text:
\section*{Introduction}

Let $k$ be a perfect field of characteristic $p>0$ and let $A$ be a
finite dimensional $k$-algebra.
For a symmetric algebra $A$ K\"ulshammer introduced in \cite{Kueldeutsch}
a descending sequence of ideals $T_n(A)^\perp$ of the centre of $A$ satisfying
various interesting properties. In \cite{HHKM} H\'ethelyi, Horv\'ath,
K\"ulshammer and Murray continued to study this sequence of ideals,
proved their invariance under Morita equivalence, and showed further
properties linked amongst others to the Higman ideal of the algebra.
In order to show these properties K\"ulshammer introduced a sequence
of mappings $\zeta_n:Z(A)\lra Z(A)$
with image being $T_n(A)^\perp$. In a completely dual procedure he
introduced mappings $\kappa_n:A/KA\lra A/KA$
again in case $A$ is symmetric. These mappings encode many representation
theoretic informations, in particular in case $A$ is a group algebra. The reader
may refer to K\"ulshammer's original articles \cite{Kueldeutsch}
and \cite{Kuelprog} for more  details. Moreover, in a recent
development it could be proved in \cite{kuelsquest} that these mappings
actually are an invariant of the derived category of $A$.
This fact could be used to
distinguish derived categories in very subtle situations, such as some
parameter questions for blocks of dihedral or semidihedral type in
joint work with Holm \cite{hz-tame}, such as very delicate questions
to distinguish two families
of symmetric algebras of tame domestic representation type by Holm and
Skowro\'nski \cite{Holmskow}, or in joint work with Holm \cite{typeL}
to distinguish derived
equivalence classes of deformed
preprojective algebras of generalised  Dynkin type, as defined by
Bia\l kowski, Erdmann and Skowro\'nski \cite{BES} (modified slightly for type $E$;
cf \cite[page 238]{periodic}) with respect to different
parameters.
Most recently in joint work \cite{stableHochschild}
with Yuming Liu and Guodong Zhou we
gave a version which is invariant under stable equivalences of Morita type.
This approach gives a link to Auslander's conjecture on the invariance of
the number of simple modules under stable equivalences of Morita type.
K\"onig, Liu and Zhou \cite{KLZ} continued the work with focus being higher Hochschild
and cyclic homology invariants.

In joint work with Bessenrodt and Holm \cite{nonsymmetric} we could get
rid of the assumption that $A$ needs to be symmetric for the definition
of the images of $\zeta_n$. We used the trivial extension algebra $\T A$
of $A$, computed the image of $\zeta_n$ for $\T A$ and interpreted the result
purely in terms of $A$. The derived invariant K\"ulshammer's ideal structure
of the centre of $A$ becomes available for any finite dimensional algebra over
perfect fields of positive characteristic.

Since the centre of the algebra is the degree $0$ Hochschild cohomology and
$A/KA$ is the $0$-degree Hochschild homology, it is natural to try to
understand the K\"ulshammer construction for higher degree Hochschild (co-)homology.
Denote by $HH^m(A)$ the $m$-degree
Hochschild cohomology of $A$ and by $HH_m(A)$ the $m$-degree Hochschild homology
of $A$.
We defined in \cite{gerstenhaber} for a symmetric algebra $A$ certain mappings
$\kappa_n^{(m);A}:HH_{p^nm}(A)\lra HH_m(A)$
in completely analogous manner as K\"ulshammer's $\kappa_n$. Moreover
we showed that $\kappa_n^{(0);A}=\kappa_n$, and proved its derived invariance as well
(cf Theorem~\ref{derivedkappacup} below).

In Definition~\ref{Definitionofkappahat} of Section~\ref{Kuelshammerfornonsymmetric} we propose a mapping
$\hat \kappa_n^{(m);A}$, denoted also $\hat \kappa_n^{(m)}$
if no confusion may occur, in analogy of $\kappa_n^{(m);A}$ for algebras which are not necessarily symmetric.
Further, we prove that $\hat \kappa_n^{(m)}$ is invariant under a derived equivalence:
If
$F:D^b(A)\simeq D^b(B)$ is a standard equivalence
then it induces coherent isomorphisms of Hochschild homology
groups $HH_m(F):HH_m(A)\lra HH_m(B)$
and we get
$HH_m(F)\circ \hat\kappa_n^{(m);A}=\hat\kappa_n^{(m);B}\circ HH_{p^nm}(F)$ for all
positive integers $m$ and~$n$.

Finally we give an elementary example in order to illustrate that these
invariants are computable and to show that they are not trivial.

The paper is organised as follows. In Section~\ref{Hochschilddefinitions} we
recall some known facts about Hochschild (co-)homology.
In Section~\ref{Kuelshammerrappel} the known constructions concerning
K\"ulshammer ideals in degree $0$ for symmetric and non
symmetric algebras, using trivial extension algebras are recalled, as
well as the $m$-degree Hochschild cohomology generalisation
for symmetric algebras.
Section~\ref{barcomplexsection} recalls that Hochschild homology is functorial,
a statement which is used at a prominent point in our construction.
In Section~\ref{Kuelshammerfornonsymmetric} finally
we provide a K\"ulshammer mapping for not necessarily symmetric algebras
and higher Hochschild homology.
We show in
Theorem~\ref{derivedinvarianceofkappahat} that the new
mapping $\hat\kappa_n^{(m)}$ is an invariant of the derived category.
Section~\ref{dualnumbers} is devoted to a detailed computation of an example.

\bigskip

{\bf Acknowledgement:} Part of this work was done in October
2007 during a visit at Beijing Normal University and Nanjing Normal University.
I wish to express my gratitude to
Changchang Xi and to Jiaqun Wei for their invitation and their kind hospitality.

\section{Setup of the basic tools}

\subsection{Hochschild homology and cohomology}
\label{Hochschilddefinitions}

Recall some well-known facts from Hoch\-schild (co-)homology. We refer to
Loday \cite{Loday} for a complete presentation of this theory.
Let $A$ be a finitely generated
$k$-algebra for a commutative ring $k$, and suppose $A$ is
projective as $k$-module.
Then for every $A\otimes_kA^{op}$-module $M$ we define
the Hochschild homology of $A$ with values in $M$ as
$$HH_n(A,M):=Tor_n^{A\otimes_kA^{op}}(A,M)$$
and Hochschild cohomology of $A$ with values in $M$ as
$$HH^n(A,M):=Ext_{A\otimes_kA^{op}}^n(A,M).$$
Often and frequently throughout the paper, for a $k$-algebra $R$ we abbreviate
$R^e:=R\otimes_kR^{op}$.
There is a standard way to compute Hochschild (co-)homology
by the bar resolution. We recall its definition in order to set
up the notations.

The bar complex $\B A$ is given by
$$(\B A)_n:=A^{(\otimes_k)^{n+1}}:=
\underbrace{A\otimes_kA\otimes_k\dots\otimes_kA}_{n+1\mbox{ \scriptsize
factors}}$$
and differential
$$d(a_0\otimes\dots a_n)=
\sum_{j=0}^{n-1}(-1)^ja_0\otimes\dots\otimes
a_{j-1}\otimes a_ja_{j+1}\otimes a_{j+2}\otimes \dots\otimes a_n.$$
One observes that $d^2=0$, that $(\B A)_n$ is a free $A\otimes_kA^{op}$-module
for every $n> 0$ and that the complex $(\B A,d)$ is
a free resolution of $A$ as a $A\otimes_kA^{op}$-module, called the bar resolution.
Hence, $$HH_n(A,M)\simeq H_n(\B A\otimes_{A\otimes A^{op}}M)\mbox{ and }
HH^n(A,M)\simeq H^n(Hom_{A\otimes A^{op}}(\B A,M)).$$

Moreover, the bar complex is functorial in the sense that whenever
$f:A\lra B$ is a homomorphism of $k$-algebras, then
$$(\B A)_n\ni a_0\otimes\dots \otimes a_n\mapsto f(a_0)\otimes\dots \otimes
f(a_n)\in \B B$$
induces a morphism of complexes $\B A\lra \B B$.

\subsection{The K\"ulshammer ideals; constructions and their Hochschild
generalisation}

\label{Kuelshammerrappel}

We recall the constructions introduced by K\"ulshammer in \cite{Kueldeutsch}
and \cite{Kuelprog} as well as their generalisation introduced in \cite{gerstenhaber}
to higher degree Hochschild cohomology and their generalisations to $0$-degree
Hochschild cohomology for not necessarily symmetric algebras.

\subsubsection{The K\"ulshammer ideal theory for algebras which are symmetric}
\label{Kuelshammeridealsknownfacts}

Let $k$ be a perfect field of characteristic $p>0$
and let $A$ be a $k$-algebra. Then
$$KA:=<ab-ba\;|\;a,b\in A>_{k-\mbox{\scriptsize space}}$$
is a $Z(A)$-module and for all $n\in \N$
$$T_n(A):=\{x\in A\;|\;x^{p^n}\in KA\}$$
is a $Z(A)$ submodule of $A$. If $A$ is symmetric with symmetrising form
$$<\;,\;>:A\otimes_kA\lra k$$ taking orthogonal spaces, since $KA^\perp=Z(A)$,
$T_n(A)^\perp$ is an ideal of $Z(A)$.

Given an equivalence of triangulated
categories of standard type (cf Section~\ref{derivedequivonhomology} below)
$D^b(A)\lra D^b(B)$ between two symmetric
$k$-algebras $A$ and $B$, Rickard showed in \cite{Ri3} that
this equivalence induces an isomorphism $Z(A)\simeq Z(B)$
and if $k$ is a perfect field, it is shown in \cite{kuelsquest} that
$T_n(A)^\perp$ is mapped by this isomorphism to $T_n(B)^\perp$.

Moreover $A$ is symmetric if and only if $A\simeq A^*:=Hom_k(A,k)$ as
$A\otimes_kA^{op}$-modules.
Since we supposed $k$ to be perfect the Frobenius map is invertible and let
$k\ni \lambda\mapsto\lambda^{p^{-1}}\in k$ be its inverse.
Denote the $n$-th iterate by $p^{-n}:=\left(p^{-1}\right)^n$.
Since $A/KA\ni b\mapsto b^p\in A/KA$ is a well-defined additive mapping,
semilinear with respect to the Frobenius mapping, one defines an
element in $Hom_k(A,k)$ by
$$a\mapsto \left(b\mapsto <a,b^{p^n}>^{p^{-n}}\right)$$
and so for every $a\in Z(A)$ there is a unique element $\zeta_n(a)\in A$
so that $$<a,b^{p^n}>=<\zeta_n(a),b>^{p^n}\forall\;b\in A.$$

Similarly, using that $Z(A)^\perp=KA$, one gets a mapping
$\kappa_n:A/KA\lra A/KA$ satisfying
$$<a^{p^n},b>=<a,\kappa_n(b)>^{p^n}\forall\;a\in Z(A), b\in A/KA.$$

In \cite{gerstenhaber} this construction was generalised to a structure on
Hochschild homology which is invariant under equivalences of derived categories.
More precisely, let $A$ be a {\em symmetric} finite dimensional
$k$-algebra for $k$
a perfect field of characteristic $p>0$.
Then, denoting by $\B A$ the bar resolution of $A$ and $A^e=A\otimes_kA^{op}$,
$$Hom_k(\B A\otimes_{A^e}A,k)
\simeq Hom_{A^e}(\B A,Hom_k(A,k))\simeq Hom_{A^e}(\B A,A)$$
and so $$Hom_k(HH_m(A,A),k)\simeq HH^m(A,A)$$
which induces a non degenerate pairing
$$<\;,\;>_m:HH^m(A,A)\times HH_m(A,A)\lra k\;.$$
By a construction analogous to the one for the pairing
$$<\;,\;>:Z(A)\times A/KA\lra k$$
one gets a mapping
$$\kappa_n^{(m);A}:HH_{p^nm}(A,A)\lra HH_m(A,A)$$
denoted simply by $\kappa_n^{(m)}$ if no confusion may occur, satisfying
$$<z^{p^n},x>_{p^nm}=\left(<z,\kappa_n^{(m)}(x)>_{m}\right)^{p^n}$$
for all $x\in HH_{p^nm}$ and $z\in HH^m(A,A)$.

