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\begin{document}

	 \title{Leibniz homology of Lie algebras as functor homology}

		      \author{Eric Hoffbeck}
		      \author{Christine Vespa}
\address{Universit\'e Paris 13, Sorbonne Paris Cit\'e, LAGA, CNRS (UMR 7539)
99 avenue Jean-Baptiste Cl\'ement,
93430 Villetaneuse,
France}
\email{hoffbeck@math.univ-paris13.fr}

\address{ Universit\'e de Strasbourg, Institut de Recherche
 Math\'ematique Avanc\'ee, Strasbourg, France. }
\email{vespa@math.unistra.fr}

\date{\today}

\begin{abstract}
We prove that Leibniz homology of Lie algebras can be described as functor homology in the category of linear functors from a category associated to the Lie operad.\\
\vspace{.3cm}
\textit{Mathematics Subject Classification: 17B55, 18D50, 17B56}
\vspace{.3cm}

\textit{Keywords}: Functor homology, operads, Lie algebras, Leibniz homology
\end{abstract}

\maketitle

\begin{minipage}{\textwidth}

\printindex
\end{minipage}

\tableofcontents

\section*{Introduction}

In the 90's, Robinson and Whitehouse defined and studied in \cite{RW2002}
$\Gamma$-homology of commutative algebras, a homology theory
suited in the context of differential graded modules over a field of positive characteristic.
Pirashvili and Richter proved in \cite{PiraR-2000} that this homology theory can be interpreted as functor homology, for functors from the category $\Gamma$ of finite pointed sets. This category can be viewed as a category associated to the commutative set operad.
For associative algebras, similar results are obtained by Pirashvili and Richter in \cite{PiraR-2002}.
In this paper, the authors interpret usual Hochschild and cyclic homology of associative algebras as functor homology. In this setting, the category $\Gamma$ is replaced by a category associated to the associative set operad.
In \cite{Liv-Richter}, Livernet and Richter give a description of $E_n$-homology of non-unital commutative algebras as functor homology. In this setting the category $\Gamma$ is replaced by a suitable category of epimorphisms related to planar trees with $n$-levels.

In all these results, homology theories are obtained as Tor functors in a category of functors, between a Loday functor and a functor $t$ playing the role of the base ring. In \cite{Pira-Hodge}, Pirashvili uses the interpretation of homology theories as functor homology to give purely homological proofs of Hodge decompositions of higher order Hochschild homology of commutative algebras.

This paper is motivated by the following natural question:
is it possible to describe homologies of Lie algebras as functor homology?

We obtain that the
Leibniz homology of Lie algebras can be interpreted as functor homology. More precisely, we prove the following theorem:

\begin{thm*}
For $A$ a Lie algebra and $M$ a $A$-module, one has an isomorphism:
$$H_*^{Leib}(A,M)\simeq Tor^{{\Sh^{Lie}}}_*(t,{\mathcal{L}}^{Lie}_{sh}(A,M))$$
where ${\Sh^{Lie}}$ is a suitable linear category associated to the Lie operad and ${\mathcal{L}}^{Lie}_{sh}(A,M)$ is a generalized Loday functor.
\end{thm*}

\vspace{1cm}
Leibniz homology was defined by Loday in \cite{Loday1} in order to understand periodicity phenomena in algebraic K-theory (see \cite{Loday2}). This homology can be defined via an explicit complex,
obtained as the non-commutative analogue of the Chevalley-Eilenberg complex and it is defined for Leibniz algebras, which are non-commutative variants of Lie algebras. In Leibniz algebras, the bracket is not required
to be anti-commutative anymore, and  the Jacobi relation is replaced by the Leibniz relation,
which can be seen as a lift of the Jacobi relation in the non-commutative context.
As Lie algebras are a particular case of Leibniz algebras, the Leibniz homology can be used to
compute homological invariants of Lie algebras.

The most striking result concerning Leibniz homology is the following Loday-Cuvier theorem \cite{Cuvier, Loday-Cyclic}: for an associative algebra $A$, the Leibniz homology of the Lie algebra $gl(A)$ is isomorphic to
the free associative algebra over the Hochschild homology of $A$.
This theorem is the non-commutative variant of the Loday-Quillen-Tsygan theorem \cite{LQ, FT}
which states that the Chevalley-Eilenberg homology of $gl(A)$ is isomorphic to
the free exterior algebra over the cyclic homology of $A$. Recall that cyclic homology is naturally isomorphic to the additive K-theory which is the analogue of K-theory obtained by replacing the general linear group $GL(A)$ by the Lie algebra $gl(A)$.

\vspace{1cm}

In order to write Leibniz homology of Lie algebras as functor homology, two objects have to be defined:
a category playing for Lie algebras the role of $\Gamma$ and a variant of the Loday functor for this category.
One main difference with the cases of commutative and associative algebras is that the operad encoding Lie algebras is not a set operad.
This requires the use of linear categories (i.e. categories enriched over the category of $\kk$-modules), instead of usual categories.

The proof of the main theorem is based on a characterization of the Tor functors. The principal difficulty is to prove that the homology vanishes on projective generators. This result is the heart of this paper. To prove it, we need to consider a linear category associated to the operad $Lie$, called ${\Sh^{Lie}}$,
with a shuffle condition, requiring some maps to preserve a part of the order. This condition is the main reason why we obtain Leibniz homology as functor homology and not Chevalley-Eilenberg homology.

The proof of the vanishing of the Leibniz homology on projective generators can be decomposed into three steps. We  begin to describe a basis of the morphisms spaces in the linear category ${\Sh^{Lie}}$. This allows us to define a filtration on the $\kk$-module associated to the complex computing the homology of projective generators. The proof of the compatibility of this filtration with the differential requires to use a basis of the operad $Lie$ and to understand its behaviour with respect to composition. Then we identify the associated graded complex with a sum of acyclic complexes. This step is heavily based on the thorough understanding of the combinatorial objects associated to the basis of the morphisms spaces in the linear category ${\Sh^{Lie}}$.

\vspace{1cm}
In a recent work \cite{Fresse}, Fresse proves that operadic homology can be obtained as functor homology, by a method different than ours.
For the commutative operad and the associative operad, Fresse recovers results of  Pirashvili and Richter \cite{PiraR-2000, PiraR-2002}.
For the operad $Lie$, Fresse obtains a description of Chevalley-Eilenberg homology of Lie algebras as functor homology over the category $\Gamma^{Lie}$. Our theorem shows that functor homology over the category ${\Sh^{Lie}}$ is naturally related to Leibniz homology of Lie algebras. The category ${\Sh^{Lie}}$ has the advantage to be more manageable than the category $\Gamma^{Lie}$ for effective computations.

\vspace{1cm}
The paper is organized as follows:
Section 1 consists in recollections on functor homology of enriched categories and on
some operadic definitions.
In Section 2 we define our category ${\Sh^{Lie}}$ and the associated Loday functor.
After some recollections of Leibniz homology, we state in Section 3 the main theorem of the paper.
The last section is devoted to the proof of the vanishing of the Leibniz homology on projective generators.

\section*{Acknowledgements}
The first author would like to thank Birgit Richter for explaining to him the usual strategy to interpret homology theories as functor homology. The authors gratefully acknowledge the hospitality of the Isaac Newton Institut in Cambridge.
The first author is supported in part by the Sorbonne-Paris-Cit\'e IDEX grant Focal and the ANR grant ANR-13-BS02-0005-02 CATHRE. The second author is supported by the ANR grant ANR-11-BS01-0002 HOGT.

\vspace{2cm}
\textbf{Notations}: The following categories will be useful throughout the whole paper.
\begin{itemize}
\item $Set$ is the category of sets with morphisms the set maps;
\item $\kkm$ is the category of modules over a fixed commutative ground ring $\kk$;
\item $\kkmg$ is the category of $\mathbb N$-graded modules over $\kk$;
\item $\Delta$ is the simplicial category, i.e. the category with objects ordered finite sets $[n]$ and morphisms order preserving maps;
\item $\Gamma$ is the skeleton of the category of finite pointed sets having as objects $[n]=\{0,1, \ldots, n\}$ with $0$ as basepoint and morphisms the set maps $f:[n] \to [m]$ such that $f(0)=0$;
\item $\Gamma^{surj}$ is the category having as objects finite pointed sets and as morphisms the pointed surjective maps;
\item $\Sh$ is  the category with objects ordered finite sets $[n]$ and morphisms pointed shuffling maps $f$, that is maps such that $min(f^{-1}(i))<min(f^{-1}(j))$ whenever $i<j$;
\item $\Sh^{surj}$ is  the category with objects finite ordered sets $[n]$ and morphisms pointed shuffling surjections;
\end{itemize}

The Lie algebras we consider in this paper are algebras over the Lie operad. In particular, in characteristic 2, a Lie bracket is just antisymmetric and we do not have in general $[x,x]=0$.

\vspace{2cm}
\section{Recollections on enriched category, functor homology and operads}
\input{Recollections}

\vspace{2cm}
\section{Functorial constructions associated to an operad}
\input{funct-constructions}

\vspace{2cm}
\section{Leibniz homology of a functor}
\input{Leibniz}

\vspace{2cm}
\section{Vanishing of the homology of the projective generators}
\input{ProofOf3}

\bibliographystyle{amsplain}

\subsection{Leibniz complex}
Let $A$ be a Lie algebra and $M$ a $A$-module.

\begin{defi} \cite{Loday-Cyclic}
The Leibniz homology of $A$ with coefficients in $M$, denoted by $HL_*(A,M)$, is the homology of the complex
$(C^{Leib}_* (A,M),d)$ where
$C^{Leib}_n (A,M)= M \otimes A^{\otimes n}$ and the differential $d: C^{Leib}_n (A,M) \to C^{Leib}_{n-1} (A,M)$ is given by:
$$d(x \otimes a_1 \otimes \ldots \otimes a_n)= \sum_{1\leq i<j \leq n } (-1)^j x\otimes a_1 \otimes \ldots a_{i-1} \otimes [a_i,a_j]\otimes \ldots \otimes \widehat{a_j} \otimes \ldots \otimes a_n$$
$$+ \sum_{1 \leq j \leq n} (-1)^j [x,a_j] \otimes a_1 \otimes \ldots \otimes \widehat{a_j} \otimes \ldots \otimes a_n.$$
\end{defi}

\begin{rem}
\begin{itemize}
\item
If we replace $x$ by the symbol $a_0$, we can notice that:
$$d(a_0 \otimes a_1 \otimes \ldots \otimes a_n)= \sum_{0\leq i<j \leq n} (-1)^j a_0\otimes a_1 \otimes \ldots a_{i-1} \otimes [a_i,a_j]\otimes \ldots \otimes \widehat{a_j} \otimes \ldots \otimes a_n$$
Note that $d^2=0$ is proved in the same way as for the Chevalley-Eilenberg complex.
\item Leibniz homology and Chevalley-Eilenberg homology are related by a map $H^{Leib}_* (A,M) \to H^{CE}_* (A,M)$ induced by the anti-symmetrisation map $C^{Leib}_* (A,M) \to C^{CE}_* (A,M)$.
\end{itemize}
\end{rem}

\begin{prop}
Let $T$ be  a linear functor from ${\Sh^{Lie}}$ to $\kkm$. The map $d: T([n]) \to T([n-1])$ defined by $d=T(\sum_{0\leq i < j \leq n} (-1)^j d_{i,j})$ satisfies $d^2=0$. Therefore the sequence:
$$\ldots \xrightarrow{d}T([n])\xrightarrow{d} T([n-1])  \xrightarrow{d} \ldots$$
is a well-defined complex, denoted by  $(C^{Leib}_* (T),d)$.
\end{prop}

\begin{proof}
The relation $d^2=0$ is obtained by exactly the same computation as $d^2=0$ for the Leibniz complex of an algebra.
\end{proof}

\begin{defi}
Let $T$ be  a linear functor from ${\Sh^{Lie}}$ to $\kkm$.
The Leibniz homology of $T$, denoted by $H^{Leib(T)}_*$, is the homology of the complex $(C^{Leib}_* (T),d)$. Leibniz homology produces a functor from left ${\Sh^{Lie}}$-modules to $\kkmg$ called $H^{Leib}_*$.
\end{defi}

\begin{rem}
If one chooses the Loday functor for $T$, the definition coincides with the previous one: $(C_*^{Leib} ({\mathcal{L}}^{Lie}_{sh}(A,M)),d)=(C_*^{Leib}(A,M),d)$.
\end{rem}

\subsection{Leibniz homology as functor homology}

Before stating our theorem, we first need to define a right ${\Sh^{Lie}}$-module, which will serve as basepoint.

\begin{defi}
The contravariant linear functor $t$ from ${\Sh^{Lie}}$ to $\kkm$ is defined by $\text{coker}\big({\Sh^{Lie}}(-,[1]) \xrightarrow{(d_{0,1})_*} {\Sh^{Lie}}(-,[0])\big)$,
where $-_*$ denotes the postcomposition and $d_{0,1} \in {\Sh^{Lie}}([1],[0])$ is defined in Definition \ref{dij}.
\end{defi}

It is easy to check that $(\text{Im } (d_{0,1})_*)[n] ={\Sh^{Lie}}([n],[0])$ for $n\geq 1$ and  $(\text{Im } (d_{0,1})_*)[0]=0$.
Therefore $t([n])=0$ for $n\geq 1$ and $t[0]={\Sh^{Lie}}([0],[0])=\kk$.

We can now state the main theorem of the paper:

\begin{thm}
Let $T$ be a left ${\Sh^{Lie}}$-module. Then
$$H^{Leib}_*(T) = Tor_* ^{\Sh^{Lie}} (t, T).$$
\end{thm}

To prove this theorem, we use the characterization of homology theories given in \ref{char-homo},
which relies on three hypotheses.

The first hypothesis ($H^{Leib}_*$ sending short exact sequences to long exact sequences) is satisfied,
as $H^{Leib}_*$ is defined via the homology of a complex.

