

\begin{ex}
If $\mf g_{\ov 1}$ contains a non-zero $x$ such that $[x,x]=0$,
then it is obvious (see also \cite[Lemma 2.2]{DS})
that the associated variety $X_\C$ of  the trivial module $\C$ is non-zero.
Therefore we may conclude that both $\pd_{\g\text{-Mod}}\C$ and $\pd_{\mc O}\C$ are infinite.
\end{ex}

\begin{rem}
It is worth pointing out that the condition $X_M=0$ is independent of the choice of the parabolic subalgebra $\fp$  (compare with Theorem \ref{thm::510}). It is natural to ask the following question:

{\bf Question:} Let $M\in \mc O$. Is it true that $\pd_{\mc O} M<\infty \Leftrightarrow X_M=0?$

This question has the affirmative answer for the general linear Lie superalgebra
$\gl(m|n)$ by \cite[Theorem 4.1]{CS}.
\end{rem}

The following lemma addresses infiniteness of projective dimension
for $\fp$-highest weight modules.

\begin{cor} \label{coro::infproj}
Suppose that there is a weight $\alpha$ of $\mf g_1$ such that $-\alpha$ is not a weight of $\Lambda^{\bullet} \mf g_{-1}\otimes U(\mf n^-_\oa).$ Then the associated variety
of every non-zero quotient of $\Delta^\fp(\la)$ is non-zero.
Consequently, every such quotient has infinite projective dimension in both $\mc O$ and
$\mc O^\fp$, for any $\la \in \Sigma_\fp^+$.

In particular, every block of $\mc O^\fp$ contains infinitely many simple objects.
\end{cor}

\begin{proof}
Assume that $M$ is a non-zero quotient of $\Delta^\fp(\la)$
of finite projective dimension.
Let $X_{\alpha} \in \g_1$ be a root vector of  root $\alpha$.

Pick a non-zero highest weight vector $v\in M$.   By Corollary \ref{cor::59}, we have $M_{X_{\alpha}}=0$. Therefore $X_{\alpha} v=0$ implies that  there is $m\in M$ such that $v=X_{\alpha}m$.
However,  $\la -\alpha$ is not a weight of $M$, a contradiction.

Suppose that $\mc O^\fp_\la$ is a block of $\mc O^\fp$ containing only finitely many simples.
Combining existence of tilting modules  \cite[Section 4.3]{Ma}
(also, see \cite[Theorem 3.5]{CCC}), the Ringel duality \cite[Section 5.4]{Ma}
(also, see, \cite[Corollary 3.8]{CCC}) and  the BGG reciprocity  \cite[Lemma 3.3]{CCC}
for $\mc O^\fp$, we may conclude that $\mc O^\fp_\la$ contains a simple tilting module.
This is a contradiction to Proposition \ref{thm::32}. This completes the proof.
\end{proof}

\begin{ex}
Let $\mf k$ be classical of type I such that there exists $d\in \mc Z(\mf k)$ and
$d$ acts on $\mf k_{1}$ as a non-zero scalar. Set $\mf g:=\mf g_\oa\oplus \mf g_{1}$.
Then every simple module in the category $\mc O$ for $\mf g$ has infinite projective dimension.
\end{ex}

We are going to apply Corollary \ref{coro::infproj} to the periplectic Lie
superalgebra $\pn$ later in Proposition \ref{prop::512}.

\section{Examples: projective dimensions of structural supermodules over $\pn$ and  $\mf{osp}(2|2n)$} \label{Sect::6}

\subsection{Example I: projective dimension of  modules for $\mf{osp}(2|2n)$} \label{Sect::projdimosp}

\subsubsection{}
The orthosymplectic Lie superalgebra $\mf{osp}(2|2n)$ is the
following subsuperalgebra of $\mf{gl}(2|2n)$:
\begin{equation}
\mf{osp}(2\vert 2n)=
\left\{ \left( \begin{array}{cccc} c &0 & x &y\\
0 & -c& v & u\\
-u^t& -y^t & a &b\\
v^t &x^t & c& -a^t \\
\end{array} \right):
\begin{array}{c}
c\in \C;\,\, x,y,v,u\in \C^{n};\\
a,b,c\in \C^{n^2};\\
b=b^t,\,\, c=c^t.
\end{array}
\right\}. \label{eq::osp}
\end{equation}
The even part $\mf{osp}(2|2n)_\oa$ of $\mf{osp}(2|2n)$ is isomorphic to
$\C \oplus \mf{sp}(n)$, and the odd part is isomorphic to $\C^2\otimes \C^{2n}$.
We refer the reader to \cite[Section 1.1.3]{ChWa12} and \cite[Section 2.3]{Mu12}
for more details.  Recall from  Section \ref{sect::321} that we denote by
$E_{ab}$ the elementary matrices. We now define
$$H_\vare:=E_{11}-E_{22},~H_i:=E_{2+i,2+i}-E_{2+i+n,2+i+n},
\quad \text{ for }1\leq i\leq n.
$$
The Cartan subalgebra $\mf h$ of $\mf{osp}(2|2n)$
is spanned by $H_\vare$ and $H_1,H_2,~\ldots, H_n$. We let
$\{\vare,\delta_1,\delta_2,\ldots,\delta_n\}$ be
the basis of $\mf h^\ast$ dual to  $\{H_\vare,H_1,H_2,\ldots, H_n\}$.
The  bilinear form $({}_-,{}_-):\mf h^\ast \times \mf h^\ast \rightarrow \C$ is given by $(\vare,\vare) =1$ and $(\delta_i,\delta_j) =-\delta_{i,j}$, for $1\leq i,j\leq n.$