\subsubsection{On derived equivalences and Hochschild (co-)homology}

\label{derivedequivonhomology}

In \cite{gerstenhaber} we studied invariance properties of these
mappings with respect to equivalences between derived categories.
Rickard showed in \cite{Ri3} that whenever $A$ and $B$ are $k$-algebras over a
field $k$, and whenever $D^b(A)$ and $D^b(B)$ are equivalent as triangulated categories, then there is a complex $X\in D^b(B\otimes_kA^{op})$ so that
$X\otimes_A^{\mathbb L}-:D^b(A)\lra D^b(B)$ is an equivalence. Such an equivalence is called of standard type and $X$ is called a two-sided tilting complex. It is shown in \cite{Ri3} that every equivalence of triangulated categories $D^b(A)\lra D^b(B)$ coincides with a suitable standard one on objects.

A tilting complex $T$ is a complex in $K^b(A)$, the homotopy category of bounded complexes of projective $A$-modules, satisfying  $Hom_{D^b(A)}(T,T[i])=0$ for all $i\neq 0$, and $add(T)$ generates $K^b(A)$ as triangulated category.
Two $k$-algebras $A$ and $B$ have equivalent derived categories $D^b(A)$ and $D^b(B)$ if and only if there is a tilting complex $T$ in $D^b(A)$ with endomorphism ring $B$. Given an equivalence $D^b(B)\lra D^b(A)$, the image of the regular $B$-module is a tilting complex.
Hence, by the above, for every tilting complex $T$ in $D^b(A)$ there is a two-sided tilting complex $X$ in $D^b(A\otimes_kB^{op})$ so that $X\simeq T$ in $D^b(A)$.

Let $F_X=X\otimes_A^{\mathbb L}-:D^b(A)\lra D^b(B)$ be an equivalence
of derived categories with quasi-inverse $Y\otimes^{\mathbb L}-$ (cf
Rickard \cite{Ri3}). If $A$ is symmetric, then
$B$ is symmetric as well (cf \cite{rogquest}
for general base rings, and \cite{Ri3} for fields) and
$$X\otimes_A^{\mathbb L}-\otimes_A^{\mathbb L}Y:
D^b(A\otimes_kA^{op})\lra D^b(B\otimes_kB^{op})$$
is an equivalence again. This equivalence induces an isomorphism
$$HH_*(F_X):HH_*(A)\lra HH_*(B)$$
made explicit in \cite{gerstenhaber}.
Recall the precise mapping, defined on the level of complexes, which will be needed later.
\begin{eqnarray*}
\B A\otimes_{A\otimes_kA^{op}}(X\otimes_BY)&\lra&(Y\otimes_A\B A\otimes_AX)\otimes_{B\otimes_kB^{op}}B\\
u\otimes(x\otimes y)&\mapsto&(y\otimes u\otimes x)\otimes 1
\end{eqnarray*}
Then $HH_*(F_X)$ is the mapping induced on the homology.

We recall
one of the main results of \cite{gerstenhaber}.
This result will be one of our main
ingredients in the proof of Theorem~\ref{derivedinvarianceofkappahat}.

\begin{Theorem}\label{derivedkappacup}\cite{gerstenhaber}
Let $A$ be a finite dimensional symmetric $k$-algebra
over the perfect field $k$ of
characteristic $p>0$. Let $B$ be a second algebra such that
$D^b(A)\simeq D^b(B)$ as triangulated categories,
and let $m\in\N$.
Then, there is a standard equivalence $F:D^b(A)\simeq D^b(B)$,
and any such standard equivalence induces an isomorphism
$HH_m(F):HH_m(A,A)\lra HH_m(B,B)$ of all Hochschild homology
groups, satisfying
$$HH_{m}(F)\circ\kappa_n^{(m),A}=
\kappa_n^{(m),B}\circ HH_{p^nm}(F)\;.$$
\end{Theorem}

We take the opportunity to mention that in \cite{gerstenhaber} in the statement of
the theorem the condition
on the field $k$ to be perfect is unfortunately missing.
However, this was a general assumption throughout \cite{gerstenhaber}.

\subsubsection{Trivial extension of an algebra; the K\"ulshammer ideal
theory in the general case}
\label{TA}

Let $k$ be a commutative ring and let $A$ be a $k$-algebra.
Then $Hom_k(A,k)$ is an $A\otimes_kA^{op}$-module by the action
$$\left((a,b)\cdot f\right)(x):=
(a\cdot f\cdot b)(x):=f(bxa)\;\;\forall a,x\in A;b\in A^{op};f\in Hom_k(A,k).$$

Now, the $k$-vector space $$\T A:= A\oplus Hom_k(A,k)$$ becomes
a $k$-algebra by putting
$$(a,f)\cdot (b,g):= (ab,ag+fb)\;\;\forall a,b\in A;f,g\in Hom_k(A,k)\;.$$
Then
\begin{eqnarray*}
A&\stackrel{\iota_A}{\lra}&\T A\\
a&\mapsto&(a,0)
\end{eqnarray*}
is a ring homomorphism. Moreover $\{0\}\oplus Hom_k(A,k)$ is a two-sided ideal
of $\T A$ and the canonical projection
\begin{eqnarray*}
\T A&\stackrel{\pi_A}{\lra}&A\\
(a,f)&\mapsto&a
\end{eqnarray*}
is a splitting for $\iota_A$; i.e. $\pi_A\circ\iota_A=id_A$.
The algebra $\T A$ is always symmetric via the bilinear form
$$\T A\otimes_k\T A\ni <(a,f);(b,g)>\mapsto f(b)+g(a)\in k$$
which induces, for $k$ a field and $A$ finite dimensional over $k$, an
isomorphism
\begin{eqnarray*}
\T A&\lra&Hom_k(\T A,k)\\
(a,f)&\mapsto&\left((b,g)\mapsto f(b)+g(a)\right)
\end{eqnarray*}
of $\T A\otimes_k\T A^{op}$-modules.

In the joint paper \cite{nonsymmetric} with Bessenrodt and Holm
it is shown that the K\"ulshammer ideals satisfy
$$T_n(\T A)^\perp=\{0\}\oplus Ann_{Hom_k(A,k)}(T_n(A))$$
for all $n\geq 1$, and therefore for all $n\geq 1$
the $Z(A)$-modules $Ann_{Hom_k(A,k)}(T_n(A))$ are invariants
under derived equivalences.

\begin{Remark} \label{findbackoriginal}
It should be noted that for algebras $A$ which are already symmetric,
via the symmetrising form $<\;,\;>$ we get an isomorphism of
vector spaces $\lambda:Z(A)\lra Hom_k(A/KA,k)$. Now, every linear form on $A$
equals a form of the shape $<z,->$ for some $z\in Z(A)$ and so
$$\lambda^{-1}(Ann_{Hom_k(A,k)}(T_n(A)))=
\{z\in Z(A)\;|\;<z,T_n(A)>=0\}=T_n(A)^\perp.$$
We find back our original result.
\end{Remark}

\subsection{Relating Hochschild homology of an algebra and of its trivial extension}
\label{barcomplexsection}

Let $R$ and $S$ be $k$-algebras and let $\alpha:R\lra S$ and $\beta:S\lra R$ be
algebra homomorphisms. Then it is well-known
(cf e.g. Loday \cite[Chapter 1, Section 1.1.4]{Loday})
that Hochschild homology is functorial, i.e. $\alpha$ and $\beta$
induce mappings
$$HH_*(\alpha):HH_*(R)\lra HH_*(S)\mbox{ and }HH_*(\beta):HH_*(S)\lra HH_*(R)$$
so that $$HH_*(id_R)=id_{HH_*(R)}\mbox{
as well as
}HH_*(\beta)\circ HH_*(\alpha)=HH_*(\beta\circ\alpha).$$
In particular in case of the mappings $\iota_A:A\lra \T A$ and $\pi_A:\T A\lra A$
from an algebra $A$ to its trivial extension $\T A$ we get induced mappings
giving a split projection
$$HH_n(\pi_A):HH_n(\T A)\lra HH_n(A)$$
and a split injection
$$HH_n(\iota_A):HH_n(A)\lra HH_n(\T A).$$

This can be defined on the level of complexes.
As seen in Section~\ref{Hochschilddefinitions}
the algebra homomorphism $\alpha: R\lra S$ induces a morphism of complexes
\begin{eqnarray*}
\B\alpha:\B R&\lra& \B S\\
(x_0\otimes\dots\otimes x_n)&\mapsto&
(\alpha(x_0)\otimes\dots\otimes \alpha(x_n))
\end{eqnarray*}
and likewise for $\beta:S\lra R$.
Also on the complex computing Hochschild homology an analogous
morphism of complexes is defined by
\begin{eqnarray*}
\B R\otimes_{R^e}R&\lra& \B S\otimes_{S^e}S\\
(x_0\otimes\dots\otimes x_n)\otimes x_{n+1}&\mapsto&
(\alpha(x_0)\otimes\dots\otimes \alpha(x_n))\otimes \alpha(x_{n+1})
\end{eqnarray*}
and likewise for $\beta.$
Its homology induces a mapping $H_*\B\alpha:HH_*(R)\lra HH_*(S)$
which is easily seen to be $HH_*(\alpha)$ obtained above.

In particular in our situation
$$H_*\B\pi_A\circ H_*\B\iota_A=id_{\B A\otimes_{A^e}A}$$
and it is clear that
$$H_*\B\pi_A=HH_*(\pi_A)\mbox{ as well as }H_*\B\iota_A=HH_*(\iota_A).$$

\medskip

We reproved the following proposition which was obtained in a much larger context and generality, and much more sophisticated methods, by Cibils, Marcos, Redondo and Solotar in \cite[Theorem 5.8]{CMRS}.

\begin{Prop}\label{directfactor} \cite[Theorem 5.8]{CMRS}
Let $k$ be a field and let $A$ be a $k$-algebra. Then the canonical embedding
$A\lra \T A$ induces a canonical embedding of
$HH_*(A)$ as a direct factor of $HH_*(\T A)$.
\end{Prop}

\section{K\"ulshammer-like Hochschild invariants for non symmetric algebras}
\label{Kuelshammerfornonsymmetric}

Let $k$ be a perfect field of characteristic $p>0$ and let $A$ be a
$k$-algebra. Recall from Section~\ref{Kuelshammeridealsknownfacts} that in case
$A$ is symmetric, we defined in \cite{gerstenhaber} mappings
$$\kappa_n^{(m)}:HH_{p^nm}(A)\lra HH_m(A)$$
satisfying
$$\left(<x,\kappa_n^{(m)}(y)>_{m}\right)^{p^n}=<x^{p^m},y>_{p^nm}\;
\forall \;x\in HH^{m}(A);y\in HH_{p^nm}(A).$$
We should mention that the pairing $<\;,\;>_m$ is defined only on the
(co-)homology of finite dimensional symmetric algebras by the isomorphism
$$
\B A\otimes_{A^e}A\simeq Hom_k(Hom_k(\B A\otimes_{A^e}A,k),k)\simeq
Hom_k(Hom_{A^e}(\B A,A),k)$$
whose homology gives an isomorphism
$$HH_*(A)\simeq Hom_k(HH^*(A),k)\;.$$
Observe that the double dual is the identity for finite dimensional vector
spaces only, and hence here we use the fact that finite dimensional algebras
have projective bimodule resolutions with finite dimensional homogeneous
components.

\begin{Def}\label{Definitionofkappahat}
Let $A$ be a finite dimensional (not necessarily symmetric) $k$-algebra.
Then put
$$\hat\kappa_n^{(m);A}:=
HH_m(\pi_A)\circ \kappa_n^{(m);\T A}\circ HH_{p^nm}(\iota_A)$$
\end{Def}

\bigskip

We shall prove now that the invariant $\hat\kappa_n^{(m)}$ is an
invariant of the derived category of $A$ in the same sense as
it was proved for symmetric algebras (cf
Theorem~\ref{derivedkappacup}).

Rickard showed in \cite{Rickard} that an equivalence between the derived categories of two $k$-algebras induces an equivalence between the derived categories of their trivial extension algebras. We need the following improvement of his result, which seems to be of interest in its own right.

\begin{Prop}\label{twosidedtrivialextension}
Let $A$ and $B$ be finite dimensional $k$ algebras and let $T$ be a tilting complex in $D^b(A)$ with endomorphism ring $B$.
\begin{enumerate}
\item
(Rickard \cite{Rickard}) Then $\T A\otimes_AT$ is a tilting complex in $D^b(\T A)$ with endomorphism ring $\T B$.
\item Let $X$ be a two-sided tilting complex in $D^b(\T A\otimes_k\T B)$ so that $\T A\otimes_AT\simeq X$ in $D^b(\T A)$. Then $A\otimes_{\T A}X$ and $X\otimes_{\T B}B$ are two-sided tilting complexes in $D^b(A\otimes_kB)$. Moreover, $T\simeq A\otimes_{\T A}X$.
\end{enumerate}
\end{Prop}

Proof. Let $T$ be a tilting complex in $D^b(A)$ with endomorphism ring $B$.
Then by
Rickard's theorem \cite[Corollary 5.4]{Rickard} the complex $\T A\otimes_AT$
is a tilting complex in
$D^b(\T A)$ with endomorphism ring $\T B$. By Keller's theorem
\cite{Kellercoderivations}
there is a two-sided tilting complex $X$ in $D^b(\T A\otimes_k\T B^{op})$
so that $_{\T A}|X\simeq \T A\otimes_AT$.