The functor $t$ has been defined in a way to satisfy the second hypothesis (concerning $H_0$).

\begin{prop}
For all left ${\Sh^{Lie}}$-modules $T$, we have
$$H_0^{Leib}(T)=t \otimes_{\Sh^{Lie}} T.$$
\end{prop}

\begin{proof}
We follow the ideas of the proof of Theorem 1.3 of \cite{PiraR-2002}.
By the definition of $t$, we have an exact sequence:
$${\Sh^{Lie}}(-,[1]) \xrightarrow{(d_{0,1})_*} {\Sh^{Lie}}(-,[0]) \to t \to 0.$$
For $T$ a left ${\Sh^{Lie}}$-module, by Proposition \ref{exact} the functor $- \underset{{\Sh^{Lie}}}{\otimes}T$ is right exact, so we deduce the following exact sequence:
$${\Sh^{Lie}}(-,[1]) \underset{{\Sh^{Lie}}}{\otimes}T \to {\Sh^{Lie}}(-,[0]) \underset{{\Sh^{Lie}}}{\otimes}T \to t\underset{{\Sh^{Lie}}}{\otimes}T \to 0.$$
By Observation \ref{projectifs}, ${\Sh^{Lie}}(-,[1]) \underset{{\Sh^{Lie}}}{\otimes}T \simeq T([1])$ and ${\Sh^{Lie}}(-,[0]) \underset{{\Sh^{Lie}}}{\otimes}T \simeq T([0])$ so:
$$t\underset{{\Sh^{Lie}}}{\otimes}T  \simeq T([0])/Im\big(T([1]) \to T([0])\big)=H_0^{Leib}(T).$$

\end{proof}

The third hypothesis consists in the vanishing of the homology of the projective generators.
The proof of this part is detailed in the remaining section of the paper.

\subsection{Sketch of the proof}
Let us first give a survey of the three steps of the proof of this proposition. In the first step, we give a description of bases of the Lie operad and of ${\Sh^{Lie}}([n],[m])$. In Lemma \ref{bases}, we identify these bases with
easy combinatorial objects (tuples, or ordered partitions into tuples).
One important point is to consider all possible values of $m$ at the same time.
This allows us to use the ordering between tuples to obtain, in Proposition \ref{filtration}, a filtration of the $\kk$-module $\oplus_m C_m^{Leib}(P_n)$. Then we show, in Proposition \ref{compatible} that this filtration is compatible with the differential of the complex $C_*^{Leib}(P_n)$. This step in based on the crucial lemma \ref{crochet}
describing the bracket of two elements of the basis of $Lie$ in terms of elements of the basis.
In the last part of the proof we identify the associated graded complex with a sum of acyclic complexes. This step is heavily based on the thorough understanding of relationships between tuples and ordered partitions into tuples. We deduce Proposition \ref{Leib-projectifs} by a classical spectral sequence argument.

\subsection{Filtration as a $\kk$-module}

\subsubsection{Basis of the Lie operad}

Recall that one can define the free symmetric operad generated by a symmetric collection, and that the notion of operadic ideal exists (cf \cite{LV}).
Therefore symmetric operads can be defined by generators and relations.

Our operad of interest in this paper, $Lie$, can then be defined by the free symmetric operad generated by the symmetric collection
$\kk \mu$ concentrated in arity $2$, where $\mu$ denotes an antisymmetric bracket, quotiented by the ideal generated by
the Jacobi relation.

Recall that the Jacobi relation can be written operadically as
\begin{equation} \label{Jacobi}
\begin{array}{c}
$\xymatrix@M=1pt@C=6pt@R=6pt{
1 \ar@{-}[ddrr] & & 2\ar@{-}[dr] & & 3 \ar@{-}[dl] \\
 & & & *{\mu}\ar@{-}[dl]  & \\
 & & *{\mu}\ar@{-}[d]  & & \\
& &  & &
}$\end{array}
=
\begin{array}{c}
 $\xymatrix@M=1pt@C=6pt@R=6pt{
1 \ar@{-}[dr] & & 2\ar@{-}[dl] & & 3 \ar@{-}[ddll] \\
 & *{\mu}\ar@{-}[dr]  & &  & \\
 & & *{\mu}\ar@{-}[d]  & &\\
& &  & &
}$\end{array}
-
\begin{array}{c}
 $\xymatrix@M=1pt@C=6pt@R=6pt{
1 \ar@{-}[dr] & & 3\ar@{-}[dl] & & 2 \ar@{-}[ddll] \\
 & *{\mu}\ar@{-}[dr]  & &  & \\
 & & *{\mu}\ar@{-}[d] & & \\
& &  & &
}$\end{array}.
\end{equation}

In the rest of the paper, all internal vertices of the trees are labelled with the bracket $\mu$,
therefore we do not write it anymore.

Methods have been developed in \cite{Eric} and \cite{DK2010} to find monomial bases with good properties with respect to an ordering.

A basis of the $Lie$ operad is given by the following planar trees (cf \cite{Eric}):

$$
\left\{ \vcenter{ \xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
1\ar@{-}[dr]& &\sigma(2)\ar@{-}[dl] \\
& \ar@{-}[dr] & & \sigma(3)\ar@{-}[dl]  \\
& & \ar@{-}[dr] & & \cdots \ar@{-}[dl] \\
& & & \ar@{-}[dr] & & \cdots \ar@{-}[dl] \\
& & & & \ar@{-}[dr] & & \sigma(n) \ar@{-}[dl] \\
& & & & & \ar@{-}[d] \\
& & & & & &
} } \  \Bigg\vert \ n \geq 1, \sigma \text{ permutation of } \{2, \ldots, n\} \right\}. $$

Note that this operadic basis corresponds to a basis of the multilinear part of the free Lie algebra on $x_1, \ldots, x_n$
given by elements of the form $[ \ldots[[x_1, x_{\sigma(2)}], x_{\sigma(3)}], \ldots, x_{\sigma(n)}]$.
(See \cite[Section 5.6.2]{Reutenauer}).

We denote the part of the basis in arity $n$ by $\mathcal B^{Lie}(n)$. We define similarly $\mathcal B^{Lie}(I)$ for $I$ a finite ordered set of cardinality $n$,
using the order preserving bijection between $I$ and $\{1, \ldots, n\}$.

We denote the whole basis by $\mathcal B^{Lie}=\bigcup_n \mathcal B^{Lie}(n)$.

\subsubsection{Explicit description of the category associated to the Lie operad}

To describe the category ${\Sh^{Lie}}$, it is enough to describe the vector spaces of morphisms and to understand the composition of elements of the basis.
We first give a $\kk$-basis for ${\Sh^{Lie}}([n],[m])$.
Recall that ${\Sh^{Lie}}([n],[m])=\underset{\alpha \in \Sh([n],[m])}{\bigoplus} Lie (\alpha^{-1}(0)) \otimes \ldots \otimes Lie (\alpha^{-1}(m))$.

Obviously, if $m>n$, the basis is empty.

If $m \leq n$, we can describe a basis $\mathcal B^{{\Sh^{Lie}}}(n,m)$ of ${\Sh^{Lie}}([n],[m])$ via shuffle maps and a tuple of elements of a Lie basis:

$$\mathcal B^{{\Sh^{Lie}}}(n,m)= \bigcup_{\alpha \in \Sh([n],[m])} \mathcal B^{Lie}(\alpha^{-1}(0)) \times \ldots \times \mathcal B^{Lie}(\alpha^{-1}(m)).$$

We also define $\mathcal B^{{\Sh^{Lie}}}(n,\mathbb N) = \bigcup_m \mathcal B^{{\Sh^{Lie}}}(n,m)$.

\subsubsection{Notations for elements of the bases}
To avoid drawing huge forests of trees, we first identify the elements of $\mathcal B^{{\Sh^{Lie}}}(n,\mathbb N)$ as some particular tuples with splittings.
We begin by defining these particular tuples and introduce some notations.

\begin{obs}
All trees appearing in $\mathcal B^{Lie}(n)$ are left combs, so they are characterized by their leaves.
An element of $\mathcal B^{Lie}(n)$ can be identified with a $n$-tuple $(1, \sigma(2), \ldots, \sigma(n))$ where $\sigma$ is a permutation of $\{2, \ldots, n\}$,
just by reading the inputs from left to right. So we have a bijection $\mathcal B^{Lie}(n) \simeq \mathfrak{S}_{n-1}$, where $\mathfrak{S}_{n-1}$ is the permutation group. In the sequel, we identify these two sets.
\end{obs}

\begin{exple}
Let $(\alpha,f)$ be the generator of ${\Sh^{Lie}}([6],[2])$ where $\alpha \in \Gamma_{sh}([6],[2])$ is given by:
$$\alpha(0)=\alpha(1)=\alpha(5)=0;\  \alpha(2)=1;\  \alpha(3)=\alpha(4)=\alpha(6)=2$$
and $f=f_0 \otimes f_1 \otimes f_2$ where:
$$f_0=Id \in \mathcal B^{Lie}(3) \simeq \mathfrak{S}_{2};\  f_1=Id \in \mathcal B^{Lie}(1) \simeq \mathfrak{S}_{0};$$
$$ f_2=\tau_{1,2} \in \mathcal B^{Lie}(3) \simeq \mathfrak{S}_{2}.$$
We can represent this morphism by the following picture:
$$
\vcenter{\xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
0\ar@{-}[dr]& & 1\ar@{-}[dl] & \\
& \ar@{-}[dr]& & 5\ar@{-}[dl]\\
& & \ar@{-}[d] \\
& & 0&}}
\
\vcenter{\xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
2\ar@{-}[ddd]\\
\\
\\
1&}}
\
\vcenter{\xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
3\ar@{-}[dr]& & 6\ar@{-}[dl] & \\
& \ar@{-}[dr]& & 4\ar@{-}[dl]\\
& & \ar@{-}[d] \\
& & 2&}}$$

\end{exple}

\begin{defi} \label{STuple}
 A $m$-split $(n+1)$-tuple is an ordered partition of $[n]$ into $m+1$ tuples:
$B_0 \cup B_1 \cup \ldots \cup B_m$
where $m \geq 0$ and $B_i=(k_{i,0}, \ldots, k_{i,\ell_i})$ satisfying two conditions:
\begin{enumerate}
 \item $0 \leq i<j \leq m \Rightarrow k_{i,0} < k_{j,0}$ ,
 \item $\forall \, 0 \leq i \leq m, \forall \,  0 \leq q \leq \ell_i, k_{i,0} < k_{i,q}$.
\end{enumerate}
To ease notations, we denote $B_0 \cup B_1 \cup \ldots \cup B_m$ by $(B_0 \mid B_1 \mid \ldots \mid B_m)$.
\end{defi}

\begin{rem}
The conditions impose that $k_{0,0}=0$.
The first condition can be read as ``the first term of a tuple is smaller than the first term of the following tuples''.
The second condition can be read as ``the first term of a tuple is smaller than the other terms of this tuple''.
\end{rem}

\begin{nota}
We denote by $STuple(n,m)$ the set of $m$-split $(n+1)$-tuples, by $STuple(n,\mathbb N)$ the set $\cup_m STuple(n,m)$
and by $Tuple(n)$ the set $STuple(n,0)$ corresponding to the set of $(n+1)$-tuples starting with $0$.
\end{nota}

\begin{exple}
 For $n=2$, the set $STuple(2, \mathbb N)$ is composed of $6$ elements:

$(0\mid1\mid2), (0\mid1,2), (0,1\mid2), (0,2\mid1), (0,1,2), (0,2,1).$

The set $Tuple(2)$ is composed of only $2$ elements: $(0,1,2)$ and $(0,2,1)$.

\end{exple}

\begin{lm} \label{condition3}
The two conditions $(1)$ and $(2)$ in Definition \ref{STuple} are jointly equivalent to the following single condition:
\begin{equation}
 \tag*{(3)} \forall \, 0 \leq i \leq j \leq m, \forall \, 0 \leq q \leq \ell_j, k_{i,0} < k_{j,q}.
\end{equation}
\end{lm}
This condition can be read as ``the first term of a tuple is smaller than the following terms''.

\begin{proof}
Each direction is easy to check.
\end{proof}

If $B_i$ and $B_{i+1}$ are two consecutives blocks of a split tuple, we denote by $B_iB_{i+1}$ their concatenation.

\begin{lm}
Let $(B_0 \mid B_1 \mid \ldots \mid B_m)$ be a split tuple, and $0 \leq i <m$.

Then $(B_0 \mid B_1 \mid \ldots \mid B_iB_{i+1} \mid \ldots \mid B_m)$
is also a split tuple.
\end{lm}

\begin{proof}
The condition (3) in the above lemma for $(B_0 \mid B_1 \mid \ldots \mid B_m)$ directly implies the condition for $(B_0 \mid B_1 \mid \ldots \mid B_iB_{i+1} \mid \ldots \mid B_m)$.
\end{proof}

We now use these tuples and split tuples to describe the bases $\mathcal B ^{{\Sh^{Lie}}}(n,m)$ and $\mathcal B ^{{\Sh^{Lie}}}(n,\mathbb N)$.