We define the set $\Phi^+_\oa$ of even positive roots  and
the set $\Phi^+_\ob$ of odd positive roots  by
\begin{align}
&\Phi^+_\oa:=\{\delta_i \pm\delta_j, 2\delta_p|~1\leq i< j <n, ~1\leq p \leq n \}.\\
&\Phi_\ob^+:=\{\vare \pm \delta_p|~1\leq p\leq n\}.
\end{align}
This gives rise to a triangular decomposition
$\mf g=\mf n^+\oplus \mf h \oplus \mf n^-$ with Borel subalgebra $\mf b:= \mf b_\oa \oplus \mf g_1.$

\subsubsection{Simple supermodules over $\mf{osp}(2|2n)$}
The aim of this subsection is to study the projective dimension of simple
supermodules over arbitrary basic classical Lie superalgebras of type I including $\mf{osp}(2|2n)$.

We recall the notation of typical weights for basic classical Lie superalgebras
of type I, see, e.g., \cite[Section 2.4]{Gor2}.
Let $X^{-}\in \Lambda^{\text{max}}\mf g_{-1}$ and
$\quad X^{+}\in \Lambda^{\text{max}}\mf g_{1}$ be non-zero vectors.
Then by   \cite[Lemma 6.5]{CM} (also, see, \cite[Section 4]{Gor1}) it follows that
\begin{align}
&X^{+} X^- = \Omega + \sum_i x_i r_i y_i, \label{eq::XYeq}
\end{align}
for some $x_i \in \Lambda(\mf g_{ -1})\backslash \mathbb{C}$,
$y_i \in \Lambda(\mf g_{1})\backslash \mathbb{C}$,
$r_i\in U(\mf g_{\oo})$ and $\Omega \in Z(\mf g_\oo)$.
A weight $\la \in \h^\ast$ is typical if its evaluation at the
Harish-Chandra projection of $\Omega$ is non-zero.
It was shown in \cite[Theorem 1]{Ka2} that the typicality of
$\la$ is equivalent to the simplicity of the Kac module of
highest weight $\la$. A more general definition of {\em typical Kac modules}
can be found in \cite[Section 6.2]{CM}.

The following proposition shows that all simple supermodules of atypical
blocks over basic classical Lie superalgebras have infinite projective dimension
in $\mc O$. This generalizes \cite[Proposition~5.14]{CS} where the case
of the superalgebra
$\gl(m|n)$ was considered.

\begin{prop} \label{thm::basicgpdsimple}
Let $\mf g$ be a basic classical Lie superalgebra of type I.
Suppose that $\la$ is atypical. Then the associated variety
$X_{L(\la)}$ is non-zero. In particular, if  $L(\la) \in \mc O^\fp$,
for some $\fp$, then $\epd_{\mc O^\fp}L(\la)=\infty.$
\end{prop}

Before we prove Proposition \ref{thm::basicgpdsimple}, we need to
recall the notion of odd reflection for basic classical Lie superalgebras,
see, e.g., \cite[Lemma 1]{PS89}, \cite[Section 1.4]{ChWa12} and
\cite[Section 3]{Mu12}, for more details.

Since $\mf g$ is contragradient,  the set of roots of $\mf g_{-1}$ can be obtained
from the set of roots of $\mf g_{1}$ by multiplying with the scalar  $-1$.
Set $d:=\text{dim}\mf g_1=\text{dim}\g_{-1}$.
By \cite[Lemma 1.30, Proposition 1.32]{ChWa12} (also, see \cite[Theorem 3.13]{Mu12}),
the positive root system given by the Borel subalgebra $\mf b_\oa\oplus\mf g_{-1}$
can be obtained from that of  $\mf b$ via a sequence of odd reflections.
Namely, we can fix an ordering of   positive odd roots
\begin{align}
&\{\alpha_1, \alpha_2, \ldots, \alpha_d\} = \Phi^+_\ob \label{eq::oddfleordering}
\end{align}
such that the following conditions are satisfied:
\begin{itemize}
\item[(1)] For any $1\leq i\leq d$, the set
$R^{i} =\Phi^+_\oa \cup \{-\alpha_1,\ldots ,-\alpha_{i}\}
\cup  \{\alpha_{i+1},\ldots, \alpha_d\}$ forms a positive root system.
\item[(2)]  We set $R^{-1}:=R^0$. Then $\alpha_i$ is a
simple root in $R^{i-1}$ and $-\alpha_i$ is a simple
root in $R_{i}$, for any $1\leq i\leq d$.
\end{itemize}

Let us fix root vectors $X_i\in \mf g_1\cap \mf g^{\alpha_i}$ and
$Y_i \in \mf g_{-1}\cap \mf g^{-\alpha_i}$, for $1\leq i \leq n$.
Also,	let $\mf b^{(i)}$ denote the Borel subalgebra corresponding
to $R^i$, namely, $\mf b^{(i)}_\oa = \mf b_\oa$ and $\mf b^{(i)}_\ob$
is generated by all root vectors of odd roots in $R^i$.
By \cite[Lemma 1.40]{ChWa12}, we have the following lemma:

\begin{lem} \label{lem::29}
Suppose that $L\in \mc O$ is a simple supermodule having $\mf b^{(i)}$-highest
weight  $\mu\in \h^\ast$. Let $v\in L$ be a $\mf b^{(i)}$-highest weight vector.
\begin{itemize}
\item[(1)] 	If $Y_{i+1} v=0$, then $v$ is a  $\mf b^{(i+1)}$-highest weight vector of $L.$
\item[(2)]  If  $Y_{i+1} v\neq 0$, then $Y_{i+1} v$  is a $\mf b^{(i+1)}$-highest
weight vector of $L$.
\end{itemize}
\end{lem}