Now,
\begin{eqnarray*}
Hom_{D^b(A)}(A\otimes_{\T A}X,A\otimes_{\T A}X)&=&
Hom_{D^b(A)}(A\otimes_{\T A}\ _{\T A}|X,A\otimes_{\T A}\ _{\T A}|X)\\
&\simeq &Hom_{D^b(A)}(A\otimes_{\T A}\T A\otimes_AT,A\otimes_{\T A}\T A\otimes_AT)\\
&\simeq &Hom_{D^b(A)}(T,T)\\
&\simeq &B
\end{eqnarray*}
as rings.
But, we know that $\T B$ acts on $A\otimes_{\T A}X$ by multiplication on the right.
Hence, we get a ring homomorphism
$$\T B\lra Hom_{D^b(A)}(A\otimes_{\T A}X,A\otimes_{\T A}X).$$
Since we have seen that the endomorphism ring of $A\otimes_{\T A}X$ is
isomorphic to $B$, the mapping
$$\T B\lra Hom_{D^b(A)}(A\otimes_{\T A}X,A\otimes_{\T A}X)$$
factorises through the canonical projection
$\T B\lra B$. Hence, the action of $\T B$ on $A\otimes_{\T A}X$
has $B^*$ in the kernel.

Now, the $\T B\otimes_k \T B^{op}$-module structure of
$Hom_{D^b(A)}(A\otimes_{\T A}X,A\otimes_{\T A}X)$ is the following.
The action of $\T B$ on the first argument gives the $\T B$-action from the left on
$Hom_{D^b(A)}(A\otimes_{\T A}X,A\otimes_{\T A}X)$ and the action of $\T B$ on the right
comes from the action of $\T B$ on the second argument. Both have the degree $2$
nilpotent ideal $B^*$ in the kernel and so, the natural action of
$\T B\otimes_k\T B^{op}$ is actually an action of $B\otimes_kB^{op}$.

Therefore,
$$Hom_{D^b(A)}(A\otimes_{\T A}X,A\otimes_{\T A}X)\simeq B$$
as $B\otimes_kB^{op}$-modules.

Hence $A\otimes_{\T A}X$ is invertible from the left in the sense that
$$Hom_{D^b(A)}(A\otimes_{\T A}X,A\otimes_{\T A}X)\simeq
Hom_{A}(A\otimes_{\T A}X,A)\otimes_A(A\otimes_{\T A}X)\simeq B$$
as $B\otimes_kB^{op}$-modules.

We still need to show that $A\otimes_{\T A}X$ is invertible from the right as well.
Since $\T A$ and $\T B$ are both symmetric, and since $X$ is a two-sided tilting
complex in $D^b(\T A\otimes_k\T B^{op})$, its inverse complex is given by its
$k$-linear dual $Hom_k(X,k)=:\check X$. We claim that $\check X\otimes_{\T A}A$
is a right inverse of $A\otimes_{\T A}X$. Indeed,
by the previous paragraph, we see that $\T B$ acts on the right of $A\otimes_{\T A}X$
via the projection $\T B\lra B$, i.e. the nilpotent ideal $B^*$ is in the kernel
of this action. Analogously, the same holds for $\check X\otimes_{\T A}A$. Hence,
the tensor product over $\T B$ equals the tensor product over $B$ only.
\begin{eqnarray*}
(A\otimes_{\T A}X)\otimes_{B}(\check X\otimes_{\T A}A)
&\simeq&(A\otimes_{\T A}X)\otimes_{\T B}(\check X\otimes_{\T A}A)\\
&\simeq&A\otimes_{\T A}(X\otimes_{\T B}\check X)\otimes_{\T A}A\\
&\simeq&A\otimes_{\T A}\T A\otimes_{\T A}A\\
&\simeq& A
\end{eqnarray*}
as $A\otimes_kA^{op}$-modules.

As a consequence $A\otimes_{\T A}X$ is a two-sided tilting complex with restriction to
the left being $T$. Moreover, it is clear that $\pi_A$ will map $X$ to
$A\otimes_{\T A}X$.
\dickebox

\begin{Theorem}\label{derivedinvarianceofkappahat}
Let $k$ be a perfect field of characteristic $p>0$, let
$A$ and $B$ be finite dimensional  $k$-algebras and suppose
that $D^b(A)\simeq D^b(B)$ as triangulated categories.
Let $F$ be an explicit standard equivalence between $D^b(A)$ and
$D^b(B)$. Then, $F$ induces a sequence of isomorphisms
$HH_m(F):HH_m(A)\lra HH_m(B)$ so that
$$HH_m(F)\circ \hat \kappa_n^{(m);A}=\hat \kappa_n^{(m);B}\circ HH_{p^nm}(F).$$
\end{Theorem}

Proof. Let $T$ be a tilting complex in $D^b(A)$ with endomorphism ring $B$ and let $X$ be a two-sided tilting complex in $D^b(\T A\otimes_k\T B^{op})$ with
$\T A\otimes_AT\simeq X$ in $D^b(\T A)$.
By Proposition~\ref{twosidedtrivialextension} we get $A\otimes_{\T A}X$ is a two-sided tilting complex in $D^b(A\otimes_kB)$.

We now use Theorem~\ref{derivedkappacup}:
$$HH_{m}(F_X)\circ\kappa_n^{(m);\T A}=\kappa_n^{(m);\T B}\circ HH_{p^nm}(F_X).$$
Multiplying with $HH_{p^nm}(\iota_A)$ from the right and with $HH_{m}(\pi_B)$ from the
left gives then
$$HH_{m}(\pi_B)\circ HH_{m}(F_X)\circ\kappa_n^{(m);\T A}\circ HH_{p^nm}(\iota_A)=HH_{m}(\pi_B)\circ\kappa_n^{(m);\T B}\circ HH_{p^nm}(F_X)\circ HH_{p^nm}(\iota_A).$$

We now claim that
$$HH_{p^nm}(F_X)\circ HH_{p^nm}(\iota_A)=HH_{p^nm}(\iota_B)\circ HH_{p^nm}(F_{A\otimes_{\T A}X})$$
and
$$HH_{m}(\pi_B)\circ HH_{m}(F_X)=HH_{m}(F_{A\otimes_{\T A}X})\circ HH_{m}(\pi_A).$$

Recall from Section~\ref{derivedequivonhomology} how $HH_{m}(F_X)$ is defined.
Then, the commutativity relation will be proven as soon as we have that
the diagrams
$$
\begin{array}{ccc}
\left(\check X\otimes_{\T A}\B(\T A)\otimes_{\T A}X\right)
\otimes_{(\T B)^e}\T B
&\lla&
\B\T A\otimes_{(\T A)^e}(X\otimes_{\T B}\check X)\\
\uar\mbox{\scriptsize $HH(\iota_B)$}&&\mbox{\scriptsize $HH(\iota_A)$}\uar\\
\left((\check X\otimes_{\T A}A)\otimes_A\B A\otimes_A(A\otimes_{\T A}X)\right)
\otimes_{B^e}B
&\lla&
\B A\otimes_{A^e}((A\otimes_{\T A}X)\otimes_{B}(\check X\otimes_{\T A}A))
\end{array}
$$
as well as
$$
\begin{array}{ccc}
\left(\check X\otimes_{\T A}\B(\T A)\otimes_{\T A}X\right)
\otimes_{(\T B)^e}\T B
&\lla&
\B\T A\otimes_{(\T A)^e}(X\otimes_{\T B}\check X)\\
\dar\mbox{\scriptsize $HH(\pi_B)$}&&\mbox{\scriptsize $HH(\pi_A)$}\dar\\
\left((\check X\otimes_{\T A}A)\otimes_A\B A\otimes_A(A\otimes_{\T A}X)\right)
\otimes_{B^e}B
&\lla&
\B A\otimes_{A^e}((A\otimes_{\T A}X)\otimes_{B}(\check X\otimes_{\T A}A))
\end{array}
$$
are commutative.

We use the canonical isomorphisms
$$(\check X\otimes_{\T A}A)\otimes_A\B A\otimes_A(A\otimes_{\T A}X)\simeq
\check X\otimes_{\T A}\B A\otimes_{\T A}X$$
and
$$\B A\otimes_{A^e}(A\otimes_{\T A}X\otimes_B\check X\otimes_{\T A}A)
\simeq \B A\otimes_{(\T A)^e}(X\otimes_{\T B}\check X),$$
which proves that we only need to show that the diagrams
$$
\begin{array}{ccc}
\left(\check X\otimes_{\T A}\B(\T A)\otimes_{\T A}X\right)
\otimes_{(\T B)^e}\T B
&\lla&
\B\T A\otimes_{(\T A)^e}(X\otimes_{\T B}\check X)\\
\uar\mbox{\scriptsize $HH(\iota_B)$}&&\mbox{\scriptsize $HH(\iota_A)$}\uar\\
\left(\check X\otimes_{\T A}\B A\otimes_{\T A}X\right)
\otimes_{B^e}B
&\lla&
\B A\otimes_{(\T A)^e}(X\otimes_{\T B}\check X)
\end{array}
$$
and
$$
\begin{array}{ccc}
\left(\check X\otimes_{\T A}\B(\T A)\otimes_{\T A}X\right)
\otimes_{(\T B)^e}\T B
&\lla&
\B\T A\otimes_{(\T A)^e}(X\otimes_{\T B}\check X)\\
\dar\mbox{\scriptsize $HH(\pi_B)$}&&\mbox{\scriptsize $HH(\pi_A)$}\dar\\
\left(\check X\otimes_{\T A}\B A\otimes_{\T A}X\right)
\otimes_{B^e}B
&\lla&
\B A\otimes_{(\T A)^e}(X\otimes_{\T B}\check X)
\end{array}
$$
are commutative.
But this is clear since the unit element of $\T B$ (and of $\T A$ resp.)
is the image of the unit element of $B$ (and of $A$ resp.)
under $\iota$.

Hence the claim is proven.

This implies now
\begin{eqnarray*}
HH_m(F_{A\otimes_{\T A}X})\circ\hat\kappa_n^{(m);A}
&=&HH_{m}(F_{A\otimes_{\T A}X})\circ HH_{m}(\pi_A) \circ\kappa_n^{(m);\T A}\circ HH_{p^nm}(\iota_A)\\
&=&HH_{m}(\pi_B)\circ HH_{m}(F_{X})\circ\kappa_n^{(m);\T A}\circ HH_{p^nm}(\iota_A)\\
&=&HH_{m}(\pi_B)\circ\kappa_n^{(m);\T B}\circ HH_{p^nm}(F_X)\circ HH_{p^nm}(\iota_A)\\
&=&HH_{m}(\pi_B)\circ\kappa_n^{(m);\T B}\circ HH_{p^nm}(\iota_B)\circ HH_{p^nm}(F_{A\otimes_{\T A}X})\\
&=&\hat\kappa_n^{(m);B}\circ HH_{p^nm}(F_{A\otimes_{\T A}X})
\end{eqnarray*}
which proves Theorem~\ref{derivedinvarianceofkappahat}.
\dickebox

\bigskip

\section{Dual numbers as an example}
\label{dualnumbers}

It should be noted that the mappings $\kappa_n^{(m)}$ are not zero in general.
The dual numbers provide an example. In this section we shall show this fact and prove moreover that $\kappa^{(n)}_m=\hat\kappa^{(n)}_m$ in this case.

\medskip

Holm computed the cohomology ring of rings $k[X]/(f(X))$ for all polynomials $f(X)$ and all fields $k$. In particular for $f(X)=X^2$ one obtains a symmetric algebra, the algebra of dual numbers, whose Hochschild homology and cohomology are isomorphic. For fields of characteristic zero Lindenstrauss \cite{Lindenstrauss} computed in general the Hochschild homology of algebras $k[X_1,\dots,X_n]/{\mathfrak m}^m$ for $\mathfrak m$ being the maximal ideal corresponding to the point $(0,0,\dots,0)$ by exhibiting an explicit projective resolution.

