text:
\begin{document}

\title{Crossed products of Calabi-Yau algebras by finite groups}
\author{Patrick Le Meur}
\address{CMLA, ENS Cachan, CNRS, UniverSud,
61 Avenue du President Wilson, F-94230 Cachan, France}
\email{patrick.lemeur@cmla.ens-cachan.fr}

\subjclass[2000]{Primary 18G20, Secondary 16W22}

\keywords{Calabi-Yau algebra; crossed product; quiver
  with potential}

\begin{abstract}
Let a finite group $G$ act on a differential graded Calabi-Yau algebra
$A$ over a field
whose characteristic does not divide the order of $G$. This note
studies when the crossed-product of $A$ by $G$ is still
Calabi-Yau. Under a compatibility condition between the action of $G$
on $A$ and the
Calabi-Yau structure of $A$, it is proved that this is indeed the
case. Similar results are proved for general constructions of
Calabi-Yau algebras such as deformed Calabi-Yau completions and
Ginzburg algebras.
\end{abstract}

\maketitle

\section*{Introduction}

Ginzburg introduced Calabi-Yau algebras in \cite{G06} as non
commutative versions of coordinate rings of Calabi-Yau varieties. They
have been object of much interest recently. For instance, in the study of
Donaldson-Thomas invariants, Calabi-Yau algebras are associated to
brane tilings (\cite{MR10}). The algebras considered there are
Jacobian algebras, that is, the $0$-th cohomology algebra of
Ginzburg algebras $\mathcal A(Q,W)$ of quivers with potentials $(Q,W)$
 (\cite[3.6]{G06}).  The Ginzburg algebras were proved
to be Calabi-Yau in \cite{K09}. Moreover, in representation theory, Calabi-Yau
algebras satisfying some finiteness condition are used to construct the generalised cluster categories
\cite{A08} which serve to categorify cluster algebras \cite{FZ02}.
Finally, it is proved in \cite{KY09} that the mutation of quivers with
potentials in the sense of \cite{DWZ08} define equivalences between the derived categories of the
associated Ginzburg algebras. Since the definition of Calabi-Yau
algebras was given in \cite{G06}, many known algebras  have been
proved to be Calabi Yau: The algebra $\mathbb{C}[x_1,\ldots,x_n]$ of
polynomials in $n$ variables is Calabi-Yau of dimension $n$; if
$X$ is a smooth affine variety with trivial canonical sheaf, then
$\mathbb{C}[X]$ is Calabi-Yau of dimension $\ts{dim}\,X$ (see
\cite[3.2]{G06}); also the Weyl algebra
$A_n=\mathbb{C}[x_1,\ldots,x_n,\frac{\partial}{\partial
  x_1},\ldots,\frac{\partial}{\partial x_n}]$ is Calabi-Yau of
dimension $2n$ (see \cite{B09}); and, finally, the
Yang-Mills algebras (\cite{CDV02}) are Calabi-Yau of dimension $3$
(\cite{HS09}).