\begin{lm} \label{bases}
\begin{itemize}
 \item The set $\mathcal B ^{{\Sh^{Lie}}}(n,m)$ is in bijection with $STuple(n,m)$. In particular, $\mathcal B ^{{\Sh^{Lie}}}(n,0)$ is in bijection with $Tuple(n)$
 \item The set $\mathcal B ^{{\Sh^{Lie}}}(n,\mathbb N)$ is in bijection with $STuple(n, \mathbb N)$.
\end{itemize}
\end{lm}

\begin{proof}

Let $(\alpha, \sigma_0, \ldots, \sigma_m)$ be an element in $\mathcal B ^{{\Sh^{Lie}}}(n,m)$ where
$\alpha \in \Sh([n],[m])$ and $\forall i \in \{0,\ldots,m\}$, $\sigma_i \in \mathcal B ^{Lie}(\alpha^{-1}(i))\simeq \mathfrak{S}_{\mid  \alpha^{-1}(i)\mid -1}$.
 We consider for all $i \in \{0, \ldots, m\}$ the ordered set $\alpha^{-1}(i)=\{k_{i,0}<k_{i,1}< \ldots<k_{i,\mid  \alpha^{-1}(i)\mid} \}$.
We associate to $(\alpha, \sigma_0, \ldots, \sigma_m)$ the ordered partition of $[n]$ into $m+1$ tuples $(B_0 \mid B_1 \mid \ldots \mid B_m)$
 where for all $i \in \{0, \ldots, m\}$, $B_i=\{k_{i,0}, k_{i, \sigma_i(1)}, \ldots, k_{i, \sigma_i(\mid  \alpha^{-1}(i)\mid})\}$.
We have $(B_0 \mid B_1 \mid \ldots \mid B_m) \in STuple(n,m)$ since condition $(1)$ in Definition \ref{STuple}
corresponds to the shuffling property of $\alpha \in \Sh([n],[m])$ and condition $(2)$ is obviously satisfied. So we define the map
$$\Psi: \mathcal B ^{{\Sh^{Lie}}}(n,m) \to STuple(n,m)$$
 by:
$$\Psi(\alpha, \sigma_0, \ldots, \sigma_m)=(B_0 \mid B_1 \mid \ldots \mid B_m).$$
This map $\Psi$ is a bijection whose inverse is given by:
$$\Psi^{-1}(B_0 \mid B_1 \mid \ldots \mid B_m)=\Psi^{-1}(k_{0,0}, \ldots, k_{0, l_0} \mid \ldots \mid k_{n,0}, \ldots, k_{n, l_n})=(\alpha, \sigma_0, \ldots, \sigma_m)$$
where for all $i \in \{0, \ldots, m\}$, $\alpha(k_{i,0})= \ldots= \alpha(k_{i, l_i})=i$ and $\sigma_i$ is the permutation
such that $k_{i,0}<k_{i,\sigma_i^{-1}(1)}<k_{i,\sigma_i^{-1}(2)}< \ldots< k_{i,\sigma_i^{-1}(l_i)}.$
The map $\alpha$ is a shuffle map by condition $(1)$ in Definition \ref{STuple}
\end{proof}

\begin{exple}
 The following basis element of ${\Sh^{Lie}}([6], [2])$:
$$
\vcenter{\xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
0\ar@{-}[dr]& & 1\ar@{-}[dl] & \\
& \ar@{-}[dr]& & 5\ar@{-}[dl]\\
& & \ar@{-}[d] \\
& & 0&}}
\
\vcenter{\xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
2\ar@{-}[ddd]\\
\\
\\
1&}}
\
\vcenter{\xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
3\ar@{-}[dr]& & 6\ar@{-}[dl] & \\
& \ar@{-}[dr]& & 4\ar@{-}[dl]\\
& & \ar@{-}[d] \\
& & 2&}}$$
corresponds to the $2$-split $7$-tuple  in $STuple(6,2)$

$$
(0,1,5 \mid 2 \mid 3,6,4)
$$
by the bijection in Lemma \ref{bases}.
\end{exple}

In the rest of the paper, we identify the set  $\mathcal B ^{{\Sh^{Lie}}}(n,m)$ with $STuple(n,m)$.

\subsubsection{Ordering and filtration }

We now define an ordering on the set $\mathcal B^{{\Sh^{Lie}}}(n,\mathbb N)$ and use it to endow $\bigoplus_m {\Sh^{Lie}}([n],[m])$ with a filtration
(note that this direct sum is actually finite).

\begin{obs}
Forgetting the vertical bars of an element of $STuple(n, \mathbb N)$ gives an element of $Tuple(n)$. This defines a surjective map $p: STuple(n, \mathbb N) \to Tuple(n)$.
\end{obs}

Using the bijection between $\mathcal B ^{{\Sh^{Lie}}}(n,\mathbb N)$ and $STuple(n, \mathbb N)$, we define $p:  \mathcal B ^{{\Sh^{Lie}}}(n,\mathbb N) \to Tuple(n)$, which is again a surjection.

We order the set $Tuple(n)$ with the usual total lexicographical ordering.

We now define an ascending filtration on $\bigoplus_m {\Sh^{Lie}}([n],[m])$

\begin{lm}
Let $u$ be an element in $Tuple(n)$. The $\kk$-modules:
$$F_u:=\displaystyle{\bigoplus_{b \in \mathcal B ^{\Sh^{Lie}}(n, \mathbb N), p(b) \geq u} \kk b}$$
define an ascending filtration of the $\kk$-module $\bigoplus_m {\Sh^{Lie}}([n],[m])$.

The associated graded object is $\displaystyle{gr\Big(\bigoplus_m {\Sh^{Lie}}([n],[m])\Big) = \bigoplus_{u \in Tuple(n)} gr_u}$
where
$gr_u$ is $\displaystyle{\bigoplus_{b \in \mathcal B ^{\Sh^{Lie}}(n, \mathbb N), p(b) = u} \kk b}$.
\end{lm}

Recall that $C^{Leib}_m(P_n)= {\Sh^{Lie}}([n],[m])$ as $k$-modules. Thus we obtain the following proposition:

\begin{prop} \label{filtration}
We obtain a filtration on $C^{Leib}_*(P_n)$ (seen as a $\kk$-module) indexed by elements $u$ in $Tuple(n)$.
\end{prop}

We have to show that this filtration is compatible with the differential. This is the point of the next subsection.

\subsection{Compatibility of the filtration with the differential}

We first need to rewrite in $\mathcal B^{Lie}$ a product of two basis elements.
This allows us to understand the image by $P_n(d_{i,j})=(d_{i,j})_*$ (where $-_*$ denotes the postcomposition) of an element in  $\mathcal B^{{\Sh^{Lie}}}(n,\mathbb N)$.
We actually do not need all the terms of the image, but only its leading term relatively to the ordering defined in the previous section.

\subsubsection{Products of elements in the basis of $Lie$}
We need to rewrite in $\mathcal B^{Lie}$ such a composite of two basis elements:
$$\xymatrix@M=0pt@H=0pt@W=0pt@R=11pt@C=11pt@!0{
i_0\ar@{-}[dr]& & i_1\ar@{-}[dl] & & j_0\ar@{-}[dr] & & j_1\ar@{-}[dl] & & \\
& \ar@{-}[dr] & & \cdots \ar@{-}[dl] & & \ar@{-}[dr] & & \cdots\ar@{-}[dl] & \\
& & \ar@{-}[dr] & & i_n\ar@{-}[dl] & & \ar@{-}[dr] & & j_m\ar@{-}[dl] \\
& & & \ar@{-}[ddrr] & & & & \ar@{-}[ddll]  \\
& & & & & & & \\
& & & & &\ar@{-}[d] & &\\
& & & & & &
}$$
where $i_0<j_0$, $i_0<i_k$ for all $1\leq k \leq n$ and $j_0 < j_k$ for all $1 \leq k \leq m$.

\begin{lm}\label{crochet}
 The decomposition of this element in the basis $\mathcal B^{Lie}$ is the following sum:
$$\sum_{k=0}^m \
\sum_{S \subset \{1, \ldots, m \}, \mid S \mid =k} (-1)^k
\vcenter{\xymatrix@M=0pt@H=0pt@R=11pt@C=11pt@!0{
i_0\ar@{-}[dr]& & i_1\ar@{-}[dl] \\
& \ar@{-}[dr] & & \cdots\ar@{-}[dl]  \\
& & \ar@{-}[dr] & & i_n\ar@{-}[dl] \\
& & & \ar@{-}[dr] & & \cdots \ar@{-}[dl] \\
& & & & \ar@{-}[dr] & & j_0\ar@{-}[dl] \\
& & & & & \ar@{-}[dr] & & \cdots\ar@{-}[dl] \\
& & & & & & \ar@{-}[d] \\
& & & & & & & & & &
}}$$
where: \\
- the inputs between $i_0$ and $i_n$ are in the same order as before,\\
- $j_0$ is in the $(n+2+k)$-th position, \\
- the inputs between $i_n$ and $j_0$ are labelled by the $j_\ell$ for $\ell \in S$ (with $\ell$ decreasing from top to bottom),\\
- the inputs below $j_0$ are labelled by the $j_\ell$ for $\ell \notin S$ (with $\ell$ increasing from top to bottom).
\end{lm}

Let us denote by $b_k^S$ the element of the basis appearing for the indices $k$ of the first sum and $S$ of the second sum.

\begin{proof}[Sketch of proof]
 We proceed by induction on $m$.

For $m=0$, there is nothing to prove.

For $m=1$, we apply the Jacobi relation.

To prove the formula for $m+1$ if $m \geq 1$, the idea is to apply the formula for $m$ to the tree where
$ \vcenter{\xymatrix@M=0pt@H=0pt@R=7pt@C=10pt@!0{ j_0\ar@{-}[dr]& & j_1\ar@{-}[dl] \\ &\ar@{-}[d] &\\&&}}$
has been replaced by $j_0$. Then we graft
$ \vcenter{\xymatrix@M=0pt@H=0pt@R=7pt@C=10pt@!0{ j_0\ar@{-}[dr]& & j_1\ar@{-}[dl] \\ &\ar@{-}[d] &\\&&}}$
back in every term where $j_0$ appears. Finally we use the Jacobi relation again, and obtain the desired formula for $m+1$. We refer the reader to the proof of  \cite[Theorem 5.1]{Reutenauer} for details.
\end{proof}

\begin{rem}
In \cite[Theorem 5.1]{Reutenauer} the description of the basis $\mathcal B^{Lie}$ corresponds to the fact that Lyndon words form a Hall set and the previous lemma corresponds to the triangularity property of the corresponding Hall basis. Notice that a careful study of the proof of  \cite[Theorem 5.1]{Reutenauer}  gives the more precise formula given in the statement of Lemma \ref{crochet}.
\end{rem}

\begin{rem}
This lemma moreover implies that the basis satisfies a property very similar to the Poincar\'e-Birkhoff-Witt property of associative algebras.
It is then possible to show that the category ${\Sh^{Lie}}$ is actually a Koszul category.
\end{rem}

\begin{exple} \label{exampled01}
 $$\vcenter{\xymatrix@M=0pt@H=0pt@W=0pt@R=13pt@C=13pt@!0{
i_0\ar@{-}[dr]& & i_1\ar@{-}[dl] & & j_0\ar@{-}[dr] & & j_1\ar@{-}[dl] & & \\
& \ar@{-}[dr] & &  & & \ar@{-}[dr] & & j_2\ar@{-}[dl] & \\
& & \ar@{-}[dr] & &  & & \ar@{-}[dr] & & j_3\ar@{-}[dl] \\
& & & \ar@{-}[ddrr] & & & & \ar@{-}[ddll]  \\
& & & & & & & \\
& & & & &\ar@{-}[d] & &\\
& & & & & &
}}
=$$

$$
\vcenter{\xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
i_0\ar@{-}[dr]& & i_1\ar@{-}[dl] \\
& \ar@{-}[dr] & & j_0\ar@{-}[dl]  \\
& & \ar@{-}[dr] & & j_1\ar@{-}[dl] \\
& & & \ar@{-}[dr] & & j_2 \ar@{-}[dl] \\
& & & & \ar@{-}[dr] & & j_3\ar@{-}[dl] \\
& & & & & \ar@{-}[d] \\
& & & & & &
}}
-
\vcenter{\xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
i_0\ar@{-}[dr]& & i_1\ar@{-}[dl] \\
& \ar@{-}[dr] & & j_1\ar@{-}[dl]  \\
& & \ar@{-}[dr] & & j_0\ar@{-}[dl] \\
& & & \ar@{-}[dr] & & j_2 \ar@{-}[dl] \\
& & & & \ar@{-}[dr] & & j_3\ar@{-}[dl] \\
& & & & & \ar@{-}[d] \\
& & & & & &
}}
-
\vcenter{\xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
i_0\ar@{-}[dr]& & i_1\ar@{-}[dl] \\
& \ar@{-}[dr] & & j_2\ar@{-}[dl]  \\
& & \ar@{-}[dr] & & j_0\ar@{-}[dl] \\
& & & \ar@{-}[dr] & & j_1 \ar@{-}[dl] \\
& & & & \ar@{-}[dr] & & j_3\ar@{-}[dl] \\
& & & & & \ar@{-}[d] \\
& & & & & &
}}
-
\vcenter{\xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
i_0\ar@{-}[dr]& & i_1\ar@{-}[dl] \\
& \ar@{-}[dr] & & j_3\ar@{-}[dl]  \\
& & \ar@{-}[dr] & & j_0\ar@{-}[dl] \\
& & & \ar@{-}[dr] & & j_1 \ar@{-}[dl] \\
& & & & \ar@{-}[dr] & & j_2\ar@{-}[dl] \\
& & & & & \ar@{-}[d] \\
& & & & & &
}}
+$$

$$
\vcenter{\xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
i_0\ar@{-}[dr]& & i_1\ar@{-}[dl] \\
& \ar@{-}[dr] & & j_2\ar@{-}[dl]  \\
& & \ar@{-}[dr] & & j_1\ar@{-}[dl] \\
& & & \ar@{-}[dr] & & j_0 \ar@{-}[dl] \\
& & & & \ar@{-}[dr] & & j_3\ar@{-}[dl] \\
& & & & & \ar@{-}[d] \\
& & & & & &
}}
+
\vcenter{\xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
i_0\ar@{-}[dr]& & i_1\ar@{-}[dl] \\
& \ar@{-}[dr] & & j_3\ar@{-}[dl]  \\
& & \ar@{-}[dr] & & j_1\ar@{-}[dl] \\
& & & \ar@{-}[dr] & & j_0 \ar@{-}[dl] \\
& & & & \ar@{-}[dr] & & j_2\ar@{-}[dl] \\
& & & & & \ar@{-}[d] \\
& & & & & &
}}
+
\vcenter{\xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
i_0\ar@{-}[dr]& & i_1\ar@{-}[dl] \\
& \ar@{-}[dr] & & j_3\ar@{-}[dl]  \\
& & \ar@{-}[dr] & & j_2\ar@{-}[dl] \\
& & & \ar@{-}[dr] & & j_0 \ar@{-}[dl] \\
& & & & \ar@{-}[dr] & & j_1\ar@{-}[dl] \\
& & & & & \ar@{-}[d] \\
& & & & & &
}}
-
\vcenter{\xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
i_0\ar@{-}[dr]& & i_1\ar@{-}[dl] \\
& \ar@{-}[dr] & & j_3\ar@{-}[dl]  \\
& & \ar@{-}[dr] & & j_2\ar@{-}[dl] \\
& & & \ar@{-}[dr] & & j_1 \ar@{-}[dl] \\
& & & & \ar@{-}[dr] & & j_0\ar@{-}[dl] \\
& & & & & \ar@{-}[d] \\
& & & & & &
}}$$
$$=b_0^{\emptyset}-b_1^{\{1\}}-b_1^{\{2\}}-b_1^{\{3\}}+b_2^{\{1,2\}}+b_2^{\{1,3\}}+b_2^{\{2,3\}}-b_3^{\{1,2,3\}}.$$
\end{exple}

Using our notations for the basis of ${\Sh^{Lie}}([n],[m])$, we can notice that the previous example can be seen as
applying $(d_{0,1})_*$ to
$(i_0, i_1 | j_0, j_1, j_2, j_3) \in {\Sh^{Lie}}([5],[1])$, which gives an element in ${\Sh^{Lie}}([5],[0])$.
And the sum of eight terms we obtain is the decomposition of  $(d_{0,1})_*(i_0, i_1 | j_0, j_1, j_2, j_3)$ in the basis of ${\Sh^{Lie}}([5],[0])$.
The sum begins with :
$$(i_0, i_1, j_0, j_1, j_2, j_3) - (i_0, i_1, j_1, j_0, j_2, j_3) - \ldots $$
We can notice that the first term of the sum is in the same stage of the filtration as the element we were starting with,
while the other 7 terms are in a higher stage of the filtration.