\begin{ex}
Let $\mf g= \mf{osp}(2|2n)$. We start with the simple system
(see \cite[Section 1.3.4]{ChWa12})
\[\{\vare-\delta_1,~\delta_2-\delta_3,~\ldots,~\delta_{n-1}-\delta_n,~ 2\delta_n\}.\]
Then, by a direct computation using \cite[Equation (1.44) of Lemma 1.30]{ChWa12},
the ordering  \eqref{eq::oddfleordering} in this case can be chosen as follows:
\begin{align}
&\vare -\delta_1,~\vare-\delta_2,~\ldots,~ \vare-
\delta_n,~\vare +\delta_n,~\vare +\delta_{n-1},~\ldots,\vare+\delta_1. \label{eq::osproots}
\end{align}
\end{ex}

Now we are in a position to  prove Proposition \ref{thm::basicgpdsimple}.

\begin{proof}[Proof of Proposition \ref{thm::basicgpdsimple}]
Let $v\in L(\la)$ and $v'\in K(\la)$  be highest weight vectors.
Since $\la$ is atypical, the kernel of the natural epimorphism
$\phi: K(\la)\rightarrow L(\la)$ is non-zero.
By \cite[Lemma 3.2]{CM}, the kernel of $\phi$ contains $Y_dY_{d-1}\cdots Y_1v'$
and so $Y_dY_{d-1}\cdots Y_1v= 0$. This means that there is
$0\leq \ell < d$ such that $Y_{\ell} Y_{\ell - 1}\cdots Y_1 v\neq 0$
and $Y_{\ell+1} Y_{\ell}\cdots Y_1 v =0$. By
Lemma \ref{lem::29}, $Y_{\ell} Y_{\ell - 1}\cdots Y_1 v$ is a
highest weight vector of $L$ with respect to both $\mf b^{\ell}$
and $\mf b^{\ell+1}$. Therefore the weight spaces corresponding to the weights
$\la-\alpha_1 +\cdots -\alpha_{\ell}\pm \alpha_{\ell+1}$ of $L(\la)$
are zero. In particular, the vector
$Y_{\ell} Y_{\ell - 1}\cdots Y_1 v\not \in Y_{\ell+1} L(\la)$.
This means that that $L(\la)_{Y_{\ell+1}}\neq 0$
and the claim  of our lemma follows from Corollary \ref{cor::59}.
\end{proof}

\subsubsection{(Dual) Verma supermodules  over $\mf{osp}(2|2n)$}

In this subsection, we assume that $\mf g=\mf{osp}(2|2n)$, for some positive integer $n$.
The following proposition shows that atypical Verma supermodules have infinite projective dimension.  We recall that $\Sigma^+_\fp$ denotes the set of highest weights of simple modules in $\mc O^\fp$.

\begin{prop}  \label{thm::pdospVerma}
Suppose that $\la\in \Sigma_\fp^+$.
Then the following conditions are equivalent:
\begin{itemize}
\item[(1)] $\epd_{\mc  O^\fp}\Delta^\fp(\la) <\infty$.
\item[(2)] $\epd_{\mc  O}\Delta^\fp(\la) <\infty$.
\item[(3)] $\epd_{\mc  O^\fp}\Delta^\fp(\la) = \epd_{\mc  O^\fp_\oa}\Delta^\fp(\la)
= \epd_{\mc  O}\Delta^\fp(\la) -2\ell(w_0^\fp)$.
\item[(4)] $\la$ is typical.
\end{itemize}
\end{prop}

\begin{proof}
We first note that the implication $(1)\Rightarrow (3)$ follows from
\cite[Corollary~3.2.5]{Co} and \cite[Lemma 6.3.2]{Co}.
Therefore we have $(1)\Leftrightarrow (3)$. From \cite[Theorem~1.3.1, Theorem 1.4.1]{Gor2}
it follows that $(4)\Rightarrow (1), (2)$.  Therefore it remains to show that, if
$\la$ is atypical, then the associated variety of $\Delta^\fp(\la)$ is non-zero.

We first define an ordering of roots of $\mf g_{-1}$
(i.e. roots in $\Phi_\ob \backslash \Phi^+_\ob$) as follows:
\begin{align}
&\alpha_1:=-\vare -\delta_1<\alpha_2:=-\vare-\delta_2<\ldots< \alpha_{n}:=-\vare- \delta_n\\&< \alpha_{n+1}:=-\vare +\delta_n
<\alpha_{n+2}:=-\vare +\delta_{n-1}<\ldots<\alpha_{2n}:=-\vare+\delta_1. \label{eq::osproots2}
\end{align}
By \eqref{eq::osp}, for each $1\leq i \leq n$, there exist non-zero
vectors $X_i \in \g^{-\alpha_i}$ and $Y_{i}\in \g^{\alpha_i}$ such that
\begin{align}
&[X_i,Y_i] = H_\vare-H_{i},\text{ for $1\leq i\leq n$},\\
&[X_{n+i},Y_{n+i}] = -H_\vare-H_{n-i+1} ,\text{ for $1\leq i\leq n$}.
\end{align}
We observe the following crucial fact:
\begin{align}
&\mf n^+_\oa Y_i \subset \bigoplus_{i< j}\C Y_j,~[X_i,Y_j] \subset \mf n^+_\oa,
\label{eq::611}
\end{align}
for any $1\leq i<j\leq 2n.$  As a consequence, we have
\begin{align}
&X_iY_{j}\cdots Y_{2n}v  =0, \label{eq::613}
\end{align} for any $1\leq i<j\leq 2n$.