Let $\ts k$ be a field and $A$ be a differential graded algebra over
$\ts k$ (dg algebra, for short). The enveloping algebra of $A$ is
 the algebra $A^e=A\otimes A^{op}$. The definition of the Calabi-Yau
property is expressed in terms of the duality
\begin{equation}
  \ts{RHom}_{A^e}(-,A\otimes A)\colon \ts{per}(A^e)\to \ts{per}((A^e)^{op})\notag
\end{equation}
denoted by $M\mapsto M^\vee$, for simplicity. Here,
$\ts{RHom}_{A^e}(-,A\otimes A)$ is the total derived functor, and
$\ts{per}(A^e)$ is the perfect derived category of $A^e$, that is, the
full triangulated subcategory of the derived category of right dg
$A^e$-modules containing $A^e$ and stable under shifts in both
directions and under taking direct summands. Following \cite{G06}, the dg algebra $A$ is
called \emph{Calabi-Yau of dimension $d$} if:
\begin{enumerate}
\item $A$ is \emph{homologically smooth}, that is,
  $A\in\ts{per}(A^e)$, and
\item there exists an isomorphism $f\colon A^\vee[d]\to A$ in the
  derived category of left dg $A^e$-modules such that $f^\vee[d]=f$.
\end{enumerate}
In the last equality, right dg $(A^e)^{op}$-modules (equivalently, left dg
$A^e$-modules) are identified with right dg $A^e$-modules using the
canonical anti-involution of $A^e$. Recall that if $A$ is an algebra
concentrated in degree $0$ (with zero differential) then condition
($1$) above means that the $A$-bimodule $_AA_A$ has a resolution of
finite length by finitely generated projective $A$-bimodules, and
condition ($2$) gives the following isomorphisms of $A$-bimodules:
\begin{equation}
  \ts H^i(A;A\otimes A)\overset{\ts{def}}{=}
  \ts{Ext}_{A^e}^i(A,A\otimes A)\simeq \left\{
    \begin{array}{ll}
      0 & \text{if $i\neq d$,}\\
      A & \text{if $i=d$}
    \end{array}
\right.\notag
\end{equation}
where $\ts H^i(A;-)$ denotes the Hochschild cohomology of $A$.
For a Calabi-Yau algebra $A$, the complex $A^\vee$ plays an important
rôle. If $\theta_A\to A^\vee$ is a cofibrant resolution (of left dg
$A^e$-bimodule), then $\theta_A$ is called the \emph{inverse dualizing
complex} of $A$ (\cite{V98}). It is unique up to quasi-isomorphism.

This text studies the actions of finite groups on dg Calabi-Yau
algebras. The motivation for this work is the following. Let $X$ be a
smooth affine Calabi-Yau variety and $G$ be a finite group acting on
$X$. Then, the quotient variety $X/G$ has singularities (if $G\neq
1$), $G$ acts on $\mathbb{C}[X]$ and $\mathbb{C}[X/G]$ is the algebra
of invariants $\mathbb{C}[X]^G$. A crepant resolution of $X/G$ is a
resolution of singularities $f\colon Y\to X/G$ such that
$f^*\omega_{X/G}\simeq \omega_Y$ (which implies that $Y$ is itself
Calabi-Yau). Such resolutions do not always exist and are not always
unique. But it is conjectured by Bondal and Orlov that given crepant
resolutions $Y_1\to X/G$ and $Y_2\to X/G$, there exists an equivalence
$\mathcal D^b(\ts{Coh}\,Y_1)\simeq \mathcal D^b(\ts{Coh}\,Y_2)$. In
some cases, these resolutions, are related to the crossed-product
algebra (or, skew-group algebra) $\mathbb{C}[X]\rtimes G$. For
example, if $X=V$ is a vector space and
$G\underset{\ts{finite}}{<}\ts{SL}(V)$, then $\mathbb{C}[X]$ is the
symmetric algebra $S(V)$ and $S(V)\rtimes G$ is a non commutative
crepant resolution of $V/G$ in the sense of \cite{V04}; also, if
$V=\mathbb{C}^3$, then $S(V)\rtimes G$ is Morita equivalent to the
$0$-th cohomology algebra $H^0(\mathcal A(Q,W))$ for some quiver with
potential $(Q,W)$, and, for any crepant resolution $Y\to
\mathbb{C}^3/G$, there is an equivalence $\mathcal D^b(S(V)\rtimes
G)\simeq\mathcal D^b(\ts{Coh}\,Y)$ (\cite[4.4]{G06} and \cite{BKR01}).