\vspace{1cm}

Let us make some additional observations about Lemma~\ref{crochet} in the general case:
\begin{enumerate}
 \item For $k=0$, the sum on subsets $S$ is composed of a single term for $S=\emptyset$.
       This term $b_0^{\emptyset}$ in the basis is identified with the tuple $(i_0, i_1, \ldots, i_n, j_0, j_1, \ldots, j_n)$, and has a $+1$ coefficient.
       This means that, if we denote $B_i=(i_0, \ldots, i_n)$ and $B_j=(j_0, \ldots, j_m)$, we obtain the equality $b_0^{\emptyset}=B_iB_j$.
       Notice that $p(b_0^{\emptyset})=p(B_iB_j)=p(B_i \mid B_j)$.
 \item For $k \geq 1$, notice that all the terms $b_k^S$ appearing in the sum have a $(n+2)$th label equal to $j_l$ for $l>0$.
       As $j_0$ is the $(n+2)$th label of $p(B_i \mid B_j)$ and $j_l > j_0$, using the lexicographical ordering in $Tuple(n)$,
       we obtain the inequality  $p(b_k^S) > p(B_i \mid B_j)$.
\end{enumerate}

\subsubsection{Compatibility of the filtration with the differential}

We now want to see how the observation from the previous example can be generalized to $(d_{i,j})_*(b)$ where $b$ is a basis element ${\Sh^{Lie}}([n],[m])$.

\begin{lm}\label{calculdiff}
Let $b=(B_0 \mid \ldots \mid B_m)$ be in $\mathcal B^{\Sh^{Lie}} (n,m)$ where $m>0$. Let $u$ denote $p(b)$.
\begin{enumerate}
\item [(i)]
If $0 \leq i <m$, then $(d_{i,i+1})_*(b)$ is in $F_u$. Moreover, in $gr_u$,  we obtain $(d_{i,i+1})_*(b)= (B_0 \mid \ldots \mid B_iB_{i+1} \mid \ldots \mid B_m)$.

\item [(ii)]
If $0 \leq i <m$ and $i+2 \leq j \leq m$, then $d_{i,j}(b)$ is in $F_u$. Moreover, in $gr_u$, we obtain $d_{i,j}(b)= 0$.
\end{enumerate}

\end{lm}

That means that $(d_{i,i+1})_*(b)$ has one term (with a plus sign) in the same stage of the filtration as $b$, and other terms in higher stages.
And $(d_{i,j})_*(b)$ for $j>i+1$ has only terms in stages of the filtration higher than the stage containing $b$, and thus vanishes in the graded object.

\begin{proof}

By Lemma \ref{crochet} we see that $(d_{i,i+1})_*(B_0 \mid \ldots \mid B_m)$ is
a sum of the term $(B_0 \mid \ldots \mid B_iB_{i+1} \mid \ldots \mid B_m)$
with the terms $\pm (B_0 \mid \ldots \mid b_k^S \mid \ldots \mid B_m)$ where $k>0$.

Using the projection to $Tuple(n)$ and the lexicographical ordering there, we notice
$u=p(B_0 \mid \ldots \mid B_iB_{i+1} \mid \ldots \mid B_m)$ and $ u < p(B_0 \mid \ldots \mid b_k^S \mid \ldots \mid B_m)$.

Thus we obtain the part (i) of the lemma.

For the second part, first notice that $(d_{i,j})_*(B_0 \mid \ldots \mid B_m)$ is a sum of terms whose smallest one is
$(B_0 \mid \ldots \mid B_iB_j \mid \ldots \mid \widehat{B_j} \mid \ldots \mid B_m)$.
But $p(B_0 \mid \ldots \mid B_iB_j \mid \ldots \mid \widehat{B_j} \mid \ldots \mid B_m)$ is larger than $p(B_0 \mid \ldots \mid B_m)$
because $k_{i+1,0} < k_{j,0}$ as $i+1<j$ by condition $(1)$ in Definition \ref{STuple}.

Thus all terms appearing in $d_{i,j}(B_0 \mid \ldots \mid B_m)$ are in $F_u$ and vanish in $gr_u$.
This concludes the proof of (ii).
\end{proof}

\begin{exple}
An example of the behaviour of $(d_{i,i+1})_*$ was already given in the observation after Example \ref{exampled01}.

Let us now consider an example of the second case: $(d_{0,2})_*$ applied to the following element of the basis of ${\Sh^{Lie}}([6], [2])$:
$$ b=
\vcenter{\xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
0\ar@{-}[dr]& & 1\ar@{-}[dl] & \\
& \ar@{-}[dr]& & 5\ar@{-}[dl]\\
& & \ar@{-}[d] \\
& & 0&}}
\
\vcenter{\xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
2\ar@{-}[ddd]\\
 \\
 \\
1&}}
\
\vcenter{\xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
3\ar@{-}[dr]& & 6\ar@{-}[dl] & \\
& \ar@{-}[dr]& & 4\ar@{-}[dl]\\
& & \ar@{-}[d] \\
& & 2&}}.$$

The computation gives a sum of four terms of the form
$$\vcenter{\xymatrix@M=0pt@H=0pt@R=13pt@C=13pt@!0{
0\ar@{-}[dr]& & 1\ar@{-}[dl] &&&&& 2\ar@{-}[dddddd]\\
& \ar@{-}[dr] & & 5\ar@{-}[dl]  &&&&\\
& & \ar@{-}[dr] & & \sigma(3)\ar@{-}[dl] &&&\\
& & & \ar@{-}[dr] & & \sigma(6) \ar@{-}[dl] &&\\
& & & & \ar@{-}[dr] & & \sigma(4)\ar@{-}[dl] & \\
& & & & & \ar@{-}[d] & & \\
& & & &  & 0 & &1
}}$$
where $\sigma$ is a bijection of the set $\{ 3, 4, 6 \}$.

As split tuples, the equality reads
$$(d_{0,2})_*(0,1,5 \mid 2 \mid 3,6,4) = $$
$$(0,1,5,3,6,4 \mid 2) - (0,1,5,6,3,4 \mid 2) - (0,1,5,4,3,6 \mid 2) + (0,1,5,6,4,3 \mid 2).$$

Clearly, when projecting on Tuples(6), the four terms of the right hand side are larger than
the term of the left hand side.
Thus in the associated graded object, the equality reduces to
$$(d_{0,2})_*(0,1,5 \mid 2 \mid 3,6,4) \cong 0.$$
\end{exple}

\begin{prop} \label{compatible}
The differential of $C^{Leib}_*(P_n)$  is compatible with the filtration.
\end{prop}

\begin{proof}
 This is a direct consequence of the previous lemma, using that the differential in  $C^{Leib}_*(P_n)$ is $d= \displaystyle{\sum_{0\leq i<j \leq n} (-1)^j (d_{i,j})_*}$.
\end{proof}

\subsection{Acyclicity of the associated graded complex}

We now identify the associated graded complex with a sum of acyclic ones.

First, let us describe the differential in the associated graded complex. From Lemma \ref{calculdiff}, we obtain
$$d_{gr}(B_0 \mid \ldots \mid B_m)= \sum_{0\leq i <m} (-1)^{i+1} (B_0 \mid \ldots \mid B_iB_{i+1} \mid \ldots \mid B_m).$$

For a fixed $u=(0, k_1, \ldots, k_n) \in Tuple(n)$, we want to describe the elements $b\in STuple(n)$ such that $p(b)=u$.
We use the definition of a split tuple with the condition (3) from Lemma \ref{condition3}, which characterizes the first terms of a block.

\begin{defi}
We call an admissible cut of the tuple $u=(0, k_1, \ldots, k_n)$ an index $i$ with $1 \leq i \leq n$ satisfying the condition
$$\forall \, 1 \leq j \leq m, j>i \Rightarrow k_j > k_i.$$
We denote by $Cuts(u)$ the set of admissible cuts of $u$.
\end{defi}

Notice that $Cuts(u)$ is non-empty as $n$ always belongs to $Cuts(u)$ (the condition of the definition is empty for this index).

\begin{exple}
 Let $u$ be $(0,2,1,5,3,4)$. The set $Cuts(u)$ is $\{2,4,5\}$.
These indices $2$, $4$ and $5$ are exactly the ranks where it is possible to split the tuple.
For instance, $(0,2 \mid 1,5 \mid 3 \mid 4)$ is the split tuple obtained by cutting as most as possible.
\end{exple}

\begin{lm}
The set of $b\in STuple(n)$ such that $p(b)=u=(0, k_1, \ldots, k_n)$ is in bijection with the power set $\mathscr P (Cuts(u))$, via
$$(B_0 \mid \ldots \mid B_m) \mapsto \{ |B_0|, |B_0| + |B_1|, \ldots, |B_0| + \ldots + |B_{m-1}| \}$$

The inverse bijection is given by
$$\{i_1, \ldots, i_m\} \mapsto (0, k_1, \ldots, k_{i_1 -1} \mid k_{i_1}, k_{i_1+1}, \ldots  \ldots \mid k_{i_2} \ldots \mid \ldots \mid k_{i_m}, k_{i_m+1}, \ldots ).$$
\end{lm}

Recall that for any non-empty finite set $E=\{e_0, \ldots, e_n\}$,
there exists an associated chain complex $C_*(E)$ whose basis is the power set $\mathscr P (E)$, a subset of cardinality $k$ giving a generator in degree $k$.
The differential $d: C_{l+1} \to C_l$ is given by $d(\{e_{i_0}, \ldots e_{i_\ell} \}) = \displaystyle{ \sum_{j=0}^\ell (-1)^j \{e_{i_0}, \ldots, \widehat{e_{i_j}}, \ldots, e_{i_\ell} \} }$.
This complex $C_*(E)$ is acyclic (for instance it can be seen as a complex associated to a simplex).

\begin{lm}
 There is an isomorphism of complexes between $(gr_u,d_{gr})$ and $(C_*(Cuts(u)),d)$.
\end{lm}

\begin{proof}
Recall that in the graded object, the part $d_{i,i+1}$ of the differential is the concatenation of the blocks $B_i$ and $B_{i+1}$, that is the removal of the $(i+1)$th cut.
The part $d_{i,i+1}$ comes with the sign $(-1)^{i+1}$, and in the differential of the chain complex $(C_*(Cuts(u)),d)$, removing the $(i+1)$th cut comes with the same sign.
\end{proof}

This isomorphim and the acyclicity of the complex $\big(C_*(Cuts(u)),d\big)$ imply the following lemma:

\begin{lm}
 The complex $(gr_u, d_{gr})$ is acyclic for  $u \in Tuple(n)$.
\end{lm}

We can now finish the proof of Proposition \ref{Leib-projectifs}.

\begin{proof}[Proof of Proposition \ref{Leib-projectifs}]
The filtration of $C_*^{Leib}(P_n)$ is bounded since $Tuple(n)$ is a finite set.
So, by the classical convergence theorem of the spectral sequence associated to a filtration (see \cite[Theorem 5.5.1]{Weibel}), this spectral sequence converges to $H_*^{Leib}(P_n)$.
The previous lemma implies that
$\displaystyle{gr C^{Leib}_*(P_n) = \bigoplus_{u \in Tuple(n)} gr_u} $ is a sum of acyclic complexes, and therefore is also acyclic. We deduce that $H_*^{Leib}(P_n)=0$
\end{proof}

\subsection{Enriched  category}
One of the standard references for symmetric monoidal categories and enriched categories is the book of Borceux \cite[Chapter 6]{Borceux2}.
Let $({\mathcal{C}}, \otimes, 1)$ be a closed symmetric monoidal category. Recall that a symmetric monoidal category ${\mathcal{C}}$ is closed when for each object $C \in {\mathcal{C}}$, the functor $- \otimes C: {\mathcal{C}} \to {\mathcal{C}}$ admits a right adjoint denoted by $[C,-]: {\mathcal{C}} \to {\mathcal{C}}$.
\begin{defi}
 A category ${\mathcal{D}}$ enriched  over ${\mathcal{C}}$ (or a ${\mathcal{C}}$-category) consists of a class $I$ (representing the objects of ${\mathcal{D}}$) and for any objects $i,j,k \in I$ an object of ${\mathcal{C}}$: ${\mathcal{D}}(i,j)$ (representing the morphisms from $i$ to $j$ in ${\mathcal{D}}$) and morphisms in ${\mathcal{C}}$
$${\mathcal{D}}(i,j) \otimes {\mathcal{D}}(j,k) \to {\mathcal{D}}(i,k) \quad \text{and} \quad 1\to {\mathcal{D}}(i,i)$$
(representing the composition of morphisms in ${\mathcal{D}}$ and the identity morphism on $i$). These structure morphisms are required to be associative and unital in the obvious sense.
\end{defi}

\begin{exple}
\begin{enumerate}
\item A category enriched over $(Set, \times, [0])$ is an usual category.
\item A category enriched over $(\kkm, \otimes, \kk)$ is a $\kk$-linear category and  a functor enriched over $(\kkm, \otimes, \kk)$ is a $\kk$-linear functor.
\item The category ${\mathcal{C}}$ can be provided with the structure of an enriched category over $({\mathcal{C}}, \otimes, 1)$ using the bifunctor $[-,-]: {\mathcal{C}}^{op} \times {\mathcal{C}} \to {\mathcal{C}}$ whose composition with the forgetful functor ${\mathcal{C}}(1,-): {\mathcal{C}} \to Set$ is just ${\mathcal{C}}(-,-): {\mathcal{C}}^{op} \times {\mathcal{C}} \to Set$.