We shall reprove parts of these statements since we will need quite detailed information about the
(co-)cycles that represent each element in the Hochschild structure.

Let $A=k[\epsilon]/(\epsilon^2)$ throughout this section and let
$k$ be a field of characteristic $p>0$.

\begin{Lemma}
$\T A\simeq A\otimes_kA$.
\end{Lemma}

Proof.
Clearly, since $A$ is commutative, also $\T A$ is commutative. Moreover, there is a $k$-basis $A=k\cdot 1\oplus k\cdot \epsilon$ and since $A$ is symmetric, the symmetrising form being $\langle x+y\epsilon,x'+y'\epsilon\rangle:=xy'+x'y$, we get a $K$-basis of $\T A=A^*\sdp A$ by
$$\T A=k\cdot (\langle 1,-\rangle,0)\oplus k\cdot (\langle \epsilon,-\rangle,0)
\oplus k\cdot (\langle 0,-\rangle,1)\oplus k\cdot (\langle 0,-\rangle,\epsilon).$$
Let $\delta:=(\langle 1,-\rangle,0)$, $\sigma:=(\langle \epsilon,-\rangle,0)$,
$\varepsilon:=(\langle 0,-\rangle,\epsilon)$ and $1:=(\langle 0,-\rangle,1)$ (remarking that this is still the unit element of the trivial extension algebra).
We verify immediately that $\varepsilon^2=\delta^2=0$. Furthermore,
$\varepsilon\cdot\delta=\delta\cdot\varepsilon=\sigma$.
Hence, $\T A$ is the quotient of the quiver algebra with one vertex and two loops by the relations saying that the two loops are nilpotent of order $2$ and that they commute. This describes exactly the algebra $A\otimes_kA$. \dickebox

\begin{Rem}
If $p=2$, then $A$ is isomorphic to the group algebra of the cyclic group of order $2$.
Then $A\otimes_kA^{op}$ is isomorphic to the group algebra of the Klein four group and
the Hochschild cohomology ring is known. Holm showed \cite[Theorem 3.2.1]{holmhabil}
that in this case
$$HH^*(A\otimes_kA^{op})\simeq A[X,Y]$$
where $X$ and $Y$ are algebraically independent of degree $1$.
\end{Rem}

For any $p>0$ we know how to determine the Hochschild homology in this more general case
as well by the K\"unneth formula. Indeed,
$$HH_*(A\otimes_kA)\simeq HH_*(A)\otimes_kHH_*(A)$$ by
the K\"unneth formula.

We have to study the injection
$$HH_*(A)\lra HH_*(\T A)\simeq HH_*(A\otimes_KA)\simeq HH_*(A)\otimes_KHH_*(A)$$
given by the isomorphism $\T A\simeq A\otimes_KA$, the injection $A\lra \T A$ and the
K\"unneth formula, as well as the projection
$$HH_*(A)\otimes_kHH_*(A)\simeq HH_*(A\otimes_kA)\simeq HH_*(\T A)\lra HH_*(A).$$

The algebra $A$ is symmetric and hence
$$Hom_k(HH_m(A),k)\simeq HH^m(A)$$
by an isomorphism induced by the symmetrising bilinear form. Therefore there
is a non degenerate pairing $$HH^m(A)\times HH_m(A)\lra K$$
as usual.

\begin{Rem}
The algebra $A\otimes A$ is the quotient of the quiver algebra with one vertex and
two loops by the relations saying that the two loops are nilpotent of order $2$ and
that they commute. Hence, we may replace any loop by a non trivial linear combination
of these two loops, completing by another linear combination of the two loops so that
the determinant of the coefficient matrix is non zero. Hence, we may suppose that the
inclusion $A\lra \T A$ is given by the inclusion $A\lra A\otimes A$ defined by
$a\mapsto 1\otimes a$. Indeed, the Hochschild homology computation by
means of the corresponding double complex does not depend on the choice
of a basis.
\end{Rem}

BACH \cite{Bach} and Holm \cite{Holmpolynomial} computed an explicit resolution of bimodules for a monogenic algebra, such as the dual numbers. Abbreviate for simplicity $A=K[\epsilon]/\epsilon^2$ (as usual) and $A^2:=A\otimes_kA$.
Then a free resolution $C$ of $A$ is periodic of period $2$ and is given by
$$C:\;\;\;(A\lla) A^2\stackrel{d_1}{\lla} A^2\stackrel{d_2}{\lla} A^2 \stackrel{d_1}{\lla} A^2\stackrel{d_2}{\lla} A^2\lla \cdots$$
where $d_1$ is multiplication by $1\otimes\epsilon-\epsilon\otimes 1$ and $d_2$ is
multiplication by $1\otimes\epsilon+\epsilon\otimes 1$.

Applying the functor $-\otimes_{A^2}A$ gives a complex
$$A\stackrel{(d_1)_\otimes}{\lla} A\stackrel{(d_2)_\otimes}{\lla} A \stackrel{(d_1)_\otimes}{\lla} A\stackrel{(d_2)_\otimes}{\lla} A\lla \dots$$
with $(d_1)_\otimes=0$ and $(d_2)_\otimes=2\epsilon$.

Now, applying the functor $Hom_{A^2}(-,A)$ gives a complex $Hom_{A^2}(C,A)$
$$\dots\lra A\stackrel{(d_1)_h}{\lra} A\stackrel{(d_2)_h}{\lra} A \stackrel{(d_1)_h}{\lra} A\stackrel{(d_2)_h}{\lra} A$$
where again $(d_1)_h=0$ and $(d_2)_h=2\epsilon$.

The K\"unneth formula  gives that the tensor product of the cohomology is the cohomology of the tensor product
$$H\left(Hom_{A^2}(C,A)\otimes Hom_{A^2}(C,A)\right)=
H\left(Hom_{A^2}(C,A)\right)\otimes H\left(Hom_{A^2}(C,A)\right).$$
We observe that
$$Hom_{A^2}(C,A)\otimes Hom_{A^2}(C,A)\simeq
Hom_{A^2\otimes A^2}(C\otimes C,A\otimes A)$$
and this is the complex computing the Hochschild cohomology of $A\otimes_kA$, and whence for $\T A$.

As usual the Hochschild (co-)homology depends on weather $p=2$ or $p>2$. Our arguments use the explicit structure of the Hochschild (co-)homology and therefore we shall need to treat these two cases separately.

\subsection{The case $p>2$}

Holm \cite{Holmpolynomial} shows that
$$HH^*(A)\simeq A[U,Z]/(Z\epsilon, U\epsilon, U^2)$$
for an element $U$ in degree $1$ and an element $Z$ in degree $2$. Hence
$HH^m(A)$ is one-dimensional, generated by $Z^n$ if $m=2n$ or $UZ^n$ if $m=2n+1$,
for $m>0$ and isomorphic to $A$ in degree $0$.

We may therefore choose for each $n\in{\mathbb N}\setminus \{0\}$ an element $z_{n}$ in $HH_{2n}(A)$ which corresponds to $Z^n$ in $HH^*(A)$ under the above isomorphism and we get $HH_{2n}(A)=kz_n$. Moreover, choose an element $uz_n\in HH_{2n+1}(A)$ which corresponds to $UZ^n$ under the above isomorphism. Hence, $HH_{2n+1}(A)=kuz_n$.

Observe that
$$\left(HH_*(A)\otimes HH_*(A)\right)^*\simeq HH_*(A)^*\otimes HH_*(A)^*\simeq
HH^*(A)\otimes HH^*(A)$$
where the first isomorphism is canonical and the second one is induced by the
isomorphism $HH_*(A)^*\simeq HH^*(A)$. Hence {\em using this chain of isomorphisms}
we choose the $k$-basis
$$x_iy_j\mbox{ in bidegree }(2i,2j)\mbox{ of }HH_{2i+2j}(A\otimes A),$$
$$x_ivy_j\mbox{ in bidegree }(2i+1,2j)\mbox{ of }HH_{2i+1+2j}(A\otimes A),$$
$$x_iy_jw\mbox{ in bidegree }(2i,2j+1)\mbox{ of }HH_{2i+2j+1}(A\otimes A),$$
$$x_ivy_jw\mbox{ in bidegree }(2i+1,2j+1)\mbox{ of }HH_{2i+1+2j+1}(A\otimes A),$$
which is the dual basis element to the corresponding monomial in $X^iY^j$, the
monomial $X^iVY^j$, the monomial $X^iY^jW$ or the monomial $X^iVY^jW$ of
$HH^{*+*}(A\otimes A)$ under that isomorphism. The $p$-th power of all these basis
elements is zero, except the element $X^iY^j$, whose $p$-th power is $X^{pi}Y^{pj}$.

The (minimal) projective resolution of $A^2$ used above can most easily be expressed as a double complex $C\otimes C$, as was shown above.

$$
\begin{array}{ccccccccccccccc}
&&A^2&\lla&A^2&\lla&A^2&\lla&A^2&\lla&A^2&\lla\\
&&\uar&&\uar&&\uar&&\uar&&\uar\\
A^2&\lla&A^2\otimes A^2&\lla&A^2\otimes A^2&\lla&A^2\otimes A^2&\lla&A^2\otimes A^2&\lla&A^2\otimes A^2&\lla\\
\uar&&\uar&&\uar&&\uar&&\uar&&\uar\\
A^2&\lla&A^2\otimes A^2&\lla&A^2\otimes A^2&\lla&A^2\otimes A^2&\lla&A^2\otimes A^2&\lla&A^2\otimes A^2&\lla\\
\uar&&\uar&&\uar&&\uar&&\uar&&\uar\\
A^2&\lla&A^2\otimes A^2&\lla&A^2\otimes A^2&\lla&A^2\otimes A^2&\lla&A^2\otimes A^2&\lla&A^2\otimes A^2&\lla\\
\uar&&\uar&&\uar&&\uar&&\uar&&\uar\\
A^2&\lla&A^2\otimes A^2&\lla&A^2\otimes A^2&\lla&A^2\otimes A^2&\lla&A^2\otimes A^2&\lla&A^2\otimes A^2&\lla\\
\uar&&\uar&&\uar&&\uar&&\uar&&\uar
\end{array}
$$

In order to view the image of $z_n\in HH_{2n}(A)$ in $HH_{2n}(\T A)$
we need to first establish the multiplicative structure of $HH^*(A\otimes A)$ in terms
of maps in the double complex.

The element $X$ is the degree $(2,0)$ mapping consisting of the identity on all
homogeneous components except on the borders of this bi-complex, and the element $Y$ is the degree $(0,2)$ mapping identity on all homogeneous components except on the
borders of this bi-complex.

The element $V$ is the degree $(1,0)$ mapping consisting of $\epsilon\otimes 1$ in all degrees, except on the borders of the bi-complex, where it is $0$.
Likewise the element $W$ is the degree $(0,1)$ mapping consisting of $\epsilon\otimes 1$ in all degrees, except on the borders of the bi-complex, where it is $0$.

We claim that the element $X^iY^j$ is represented by the degree $(2i,2j)$ mapping
consisting of the identity on all homogeneous components except on the borders of
this bi-complex. The proof of this statement is an easy induction on $i$ and $j$.
Actually, the composition of a morphism of bi-complexes of degree $(2i,2j)$ of the
given shape by a morphism of complexes of degree $(2,0)$, or $(0,2)$ respectively,
corresponds to a morphism of bi-complexes of degree $(2(i+1),2j)$, or $(2i,2(j+1))$
respectively. This corresponds to the cup product of $X^iY^j$ with $X$, or $Y$
respectively.

The element $X^iVY^j$ is represented by the degree $(2i+1,2j)$ mapping
consisting of the mapping $(\epsilon\otimes id)\otimes (id\otimes id)$
on all homogeneous components except on the borders of this bi-complex. Likewise
the element $X^iY^jW$ is represented by the degree $(2i,2j+1)$ mapping
consisting of the mapping $(id\otimes id)\otimes (\epsilon\otimes id)$
and the element $X^iVY^jW$ is represented by the degree $(2i+1,2j+1)$ mapping
consisting of the mapping $(\epsilon\otimes id)\otimes (\epsilon\otimes id)$.