This motivates the following question:
\begin{center}
($\mathcal Q$): \emph{Let $A$ be a dg Calabi-Yau algebra over
a field  $\ts k$ and $G$ be a finite group acting on $A$ by
  dg automorphisms such that $\ts{car}(\ts k)$ does not divide $\sharp
  G$. Is the crossed-product $A\rtimes G$ a Calabi-Yau
  algebra?}
\end{center}
Recall the following result closely related to ($\mathcal Q$): Given
an algebra $A$ (concentrated in degree $0$), if $A$ has the Van den
Bergh duality (that is, $\ts{Ext}^i_{A^e}(A,A\otimes A)$ is $0$ is
$i\neq d$ and is an invertible $A$-bimodule if $i=d$), then so does
$A\rtimes G$ (\cite[Prop. 7.1]{AFLS02}, see also \cite{F05} for a
generalisation to crossed-products of algebras by Hopf algebras). The
main result of this text is the following. Recall that the definition
of the Calabi-Yau property involves a quasi-isomorphism
$f\colon \theta_A[d]\simeq A^\vee[d]\to A$.
\begin{thm}
\label{thm}
  Let $\ts k$ be a field and $A$ be a dg algebra over $\ts k$. Let $G$
  be a finite group whose order is not divisible by $\ts{car}\,k$ and
  assume that $G$ acts on $A$ by dg automorphisms.
  \begin{enumerate}
  \item Up to an adequate choice of the inverse dualizing complex
    $\theta_A$ of $A$, the group $G$ acts on $\theta_A$ by
    automorphisms of dg modules and the inverse dualizing complex of
    $A\rtimes G$ is:
    \begin{equation}
      \theta_{A\rtimes G}=\theta_A\rtimes G\,.\notag
    \end{equation}
  \item If $A$ is Calabi-Yau of dimension $d$ and if the
    quasi-isomorphism $f\colon \theta_A[d]\simeq A^\vee[d]\to A$ is
    $G$-equivariant, then $A\rtimes G$ is Calabi-Yau of dimension $d$.
  \end{enumerate}
\end{thm}

When $A$ is concentrated in degree $0$, part ($1$) was proved in
\cite[Thm. 17, Ex. 21]{F05} and it is easy to check that part ($2$)
follows from the same result (see also \cite{AFLS02}).
Unfortunately, the proofs in \cite{AFLS02} and \cite{F05} are based on
Stefan's spectral sequence on Hochschild cohomology for Hopf-Galois
extensions (\cite{S95}) so their proof cannot apply directly to dg
algebras not concentrated in degree $0$ (for
instance, all Calabi-Yau algebras used to construct generalised
cluster categories). Therefore, in order to prove
the above theorem, different techniques are needed. Recently,
Witherspoon introduced methods to study the Hochschild cohomology ring
of the crossed-products of algebras by finite groups (\cite{W04}). It
appears that the techniques introduced there can be adapted to dg
algebras and applied to prove Theorem~\ref{thm}.

The text is organised as follows. Section~$1$ fixes notation and all
the definitions. Section~$2$ proves Theorem~\ref{thm}. Section~$3$
studies ($\mathcal Q$) for dg Calabi-Yau algebras arising from a general
construction by Keller and called deformed Calabi-Yau completion
(\cite{K09}): Given a homologically smooth dg algebra $A$, an integer
$d\geqslant 3$ and $c\in \ts{HH}_{d-2}(A)$ (the Hochschild homology of
$A$ with coefficients in $A$), the construction defines a dg Calabi-Yau algebra of dimension
$d$ and denoted by $\Pi_d(A,c)$. This section proves that if $G$ acts
on $A$ and if $c\in\ts{HH}_{d-2}(A)^G$, then $G$ acts naturally on
$\Pi_d(A,c)$ and $\Pi_d(A,c)\rtimes G$ is itself a deformed Calabi-Yau
completion and, therefore, is Calabi-Yau. Section~$4$ applies the
results of Section~$3$ to crossed products of Ginzburg algebras
by finite groups. In particular, sufficient conditions are given for
the crossed-product of a Ginzburg algebra by a finite group to have a
derived category equivalent to that of a Ginzburg algebra.
 Finally, Section~$5$ comments the equivariance
hypothesis of Theorem~\ref{thm}, it appears that this hypothesis is of
geometric nature.

\section{Definitions and notation}
\label{sec:not}
Throughout the text, $\ts k$ will denote a field.
\subsection{Differential graded algebras and their modules}