\end{enumerate}
\end{exple}

\begin{defi}
Let ${\mathcal B}$ and ${\mathcal{D}}$ categories enriched over ${\mathcal{C}}$. A functor enriched over ${\mathcal{C}}$ (or a ${\mathcal{C}}$-functor) from ${\mathcal B}$ to ${\mathcal{D}}$, $F:{\mathcal B} \to {\mathcal{D}}$ consists of an object of ${\mathcal{D}}$, $F(B)$ for every object $B \in {\mathcal B}$ and of morphisms in ${\mathcal{C}}$:
$$F(B,B'): {\mathcal B}(B,B') \to {\mathcal{D}}(F(B), F(B'))$$
for every pair of objects $B,B' \in {\mathcal B}$ that are associative and unital.
\end{defi}

\begin{defi}
A natural transformation enriched over ${\mathcal{C}}$, $\sigma: F \to F'$, between two functors enriched over ${\mathcal{C}}$, $F,F': {\mathcal B} \to {\mathcal{D}}$ consists in giving, for every object $B \in {\mathcal B}$, a morphism:
$$\sigma_B: 1 \to {\mathcal{D}}(F(B), F'(B))$$
in ${\mathcal{C}}$,
for every object $B$ of ${\mathcal B}$, that satisfy obvious commutativity conditions for a natural transformation.
\end{defi}

We denote by ${\mathcal{C}}\text{-Nat}(F,F')$ the object of natural transformations enriched over ${\mathcal{C}}$ between $F$ and $F'$.

\begin{thm}[\textbf{Enriched Yoneda lemma}]
Let ${\mathcal B}$ be a small category enriched over ${\mathcal{C}}$. For every object $B \in {\mathcal B}$ and every functor enriched over ${\mathcal{C}}$, $F: {\mathcal B} \to {\mathcal{C}}$, the object of natural transformations enriched over ${\mathcal{C}}$ from ${\mathcal B}(B,-)$ to $F$ exists and there is an isomorphism in ${\mathcal{C}}$:
$${\mathcal{C}}\text{-Nat}({\mathcal B}(B,-),F) \simeq F(B)$$
which is natural both in $F$ and in $B$.
\end{thm}

\subsection{Functor homology}

\begin{defi}
For ${\mathcal{D}}$ a ${\mathcal{C}}$-category and $F: {\mathcal{D}}^{op} \to {\mathcal{C}}$ and $G: {\mathcal{D}} \to {\mathcal{C}}$ a pair of ${\mathcal{C}}$-functors, the enriched tensor product of $F$ and $G$, $F \underset{{\mathcal{D}}}{\otimes}G$ is the coequalizer
$$\xymatrix{
\underset{d,d' \in D}{\coprod} F(d') \otimes {\mathcal{D}}(d,d') \otimes G(d) \ar@<+0.4ex>[r]^-{U} \ar@<-0.4ex>[r]_-{B}&\underset{d \in D}{\coprod} F(d) \otimes G(d) \ar[r] & F \underset{{\mathcal{D}}}{\otimes}G
}$$
in ${\mathcal{C}}$ where the map $U$ is induced by the composite
$$F(d') \otimes {\mathcal{D}}(d,d') \otimes G(d) \xrightarrow{1 \otimes F(d,d')} F(d') \otimes {\mathcal{C}}(F(d'), F(d)) \otimes G(d) \xrightarrow{ev \otimes 1} F(d) \otimes G(d) $$
where $F(d,d')$ is the morphism in ${\mathcal{C}}$ given because $F$ is a ${\mathcal{C}}$-functor and $ev$ is the evaluation map. The map $B$ is induced by a similar composite  with $G$ in place of $F$.
\end{defi}
Recall that the evaluation map is the counit of the adjunction on ${\mathcal{C}}$ of the monoidal product with the internal-hom.

For a category ${\mathcal{D}}$ enriched over ${\mathcal{C}}$, we call left ${\mathcal{D}}$-modules covariant ${\mathcal{C}}$-functors from ${\mathcal{D}}$ to  ${\mathcal{C}}$ and right ${\mathcal{D}}$-modules contravariant ${\mathcal{C}}$-functors from ${\mathcal{D}}$ to ${\mathcal{C}}$. Let ${\mathcal{D}}$-mod (resp. mod-${\mathcal{D}}$ ) the category of left (resp. right) ${\mathcal{D}}$-modules.
If ${\mathcal{C}}$ is an abelian category, the categories ${\mathcal{D}}$-mod and mod-${\mathcal{D}}$ are abelian. In the sequel ${\mathcal{C}}$ is an abelian category.

\begin{prop} \label{exact}
The bifunctor $-\underset{{\mathcal{D}}}{\otimes}-: \text{mod-}{\mathcal{D}} \times {\mathcal{D}}\text{-mod} \to {\mathcal{C}}$ is right exact with respect to each variable.
\end{prop}

We have the following characterization of homology theories in the enriched setting:

\begin{prop} \label{char-homo}
Let ${\mathcal{D}}$ be a ${\mathcal{C}}$-category and $G$ a right ${\mathcal{D}}$-module.
 If $H_*$ is a functor from ${\mathcal{D}}$-mod to ${\mathcal{C}}$ such that
\begin{enumerate}
 \item $H_*$ sends short exact sequences to long exact sequences,
 \item for all $F \in {\mathcal{D}}$-mod we have a natural isomorphism $H_0(F) \simeq G\otimes_{\mathcal{D}} F$ ,
 \item $H_i(F)=0$ for all projective ${\mathcal{D}}$-modules $F$ and $i>0$,
\end{enumerate}
then, for all $F \in {\mathcal{D}}$-mod, we have a natural isomorphism
$$H_i(F)\simeq Tor_i^{\mathcal{D}}(G,F).$$
\end{prop}

In the rest of the paper, all our categories will be enriched over  $(\kkm, \otimes_\kk, \kk)$.

\subsection{Algebraic operads}
A good reference for operads is the book of Loday and Vallette \cite{LV}.
We denote by $OrdSet$ the category of finite ordered sets (with order-preserving bijections as morphisms)
and by $Fin$ the category of finite sets (with bijections as morphisms).

\begin{defi}
A collection (resp. symmetric collection) is a contravariant functor from the category $OrdSet$
(resp. $Fin$) to the category $\kkm$.
\end{defi}

For a collection $P$, we denote $P([n-1])$ by $P(n)$ for $n \geq 0$.

Let $P$ and $Q$ be two symmetric collections. One defines their symmetric composition $P \circ Q$ by

$$(P \circ Q)(n) = \bigoplus_{m} P(m) \otimes_{\Sigma_m} \left( \bigoplus_{\alpha \in Set([n-1],[m-1]) } Q(\alpha^{-1}(0)) \otimes \ldots  \otimes Q(\alpha^{-1}(m-1)) \right).$$

Let $P$ and $Q$ be two collections.

One defines their nonsymmetric composition $P \circ Q$ by
$$(P \circ Q)(n) = \bigoplus_{m} P(m) \otimes \left( \bigoplus_{\alpha \in \Delta([n-1],[m-1])} Q(\alpha^{-1}(0)) \otimes \ldots  \otimes Q(\alpha^{-1}(m-1)) \right).$$

When $P(0)=Q(0)=0$, one defines their shuffle composition $P \circ_{sh} Q$ by
$$(P \circ_{sh} Q)(n) = \bigoplus_{m} P(m) \otimes \left( \bigoplus_{\alpha \in \Sh([n-1],[m-1])}Q(\alpha^{-1}(0)) \otimes \ldots  \otimes Q(\alpha^{-1}(m-1)) \right).$$

In the rest of the paper, unless otherwise stated, all our collections will be reduced, that is $P(0)=0$.
This hypothesis will be required to get a functor from symmetric operads to shuffle operads.

\begin{rem}
Note that
$$(P \circ_{sh} Q)(n) \simeq \bigoplus_{m} P(m) \otimes \left( \bigoplus_{\alpha \in \Sh^{surj}([n-1],[m-1])}Q(\alpha^{-1}(0)) \otimes \ldots  \otimes Q(\alpha^{-1}(m-1)) \right)$$
since $Q(0)=0$.
\end{rem}

\begin{rem}
Note that the previous definitions and remark can be extended to a symmetric monoidal category $({\mathcal{C}}, \otimes, 1)$ having an initial object $0$ satisfying $C \otimes 0 \simeq 0 \simeq 0 \otimes C$, $\forall C \in {\mathcal{C}}$. This is the case for $(Set, \times, [0])$, the initial object being $\emptyset$.
\end{rem}

\begin{defi}
A linear nonsymmetric (resp. shuffle) operad is a monoid $P$ in the category of collections equipped with the nonsymmetric (resp. shuffle) composition.
A linear symmetric operad is a monoid $P$ in the category of symmetric collections equipped with the symmetric composition.
\end{defi}

An algebra $A$ over an operad $P$ is a $\kk$-module endowed with an action of $P$, that is there are maps
$\theta : P(n) \otimes A^{\otimes n} \to A$ compatible with the monoid structure of $P$
(and with the symmetric group action in the symmetric case).

For $P$ a symmetric operad, an $A$-module $M$ is a $\kk$-module endowed with $M \oplus A$ has a $P$-algebra structure extending the
structure on $A$ and satisfying
$$\theta (P(n) \otimes (M^{\otimes k} \otimes A^{\otimes n-k}))=0 \textrm { if } n \geq k \geq 2.$$

\begin{exple} \label{uCom}
\begin{enumerate}

\item
Usual commutative algebras can be seen as algebras over a symmetric operad called $Com$,
determined by $Com(n)=\kk$ for all $n\geq1$ and $Com(0)=0$, and equipped with the obvious composition.
The notion of $A$-modules for $A$ a commutative algebra is the usual one.
The case of unital commutative algebras can be dealt using the symmetric operad called $uCom$,
determined by $uCom(n)=\kk$ for all $n\geq 0$ and equipped with the obvious composition.

\item Lie algebras can be seen as algebras over a symmetric operad called $Lie$.
This operad is generated by a symmetric bracket which satisfies the Jacobi relation.
The notion of $A$-modules for $A$ a Lie algebra is the usual one.

\end{enumerate}
\end{exple}

There exists a forgetful functor $(-)^{sh}$ from symmetric collections to nonsymmetric collections.
This functor allows us to make a direct link between the notion of symmetric operads and the notion of shuffle operads because of the following lemma,
proved by Dotsenko and Khoroshkin:
\begin{lm} \cite[Proposition 3]{DK2010}
The forgetful functor $(-)^{sh}$ is monoidal.
\end{lm}

Therefore this functor induces a forgetful functor from symmetric operads to shuffle operads.

\begin{exple}
The usual $Lie$ symmetric operad induces a linear shuffle operad.
\end{exple}

The fundamental property we will use later in the paper is the following:
For ${\mathcal{P}}$ a symmetric operad, the $\kk$-modules ${\mathcal{P}}(n)$ and ${\mathcal{P}}^{sh}(n)$ are the same. For instance they have the same bases.
This observation also implies that for a ${\mathcal{P}}$-algebra $A$, the maps $\theta : P(n) \otimes A^{\otimes n} \to A$ defining
the algebra structure can be seen as maps $P^{sh}(n) \otimes A^{\otimes n} \to A$. The same holds for maps
defining a $A$-module structure on $M$.

\subsection{Enriched category associated to an operad}

The notion of enriched category associated to an operad is quite classical, already appearing in a paper of May and Thomason \cite{MayT}.
We recall the classical definition and then extend it to the shuffle context.

\begin{defi}
\begin{itemize}
\item The linear category of pointed operator $\Gamma^{\mathcal{P}}$ associated to a reduced \textbf{symmetric} linear operad ${\mathcal{P}}$ is the linear category whose objects are pointed finite sets $[n]$ and such that:
$$\Gamma^{\mathcal{P}}([n],[m])=\underset{\alpha \in \Gamma([n],[m])}{\bigoplus}{\mathcal{P}}( \alpha^{-1}(0) ) \otimes \ldots \otimes {\mathcal{P}}( \alpha^{-1}(m)).$$
Composition of morphisms is prescribed by symmetric operad structure maps in ${\mathcal{P}}$.
\item The linear category of operator $\Sh^{{\mathcal{P}}}$ associated to a reduced \textbf{shuffle} linear operad ${\mathcal{P}}$  is the linear category whose objects are pointed  finite sets $[n]$ and such that:
$$\Sh^{{\mathcal{P}}}([n],[m])=\underset{\alpha \in \Sh([n],[m])}{\bigoplus}{\mathcal{P}}( \alpha^{-1}(0)) \otimes \ldots \otimes {\mathcal{P}}( \alpha^{-1}(m)).$$
Composition of morphisms is prescribed by shuffle operad structure maps in ${\mathcal{P}}$.
\end{itemize}
\end{defi}

\begin{rem}
Note that for a symmetric operad ${\mathcal{P}}$ satisfying ${\mathcal{P}}(0)=0$, we have
$$\Gamma^{\mathcal{P}}([n],[m])\simeq \underset{\alpha \in \Gamma^{surj}([n],[m])}{\bigoplus}{\mathcal{P}}( \alpha^{-1}(0) ) \otimes \ldots \otimes {\mathcal{P}}( \alpha^{-1}(m))$$
and
$$\Sh^{{\mathcal{P}}}([n],[m])\simeq \underset{\alpha \in \Sh^{surj}([n],[m])}{\bigoplus}{\mathcal{P}}( \alpha^{-1}(0)) \otimes \ldots \otimes {\mathcal{P}}( \alpha^{-1}(m)).$$

\end{rem}
For a shuffle operad ${\mathcal{P}}^{sh}$ coming from a symmetric operad ${\mathcal{P}}$, we ease notation by abbreviating $\Sh^{{\mathcal{P}}^{sh}}$ to $\Sh^{{\mathcal{P}}}$.

Note that $\Sh^{\mathcal{P}}([n],[0])= {\mathcal{P}}^{sh}([n])= {\mathcal{P}}([n])={\mathcal{P}}(n+1)$.

\begin{exple}
We explain graphically how the composition works in $\Gamma^{\mathcal{P}}_{sh}$ for ${\mathcal{P}}$ a shuffle operad.