Let $\displaystyle\rho : =-n \vare+ \sum_{1\leq i\leq n} (n-i+1) \vare_i$ be the Weyl
vector, i.e., $\rho$ is the half of the sum of roots in $\Phi_\oa^+$ minus the half of
the sum of roots in $\Phi^+_\ob$. Since $\la$ is atypical, we have
\begin{displaymath}
\prod_{\alpha\in\Phi^+_{\bar{1}}}(\la+\rho,\alpha) =0,
\end{displaymath}
by \cite[Subsection~4.2]{Gor1}. Therefore there exists $\alpha = \vare+c\delta_i$
such that $(\la +\rho ,\alpha) =0$ and $c=\pm 1.$ We let $v$ be a highest
weight vector of $\Delta^\fp(\la)$. Also, we consider the element
$\displaystyle\la = \la_\vare \vare+\sum_{1\leq i\leq n} \la_i \delta_i$.

Suppose that $c=1$. By \eqref{eq::613} and direct computation, we
get $X_iY_{i+1}\cdots Y_{2n}v =0$ and
\begin{align*}
&X_iY_iY_{i+1}\cdots Y_{2n}v \\
&= (H_\vare -H_i)Y_{i+1}\cdots Y_{2n}v \\
&= (\la_\vare -\la_i-2n+i -1) Y_{i+1}\cdots Y_{2n}v \\
&=(\la+\rho, \alpha)Y_{i+1}\cdots Y_{2n}v =0.
\end{align*}
Since the weight subspace of $\Delta^\fp(\la)$ of weight
$\displaystyle\la-\sum_{p\leq q\leq 2n}\alpha_q$ is one-dimensional and spanned by
$Y_{p}Y_{p+1}\cdots Y_{2n}v$, for any $1\leq p\leq 2n$, we may conclude
that $Y_{i+1}\cdots Y_{2n}v \notin X_i\Delta^\fp(\la)$, for otherwise
$Y_{i+1}\cdots Y_{2n}v=0$, which is a contradiction. Consequently,
$X_{\Delta^\fp(\la)}\neq 0$. Hence, in this case, the fact that
the associated variety of $\Delta^\fp(\la)$ is non-zero
follows from  Corollary~\ref{cor::59}.

Suppose that $c=-1$. Set $\ov i:=n-i+1$, for each $1\leq i\leq n$.
Similarly, using \eqref{eq::613} and direct computation, we
obtain $X_{n+\ov i}Y_{n+\ov i +1 +1}Y_{2n}v=0$ and
\begin{align*}
&X_{n+\ov i}Y_{n+\ov i}Y_{n+\ov i+1}\cdots Y_{2n}v \\
&= -(H_\vare +H_i)Y_{n+\ov i+1}\cdots Y_{2n}v \\
&=(-1) (\la_\vare +\la_i -i+1) Y_{n+\ov i+1}\cdots Y_{2n}v \\
&=-(\la+\rho, \alpha)Y_{n+\ov i+1}\cdots Y_{2n}v =0.
\end{align*}
As above, this implies that $Y_{n+\ov i+1}\cdots Y_{2n}v \notin X_{n+\ov i}\Delta\fp(\la)$
and yields  $X_{\Delta^\fp(\la)}\neq 0$. Again, the fact that
the associated variety of $\Delta^\fp(\la)$ is non-zero
now follows from  Corollary \ref{cor::59}.
This completes the proof.
\end{proof}

\begin{cor}
Suppose that $\la\in \Sigma_\fp^+$.
Then the following are equivalent:
\begin{itemize}
		\item[(1)] $\epd_{\mc  O^\fp}\nabla^\fp(\la) <\infty$.
\item[(2)] $\epd_{\mc  O}\nabla^\fp(\la) <\infty$.
\item[(3)] $\epd_{\mc  O^\fp}\nabla^\fp(\la) = \epd_{\mc  O^\fp_\oa}\nabla^\fp(\la) = \epd_{\mc  O}\nabla^\fp(\la) -2\ell(w_0^\fp)$.
\item[(4)] $\la$ is typical.
\end{itemize}
\end{cor}
\begin{proof}

The implication $(1)\Rightarrow (3)$ follows from
\cite[Corollary~3.2.5]{Co} and \cite[Lemma 6.3.2]{Co}. Therefore the statements in  $(1)$ and $(3)$ are equivalent.

	We recall that $\mf g= \mf{osp}(2|2n)$ is  one of the contragredient Lie superalgebras in \cite[Theorem 5.1.5]{Mu12}. Therefore $\mc O^\fp$ admits a simple-preserving duality (see, e.g., \cite[Section 13.7]{Mu12}) which sends dual Verma supermodules to Verma supermodules. By \cite[Theorem 3]{Ma}, it follows from Proposition \ref{thm::pdospVerma} that the statements in $(1),(2)$ and $(4)$ are equivalent.