The cup product on $HH^*(A)$ is similar: $Z$ corresponds to the degree $2$ mapping
being the identity on all homogeneous of $C$, except the degree $0$ and degree $1$
component, where the mapping clearly is $0$. Now, again by an analogous induction
as in the bi-complex case, the element $Z^n$ corresponds to the degree $2n$ mapping
being the identity on all homogeneous of $C$, except the degrees up to $2n-1$ component,
where the mapping clearly is $0$. The element $U$ corresponds to the degree $1$ mapping
$\epsilon\otimes 1$ in all degrees except the degree $0$, where it is $0$. The cup
product with $Z^n$ is just an additional shift in degree, the cup product of $U$
with $U$ is $0$, since $(\epsilon\otimes 1)^2=0$.

We now need to compare this mapping to the dual of the Hochschild homology side.
Hence, we dualise the bi-complex and the cocycle representing $X^iY^j$. This just reverses the arrows, identifying again $A^*$ with $A$. The dual of the cocycle $X^iY^j$ can also be obtained by first applying $Hom_{A^2\otimes A^2}(-,A^2)$
to the double complex and then $K$-dualising the result. We obtain the double complex $D_{HH(A^2)}$.
$$
\begin{array}{ccccccccccccccc}
&&.&&.&&.&&.&&.&\\
&&\uar&&\uar&&\uar&&\uar&&\uar\\
.&\lla&A^2&\stackrel{0}{\lla}&A^2&\stackrel{}{\lla}&A^2&\stackrel{0}{\lla}&A^2&
\stackrel{}{\lla}&A^2&\lla\\
&&\uar 0&&\uar 0&&\uar 0&&\uar 0&&\uar 0\\
.&\lla&A^2&\stackrel{0}{\lla}&A^2&\lla&A^2&\stackrel{0}{\lla}&A^2&\lla&A^2&\lla\\
&&\uar &&\uar&&\uar &&\uar&&\uar \\
.&\lla&A^2&\stackrel{0}{\lla}&A^2&\stackrel{}{\lla}&A^2&\stackrel{0}{\lla}&
A^2&\stackrel{}{\lla}&A^2&\lla\\
&&\uar 0&&\uar 0&&\uar 0&&\uar 0&&\uar 0\\
.&\lla&A^2&\stackrel{0}{\lla}&A^2&\lla&A^2&\stackrel{0}{\lla}&A^2&\lla&A^2&\lla\\
&&\uar&&\uar&&\uar&&\uar&&\uar
\end{array}
$$
which now represents exactly the complex computing Hochschild homology of
$A\otimes A$.
Half of the arrows represent the $0$-morphism, the rest, without a side- or superscript in the diagram, represent the mapping $2\epsilon$. The cycle representing $X^n$ is the $2n$-th mapping identity in the upper line.
The complex computing the Hochschild homology of $A$ is simpler:
$$C_{HH(A)}:\;\;A\stackrel{0}{\lla} A\lla A\stackrel{0}{\lla} A\lla A\stackrel{0}{\lla} \cdots$$
with the same convention on non superscribed arrows.

The injection $A\lra A\otimes A\simeq \T A$ maps $a\mapsto 1\otimes a$ and the  projection $A\otimes A\lra A$ maps $a\otimes b\mapsto ab$. Hence, the projective resolution $P_\bullet$ of $A$ as $A^2$-modules maps to the projective resolution $P_\bullet\otimes P_\bullet$ of $A^2$ as $A^2\otimes A^2$-modules as $x\mapsto 1\otimes x$. The second degree component is $0$, except in degree $0$, where it is constantly $1\otimes 1$. Hence, the Hochschild homology complex
of $A$ injects by the identity into the first line of the Hochschild homology (double-) complex of $A^2$.
In other words, the injection produces the mapping of complexes from
$C_{HH(A)}$ to the first line of the bi-complex $D_{HH(A^2)}$ by the mapping $A\ni a\mapsto 1\otimes a\in A\otimes A$ on the level of each homogeneous component.

Likewise, the projection produces a mapping of complexes in the inverse order, which compose to the identity on $C_{HH(A)}$.
Hence $z_n$ is mapped to $x_n$ for all $n>0$ and $z_nu$ is mapped to $x_nv$ for all $n>0$.

Finally, since $p$ is odd, $p^nm$ is odd if and only if $m$ is odd.
We proved the following

\begin{Prop} Let $k$ be a perfect field of characteristic $p>2$. Then
$\hat\kappa_n^{(m),k[\epsilon]/\epsilon^2}=\kappa_n^{(m),k[\epsilon]/\epsilon^2}$
for all $n$ and $m$.
Moreover, $\kappa_n^{(m)}(z_{p^nm})=z_m$ and $\kappa_n^{(m)}(z_{p^nm-1}u)=0$.
\end{Prop}

\subsection{The case $p=2$}

The case of even characteristic is completely analogous, with a few exceptions.
The generator $Z$ of $HH^*(A)$ is in degree $1$, $Z$ is not annihilated by $\epsilon$,
and the differentials for the homology and the cohomology complex are all $0$.
The Hochschild cohomology ring does only contain
nilpotency coming from the centre. The rest is immediate.

\begin{Prop}
Let $k$ be a perfect field of characteristic $2$. Then
$\hat\kappa_n^{(m),k[\epsilon]/\epsilon^2}=\kappa_n^{(m),k[\epsilon]/\epsilon^2}$
for all $n$ and $m$.
Moreover, $\kappa_n^{(m)}(z_{p^nm})=z_m$ for all $m$ and $n$.
\end{Prop}

{\bf Proof.} The proof is a straightforward analogue
of the proof in the $p>2$-case. \dickebox

redpajama_set_name: RedPajamaArXiv


In episode 67 of The Podcast of Oa co-hosts Bill Giancoli and Myron Rumsey discuss the start of the latest Green Lantern story arc, Uprising. Along with parts 1 and 2 of Uprising that took place in Green Lantern #31 and Green Lantern Corps #31, conversation turns to the thirty first issues of Green Lantern: New Guardians and Red Lanterns as well as the second issue of Cullen Bunn's Sinestro series which includes the long awaited showdown between Sinestro and Arkillo. Was it all that fans hoped it would be, or did it leave readers wanting more?
Myron and Bill share listener feedback including comments and results from the Facebook poll that asked listeners if they were Team Kyle or Team Hal when it came to Carol Ferris.


Molly Wright Steenson is a designer, author, professor, and international speaker whose work focuses on the intersection of architecture, design, and artificial intelligence. She is the author of Architectural Intelligence: How Designers and Architects Created the Digital Landscape (MIT Press, 2017), which tells the radical history of AI's impact on design and architecture and how it poured the foundation for contemporary digital design. Molly is an associate professor at Carnegie Mellon University's School of Design, with a courtesy appointment in the School of Architecture. Previously, she was an assistant professor in the School of Journalism and Mass Communication at the University of Wisconsin-Madison, adjunct faculty at Art Center in Pasadena, CA, and an associate professor at the Interaction Design Institute Ivrea in Italy. Molly cut her teeth on the web in 1994 and has since worked with groundbreaking studios, consultancies, and corporations. She holds a PhD in Architecture from Princeton University and an M.E.D from the Yale School of Architecture.


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Submitted by matt on Wed, 11/01/2017.
What do life insurance product developers have on their minds? A recent study group gave us some clues.


Do you remember my previous post about Maybank Go Ahead Challenge 2015 (#MGAC2015)? The global competition had been ongoing for months, and last week marked the final week of the top 60 challengers from around the world to accept final round of challenges.
As part of the challenges, the global finalists of #MGAC2015 took over Bukit Bintang with creative projects and activities for the general public, through their CSR fundraising initiative dubbed Season Of Smile (#SOS). All the funds raised would go into different charities.
10 groups of finalists had set up their own booth along Bukit Bintang, namely in front of Pavilion KL, Starhill Gallery and Lot 10 shopping malls. Kudos to their strong perseverance and strong spirits, they were promoting for the good cause regardless of rain or shine!
Let's take a picture with the cute mascot!
and her colleagues were there to check out this meaningful campaign too.
Among the 10 booths that we had visited, "Amsanna" was one of my favourite because of their great services and friendliness. The station was transformed into a beauty parlour where I managed to get my nail manicure fix while Carmen went for eyebrow shaping and hand massage. I guess a nice pampering session certainly makes girls happy!
These #MGAC2015 finalists really deserved a pat on the shoulder! They are really creative when it comes to creating CSR projects. "MR Mary & Ross" shopping kiosk allowed us to play a shopping game by matching the shopping bag price to a targeted amount.
For instance, Carmen and I were given 30 seconds to grab anything in store that can accumulate to the range between RM49 to RM51. If we succeed in this challenge, then we would be able to bring everything we picked home.
Haha, but neither of us were grocery shopping queen. Luck was not on our side, and our total amount was RM52.63. Anyway it was alright, because we managed to donate RM20 to the charity homes!
After shopping, it was time to treat our hungry tummy. Thanks to one of the booths, "Finnie", Milkcow soft serve ice cream came perfectly at the nick of time as a late afternoon dessert.
As we strolled along the street, more and more food were coming our way. The brownies and cupcakes from different teams were all selling like hot cakes! I could see it was indeed a great effort from the MGAC challengers to prepare for this event within such a short timeframe.
"Tom's Food: Best Hotdogs and Pizzas in Town"
On the other hand, the "Hamilton" team spread happiness by offering selfie and instant photo printing services for just RM5! Well, how often do you get instant prints like this? It's super worth it right!
Thanks for being the smiley agents and painting the town yellow!
"Shroom"--- Another unique idea I've seen, combinning both Uber cab and AirBNB accommodation services into one.
Look at those angels! I can feel smiles are everywhere and love is in the air.
Despite of being just a short one-day Season Of Smile (#SOS) fundraising event, it was great to see how hard everyone in MGAC2015 had worked to make a difference in other's lives. With this, MGAC become more than just a global challenge platform, but an avenue for everyone to become a winner because they have hearts of champions.
Now that MGAC2015 has come to an end, and congratulations to the winning teams who walked away with amazing cash prizes and global internship opportunities.
Anyway, I hope to see you again in Maybank Go Ahead Challenge 2016.
Don't forget to check out www.facebook.com/roaringcareers for latest updates!


text:
John-F.-Kennedy-Institut/
Graduate School of North American Studies (GSNAS)
(Post-) Doktoranden
Doktorandinnen und Doktoranden
Sabine Isabell Engwer
Sabine Engwer
Doktorandin
Lansstraße 5-9
engwer@gsnas.fu-berlin.de
Writing the African American Nation - Notions of Folk, Volk, and Nation in the Literature of the Nadir (Dissertationsprojekt)
Dissertation in Literatur
Mentoring Team:
First supervisor: Prof. Florian Sedlmeier
Second supervisor: Karl-Heinz Ickstadt
Third supervisor: Prof. Carla Peterson
Sabine Isabell Engwer has been a doctoral candidate at the Graduate School of North American Studies since the fall of 2012. Her dissertation project focuses on African American literature from the "Nadir" period.
The Nadir spans the era from the end of Reconstruction in the 1870s to the early years of the twentieth century. This period was characterized by a severe backlash against the newly-freed African American population, the institutionalized racism of the "Jim Crow laws," and a significant rise in lynchings in the South of the United States.
For African American writers at the time, this climate of repression led to a decidedly political agenda for their literary work. The battle for equality, political enfranchisement, and against the ubiquitous racism they experienced, informed their writing significantly. Black authors dealt with the political realities of the Nadir in a number of ways – literary reactions ranged from the disillusioned portrayal of racism in Charles Chesnutt's The Marrow of Tradition (1901), to Sutton Grigg's bold suggestion of a separate African American nation within the United States in Imperium in Imperio (1899). A central component of this project will be the analysis of the oeuvre of Pauline Elizabeth Hopkins (1859-1930), a leading African American journalist, editor and novelist, and one of the few influential black women writers of the Nadir, whose phantasmatic novel Of One Blood (1903) even has its African American protagonist come to political power as the long-awaited king of a mythical subterranean civilization in Ethiopia.
The dissertation will critically examine these and other African American texts of the period in their historical context, in order to come to a deeper understanding of their underlying political agendas and the Nadir's prevalent notions of "folk," "nation" and "nationality" at large.