Let $(\alpha,f)$ be the generator of $\Gamma^{\mathcal{P}}_{sh}([9],[5])$ where $\alpha \in \Gamma_{sh}([9],[5])$ is given by:
$$\alpha(0)=\alpha(4)=0; \ \alpha(1)=1;\ \alpha(2)=\alpha(3)=\alpha(7)=2;\ $$
$$\alpha(5)=3;\ \alpha(6)=\alpha(9)=4;\ \alpha(8)=5$$
and
$$f=f_0 \otimes f_1 \otimes f_2 \otimes f_3 \otimes f_4 \otimes f_5$$
 where $\forall i \in\{0,1,2,3,4,5\}\ f_i \in {\mathcal{P}}(\alpha^{-1}(i))$. We can represent this morphism by the following picture:
$$ \vcenter{\xymatrix@M=1pt@H=0pt@R=10pt@C=14pt@!0{
 0 \ar@{-}[ddr] & & 4 \ar@{-}[ddl]& & & 1\ar@{-}[dd] &  & & 2\ar@{-}[ddrr] & & 3\ar@{-}[dd] & & 7\ar@{-}[ddll] & & & 5\ar@{-}[dd] & & & 6\ar@{-}[ddr] & & 9\ar@{-}[ddl] & & & 8 \ar@{-}[dd] \\
 &\\
 & f_0\ar@{-}[dd] & & & & f_1 \ar@{-}[dd]& & & & & f_2\ar@{-}[dd] & & & & & f_3\ar@{-}[dd] & & & & f_4 \ar@{-}[dd]& & & & f_5 \ar@{-}[dd]& \\
 &\\
 &0&&&&1&&&&&2&&&&&3&&&&4&&&&5\\
}} $$
Let $g$ be the generator of $\Gamma^{\mathcal{P}}_{sh}([5],[2])$ represented by the following picture:
$$ \vcenter{\xymatrix@M=1pt@H=0pt@R=10pt@C=14pt@!0{
 0 \ar@{-}[ddr] & & 3 \ar@{-}[ddl]& & & 1\ar@{-}[dd] &  & & 2\ar@{-}[ddrr] & & 4\ar@{-}[dd] & & 5\ar@{-}[ddll] \\
 &\\
 & g_0\ar@{-}[dd] & & & & g_1 \ar@{-}[dd]& & & & & g_2\ar@{-}[dd] \\
 &\\
 &0&&&&1&&&&&2\\
}} $$

To compose them, we use the composition of the shuffling maps $\alpha \in \Gamma_{sh}$ and the shuffle composition of the operad ${\mathcal{P}}$.

We obtain the following element in $\Gamma^{\mathcal{P}}_{sh}([9],[2])$, where the dotted lines shows where the shuffle composition of ${\mathcal{P}}$ has to be done.

$$ \vcenter{\xymatrix@M=1pt@H=0pt@R=10pt@C=14pt@!0{
0\ar@{-}[ddr] && 4\ar@{-}[ddl] && 5\ar@{-}[dd] &&& 1\ar@{-}[dd] &&& 2\ar@{-}[ddrr] && 3\ar@{-}[dd] && 7\ar@{-}[ddll] && 6\ar@{-}[ddr] && 9\ar@{-}[ddl] && 8\ar@{-}[dd] \\
&\\
& f_0 \ar@{.}[ddrr] &&& f_3\ar@{.}[ddl] &&& f_1 \ar@{.}[dd]&&&&& f_2\ar@{.}[ddrrrrr] &&&&& f_4\ar@{.}[dd] &&& f_5\ar@{.}[ddlll] \\
&\\
&&& g_0\ar@{-}[dd] &&&& g_1\ar@{-}[dd] &&&&& &&&&& g_2\ar@{-}[dd] \\
&\\
 &&&0&&&&1&&&&&&&&&&2&&&&&&&\\
}}$$

This element can also be written as

$$ \vcenter{\xymatrix@M=1pt@H=0pt@R=10pt@C=14pt@!0{
0\ar@{-}[ddrr] && 4\ar@{-}[dd] && 5\ar@{-}[ddll] &&& 1\ar@{-}[dd] &&&
2\ar@{-}[ddrrrrr] && 3\ar@{-}[ddrrr] && 6\ar@{-}[ddr] && 7\ar@{-}[ddl] && 8\ar@{-}[ddlll] && 9 \ar@{-}[ddlllll]  \\
& \\
&& h_0\ar@{-}[dd] &&&&& h_1\ar@{-}[dd] &&&&& &&& h_2\ar@{-}[dd] \\
& \\
 &&0&&&&&1&&&&&&&&2&&&&&&&&&\\
}}$$
where the element $h_2 \in {\mathcal{P}} (\{2,3,6,7,8,9\})$ is given by the composition $g_2 \circ (f_2, f_4, f_5)$
associated to the shuffling map $\alpha: \{2,3,6,7,8,9\} \to \{0,1,2\}$ defined by $\alpha^{-1}(0)=\{2,3,7\}$, $\alpha^{-1}(1)=\{6,9\}$ and
$\alpha^{-1}(2)=\{8\}$. The elements $h_0$ and $h_1$ are defined in a similar way.

\end{exple}

For ${\mathcal{P}}=Lie$, there are some particular maps in $\Gamma^{Lie}$ and $\Sh^{Lie}$ which are central in our later constructions.

\begin{defi} \label{dij}
For every $n \geq0$ and for $0\leq i <j \leq n+1 $,
there is one element in  $\Sh^{Lie}([n+1],[n])$ called $d_{i,j}$ associated to the single map $f \in\Sh ([n+1],[n])$ sending $i$ and $j$ to $i$,
where the element in $Lie(1)$ is the identity and the element $\mu$ in $Lie(2)$ is the $Lie$ bracket.
$$ \vcenter{\xymatrix@M=2pt@H=0pt@R=16pt@C=14pt@!0{
 0\ar@{-}[dd]& & 1\ar@{-}[dd] & & \ldots\ar@{-}[dd] & & i\ar@{-}[dr] & & j\ar@{-}[dl]
& & i+1\ar@{-}[dd] & & \ldots\ar@{-}[dd] & & \hat j&  & \ldots\ar@{-}[dd] & & n+1\ar@{-}[dd] \\
& &  & &  & & & \mu\ar@{-}[d] & & &  & &   & & &  &  & &  \\
0& & 1 & & \ldots & & & i & & & i+1 & & \ldots  & & &  & \ldots & & n  }} $$
\end{defi}

\begin{rem}
These elements $d_{i,j}$ are actually generators (via linear combination and composition) of $\bigoplus_{n,m}{\Sh^{Lie}}([n],[m])$.
\end{rem}

\begin{rem} \label{linear}
If $S$ is a symmetric set operad, we consider the corresponding linear symmetric operad $\kk S$ defined as $\kk S(n):=\kk[S(n)]$ with $\kk[-]$ denoting the free $\kk$-module. The $\kk$-linearization of a category ${\mathcal{C}}$ , which will be denoted by $\kk[{\mathcal{C}}]$, is the $\kk$-linear category which has the same objects as ${\mathcal{C}}$ and whose sets of morphisms $\kk[{\mathcal{C}}](c,c')$ are the $\kk$-modules $\kk[{\mathcal{C}}(c,c')]$ generated by the sets of morphisms of ${\mathcal{C}}$.

It is easy to check that $\Gamma^{\kk S} = \kk [\Gamma(S)]$, where $\Gamma(S)$ denotes the category associated to $S$.

For example, $\Gamma^{uCom}=\kk[\Gamma]$ and $\Gamma^{Com}=\kk[\Gamma^{surj}]$, where $Com$ and $uCom$ are the symmetric linear operad defined in Example \ref{uCom}.

Furthermore, for $F: {\mathcal{C}} \to \kkm$ and $G: {\mathcal{D}}^{op} \to \kkm$  and $\kk[F]: \kk[{\mathcal{C}}] \to \kkm$ and $\kk[G]: \kk[{\mathcal{D}}^{op}] \to \kkm$ their linearizations, we have:
$$Tor^{{\mathcal{C}}}(F,G) \simeq Tor^{\kk[{\mathcal{C}}]}(\kk[F], \kk[G]).$$
\end{rem}

\begin{obs} \label{projectifs}
Let ${\mathcal{P}}$ be a symmetric operad.
By the enriched Yoneda lemma, the representable functors $\Gamma^{\mathcal{P}}([n],-)$ for $n \geq 0$ are projective generators of the category of $\kk$-linear functors from $\Gamma^{\mathcal{P}}$ to $\kkm$.
Similarly, the representable functors $P_n= \Gamma^{\mathcal{P}}_{sh}([n],-)$ for $n \geq 0$ are projective generators of the category of linear functors from $\Gamma^{\mathcal{P}}_{sh}$ to $\kkm$.

Notice that for $G$ a $\kk$-linear functor from $(\Gamma^{\mathcal{P}})^{op}$ to $\kkm$ and $F$ a $\kk$-linear functor from $\Gamma^{\mathcal{P}}$ to $\kkm$, we have:
$$G \underset{\Gamma^{\mathcal{P}}}{\otimes} \Gamma^{\mathcal{P}}([n],-) \simeq G([n]); \qquad \Gamma^{\mathcal{P}}(-,[n]) \underset{\Gamma^{\mathcal{P}}}{\otimes} F \simeq F([n]).$$
\end{obs}

\subsection{Loday functor}

We recall the definition of the Loday functor, appearing in \cite{Loday-Cyclic}.
This functor appears in various contexts (associative or commutative algebras, pointed or unpointed sets)

Let $A$ be a unital commutative algebra and $M$ an $A$-module.
The Loday functor ${\mathcal{L}}(A,M): \Gamma \to \kkm$ is defined by
${\mathcal{L}}(A,M)([n])= M \otimes A^{\otimes n}$
 and for $f$ an element of $\Gamma([n],[m])$
the induced map $f_*:{\mathcal{L}}(A,M)([n]) \to {\mathcal{L}}(A,M)([m])$ is given by
$$f_*(a_0 \otimes a_1 \otimes \ldots \otimes a_n)=b_0 \otimes \ldots \otimes  b_m$$
where $b_i= \underset{j \in f^{-1}(i)}{\prod} a_j$ or $b_i=1$ if $f^{-1}(i)=\emptyset$.

We first extend this definition for algebras over a symmetric operad,
and then see how it can be adapted in the shuffle context.

Let ${\mathcal{P}}$ be a symmetric operad, $A$ be a ${\mathcal{P}}$-algebra and $M$ a $A$-module. Let us call $\theta$ the structure maps.

\begin{lm}
There exists a linear functor:
$${\mathcal{L}}^{\mathcal{P}}(A,M):\Gamma^{\mathcal{P}} \to \kkm$$
defined on an object $[n]$ by:
$$\mathcal{L}^{\mathcal{P}}(A,M)([n])=M \otimes A^{\otimes n}$$
and on a generator $(\alpha,f)$ of $\Gamma^{\mathcal{P}}([n],[m])$ where $\alpha: [n] \to [m]$ and $f=f_{0} \otimes \ldots \otimes f_m$ (with $f_i \in {\mathcal{P}}(\alpha^{-1}(i))$), by:
$$\mathcal{L}^{\mathcal{P}}(A,M)(\alpha,f)(a_0 \otimes a_1 \otimes \ldots \otimes a_n)=b_0 \otimes \ldots \otimes  b_m$$
where $b_i=\theta ( f_i \otimes \underset{j \in \alpha^{-1}(i)}{\bigotimes} a_j)$.

We extend the definition on a morphism in $\Gamma^{\mathcal{P}}([n],[m])$ by linearity.
\end{lm}

\begin{proof}
We need to prove first that $\mathcal{L}^{\mathcal{P}}(A,M)(\alpha,f)$ is well-defined, and then show the functoriality.

First, note that $b_0$ is in $M$ because $a_0$ is in $M$, the other $a_j$'s are in $A$ and because $\alpha(0)=0$. Moreover, the linearity is obtained by construction.

Concerning the functoriality, recall that the composition in $\Gamma^{\mathcal{P}}$ is defined by the operadic composition.
Recall also that the structure maps $\theta$ (for the module $M$ and the algebra $A$) are required to be compatible with the operadic composition.
This requirement gives exactly $\mathcal{L}^{\mathcal{P}}(A,M)(id,1_{\mathcal{P}} \otimes \ldots \otimes 1_{\mathcal{P}})=id$ and $\mathcal{L}^{\mathcal{P}}(A,M)((\alpha,f)(\beta,g)) = \mathcal{L}^{\mathcal{P}}(A,M)(\alpha,f)\mathcal{L}^{\mathcal{P}}(A,M)(\beta,g)$.
\end{proof}

\begin{defi}
The functor ${\mathcal{L}}^{\mathcal{P}}(A,M): \Gamma^{\mathcal{P}} \to \kkm$ is called the symmetric Loday functor
associated to the ${\mathcal{P}}$-algebra $A$ and the $A$-module $M$.
\end{defi}

\begin{rem}
For $A$ a unital commutative algebra and $M$ an $A$-module, we have ${\mathcal{L}}^{uCom}(A,M)=\Bbbk [{\mathcal{L}}(A,M)]$ where $uCom$ is the symmetric operad defined in Example \ref{uCom}.
\end{rem}
\begin{rem}
For $A$ a commutative algebra and $M$ a $A$-module, by \cite{Pira-Hodge} and \cite{PiraR-2000} we have
$$H_*^{\Gamma}(A,M) \simeq Tor^{\Gamma}(t, {\mathcal{L}}(A,M))$$
where $H_*^{\Gamma}(A,M)$ is the $\Gamma$-homology of $A$ with coefficients in $M$ and $t: \Gamma^{op} \to Ab$ is the cokernel of $\Gamma(-, [2]) \to \Gamma(-, [1])$. By the previous remark and Remark \ref{linear} this result can be rephrased in linear categories as:
$$H_*^{\Gamma}(A,M) \simeq Tor^{\Gamma^{uCom}}(\kk[t], {\mathcal{L}}^{uCom}(A,M)).$$
\end{rem}

\bigskip

Let ${\mathcal{P}}$ be a symmetric operad, $A$ be a ${\mathcal{P}}$-algebra and $M$ a $A$-module.
Recall that ${\mathcal{P}}^{sh}$ is the shuffle operad associated to ${\mathcal{P}}$.
Let us call $\theta$ the structure maps.