	\end{proof}

\subsection{Example II: projective dimension of  modules for $\pn$} \label{Sect::projdimpn}

In this section, we assume that $\mf g=\pn$, for some positive integer $n$.
It is known from \cite[Section~ 5.5]{CCC} that  every parabolic category
is equivalent to the parabolic category given
by a parabolic subalgebra  $\mf p$ with $\mf p_\ob=\mf  g_1$.
We recall that  $\Sigma^+_\fp$ denotes the set of highest weights of simple modules
in $\mc O^\fp$.

\begin{prop} \label{prop::512}
Suppose that $M\in \mc O^\fp$ is a non-zero quotient of $\Delta^\fp(\la)$,
for some $\la \in \Sigma^+_\fp$.
Then the associated variety $X_M$ is non-zero. In particular,
$$\epd_{\mc O}M =\epd_{\mc O^\fp}M =\infty.$$
In particular, the modules $L(\la),~K(\la)$ and $\Delta^\fp(\la)$
all have infinite projective dimension in both $\mc O$ and $\mc O^\fp$.
\end{prop}

\begin{proof}
We may observe that the element $2\vare_n$ is a root of $\mf g_1$ but $-2\vare_n$
is not a weight of $\Lambda^{\bullet} \mf g_{-1}\otimes U(\mf n^-_\oa)$.
The assertion of the proposition now follows directly from Corollary~\ref{coro::infproj}.
\end{proof}

We recall from \cite{Se02} that a weight
$\displaystyle\la =\sum_{1\leq i\leq n}\la_i\vare_i\in \h^\ast$ is called {\em atypical}  if
\[  \prod_{1\leq i\neq j \leq  n}(\la_i -\la_j +j-i-1) =0. \]
We describe the projective dimension of costandard objects in the  following proposition.

\begin{thm}
Let  $\la \in \Sigma^+_\fp$.  Then the following
conditions are equivalent:
\begin{enumerate}
\item $\epd_{\mc O^\fp} \nabla^\fp(\la) <\infty.$
\item $\epd_{\mc O^\fp} \nabla^\fp(\la) =\epd_{\mc O_\oa^\fp} \nabla^\fp_\oa(\la).$
\item $\la$ is typical.
\end{enumerate}
\end{thm}

\begin{proof}
From \cite[Corollary 3.2.5]{Co} and \cite[Lemma 6.3.2]{Co}  it follows that
$$\pd_{\mc O^\fp}\nb^\fp_{\mc O^\fp}(\la) = \pd_{\mc O^\fp_\oa}\nb^\fp_\oa(\la), $$
if $\pd_{\mc O^\fp}\nabla^\fp(\la)<\infty.$ Therefore it remains to show that
$\pd_{\mc O^\fp}\nb^\fp(\la)<\infty$ if and only if $\la$ is typical.

We first suppose that $\la$ is typical. Then, by \cite[Theorem 4.6]{CP}, we have
\[(T^\fp(\la): \nabla^\fp(\mu)) = (T^\fp_\oa(\la): \nabla^\fp_\oa(\mu)),\]
for any  $\mu\in \Sigma_\fp^+$.
We start by fixing $\la$ such that $T^\fp(\la) = \nabla^\fp(\la)$. That is,
let $J^\fp_\la$ be the  intersection of the set of shortest representatives
in $W^\fp \backslash W$ and the set of longest representatives in $ W/W_\la$.
We choose $\la$ such that it is  a  minimal element in $J^\fp_\la\cdot \la$.
We want to show that $\pd_{\mc O^\fp} \nabla(\mu) <\infty$,
for any $\mu \in J^\fp_\la \cdot \la$.

We first note that
\begin{align}
&\pd_{\mc O^\fp}\nabla^\fp(\la) =\pd_{\mc O^\fp} T^\fp(\la) =
\pd_{\mc O_\oa^\fp} T^\fp_\oa(\la) = \pd_{\mc O_\oa^\fp} \nabla^\fp_\oa(\la), \label{eq::55}
\end{align}
by Theorem \ref{thm::32}.
Assume that $x\in J^\fp_\la$ is such that $\pd_{\mc O^\fp}\nabla(y\cdot \la)<\infty,$
for any $y\in J^\fp_\la$ with $x\cdot \la> y\cdot \la.$ By \cite[Theorem 3.5]{CCC},
there is a submodule $K$ of $T^\fp(x\cdot \la)$ such that we have the short exact
sequence
\[0\rightarrow K\rightarrow T^\fp(x\cdot \la) \rightarrow \nabla^\fp(x\cdot \la) \rightarrow  0,\] where $(K:\nabla^\fp(z\cdot \la))>0$    only if $z\in J^\fp_\la$  and  $x\cdot \la >z\cdot \la$.
From our assumption, we have $\pd K<\infty$.
Since $\pd_{\mc O^\fp} T^\fp(x\cdot \la)<\infty$, it  follows
that $\pd_{\mc O^\fp} \nabla^\fp(x\cdot \la)<\infty$.
This completes the proof of the implication $(3)\Rightarrow (1).$

Before  proving the direction  $(1)\Rightarrow (3)$, we make the following observations.

For any positive odd root $\alpha = \vare_i +\vare_j$, we let $X_\alpha\in \g_1$ be a
root vectors for $\alpha$. If $\alpha =\vare_i +\vare_j$ with $i< j$, we can pick root
vectors  $Y_{\alpha}\in \g_{-1}$ for odd roots $-\alpha$  such that
$[Y_{\alpha}, X_{\alpha}] =H_i-H_j$.