redpajama_set_name: RedPajamaCommonCrawl


text:
Rafale deal: BJP slams Rahul Gandhi over Supreme Court order, CAG report; demand apology
The Bharatiya Janata Party (BJP) came down heavily on the Congress party and its chief Rahul Gandhi after the CAG report on the Rafale fighter jet deal was tabled in the Parliament on Wednesday. The saffron party demanded an apology from Gandhi after the CAG said in a report that the deal was 2.86 per cent cheaper than the UPA's 2007 offer. The Narendra Modi-led party also said Rahul Gandhi's "lies" on the Rafale aircraft deal have been exposed.
from IndiaTV India: Google News Feed http://bit.ly/2TQvERi


text:
Bihar Board Class 12 Solutions
Bihar Board 12th Objective Questions
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Bihar Board Class 11 English Notice Writing
May 2, 2020 May 2, 2020 by Priyanshu
Bihar Board Class 11th English Book Solutions Notice Writing
1. Your school is organising a tour to Mumbai and Goa during the winter vacation. Write a notice giving detailed information to the students. You are Madhav/Madhavi, School Pupil Leader, Ram Mohan Roy Seminary.
Feb, 20
Ram Mohan Roy Seminary
Tour to Mumbai and Goa
Our school is organising a tour to Mumbai and Goa during the coming winter vacation. It will be a 10-day tour in which only students of the Senior Secondary classes are allowed to Join. The expenses for traveling, boarding and lodging shall be around Rs. 3000/-(three thousand only) per head. Those who want to join this tour should give their names to the undersigned by the length of this month. A non-objection note by the parents is a must.
RAUSHAN
Pupil Leader
2. You lost your wallet containing your valuable documents (passport, your certificates and a bunch of keys) while traveling by Puri Express from Patna to Puri. Write a notice for publishing in 'The Times of India' (Patna) under 'Lost and Found' column. You are Naresh of station Road, Patna.
Lost my wallet of black colour containing some valuable documents, including my Board's certificates of All India Senior Secondary Examination, passport and a bunch of keys on 25th June, 200 while traveling by Puri Express from Patna to Puri. The finder will be suitably rewarded.
Contact: Naresh,
Station Road,
Patna.
3. You are Pushkar, Darbhanga, As the Secretary of the Social Service league of your school, you have organized a cultural benefit show in aid of mentally handicapped children of your town Darbhanga. Write a notice in not more than 50 words for your school notice-board, giving necessary information about the program.
Social Service League
St. Secondary School, Darbhanga
You will be glad to know that the Social Service League of Sr. Secondary School is organizing a Cultural Benefit Show in aid of mentally handicapped
children of Darbhanga on 20th April 20 Entry is by tickets. The League will also distribute Donor Cards of Rs. 1000 and Rs. 500 to some distinguished industrialists and traders willing to patronise the benefit show. For tickets and Donor cards contact the undersigned on the working days from 10 a.m. to 3 p.m. The proceeds of the show will be donated to the Secretary, Society For Mentally Handicapped Children, Darbhanga.
Magic Show: 6 p.m.
Folk Dances: 6.30 p.m.
Dance Drama (Chandalika of Tagore): 7 p.m.
Puppet Show: 7.45 p.m.
Venue-Football Ground, Sr. Secondary School, Darbhanga
4. You are the Secretary of ABC Colony Welfare Association, Patna. Write a notice to be circulated to all the residents of the colony informing them that there will be no water supply in your colony on 24th and 25th of June, 20 due to maintenance work.
Welfare Association
ABC Colony, Patna
All the residents of ABC Colony, Patna are informed that there will be no water supply in the colony on 24th and 25th June 2009. Maintenance work has necessitated this step. The residents are advised to make necessary arrangements in advance.
Inconvenience is regretted.
S.N. Rao
Secretary 20th June, 20
5. You are the Principal, Senior Secondary School, Patna. Your school is introducing Commerce Section for XI and XII for the first time from July 2009. Draft a proper 'Admission Notice' for publication in a newspaper.
Senior Secondary School, Patna
The school has decided to introduce Commerce Classes for XI and XII from July 15, 2020, There are only 100 seats for each class. Only students getting 60% or more marks in XI and XII need apply. The school will hold an Entrance Test on 5th July 2007 a 9 a.m. at the campus, Prospectus (Rs. 30/-) can be had from the school office on the working days.
Sr. Secondary School
6. Your school has just completed 25 years of its meritorious service to society. The Students' Council of your school has decided to celebrate its Silver Jubilee. As President of the Council, write a notice in not more than 50 words for the students of your school, informing them about the Councils decision and seeking their co-operation for the success of the proposed Silver Jubilee Celebrations.
Silver Jubilee Celebrations
It gives me great pleasure to inform all the students that our school has just completed 25 years of its meritorious service to society. On this happy. The occasion, the Students; Council has decided to celebrate the school's Silver Jubilee with great pomp and circumstance. The kind co-operation of all of you is sought to make the celebrations a success. Please contact the undersigned with your suggestions.
D.A.V. College, Siwan
7. The Student's Council of your school has organised an excursion to Ajanta and Elora for the students of class XII during Winter Breaks. As President of the Council writes a notice in not more than 50 words telling the students about this excursion and inviting their names for joining it.
Excursion to Ajanta and Ellora
8. Yor the students of class XII, the Students' Council or the school has organised an excursion to the Ajanta and Elora caves near Aurangabad in Maharashtra, during the coming Winter Break. It will be two days' excursion and will cost about Rs. 1000/- each. Those who are willing to join should give their names to the undersigned by the tenth of this month.
Amardeep
Patna High School, Patna
8. You are Robit/Rashi. As the Secretary of the Cultural Club of your school, you have organized a Culural Evening as a thanks giving programme on the last day of your school. Write a notice for your school notice board, giving necessary information about this event in not more then 50 words.
The cultural club of our school is holding a thanks giving programme on the excellent performance of the club during the year. The programme will be held on Saturday the 7 October, the last day of the school. All are coordially invited to attend the function in the school hall. The function will begin with 'Yajan' at 8 a.m. in the morning.
9. You are Pushpak/Pooja. As Secretary of the Social Service League of your school, you have organized a cultural benefit show in aid of mentally handicapped children of your town, Vijayawada. Write a notice in not more than 50 words for your school notice-board, giving necessary information about the programme.
Social Service League . Patna Collegiate, Patna
The Social Service League of the school is holding a variety show in aid of, mentally handicapped children of our town. The show will be held of the 15th of April at 7 p.m. in the school hall. Many renowned artists of the state have been invited to take part in it. Tickets for the show can be had from the league office.
S.S. League
10. You are Rahul/Rashmi. As President of the Literary Club of your school you have organized an inter-school debate competition on the occasion of the silver Jubilee Celebrations of your school. Write a notice in about 50 words, informing the students of your school about the competition.
26 Feb. 20.
On the occasion of the Silver Jubilee Celebration of our school, the Literary Club of the school is holding an inter-school debate competition on the 28th of the month. The competition will be held in the school hall and will begin at 11 a.m. About fifteen teams are expected to take part in the competition. All are invited to attend.
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Bihar Board Class 11 English Report Writing
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Bihar Board Class 12 Chemistry Solutions Chapter 16 दैनिक जीवन में रसायन
Bihar Board Class 12 Chemistry Solutions Chapter 15 बहुलक
Bihar Board Class 12 Chemistry Solutions Chapter 14 जैव-अणु
Bihar Board Class 12 Chemistry Solutions Chapter 13 ऐमीन
Bihar Board Class 12 Chemistry Solutions Chapter 12 ऐल्डिहाइड, कीटोन एवं कार्बोक्सिलिक अम्ल
Bihar Board Class 12 Chemistry Solutions Chapter 11 ऐल्कोहॉल, फ़िनॉल एवं ईथर
Bihar Board Class 12 Chemistry Solutions Chapter 10 हैलोऐल्केन तथा हैलोऐरीन
Bihar Board Class 12 Chemistry Solutions Chapter 9 उपसहसंयोजन यौगिक
Bihar Board Class 12 Chemistry Solutions Chapter 8 d एवं f-ब्लॉक के तत्त्व
Bihar Board Class 12 Chemistry Solutions Chapter 7 p-ब्लॉक के तत्त्व
Bihar Board Class 12 Chemistry Solutions Chapter 6 तत्त्वों के निष्कर्षण के सिद्धान्त एवं प्रक्रम
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text:
import {Request} from './request';
import moment from 'moment';
import qs from 'qs';
import {expect} from 'chai';
import {cloneDeep, castArray, set, get, find, isEmpty, range} from 'lodash';

export default class Query {
  constructor(endpoint, options) {
    this._endpoint           = endpoint;
    this._options            = cloneDeep(options);
    this._options.itemFilter = isEmpty(this._options.itemFilter) ? [] : castArray(this._options.itemFilter);
  }

  setPage(page) {
    expect(page, 'pages').to.be.within(1, 100);

    set(this._options, 'paginationInput.pageNumber', page);

    return this;
  }

  setEntriesPerPage(entries) {
    expect(entries, 'entries').to.be.within(1, 100);

    set(this._options, 'paginationInput.entriesPerPage', entries);

    return this;
  }

  setFilter(name, value) {
    const currentValue = this.getFilter(name);

    currentValue
      ? currentValue['value'] = value
      : this._options.itemFilter.push({name, value});

    return this;
  }

  getFilter(name) {
    return find(this._options.itemFilter, {name});
  }

  setEndTimeTo(time) {
    this.setFilter('EndTimeTo', time.toISOString ? time.toISOString() : time);

    return this;
  }

  setEndTimeFrom(time) {
    this.setFilter('EndTimeFrom', time.toISOString ? time.toISOString() : time);

    return this;
  }

  /**
   * Splits query into smaller chunks;
   * @param {Number} parts
   * @returns {Request}
   */
  split(parts) {
    const endTimeFrom = get(this.getFilter('EndTimeFrom'), 'value'),
          endTimeTo   = get(this.getFilter('EndTimeTo'), 'value');

    expect(endTimeFrom, 'EndTimeFrom').to.exists;
    expect(endTimeTo, 'EndTimeTo').to.exists;

    const chunks    = range(0, parts);
    const chunkSize = moment(endTimeTo).diff(endTimeFrom) / parts;

    return chunks.map(n => {
      const start = moment(endTimeFrom).add((n * chunkSize), 'ms'),
            end   = moment(start).add(chunkSize, 'ms');

      return () => new Query(this._endpoint, this._options).setEndTimeFrom(start)
                                                           .setEndTimeTo(end);
    });
  }

  invoke() {
    const queryString = qs.stringify(this._options, {delimiter: '&'});

    return new Request(this._endpoint + '?' + queryString, {json: true});
  }
}

redpajama_set_name: RedPajamaGithub


Ransomware is a type of malware that infects your computer, encrypts your data and refuses to give you the key unless you pay its makers a sum of money.
According to the Palo Alto Networks security company, the first fully functional ransomware that operates on Apple's OS X has been discovered and named "KeRanger". There goes the myth that there is no malware aimed at Apple's personal computers!
Transmission BitTorrent client installer for OS X was infected just a few hours after installers were initially posted on March 4 and anyone who downloaded Transmission 2.90 around that date may have infected their OS X machine with the KeRanger malware. Soon after the infection was discovered, Transmission released a new version of its client, Transmission 2.92, which should be malware-free.
Two KeRanger-infected Transmission installers were signed with an Apple-issued certificate but it's not clear how the malware-infested installers ended up on Transmission's website. The developer listed this certificate is a Turkish company with a different id than the developer ID used to sign previous versions of the Transmission installer.
The certificate was later revoked by Apple, so trying to run an infected version of Transmission will result in a warning dialog being shown that states: "Transmission.app will damage your computer. You should move it to the Trash." Or "Transmission can't be opened. You should eject the disk image." In any case if you see these warnings, you should follow Apple's instruction to avoid being affected.
KeRanger is not the first OS X ransomware. Mabouia was the first ransomware in the world targeting OS X but it was not fully functional, being more of a proof of concept that ransomware can be applied to Apple.