\begin{lm}
There exists a linear functor:
$${\mathcal{L}}^{\mathcal{P}}_{sh}(A,M):\Gamma^{\mathcal{P}}_{sh} \to \kkm$$
defined on an object $[n]$ by:
$$\mathcal{L}^{\mathcal{P}}_{sh}(A,M)([n])=M \otimes A^{\otimes n}$$
and on a generator $(\alpha,f)$ of $\Gamma^{\mathcal{P}}_{sh}([n],[m])$ where $\alpha: [n] \to [m]$ and $f=f_{0} \otimes \ldots \otimes f_m$ (with $f_i \in {\mathcal{P}}_{sh}(\alpha^{-1}(i))$), by:
$$\mathcal{L}^{\mathcal{P}}_{sh}(A,M)(\alpha,f)(a_0 \otimes a_1 \otimes \ldots \otimes a_n)=b_0 \otimes \ldots \otimes  b_m$$
where $b_i=\theta ( f_i \otimes \underset{j \in \alpha^{-1}(i)}{\bigotimes} a_j)$.
We extend the definition on a morphism in $\Gamma^{\mathcal{P}}_{sh}([n],[m])$ by linearity.
\end{lm}

\begin{proof}
The proof is mostly the same as in the symmetric case. The only difference comes from the shuffle context.
Recall that the structures maps in the symmetric context induce structures maps in the shuffle context.
These induced maps are compatible with the shuffle operadic composition.
\end{proof}

\begin{defi}
The functor ${\mathcal{L}}^{\mathcal{P}}_{sh}(A,M): \Gamma^{\mathcal{P}}_{sh} \to \kkm$ is called the shuffle Loday functor
 associated to the ${\mathcal{P}}$-algebra $A$ and the $A$-module $M$.
\end{defi}

redpajama_set_name: RedPajamaArXiv


text:
.PHONY: clean-pyc clean-build docs

server:
	@python manager.py run

# upgrade:
# 	@alembic upgrade head

# staticdir = june/public/static
# assets:
# 	@$(MAKE) -C assets build

# static:
# 	@$(MAKE) -C assets compile
# 	@uglifyjs ${staticdir}/app.js -m -o ${staticdir}/app.js

# translate
babel-extract:
	@python setup.py extract_messages -o messages.pot

# language = zh
# i18n = june/translations
# babel-init:
# 	@python setup.py init_catalog -i messages.pot -d ${i18n} -l ${language}

# babel-compile:
# 	@python setup.py compile_catalog -d ${i18n}

# babel-update: babel-extract
# 	@python setup.py update_catalog -i messages.pot -d ${i18n}

# Common Task
clean: clean-build clean-pyc
	@rm -fr cover/

clean-build:
	@rm -fr build/
	@rm -fr dist/
	@rm -fr *.egg-info

clean-pyc:
	@find . -name '*.pyc' -exec rm -f {} +
	@find . -name '*.pyo' -exec rm -f {} +
	@find . -name '*~' -exec rm -f {} +

lint:
	@flake8 fbone
	@flake8 tests

test:
	@nosetests -s

coverage:
	@rm -f .coverage
	@nosetests --with-coverage --cover-package=june --cover-html

redpajama_set_name: RedPajamaGithub


For the uninitiated, TEDx is a self-organized, supremely local community program that allows individuals and groups to discuss ideas to, essentially, change the world. Make it a better place. Stimulate dialogue. Speakers are given a set amount to speak (usually around 15 minutes to 20 minutes), and it's recorded.
TEDx was born from TED, a nonprofit organization dedicated to Ideas Worth Spreading. There are more than 900 TEDTalks available in a variety of languages (although subtitled in English).
My all-time favorite is a talk by Simon Sinek, a guy who reminds me of one of my college professors from Radford University in Virginia.
In his talk, How Great Leaders Inspire Action, Sinek discusses companies like Apple and leaders including Martin Luther King in a very scientific way, but keeps the content accessible and inspirational. Parts of it just gives you goosebumps.
I practically hung on every word. And then I showed it to my husband.
So without further ado, here are some of my favorite TEDx videos. I hope you enjoy them.
How Great Leaders Inspire Action by Simon Sinek (TEDxPugetSound).
How to Build a School in 3 Hours by Taylor Conroy (TEDxJuanDeFuca). Conroy offers tips on creative fundraising and getting his friends and others involved for charity and building schools.
Gay Marriage Rights Inadequacies in America by Nadine Smith (TEDxTampaBay). Smith, who is the executive director of Equality Florida, gives a powerful talk about marriage and the legal jeapardies of gay marriage.
The Power of Vulnerability by Brene Brown (TEDxHouston). Brown, who is a longtime researcher, discusses human connection under excruciating vulnerability.
Building a Personal Brand by Jacob Cass (TEDxCMU). Aussie Cass shares a pretty rough story on getting and losing his job in New York and then using his blog and social networking to find a new opportunity.
Are We Born to Run? by Christopher McDougall (TEDxPennQuarter). The runner in me finds McDougall's talk really fascinating.
Mike Rowe celebrates dirty jobs.
Christine Cube is a media relations manager for PR Newswire and freelance writer in San Diego, Calif. You can follow her @cpcube.


text:
Hydrogen-Powered Land Rover Defenders To Start Trials From End Of 2021
Chandrutpal | 23/06/2021 - 21:00
The latest-generation Land Rover Defender is going to serve as the testbed for Jaguar Land Rover's fuel-cell powertrain technology, with the manufacturer expected to start testing a hydrogen-powered..
Future Jaguar Land Rover Models To Feature Anti-COVID Filter
Yatharth Chauhan | 17/03/2021 - 11:46
Jaguar Land Rover is working with Panasonic to develop an air filter that nukes up to 97 per cent of viruses and airborne bacteria. Basically, the upcoming filtration system will work as a face mask..
Jaguar F-Pace SVR Facelifted For The International Market
Saptarshi | 07/12/2020 - 19:20
Jaguar has given the F-Pace SVR a mid-life facelift for the international market. The performance SUV gets few tasteful cosmetic updates, some new tech on the interiors, but more importantly, the..
Jaguar Land Rover Develop New Contactless Touchscreen Called Predictive Touch
Karan | 23/07/2020 - 16:00
As the Global pandemic continues, the World is adapting to this new age in various sorts of ways. While some innovate in personal protection equipment, car manufacturers are innovating in other ways...
Jaguar Land Rover looking forward to manufacturing in India
Shoeb Kalania | 14/06/2012 - 09:00
The voices are growing louder with each passing day. Jaguar-Land Rover (JLR), owned by Tata Motors, is very eager to setup a manufacturing facility in India. It could come up either in the state of..
Land Rover considers a BMW X1 rival
Kaustubh Shinde | 13/03/2012 - 05:24
The list of contenders that will rival the mighty BMW X1 just keeps on growing. Audi is already out with its Q3 model. Mercedes is planning a GLC crossover which will debut in 2014. Volkswagen is..
Jaguar shifts brand baseline, claims 'It's Alive!' [Video]
Tata Motors revival of Jaguar brand is one of the epic corporate saga of the last decade. Back in 2008, when Tata Motors bought JLR, Jaguar was almost in the dumps. But with the introduction of new..
Tata Motors to double investments in Jaguar-Land Rover
Tata Motors has indeed proved to be the 'Knight in Shining Armor' for the damsel in distress Jaguar-Land Rover. Back in 2008, when Tata Motors agreed purchase the British marquee from Ford, dark..

redpajama_set_name: RedPajamaCommonCrawl


Hi! Sign in to let us know how Tony's Pizza Ii was?
by millerltchik at Citysearch Nov 10 .
The closest I have found to true New York Style pizza. You have to try it... it's the BEST!! The famous homemade wing sauce makes their Buffalo wings "To Die For"! Great service and a very clean atmosphere.
If you've been to or used Tony's Pizza Ii, leave a review.


Deconstructing the consulting contract is done when the client or agency wants to further evaluate the agreement that they will work into specifically when it comes in the consulting part. This is usually done with a legal and professional adviser to check the validity and authenticity of the contract.
It involves further examination and breakdown of the content of the consultancy contract like any other contract involves the following parts: introduction of the contract, definition of the two parties, relationship of parties, scope of work, payment, termination, term and code ownership. Also in some cases there is a termination clause in the contract when things don't come smoothly between the two parties.
All details and terms should be further explained and agreed to by the client and the agency to make sure that both of them are aware of their duties and responsibilities in the consultancy contract. With all of this a final contract and signs of the two parties will seal the deal and the agreement will start will start on a specified date.
Contract for professional consulting services is an agreement between a client or an agency/ person whose expertise is needed when it comes to having suggestion or opinion in a given matter like in business, law, finance, service, engineering and other fields that require a professional advice.
This contract is done for the purpose of having an educated guidance when it comes to a person or company's decision when it regards to a particular matter, the agency/person finds better solution or strategy to problems and to have a better allocation of resources to come up with a better output. It is short service and the contract is done when there is no longer need for the service of the agency/person.
There are many types of consultancy service especially in the fields of business, the contract tells that the agency will only direct and suggest to the client to come up with a better decision or plan but the final say will come to the client themselves.


text:
Icelandic Voters Reject "Icesave" Again, Effectively Telling UK and Netherlands Banks "Go to Hell"; Iceland's Common Sense Stance
Posted by Mish | April 9, 2011 10:18:00 | Uncategorized
Congratulations to Icelandic voters who once again told the UK and Netherlands where to go. Please consider Iceland Rejects Icesave Debt Deal
Icelandic voters appeared Sunday to have rejected a government-approved deal to repay Britain and the Netherlands $5 billion for their citizens' deposits in the failed online bank Icesave.
Partial results of a national referendum suggested the "no" side had gained more than half the votes — a reflection of enduring anger over the economic havoc wrought by Iceland's risk-taking bankers.
Full results were not due until later Sunday. With partial results in from all six of Iceland's constituencies, the no side had almost 57 percent of the votes and the yes camp just over 43 percent.
"This is of course a disappointing result," said Prime Minister Johanna Sigurdardottir.
Icelanders overwhelmingly rejected a previous deal in a referendum last year. The government hoped a "yes" vote on an improved offer passed by parliament would finally resolve a dispute that has caused friction among the three countries and complicated Iceland's recovery from its economic collapse in 2008.
The dispute has grown acrimonious, with Britain and The Netherlands threatening to block Iceland's bid to join the European Union unless it is resolved.
"Taxpayers should not be responsible for paying the debts of a private institution," said Sigriur Andersen, a spokeswoman for the Advice group, which opposes the agreement. "I think that sends the wrong message onto the market, and sets a wrong precedent."
Icelandic voters want no part of "Icesave". Even the name "IceSave" is preposterous. Iceland was save by the fact voters rejected "Icesave". Icelanders would have been debt-slaves for decades had they accepted the original terms.
How many times do citizens have to say no? Hopefully voters give Prime Minister Johanna Sigurdardottir a well deserved boot in the next election.
Moreover, Iceland needs to rethink why it would want to be part of the Eurozone in the first place. I suggest Iceland put the Euro to another vote.
Finally, I am really disappointed in the wimps in Ireland. They should put Ireland's "reverse bailout" to a vote as well. I can guarantee the results in advance.
Banks that make stupid loans should suffer for them, not taxpayers. So far, Iceland is the only country that has taken this common-sense stance.


I have been absolutely off my feet for the last couple of weeks with the day job so the blog has taken a back step, not helped by the fact when you get home it's dark. So outfit posts are a challenge, and when it's raining all weekend it's an extra delay. So yesterday mid shoot I managed to drag a friend outside (in the brief moment it wasn't raining) and take a couple of pics of some new pieces I've picked up recently.
First up is a lovely cropped knit from Reiss, I'd been after for ages and am so happy I finally tracked one down. I'd had my eye on much more expensive versions, so this one for £120 was an absolute steal. I can also confirm it is super warm, and has pockets (an essential for me, if nothing else). My tee is an old favourite from Topshop, I've had it for years... It has little Eiffel Towers on, I found one on eBay if anyone wants to get their hands on one here.
My Chelsea boots are from Hush, who I'm working on a little something with so watch this space. I've been buying their knitwear and pyjamas for years so was thrilled when the opportunity came up. The boots are a gorgeous shade of blue which works so well with denim and pretty much everything I own - plus, they make my ridiculous tiny feet look normal sized which is a "big" one for me as I hate my tiny excuses for hooves.
On to the jeans, which are from a brand called AND Jeans. They're new to the market, and have been designed for busy girls like me. They spent years looking at how jeans wear over time, where they rip, tear, etc - and created a collection that will stay looking good for longer. Their model is the girl from the Blurred Lines video, so you can even imagine you're as hot as her in them (and you might find it encourages your boyfriend to be enthusiastic about your purchase, I speak with experience. They come up true to size and stay the same size on wear 1 as they do wear 5. Very happy.
That knit is amazing Anna, what a pretty hue. I just love how you (PIN &) create simple effortless looks using classic pieces that can be worn in so many different ways! I'm sure I've said it before, but you really remind me of Olivia P!!
LOVE that jumper! You are a total stunner, and that's a fact.


Large bohemian style gypsy hoop earrings with geometric mandala design and lotus petals. Perfect festival accessories in a hippy style.
The Lotus petals represent the beauty that comes out of the darkness as Lotus flowers always grow in muddy waters. It is a special sacred symbol in the Indian vedic tradition.
These need to be polished occasionally to keep them shined up.