Define a total ordering $\preceq$ on the set $\Phi_\ob^+:=\{\vare_i+\vare_j|~1\leq i<j\leq n\}$ by letting $$\vare_i +\vare_{j+1}\prec \vare_i+\vare_{j},~2\vare_{i}\prec \vare_{i-1} +\vare_n,$$
for $1< i\leq n$ and $1\leq j<n$. For any $1\leq i<j\leq n$ and any subset $$I\subset \hat I:=\{\vare_i+\vare_t|~i\leq t\leq j-1\},$$ we define
\begin{align*}
&S_{i,j,I}:=\{\vare_s+\vare_t|~s<t,~1\leq s\leq i-1,~1\leq t\leq j\}\cup I.
\end{align*}

Let $\la\in \h^\ast$ and  $v_\la \in \nabla^\fp_\oa(\la)\subset \nb^\fp(\la)$
be a highest weight vector. Set
$$X^{i,j,I}:=\prod_{\alpha \in S_{i,j,I}}X_\alpha\otimes v_\la \in \nb^\fp(\la).$$
We observe that
\begin{enumerate}[(a)]
\item 	For any $\alpha \in \Phi_\ob^+,$ we have
$$[X_\alpha,\mf n_\oa^+] \subseteq  \bigoplus_{\beta\in\Phi_\ob^+, ~\alpha \prec \beta}\C X_\beta.$$  \label{eq::1}
\item $[Y_{\vare_i+\vare_j}, X_{\vare_k+\vare_i}]\in \mf n_\oa^+$
has weight $\vare_k-\vare_j$, for $1\leq k\leq i$.
\item $[Y_{\vare_i+\vare_j}, X_{\vare_i+\vare_k}]\in \mf n_\oa^+$
has weight $\vare_k-\vare_j$, for $i<k<j$.
\item $[Y_{\vare_i+\vare_j}, X_{\vare_k+\vare_j}]\in \mf n_\oa^+$
has weight $\vare_k-\vare_i$, for $1\leq k<i$.  \label{eq::4}
\end{enumerate}
As a consequence, from \eqref{eq::1}-\eqref{eq::4} above, it follows that
\begin{align}
&Y_{\vare_i +\vare_j} X^{i,j,I}=0. \label{eq::k}
\end{align}

We can now proceed with the proof of $(1)\Rightarrow (3)$. Suppose that $\la$ is
integral and atypical with $\la_i -\la_j +j-i=\pm 1$, for some $i<j$, satisfying
$\pd\nb^\fp(\la)<\infty$. Then, by Corollary~\ref{cor::59}, the associated variety
$X_{\nabla^\fp(\la)}$ is zero.

We have to consider the following two cases:
       \vskip 0.2cm

{\bf Case 1}.
Suppose that $\la_i -\la_j +j-i+1=0$.

Let $x:=X^{i,j,\hat I}$ and $y:=Y_{\vare_i+\vare_j}$.
Then   $X_{\nabla^\fp(\la)}=0$ implies that $\nb^\fp(\la)_y=0$.
Therefore  $x\in y\nb^\fp(\la)$.

Assume that $m\in \nb^\fp(\la)$ is a weight vector such that
$yx=0$ and $x=ym$. Then $m \in  \C X_{\vare_i+\vare_j}X^{i,j,\hat I}.$
This implies that
\[x=ym\in \C Y_{\vare_i+\vare_j}X_{\vare_i+\vare_j}X^{i,j,\hat I} =
\C(H_i -H_j)X^{i,j,\hat I} = \la_i -\la_j +j-i+1 =0,  \]
a contradiction.
\vskip 0.2cm

{\bf Case 2}.
Suppose that $\la_i -\la_j +j-i-1=0$.
Set $I:= \hat I \backslash \{2\vare_i\}$.

Let $x:=X^{i,j, I}$ and $y:=Y_{\vare_i+\vare_j}$.
Then     $X_{\nabla^\fp(\la)}=0$ implies that $\nb^\fp(\la)_y=0$.
Again, we have  $x\in y\nb^\fp(\la)$.

Assume that $m\in \nb^\fp(\la)$ is a weight vector such that $yx=0$ and $x=ym$.
Observe that
$$(H_i-H_j)m = (\la_i-\la_j +j-i-1)m=0,$$
and
$$m =   X^{i,j,\hat I}\otimes m_j+ \sum_{i\leq t< j}
X^{i,j,\hat I\backslash \{\vare_i+\vare_t\}\cup \{\vare_i+\vare_j\}}\otimes m_t,$$
for some $m_j, m_t \in \nb_\oa(\la).$
By \eqref{eq::1}-\eqref{eq::4}, for $i\leq t <j$, we have
$$yX^{i,j,\hat I}\otimes m_j =0,$$
$$y X^{i,j,\hat I\backslash \{\vare_i+\vare_t\}\cup \{\vare_i+\vare_j\}}\otimes m_t
=(H_i-H_j)X^{i,j,\hat I\backslash \{\vare_i+\vare_t\}}\otimes m_t=0.$$
The obtained contradiction completes the proof.
\end{proof}