The Mooste Overlay performance is an art project by MoKS artist in residence Geoff Robinson. Geoff is a sound artist and has recorded sounds in two buildings in Mooste: the Cultural House hall and a large warehouse at the rear of Mooste Manor (in which a log house is being built). He is interested in the acoustic relationship between these two building and their different functions.
For the performance Geoff will be playing sound recordings from one building into the other building. He has invited musicians from the Mooste Folk School to play single note sounds with their instruments to explore the acoustics of each building. The performance aims to create an overlay of the hall and warehouse through recorded and acoustic sounds.


text:
OMG !!! & Other Popular Diets
A well written article by Emily Jupp on June 12, 2012 for The Independent, in which she pursues the challenge of the OMG Diet and attentively documents her journey, discusses what the facts are, and her overall impression about this latest diet. Thank You Emily for allowing me to share your story!
"You've heard of Dukan, scoffed fat on Atkins, fainted from drinking only honey and lemon water and lost friends from the noxious side-effects of the cabbage diet. Now there's a new one, this time designed to inspire jealousy and help you to "get skinnier than all your friends". Six Weeks to OMG comes out in print next month. It has knocked the Dukan Diet off the top of the iTunes diet book chart and a subsidiary of Penguin has bought it for a seven-figure sum.
The book, by British sports scientist and personal trainer Paul Khanna, who goes by the pen name Venice A Fulton, offers controversial tips, such as skip breakfast, take cold baths and drink black coffee. But for short-term weight loss, can the OMG diet really help you shed pounds? Consultant Dietitian and Sports Nutritionist Linia Patel, Dr Christian Jessen, the presenter of the Bafta award-winning Embarrassing Bodies series, and the celebrity fitness trainer Jay Darrell Ingleton explain the science behind Fulton's unconventional theories. I've also given Fulton's tips a try, in a (highly unscientific) experiment, to see how tough they are to stick to.
Skip breakfast……What do I do for energy?
Fulton says skipping breakfast will help to stop you becoming a "fadult" (that's a fat adult to you and me). "Breakfast is for wimps!", he writes. He recommends exercising first thing in the morning on an empty stomach and then only eating once three hours have passed, so that your body is forced to burn stored fat rather than food.
The expert view: "Skipping breakfast is a weight-loss disaster," says Patel. "When you wake up your body produces cortisol, a stress hormone that helps store fat. If you don't eat, you are putting your body under more strain and stress and therefore more likely to gain weight around the middle. Also, you will want sugary and fatty things when you're really hungry, which add to weight gain."
My experience: I skipped breakfast and still had enough energy to run up the stairs on the underground (my "workout"). It didn't affect my concentration and I ate breakfast at 10am. It was easy, but I consumed exactly the same amount that I usually eat, so I doubt it made much impact on my weight.
Drink black coffee before exercise
Fulton says the caffeine in a cup of black coffee will help to boost your exercise routine: "It's nature's cheap and secret potion," he writes. But he warns against adding any milk or sugar to your drink, because "either would tell your body that food's arriving," which, in theory, will prevent it from burning its fat stores.
The expert view: Dr Christian Jessen says coffee can help, but there's no secret trick to it. "After 12 hours or more of filming, mustering the energy to get to the gym can be hard," he says. "Coffee gives me the energy to just about squeeze in a workout. It's a stimulant and that's what I use it for, but it doesn't make your workout more efficient." He also warns that liquids contain calories too. "People forget fluids such as the Starbucks coffee mocha creamy coffee thing can be high in calories. Adding milk and sugar to coffee will add to your intake. Remember your hard work in the gym is lessened by a highly calorific drink beforehand."
My experience: It made me feel more energetic, which probably helped to psyche myself up for the gym.
Take cold baths….How Cold?
Fulton recommends taking baths of between 15C and 20C to "help you lose calories fast". He says the cold boosts adrenaline, which he says is particularly good for burning the fat stored in women's thighs.
The expert view: Ingleton says that a lot of athletes will take ice baths to help their muscles recover. "When athletes do it after a workout it gets blood to the muscles quicker. It has to be freezing otherwise it won't work, but the author is not saying go to that extent. Also I don't think everyone will be working out like a top body-builder. For normal people, it probably won't make much difference to weight-loss."
My experience: I opt for a cold shower, which is horrible. It feels like my body is in shock and makes my heart race. Afterwards, my skin looks bright and glowing and my clothes feel really warm, but the results don't outweigh the awfulness. I couldn't stick to it every day for six weeks.
OMG DIET says, "Eat less fruit" …..REALLY?
Fulton says you should eat as little as one piece of fruit a day during the diet period, although after the six weeks are up he encourages more balance in your diet.
"Fruit can be a cheeky devil," he says, because it contains fructose, which can turn to fat if it isn't burnt off.
The expert view: "It's all about calories," says Patel. "We recommend you eat five pieces of fruit and veg a day, but if you want to lose weight have more vegetables than fruit, as fruit is more calorific. It's ridiculous to just eat one piece of fruit a day for six weeks, because you're losing out on essential vitamins and minerals."
My experience: I replaced fruit with colourful salads and felt fine, but I missed having a late afternoon snack and bought a coffee loaded with honey instead of fruit – Fulton wouldn't approve.
Swap Vegetables for Coke….. WHY?
Fulton says eating unhealthily in the short-term is OK if it makes you thin. "[For some people] to get skinny", he writes,"it makes no difference whether they get their daily carbs from cans of Coke or from broccoli… for six to 12 weeks there's no harm in using this knowledge to get super-skinny."
The expert view: Patel says: "Broccoli is a fibrous carb that helps digestion and gives you nutrients. If you lay off fibrous carbs like bread, pasta and vegetables then you will be constipated with no energy."
My experience: I'm far too fond of vegetables and regular bowel movements to attempt this one.
The Truth about Duelling diets
If you don't like the sound of OMG, there are other diets that also promise fast weight-loss, but Dr Jessen warns that we shouldn't expect miracles: "Coming off the diet and coming back to normal is when the weight can pile back on. Dieting should be about taking a healthy, sensible approach that will work in the long-term. Don't try to cut everything out at once. Life is about enjoyment, too, and that's why you work hard, so you can indulge in a cheeky cake in the afternoon if you fancy."
The Headspace Diet:
10 Days to Finding Your Ideal Weight (Hodder & Stoughton). This is where you think deeply about every mouthful you eat and therefore, the theory goes, consume less.
"Being mindful while you eat is great, as most people tend to eat until they are stuffed," Patel says. "It takes about 20 minutes for the stomach to signal to the brain its full – but most of us can polish off two platefuls in less than 10 minutes!" But she warns that this diet has its risks. "Eating a tiny portion each meal, you would risk not getting in enough nutrients to keep you healthy."
The Baby Food Diet
This involves just eating jars of baby food in place of your normal, adult-size meals.
"Baby food is full of nutritious vitamins, which are easily digestible," Ingleton says. But he warns it's hard to feel satisfied with the lack of texture, and it's easy to go over the three-to-four jar allowance and overeat.
The Raw Food Diet
"The raw food diet contains fewer trans fats and saturated fats than general Western diets and it can lead to weight loss and may give you more energy," Ingleton says. He warns that there is a risk that the diet lacks some essential vitamins and fatty acids, so you would need to take vitamin supplements.
The Blood Group Diet
Devised by Peter J D'Adamo, this is when you cut out dairy, meat, wheat or grains depending on your blood type.
"There is no reasonable scientific basis for this," Ingleton says. Patel agrees: "You risk getting an unbalanced diet if you are cutting out a whole food group."
Like the Atkins diet, the Dukan diet involves eating mostly protein for a few weeks and then gradually introducing other foods.
"It's a high-protein diet, so while it might work for some to kick-start weight loss it won't work for all, as it's hard to cut out carbs," Patel says. "You risk having low energy.""
Posted by Kimberly Crocker in Diet, Food Facts
Tagged: Diet, Food Facts
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Pleasure Pier in Galveston is Hiring 600 People!
Galveston's new Pleasure Pier, opening in May 2012, will include 16 rides, a full-service restaurant, midway games, shopping and live entertainment. A $60 million project of Landry's Inc., the Pleasure Pier will be the only attraction of its kind on the Texas Gulf Coast and they are hiring now!
According to a company representative, they will be hiring approximately 600 people to fill both full and part-time positions. The venue will be open 363 days a year, so while they are certainly gearing up for a great summer vacation season, positions aren't seasonal in nature.
The company is currently accepting applications Monday through Friday from 9:00 a.m. to 4:00 p.m. at the San Luis Hotel & Resort's HR Office. They will be also hosting application events on April 11 from 2:00 p.m. to 6:00 p.m. AND on April 28 from 9:00 a.m. to 2:00 p.m. at The San Luis Hotel at 5222 Seawall Blvd. in Galveston. For more information, visit http://www.pleasure-pier.com/.
For more food and beverage positions available in Houston, visit The Houston Chronicle's job search site here.


text:
Texas Health Commissioner: Seiko plans to expand nationwide to 100 locations, including Houston
AUSTIN, Texas — Health Commissioner Mike Seiko said Friday the company plans to add more than 100 locations in the Houston area to expand its reach.
In addition to expanding Seiko's footprint in Houston, Seiko also plans to open its first retail store in the city in 2019, and to offer healthcare related services in the area.
"We are working very closely with the Texas Department of State Health Services, and they are committed to bringing these businesses here and providing quality care to the people of Texas," Seiko told reporters Friday.
The company's announcement came as the Houston Chamber of Commerce and Houston Independent School District announced plans to build a new, state-of-the-art high school.
The Houston Chamber announced the plans to the Houston Chronicle.
Seiko told the Chronicle that the new school will be called "Seiko High."
The plan to open a new high school in Houston comes after the Houston Health Department said it planned to build an entire facility to house the company's facilities.
As a result, the state of Texas said it is suspending the work permit process for the company, which currently employs 1,700 workers in Texas.
With this news, we are also suspending the process for construction of a second high school and are working closely with Texas to help them with that, Seikos said in a statement.
He said the company will begin hiring for its new campus, which is expected to open in 2019.
This announcement comes just a few months after the Texas Health Department suspended the work permits for Seiko, which employs more than 2,000 workers in the state.
Tags:btx global logistics, ceva logistics houston, logistics health, logistics planning, seko logistics


text:
Rambone XXX: A DreamZone Parody ist eine Porno-Parodie aus dem Jahr 2013 über den Film Rambo.

Handlung
Rambone, die den Charakter Rambo aus dem Originalfilm widerspiegelt, driftet in eine kleine Stadt und trifft einen feindlichen Sheriff, der sie aus der Gemeinschaft verbannen will. Es kommt zu sexuellen Handlungen und Kampf zwischen den Sheriffs und Rambone.

Szenen
 Szene 1. Brianna Brooks, Tommy Pistol
 Szene 2. Kendall Karson, Seth Gamble
 Szene 3. Christy Mack, Derrick Pierce
 Szene 4. Bonnie Rotten, Ryan McLane

Produktion und Veröffentlichung

Produziert wurde der Film  von Dream Zone Entertainment und vermarktet wird er von Vantage Distribution auf DVD. Regie führte Jordan Septo. Erstmals wurde der Film 29. Oktober 2013 in den Vereinigten Staaten veröffentlicht.

Nominierungen
 Inked Awards, 2014
 Nominee: DVD of the Year

 Nightmoves, 2014
 Winner: Best Parody: Drama

 Nightmoves Fan Awards, 2014

 XBiz Awards, 2015
 Nominee: Best Scene – Parody Release, Bonnie Rotten, Ryan McLane

Weblinks

Einzelnachweise

Porno-Parodie
US-amerikanischer Film
Filmtitel 2013

redpajama_set_name: RedPajamaWikipedia


text:
Commercial Lease Dispute & Resolution Team
Public Agency
Community and CSR
B Corp and Sustainability
Legal Netlink Alliance
Wendel Rosen Boosts Land Use Practice With Addition of Patrick Tuck
By: Wendel Rosen | 10/02/2019
Wendel Rosen LLP, the largest law firm in the East Bay representing businesses throughout the Bay Area and the state, welcomes attorney Patrick Tuck as an associate in the firm's Land Use and Litigation practices.
Tuck advises and defends private and public clients on a broad spectrum of land use and environmental law issues. He is experienced in a range of matters including, but not limited to, California Environmental Quality Act (CEQA) compliance and litigation, water rights disputes, property acquisition and development, and property entitlement procedures, including obtaining local government approvals. He has successfully counseled and defended many public entity clients, including multiple California municipalities, school districts, and police departments. During law school, Tuck clerked for the U.S. Department of Justice's Environment & Natural Resources Division and interned with the San Francisco City Attorney's Land Use & Environment Team. He joins the firm from Bertrand, Fox, Elliot, Osman & Wenzel's San Francisco office.
Tuck earned his J.D., magna cum laude, from the University of San Francisco School of Law (2015), where he served on the USF Law Review Editorial Board. He earned his B.A. in International Relations from Tufts University (2002).
Patrick Tuck
Land Use , Litigation
© 2021 Wendel Rosen LLP
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Description about my music education organization's philosophy, experience, community, curriculum, and faculty:Accelerando believes that teaching music is not just a profession, but a passion. The strongest teacher is one who is not only accomplished on his/her instrument, but who is also trained in HOW to teach, and loves doing so. Katie Ribik is an accomplished pianist/guitarist with 20 years playing experience. She has a BA in Music from UVA & a Music Ed Teaching Certificate from Montclair University.