Hey everyone, I want to sign up for a class with a coach but my Wifi connection is not always stable because I'm living basically in the middle of nowhere and helping my family out until the end of the month. Are coaching sessions done over Skype? Telephone? A portal within PrepLounge? I am just wondering what kind of Wifi connection strength I need. Thank you.
All are feasible options and the way you choose for meeting depends on dialogue between you and the coach. In your case, I would recommend phone but if you choose a coach from another country, roaming fees WILL make you wish you invested in a much better WiFi. In some cases, overseas phone calls for an hour can cost more than coaching fee itself. Thus, be careful.
We do most sessions over Skype, though Google Hangouts or even WhatsApp are also options for example. My strong preference is always to have the visual, so I can give feedback on your body language if relevant: As a consultant, you will need to inspire confidence, and a weak body language / worried demeanor will work against you.


text:
Sayer's Sister
Mark and Steph Sayers have a new little sister in the heart of Northbridge in WA.
After the couple opened Sayers Food Leederville in 2006, they waited for the right time and location before opening Sayers Sister in August 2012.
"Sayers Sister has gone off with a time bomb," says Owner Mark. "Six years ago I didn't realise [coffee] would turn into something so huge. [This industry] is about trying to incorporate completeness and an understanding of coffee as well."
Mark and his wife Steph have taken an involved approach to their coffee business, travelling around the world to coffee origins, going direct to farmers for their produce and even planting coffee trees in Bali.
"We are very passionate about what we do and there's a passion in us delivering that as well," says Mark.
The café is driven towards a mixed crowd and teaching the next generation to appreciate coffee. "We think cafés are a modern-day community centre and a good meeting place that supplies good coffee and food," says Mark. "We're trying to inspire young people to be more passionate about coffee – and many already are and that's full of encouragement."
The café interior takes a recycled approach with reused timbers and images of Mark's coffee travels displayed on the walls. "We want people to see what we do and why we do it," Mark says.
Sayer's Sister aims to push unique and one- off coffee offerings to its WA customers using Five Senses Coffee. The houseblend is a combination of Brazil Santo Antonio, Balinese Kintamani and Indian Attikan Estate.
Single origins are also available and change frequently.
"Coffee brings a whole sense of well- being. You build a product, and sell it to the best of your ability," says Mark.
Food is also important to accompany good quality coffee at Sayers Sister. Gnocchi with sweet corn, zucchini and preserved lemon is a house favourite.


text:
My Writing Projects
Occult Pop Culture
David Bowie's Blackstar: Possible Occult Interpretations
The Occult Framework of Pee-wee's Big Adventure
Frank Herbert's Dune and Modern Conspiracy Mythology
Superman: Man of Steel, God of Flesh
Duncan Trussell: A Perpetually-Blossoming Blood Lotus Found in a Gray World of Endless Death
The Mythology of SAW and the New Religion of John Kramer
Tolkien Wrote Post-Apocalyptic Fiction
Pacific Rim: Crucifying Robo-Christ
Christian Symbolism in the Videogame Infamous
Mass Effect Obsession
Dragon Age Is the Future of Mass Effect
Mass Effect: Planet of the Reapers
Mass Effect 4: Indoctrination Theory
Mass Effect: Rio de Janeiro Theory
The Ghost: Outcome of Mass Effect 3's Synthesis Ending
Big Goddamn Hero: Outcome of Mass Effect 3's Control Ending
The Crush: Outcome of Mass Effect 3′s Destroy Ending
The End of Mass Effect 3: Yet Another Blowhard Gives His Opinion
Hell: An RPG Maker Game
Game Idea: BEAT UP!
I Had an Epiphany About Open-World Games
The Controversy over Wild Kings
Review: Infinite Space (DS)
Top Ten Games Never Made by Historical Figures: 10 and 9 by Kyle B. Stiff
Top Ten Games Never Made by Historical Figures: 8 and 7 by Kyle B. Stiff
Fake Game Reviews: SAW: Everlasting Love, Blame!, and Let's Take a Shower (a Wii exclusive)
Valkyrie Profile: Covenant of the Plume (DS)
I Want the Blood of Heroes!
Metal Gear: Peace Walker
Kyle B. Stiff's Favorite Movie: Conan the Barbarian
Cloud Atlas: Formula for a Turd
A Song of Hot and Cold: The End of Game of Thrones
Arrested Development Season 4: The Ostrich Cries in Anguish from Deepest Darkness
The Key to Unlocking Iron-Man 3: Is This Even a Review?
Miyazaki's Ponyo
Shigurui: Death Frenzy review
Mamoru Oshii's The Sky Crawlers
The Weird Stuff
Which Actor from the Distant Future Are You?
Canada's Former Defense Minister, Paul Hellyer, Embraces Insanity
Will My Masturbation Marathon Enhance Humanity's Position in the Galactic Community?
Gandalf's Advice on the Subject of Aliens
War of the Worlds: Lolwut???
The Diploma Annulment Wars
Twilight of the Dark Master
kylebstiff
Created for a specific purpose by a being beyond comprehension.
by Kyle B. Stiff
Ponyo! She's a magical little fish-girl who lives in an undersea wonderland and ends up in a storybook version of our world. She's amazing, she meets interesting people straight out of Pee Wee's Big Adventure, she has the Christlike ability to walk on water, heal the sick, and control the sea, she believes her father is an evil wizard, and she can't stop thinkin' about that ham. Oh, Ponyo!
The story is… damn, was there a story? There's an antagonist who is not necessarily a villain, and a goal introduced late in the movie which is not all that difficult to accomplish. This is not necessarily a flaw; one of Miyazaki's great masterpieces, My Neighbor Totoro, also did not have a strong narrative. In fact, the lack of narrative structure makes for a lot of fun unpredictability. How can you predict what will happen in a film that's about a bunch of unique characters interacting in a strange world?
I cannot stress this enough: Ponyo does not have what is traditionally considered a "storyline". It's more a state of mind. How many meaningful dreams have you had that followed a traditional narrative structure? How many enlightened Buddhas opened the thousand-petaled lotus of metaconsciousness by conforming to rigid notions of what is and is not acceptable? How many blood-spattered Einherjar passed through one of the 540 gates of Valhalla by nitpicking an experience, rather than accepting what the fates had to offer? Not a lot, right? In just the same way, Ponyo has no real storyline – but all the same, it makes for an incredible experience.
Let's say you've never seen a Miyazaki film. Is Ponyo for kids?
I can't imagine any child understanding the appeal of a magical world of pleasant character interactions and a vision of an era fueled by joy. To me, that seems like something only an adult would care to see, because an adult knows that the minute he leaves the movie theater he's going to enter into a world where anything can be taken from him at any time, where inhuman monsters drunk on control dictate the course of human potential and the finishing prize is a face full of maggots and an eternal sleep. But children, on the other hand, are naturally amoral and pretty oblivious when it comes to the horror of the everyday. As a child who grew up on Robocop andConan the Barbarian, all I ever wanted to see were violent displays of power and justice meted out by vigilantes, and I was by no means an abnormal case. Even "Sesame Street" was only just barely tolerable because the Cookie Monster was always making life difficult for those around him.
As such, Ponyo, a work of childish wonder, is most definitely for adults.
Can Ponyo Make You F*E*E*L ?
Is it touching? Is it moving?! Yes and yes. How many times did it make me want to cry? Only twice. But wait: How many movies actually succeed at making me want to cry? Not many; Ponyomade its way onto a very short list of movies that broke through my formidably protective AT Field, navigated through the freezing meat-locker labyrinth of my soul, and bitch-slapped the sensitive child within that still believes in things like fairness and hope.
This is no small feat considering I've seen many, many Miyazaki movies, and I'm aware of his usual tactics and finishing moves. Maybe more so than I should be, as I was pretty bored duringHowl's Moving Castle and came close to writing him off. Fortunately, Miyazaki proved with Ponyo that it is possible for experience and ingenuity to offset the creative boundaries of one director's admittedly limited set of tools.
I don't know much about Miyazaki's life before animation. Judging from the gentle, good-natured themes in his work, I can only assume he's seen more horror than any single man was ever meant to see. To him, suffering must be so real that he can't imagine making it a part of his art. He's seen things like children's hospitals filled with landmine victims. Things like insane asylums where the orderlies forget the line between us and them and vent their most sadistic cravings on those placed in their care while the doctors turn a blind eye. Or maybe small towns that lost their local factory, their only real source of income, and in desperation turned to trafficking in humans, selling their children to the highest bidder. Or maybe he's just hypersensitive to commonplace horror, like the prison bars of job and family, or the collection of an individual's personal failures that fit around the neck like a noose that grows tighter and tighter with each passing day, but there's never any jerk and crack! to release the victim from his misery. I suspect that Miyazaki's seen the real thing. He's not a product of the culturally sanitized suburban scene, which would explain how he can make his animations so happy and so tender without making them asinine and irrelevant.
Dub vs. Sub
Should you wait for the subtitled release and watch it alongside the cultural elites, or give up and watch the dubbed version alongside the rest of the mindless herd?
If your Nihon-go is rusty, by all means, watch the English dub. It's fine. Ten years ago I would have insisted on subtitles; these days, I'm not even really clear on why the Anti-Dubbing Elite still insist on their continued existence. I can't even remember the last bad dub I saw!
Ponyo has taken some heat because of the English voice acting. The two main characters, Ponyo and her boyfriend Sosuke, were voiced by… nevermind, I won't mention their names and make them that much more popular, plus I'm not exactly sure who they are. I think they're some of those kids that Disney clones every few years so as to have a new batch of celebrity sensations; plenty of little kids find them appealing, but then there's a lot of twenty-somethings who scorn them because of some dramatized rejection of childhood that goes hand in hand with being twenty-something. I say: Don't worry about it. Little Ponyo is delightfully voiced, and sounds as if she's intensely happy to be exploring a new world, and her little man Sosuke gives a solid performance that is heart-breaking on more than one occasion.
Verdict: Pastel paint rather than CG. A glimpse into an alternate world, rather than a story with chapters, plot, or development. A beast genetically altered for charm enhancement; a spell cast by a wiggly smile. Definitely worth seeing. If you dig Ponyo, check out Miyazaki's My Neighbor Totoro for further glimpses into the anti-story joyworld, or Princess Mononoke to see a story chock full of human conflict without any truly evil villain.
Hey readers! If you liked this post, you should check out some of my books. I've got an epic series called Demonworld, which is equal parts Mad Max and Lord of the Rings (think "science fantasy"), and a much-loved gamebook series called Heavy Metal Thunder which is currently a hyperlinked Kindle book but will be a fancy phone app any day now.
6 responses to "Miyazaki's Ponyo"
Kyle, you speak my mind with almost every sentence you write. If we both weren't heterosexual losers, I'd ask you out on a date!
my neighbor totoro bags says:
I actually first of all watched the My Neighbor Totoro Film when
I was a teen and I have ordered it for my kids along with a
giant Totoro plush toy.
kylebstiff says:
I was a teen when I first saw it, too. So while my eyes were opening for the first time to the horrors of PENAL COLONY: EARTH, at least I had Totoro there to comfort me. I never had a Totoro plush toy or backpack or anything like that, I had to carry all my crap around in a plastic grocery bag, but I was small so I was able to fit the handles around my shoulders and wear it like a thin shapeless backpack.
Alex Daltin says:
Such a beautiful review. Now i have more ideas for baby photography and maternity photos.
Thank you Alex! Tackle the world with your inspiration! Show no mercy! Well, except to the babies of course.
Trinity craft says:
I love Ponyo she was so good movie to watch
The End of Death Stranding: Occult Bluepill Black Magic vs. Big Jesus 5.56 MAGA Rounds
Home Alone's Kevin McCallister Is a Right-Wing Bigot!
Hideo Kojima Saves America with Bluepills in Death Stranding
I Was Shot to Death While Watching JOKER! 10/10 Would Get Shot Again
Rambo V: Last Boomer


text:
Q: Chaining promises for angular-translate As someone still new to Angular I am struggling with a special promise setup to work with angular-translate.
I have a working CustomLoader for angular-translate in place which is working fine. But I have to call the CustomLoader twice (with a $translate.refresh() ) and based on an external condition the CustomLoader should either load a fixed preset of data or the same preset but merged with the result of an external backend call.
This is what I got so far:
(function () {
  'use strict';

/**
 * TranslationCustomLoader
 */
angular
   .module('myApp')
   .factory('TranslationCustomLoader', TranslationCustomLoader);

function TranslationCustomLoader($q, ENV, $http, ExternalService) {

return function (options) {

  var translations_en = {
    MYKEY1: 'Value1',
    MYKEY2: 'Value2',
    MYKEY3: 'Value3'
  };

  var translations_de = {
    MYKEY1: 'Wert1',
    MYKEY2: 'Wert2',
    MYKEY3: 'Wert3'
  };

  var deferred = $q.defer(), translations;

// Standard Dataset for Translations based on options.key (= language key)

  if (options.key === 'en') {
    translations = translations_en;
  } else {
    translations = translations_de;
  }

 // If some condition is met call external backend and retrieve additional data

  if (ExternalService.isCondition()) {
    $http({
      method: 'GET',
      params: {
        lang: options.key
      },
      url: ENV.apiEndpoint + 'api/v1/translations'
    }).success(function (data) {

 // Now replace/add data in variable 'translations' by external result
 // External result could be: [MYKEY2: 'AnotherValue', MYKEY4: 'AdditionalValue']

      angular.forEach(data, function (value, key) {
        translations[key] = value;
      });

// Return 'merged' promise

      deferred.resolve(translations);
      return deferred.promise;
    })
  } else {

// Return 'standard' promise

    deferred.resolve(translations);
    return deferred.promise;
  }
   }}})();

The CustomLoader is called,
in case that the external condition is false the standard promise is returned and successfully resolved by $translate.
in case that external condition is true, I can see the external backend is called, I even see the results, but I got an 'cannot read .then of undefined' as it seems the promise is not correct.
Can anybody help me to fix this? I tried a lot already (including some desperate $q.all, .when etc.).
Using the partial loader (which is offered by angular-translate) isn't an alternative unfortunately.

A: You're not returned anything from your function when isCondition() is true.
You're also using success(), which is deprecated, and doesn't allow chaining promises.
And if the http call fails, you never reject your promise. That, again is solved automatically by using chaining.
The code should rather look like
function (options) {

  var translations_en = {
    MYKEY1: 'Value1',
    MYKEY2: 'Value2',
    MYKEY3: 'Value3'
  };

  var translations_de = {
    MYKEY1: 'Wert1',
    MYKEY2: 'Wert2',
    MYKEY3: 'Wert3'
  };

  var translations;

  // Standard Dataset for Translations based on options.key (= language key)

  if (options.key === 'en') {
    translations = translations_en;
  } else {
    translations = translations_de;
  }

 // If some condition is met call external backend and retrieve additional data

  if (ExternalService.isCondition()) {
    return $http({
      method: 'GET',
      params: {
        lang: options.key
      },
      url: ENV.apiEndpoint + 'api/v1/translations'
    }).then(function (response) {

      // Now replace/add data in variable 'translations' by external result
      // External result could be: [MYKEY2: 'AnotherValue', MYKEY4: 'AdditionalValue']

      angular.forEach(response.data, function (value, key) {
        translations[key] = value;
      });

      // Return 'merged' promise

      return translations;
    });
  } else {
    return $q.when(translations);
  }
}

redpajama_set_name: RedPajamaStackExchange