redpajama_set_name: RedPajamaArXiv


How many of us who know the Gede think about their fashion style? These infamous spirits of death and decay are so in-your-face with their sass, their salacious talk, their spot-on mimicry of our carnal secrets that the chic elements of their persona often slip by, like weird scraps of cemetery flotsam.
Schmooze with your fellow Gede over food and drink.
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3. After everybody has embraced the idea, when you mention it they will mock you and say: "Old hat, why not propose something original !!!". < now you walk way behind the music>.
So, the message of Arthur was in brief: You will be criticised before, during and after your invention, so DO NOT SEEK OR ASK FOR APPROVAL ! Especially not by parties with a vested interest !! Remember, they are the turkeys for the next Xmas dinner.
Rise above the need for Approval! Do whats right even if its unpopular. One's worth is independent of other's opinion.
So have courage and just go ahead and DO develop new idea's.
2. There is no demand for that.
3. That is something totally new.
4. We tried that before already without success.
5. We did not try that before.
6. That is not our responsibility.
7. That is a much too drastic change.
8. That is not scientific.
9. That is not in the agreed plan.
10. Let others try that first.
11. That makes other products of us obsolete.
12. That is not our problem.
13. Why change, things work well the way they are.
14. You are way too far in front of the troops.
15. Lets work with what we have.
16. If that is really a good idea, somebody else would have thought of that.
17. You will not make it to get that done.
18. We will make fools of ourselves.
19. Ok but can you first go over the calculations again.
20. Good idea but we are not going to do that.
21. We are happy to keep this complex beast flying as it is, so do not touch anything.
22. Can you guarantee that it will be a success?
23. Since when are you a specialist in that field?
24. We can't explain this to our shareholders.
26. The system can not support this.
27. That is going to be too costly.
28. Spare parts for this are not available.
29. The client will never accept this.
30. We do not have time for that.
31. There is no budget for that.
33. We are too small for that.
34. Brussels will not allow that.
37. We are not ready for that yet.
38. This will have totally unforeseen effects.
39. Management won't buy this.
40 Let's keep this in the back of our minds.
41. That will never work here.
42. What we have been doing for years is like this, and we will not change that an inch.
43. We do not dare to do this.
44. That is not allowed.
45. That is not in line with our policy / strategy.
46. That is a good idea for in the future.
47. (in the UK) That is a very interesting idea. (meaning: it's rubbish, put it where the sun does not shine).
48. I will come back on this soon. Put it on the agenda after the holidays when things are more quiet.
49. It is not new, what value does it add??
50. Why do you tell us now at this rather late stage?
A PowerPoint sheet with most of them are here 47ideakillers.
So let the Dilbert Bosses try one or more of the above on you. Think for yourself and be brave. Just start and DO IT anyway, without their approval, without subsidies or big resources. The best inventions I made where done low key with the help of trainees or students, far away from the approved R&D projects. And without exception, the projects which my boss told me 'to stop immediately' became huge successes :-).
I can assure you, the real hallmark of great idea's is when you test them on your peers and friends a they are amazed and want it for themselves. And then you can measure how viral it is and how soon it will develop into a disruptive innovation that grows exponentially (an explosion, which has a short and constant doubling time in the number of users).
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8 Responses to Don't let these 50 'idea killers' stop you, so Do Not Ask for Approval !!!
Just DO it, and you will find that all the nay-sayers will shut up and that nobody will dare stop you.
and it will work !!!
Reason 34. Brussels will not allow it. Seems this may be a bigger reason than anything else on that list as to "why" someone may back away these days. Not only Brussels, but every capitol in EU, STARTING IN Germany would line up to take a swipe and defund anything that did not enter in VIA Brussels, etc. This is a post-Brexit world. End of any research grants or subsidies for that corporation from EU.
Sorry Jeff, every time they tell me this, it is an invention they hide behind which will stop the naive politician, since it sounds impressive. I always ask in such cases "who in Brussels, from which department of the EU Commission?" And I offer to call that person together. the fact is that most EU commision depatments want to encourage innovations and new idea's. Sure there are well oiled lobbys in Brussels of telco's and cabletv providers who want to protect their vested interests, but so they are in Washington DC.
Frankly, Jaap, I've always had a pretty good view of EU Commissions and Departments because it actually seemed they were innovative and encouraged new ideas. Like any font of Government, I'm sure there are as equally well oiled ;lobbyists in Washington DC as Brussels, in fact a whole lot more that protect their vested interests at all cost in DC.
Jeff, The real underlying reason for the reluctance of managers to approve your plans is not only that your proposed improvement may harm the present plant, sales and value chains. They hate the idea that your plan might CHANGE their running of the department and that it would show that their plan is not complete and that they are NOT IN CONTROL. So they fire excuses at you.
They hate 'disruptive innovations' in which they are disrupted. Would't you? In Indonesia a group of worried parents asked me not to lecture about Internet, since the spread of cellphones already ruined the authority of parents and the elders in the villages, who since centuries knew what was good for the children. Now their 'wisdom' was challenged by the young brats who could simply call eachother for advice.
The most conservative of them all are the old pioneers who once where the rebels and are now surprised to find that even younger kids can come up with brilliant new ideas the work !!
So the real problem is the constant process of bursts of improvements that cause CHANGE and UNCERTAINTIES. Everybody hate these two !! Because it can mean a better or worse situation for them, so most people want to stick to the status quo and hates surprises.
Logically, Power Structures always resent being superceded, but as you say, improvements always happen, sooner or later. Disruptives of any kind can be your weapon of change choice.


text:
Карликовый тритон () — вид тритонов из рода Hypselotriton отряда хвостатых земноводных.

Историческое свидетельство
В 1008 году один китайский автор отметил, что небольшие чёрные саламандры могут быть обнаружены среди водяных растений в прудах. Это животное имеет такую же голову, как ящерица, тёмно-красный живот, четыре ноги и пять пальцев. Тот же автор сообщает, что животные обитают в высокогорном озере Тяньце на вершине горы Лучан, где они считаются потомками небесного дракона. Помимо прочего, крестьяне считали, что эти животные обладают способностью вызывать дождь.