I was astonished this morning and bursting with pride. God it's good to be a Spurs fan for moments like these. In comments received on this site and seen on Twitter, Harry Redknapp and Peter Crouch take a fair amount of criticism, yet last night both provided evidence of their unquestioned ability.
Harry was brave. Very brave. In his team selection he went for Pienaar over Kranjcar and continued with Sandro in midfield. Having scored two goals in his last two appearances dropping Niko was a big decision but not as big as continuing with the Brazilian protégé.
But they both came off. In the same way that Crouchy benefits from the space and pace of the Champions League games so it was for Sandro and he put in a very good performance in the centre of the midfield. I did think that Pienaar was our most disappointing player on the night but enough of negatives when there was so much worthy of superlative on show.
In all honesty this was one of the best Spurs performances I have ever seen. Our control of the game from the off and professionalism throughout the 96 minutes played made this a thoroughly deserved victory. Yes I've seen periods when we've overrun a side and scored more goals but to play with such maturity and patience throughout is hard to recollect.
At half time all it needed was a goal to round off an excellent start to the match and there was the worry that AC Milan had to be better in the second period. They did step up their game after the break but it also looked like they had been shown a DVD of late sixties/early seventies Italian football at half time for there were a number of disgraceful moments in the second half none more so than Flamini's appalling challenge on Corluka which would have warranted an immediate red card at home.
Quite how 5 officials could not decide that defies logic. The way he launched himself through the challenge made it close to assault. Charlie has played well since his return to the side and apart from one slip in the first half had been doing extremely well linking up with Lennon down the right.
For me Wilson Palacios was outstanding. As a regular at White Hart Lane I've seen the best and worst of the man. Great in the challenge but a worry with the ball. His confidence has been growing and last night he stamped his authority upon the game, won everything and made legends like Seidorf know he was playing. Best of all he was comfortable in possession; he never dived in or took people out with late challenges. It was great to watch.
What really sealed it for me was the way in which the central midfield players assisted the defence. All too often we've got turned over because we don't defend properly all over the field but last night was a lesson for future matches. And even when Milan upped the tempo we had men in position to get the ball away.
Behind them no injury or slip ups from Gomes, on his birthday, and a couple of great saves to boot with Gallas adding a spectacular overhead kick following one in getting the ball clear. Dawson was strong exacting a variety of different headers, Gallas calm and assured and Woodgate, a massively brave substitution from Harry, collected whilst at left back Ekotto played well at times running the whole line by himself.
Rafa had a lively 60 minutes dropping deep to pick up the ball and working to keep the ball in our possession and was unlucky on two occasions, firstly forcing a decent save from the reserve keeper and then narrowly wide with a sublime chip. He was replaced by Modric who was lively and played the killer ball to Lennon that led to the goal.
Lennon had a good night. Lively in the first half you could say his final ball lacked but for me he comes into his own in the second half when the game offers space on the break. At half time I fancied this might be so as in the Liverpool game and so it proved. Thankfully he got past the last man – who tried to take him out by the waist – before squaring for Crouch to slot home. He has played well recently but still sits in Bale's shadow!
Finally, Peter Crouch. The much maligned Peter Crouch. The man who performs well in the Champions League and international level and scored our winner in our last big 0-1 win at Manchester City to get us into the Champions League in the first place. Frankly, they could not handle him and he provided us with an outlet to aim at all night. He took the goal well for it was missable to send the gathered Spurs fans into raptures.
And so at half time in the tie we are ahead with a valuable away goal. If we take the game to them at White Hart Lane we can go through but make no mistake this tie is not yet over. But that's not for today for this was an astounding victory and fully deserved.
I watched it in a pub in Exeter on my own but in touch by text and twitter. I telephoned the boy on the way back to the hotel and heard Harry in the box with the Sky team with Spurs fans singing in the background. I'll remember that for a long time. When he first started supporting Spurs we were to be fair bordering on rubbish and he's seen all the improvement and a fair bit of adversity along the way. Wins like last night don't come along every day and you have to savour them when they do.
Where from : The Lane.
Maybe all of Europe has got this message after last night's victory. Harry, thanks for that.
A tenth anniversary party – not mine I might add – prevented me from watching a feed of the Blackburn Rovers game but I did manage to pick up the Sky Football First highlights. On route to said event I had picked up the news about Luka Modric's appendix operation and after feeling more concerned then ever about our chances wondered whether Rafa would drop back and we'd play two strikers up front or put in Kranjcar or Pienaar.
In the end Pienaar was also injured and so Lennon again played on the left and Rafa went out right allowing the fit again Palacios to come into the centre with Jenas. Not the midfield we would have predicted but maybe one that was suited to the occasion. Crouch came in up front to partner Defoe and on three minutes rose ahead of Samba and headed Rafa's perfectly hit cross from deep beyond Robinson to put us ahead.
After that our chances were few and far between but we were able to stem the tide for most of the game and when we didn't Gomes was playing well and making a string of brilliant second half saves. We had two decent chances in the second half both created by Lennon. The first involved a tremendous break which saw him sprint from box to box before laying the ball to Defoe who finding the ball on his left foot allowed Robinson to make a save.
The second involved a tremendous shot from Lennon after cutting inside and forcing Robinson into an excellent save following a deflection from a Blackburn player.
In defence Charlie returned and received a yellow card after he got caught out and tugged his man to the floor. He was also unlucky when he was adjudged to have brought his man down on the edge of the box when it was clearly not a foul. After the pounding at Fulham his more defensive outlook maybe benefitted the side and at left back Benny appeared to holding himself back.
However, from what I saw Billy Gallas looked outstanding. Having taken back the armband he threw himself into the game and one brilliant sliding tackle in the second half epitomised his performance. He picked up a knock in that challenge but with us being down to bare bones on the bench he continued and along with Bassong helped keep a clean sheet despite a few dodgy moments near the end.
How this might have been the result of JJ's knee is open to debate but a clean sheet is a clean sheet and once again Crouchy is on the score sheet but JD is not. He gave way to Sandro who saw the end of the game. Wanting to impress the young man crossed the ball late in the game conceding possession and allowing Blackburn to attack us once again – or so the boy told me who watched the latter stages on a feed.
So Gomes played well, we kept a clean sheet, the makeshift midfield worked on the night, a striker scored and Woody made the bench. And the biggest thing for me is points in Lancashire! I've moaned about our performances in the North West for most of the season and to prove my point I've included a map of the UK showing our points tallies so far (click link to load). It is unfortunate that some of the best sides are from the North West but so too are some of the middle and lower tier sides and all too often we let ourselves down on our travels to their stadiums.
As a consequence last night was a bit of a triumph. It wasn't a great performance but does it always need to be when we get three points and Manchester City does not!


For new vehicles the best way of proving the vehicles meet the required standards is with a statement of compliance. In most cases the manufacturer or their agents will supply this automatically. But if you're importing your vehicle from any source other than the manufacturer or their agents, you'll need to ensure you get the necessary evidence from the recognised authorities.
In most cases, if you're sourcing your new vehicle directly from the manufacturer or their agents, they'll provide you with all the information you need automatically. Usually this will be in the form of a statement of compliance, though it may be in the form of a compliance plate. See plate examples from Australia, Europe and the United States.
You may also need test certificate information that shows it was manufactured to meet the required standards. Entry certifiers can advise whether you need this.
The manufacturer's representative must provide documentation that proves the vehicle was manufactured to meet an applicable emissions standard. Learn more about the emissions requirement.
The evidence you supply to show compliance, eg a statement of compliance, must include proof that the vehicle was manufactured to meet an applicable emissions standard.
The manufacturer's representative must provide fuel consumption information for new vehicles to the Motor Vehicle Register in the form set out in the Land Transport Rule: Fuel Consumption Information 2008.
Most of the manufacturer's authorised agents have the authority to carry out the entry certification so you can go straight to step 4.
Private or parallel imports are new vehicles that you buy from a source other than an authorised manufacturer's representative.
you are the legal owner of the vehicle.
You can take your vehicle to the same entry certifiers as for used vehicles. They'll be able to assess your vehicle's compliance with emissions standards based on their experience of similar used models of your vehicle.
You can use the online tool(external link) to provide the fuel consumption compliance information. You'll then receive a fuel consumption certificate which you take to the entry certifier.
If your fuel consumption certificate says consumption is 'unknown', this is still acceptable evidence to give to the entry certifier.


MCE Coaching as part of a learning solution aims to enhance the learning process, increase self-awareness and stimulate development planning.
It also contributes to the transfer of learning by improving leadership effectiveness and impact. By identifying and focusing on specific skills, coaching helps transform the individual's abilities and produces a rapid shift in behaviours in the workplace.
MCE has a network of experienced coaches and mentors who can provide a flexible range of options for you and your people as part of a learning solution.
We offer a balanced range of coaching or mentoring support that is flexible and connected.
This is an in-depth look with you into your strengths, qualities and preferences, helping you work out how to be your best. We use a psychometric assessment to add to the deep conversation with your coach, so you can discover more about your own potential.
A Deep Discovery session takes up to half a day to explore your profile and preferences.
This is included in some of our training programmes, however, it can also be booked independently as a 'stand-alone' discovery.
Our programme coaching sessions are one-hour personal calls from a coach. You might have taken a deep discovery session and want to work on specific personal goals, or you might want to work with a coach on your own personal development agenda.
We recommend between 3 and 6 initial coaching calls and then we review outcomes with you.
Middle and Senior Leaders often benefit from a longer coaching relationship that focuses on developing specific leadership behaviours or thinking. Our Executive Coaches usually work with you on your personal style as a leader, challenging you to examine your beliefs and approach to your leadership, and helping you to create a style that will build your business.
Some people want to get advice, guidance and knowledge on how to do a job. You are probably in a new role, or about to move into a new role, and are unsure about what you need to know, or how to find out what you need to know. A mentoring style of relationship is often the most suitable. This relationship is more focused on giving you job knowledge and information than in the coaching relationship. We will help you decide what style of relationship is best for you, and which member of our team can offer you a coaching and mentoring relationship.
We are asked to help a work-team to develop a specific set of group skills, or action learning on a specific workplace issue. This is not linked to a training programme, but is a free-standing group coaching relationship to help a group of managers or employees work on a group related issue or need.
Coaching or Mentoring? Which Is the Right Solution for You?
Coaching is helping you to achieve your own potential and become a balanced and successful person. A coach will know how to get you to your very best.
A coach is someone who can help you achieve your best through new ways of thinking and behaving. Our coaches often work with you to overcome any barriers or beliefs you possess, and that you want to change. Coaching is person centred and person focused.
Mentoring is helping a person to prepare for the next career or job, through guidance, advice, sharing business knowledge, or helping you navigate your business or industry.
Our coaches are focused on helping you to find and use your potential today, and work on your current personal development. Our mentors will be focused on helping you with business knowledge, management skills, and how to navigate your business world.
Our experience tells us that sometimes you need a coach, sometimes you need a mentor, and sometimes you need someone who is both.
Our aim is to help you succeed, and that could be as your coach, or as your mentoring, or as both!
The focus is on helping you achieve your best through new ways of thinking and behaving.
Our coaches often work with you to overcome any barriers or beliefs you possess, and that you want to change.
A coach will know how to get you to your very best.
Coaching is person centred and person focused.
The aim is to help you achieve your own potential and become a balanced and successful person.
Our coaches will know about you!
Our mentors usually stay in contact with you as you move into a new or challenging role for a period of months.
Our mentors will share their expert business knowledge, share experiences and scenarios, and challenge you to apply new ideas or skills at work.
A mentor will know about your industry and your work.
The aim is to give you knowledge to be more effective in your career and fill gaps in your knowledge.
Our mentors will know about your business!
The SOLVE coaching process is our unique model. At MCE, we aim to enhance organisational effectiveness with a behavioural change. Whenever possible, we support our training programmes with group coaching or individual coaching. We also offer individual or group/ team coaching services, independently of any training programme. SOLVE can achieve this. It consists of 5 stages or steps, which may or may not happen in a sequential order.
Surface – the goal of the session and the broader development objectives. Surfacing is also the ability of the coach to reveal what might not be visible to the coachee.
Observe – spoken and unspoken messages related to emerging topics, issues, concerns and barriers experienced by the coachee. Our coaches observe without personal judgments.
Listen – to help the coachee to formulate clearly the true essence of the matter in hand.
Verify what has been shared can be validated. Discover what could be forgotten and which might be as important.
Enable – a concrete outcome to work on practically and enable the coachee to succeed.