Описание
Карликовые тритоны — небольшие хвостатые амфибии длиной 6 — 10 см. Цвет может колебаться от тёмно-коричневого до чёрного. Цвет живота — от оранжевого до алого цвета с многочисленными чёрными пятнами. На спине и особенно верхних конечностях могут быть оранжевые пятна.

Мужские особи более мелкие, чем самки. Помимо этого половой диморфизм проявляется также в форме хвоста, который у самцов более короткий с развитым плавником. Как и у большинства тритонов, самцы в период размножения имеют распухшую клоаку.

Ареал
Эндемик юго-восточного Китая. Ареал этих тритонов охватывает китайские провинции Цзянсу, Чжэцзян, Цзянси, Аньхой и Хубэй. Естественной средой существования амфибий являются горные водоёмы, небольшие ручьи, затопленные поля в горных долинах, холодные пруды, колодцы, террасные поля и канавы.

В последнее время данный вид является одним из популярных террариумных животных.

Примечания

Ссылки
 Cynops orientalis on caudata.org (англ.)
 Cynops orientalis on Salamanders of Old World
 Cynops orientalis on AmphibiaWeb (англ.)

Настоящие саламандры
Животные, описанные в 1873 году
Земноводные Юго-Восточной Азии
Эндемики Китая

redpajama_set_name: RedPajamaWikipedia


text:
Where in the World is Jessica Nabongo?
The record-setting jetsetter is never in one place for long, and she wants the world to tag along on her journey.
Profile by Steve Root
There are people who love to travel, and then there's Jessica Nabongo, the first documented Black woman to visit every country in the world, which is 195 by the United Nations count. And that's a lot of passport stamps.
As a creator, Nabongo has parlayed her insatiable wanderlust into a cottage industry whose subsidiaries include content on major social media outlets; The Catch Me If You Can blog; a soon-to-be-released travel memoir by the same name, published by National Geographic; a members-only newsletter of luxury travel experiences; an online bazaar called The Catch that sells a curated collection of artisanal goods sourced from her travels … can her own fleet of Nabongo Air jumbo jets be far behind?
Bit by the international travel bug at the age of 4, the Detroit-born Ugandan American has scratched that itch ever since. "My parents loved to travel, and fortunately they sometimes took me and my two sisters with them," says Nabongo, "so I was exposed to the world." And whether they were visiting her mother's Ugandan village with no electricity or running water or enjoying a high-end resort vacation, she says it was always without value judgments, which has informed her open-minded and wide-eyed worldview today. "Everything just is," she says. "Nothing's better or worse. It's just different."
Jessica's 5 tips for creators
After graduating with a major in English literature and a double minor in advertising-communications and Africana studies from St. John's University in New York, Nabongo initially worked for a pharmaceutical company, but always felt the call of the world. "After two years, I quit that corporate life, cut my hair and went to teach English in Japan. It was a pivotal moment that set my life off to what it's become. I had never lived abroad, but I thought, 'I'm going to do this for three years.' That became seven, and during that time I traveled to more than 40 countries."
Not that it's been nonstop gallivanting. There was a period of more study when she got her Masters in Economic Development from the London School of Economics, followed by several years working for the United Nations in the Food and Agriculture Organization and consulting for the International Atomic Energy Agency.
Still, travel beckoned, and Nabongo fueled her quest with an entrepreneurial spirit. She launched an agency catering to Black travelers, and by running hosted trips, Nabongo's checklist of visited countries expanded greatly. "It was while in Bali—country 60—in 2015 that I decided in earnest that I was going to go to every country in the world by the time I'm 35," says Nabongo. She accomplished that goal on October 6, 2019, fittingly on her late father's birthday, with a visit to the Republic of Seychelles.
And while her journey has been well-documented on Facebook, Instagram and TikTok, it was largely self-funded. "I was not getting paid to go to those countries," Nabongo says. "I don't do free trips. I don't want to be obligated to create content. It's always been the purest form of me visiting a country."
Nearly half of her traveling (89 of 195 countries) has been done alone. "I prefer traveling with friends, but I love traveling solo because it makes me engage with the country more. I'm more likely to talk to the locals. I feel like I get a deeper connection to a place.
Sure, there are caveats. Islamophobia is real. Homophobia is real. There are challenges. But you're gonna let the possibility of racism keep you from doing something? That's crazy! And even if something happens, keep pushing. I'm not gonna let one racist person steal my joy."
Joy is certainly not lacking in Nabongo's world. She loves what she does and has plans for plenty more. "There's a TV show in development, I have ideas for film, a cookbook. But more than anything, I want to do a children's book series based on what I know now about the world that," she says with a laugh, "I hope turns into a Disney ride."
Nabongo's followers find her indomitable spirit infectious. "Everyone wants women to be afraid to go into the world alone. I am not. I've learned that most people are good. I move with positive energy, and that is a magnet for all these wonderful, amazing humans that are brought into my life.
A lot of Black people ask, 'which countries are safe for Black people?' I hate that we live in a world that makes people feel like they have to question, 'can I go? Is it open to me?'
I hope that what people get from my journey is that the world is all of ours, it belongs to all of us, and we can go, we can play, we can meet, and we can see every place in the world that we want to see."
Jessica Nabongo is an Ugandan-American travel blogger and brand influencer. She is the first black woman to have travelled to every country in the world. Her book, "The Catch Me If You Can", goes on sale on June 14. You can pre-order it on Amazon or your preferred retailer.
Follow Jessica:
Heavy Manners Library: A Creator's Oasis for Thinkers, Artists, and Doers
Anela Malik In The House
Andrea Russett Masters Multiple Mediums

redpajama_set_name: RedPajamaCommonCrawl