Tensor product over $\ts k$ is denoted by $\otimes$. The reader is
referred to \cite{K06} for a background on differential graded
algebras (dg algebras, for short). All dg algebras are over $\ts k$
and with differential of degree $1$. Let $A$ be a dg algebra. Its
\emph{opposite algebra} has $A$ as underlying complex of vector spaces
and its product map is defined by $A^{op}\otimes A^{op}\to
A^{op},\,a\otimes b\mapsto (-1)^{\ts{deg}(a)\ts{deg}(b)}ba$, for
homogeneous $a,b\in A$ (and $ba$ is the product in $A$). If $B$ is
another dg algebra, then the graded vector space $A\otimes B$ is a dg
algebra with product defined by $(a\otimes b).(a'\otimes
b')=(-1)^{\ts{deg}(a')\ts{deg}(b)}aa'\otimes bb'$ and differential
defined by $d(a\otimes b)=d(a)\otimes b+(-1)^{\ts{deg}(a)}a\otimes
d(b)$, for homogeneous $a,a'\in A$ and $b,b'\in B$. In particular, the
\emph{enveloping algebra} of $A$ is the dg algebra $A^e=A\otimes
A^{op}$. Note that the map $A^e\to (A^e)^{op}$ defined by $a\otimes b\mapsto
(-1)^{\ts{deg}(a)\ts{deg}(b)}b\otimes a$, for homogeneous $a,b\in A$,
is an isomorphism of dg algebras.

The category of (right) \emph{differential graded (dg) $A$-modules} is
denoted by $\mathcal C(A)$: Its objects are the right dg $A$-modules
(with differential of degree $1$) and its morphisms are the
homogeneous of degree $0$ morphisms of graded $A$-modules commuting
with the differentials. The suspension in $\mathcal C(A)$ is denoted
by $M\mapsto M[1]$. For convenience, $\mathcal C(A^{op})$ will be
considered as the category of left $A$-modules: If $M$ is a right dg
$A^{op}$-module and $a\in A$, $m\in M$ are homogeneous, then $a.m$ is
defined as $(-1)^{\ts{deg}(a)\ts{deg}(m)}m.a$. Similarly, $\mathcal
C(A\otimes B^{op})$ will be considered as the category of dg
$B-A$-bimodules (or, simply, $A$-bimodules, if $A=B$): If $M$ is a
right dg $A\otimes B^{op}$-module and
$a\in A$, $b\in B$, $m\in M$ are homogeneous, then $b.m.a$ is defined
as $(-1)^{\ts{deg}(b).(\ts{deg}(m)+\ts{deg}(a))}m.(a\otimes b)$. Note
that the canonical isomorphism $A^e\to (A^e)^{op}$ induces an
isomorphism between $\mathcal C(A^e)$ and $\mathcal C((A^e)^{op})$.

For example, $A$ is a dg $A^e$-module for the action defined by
$x.(a\otimes b)=(-1)^{\ts{deg}(b)(\ts{deg}(a)+\ts{deg}(x))}bxa$ for
homogeneous $x,a,b\in A$. Therefore, it is also a left dg
$A^e$-module, for the action defined by $(a\otimes
b).x=(-1)^{\ts{deg}(b)\ts{deg}(x)}axb$. Also, the product in $A^e$
endows $A\otimes A$ with a structure of $A^e-A^e$-bimodule.

Let $L\in\mathcal C(A)$ and $M\in\mathcal C(A\otimes B^{op})$. Then
$\mathcal Hom_A(L,M)$ will denote the left dg $B$-module defined as
follows: Its component of degree $n\in\mathbb{Z}$ is the vector space
of homogeneous of degree $n$ morphisms $f\colon L\to M$ of graded
$A$-modules; the differential of $f$ is $d_M\circ f-(-1)^nf\circ d_L$
(if $d_L$ and $d_M$ denote the differentials of $L$ and $M$,
respectively); and the action of $b\in B$ on $f$ is $b.f\colon
x\mapsto b.f(x).1$. If $M\in\mathcal C(A\otimes B^{op})$ is fixed,
this defines the functor $\mathcal Hom_A(-,M)\colon\mathcal C(A)\to
\mathcal C(B^{op})$. The associated \emph{right total derived functor}
is denoted by $\ts{RHom}_A(-,M)\colon\mathcal D(A)\to \mathcal
D(B^{op})$. Recall that it may be computed on $L\in\mathcal D(A)$ as
follows: Take a cofibrant resolution $X\to L$ in $\mathcal C(A)$, then
$\ts{RHom}_A(L,M)=\mathcal Hom_A(X,M)$.

\subsection{Duality for bimodules, inverse dualizing complex and the
  Calabi-Yau property} Let $A$ be dg algebra. The total
right derived functor $\ts{RHom}_{A^e}(-,A\otimes A)\colon\mathcal
D(A^e)\to \mathcal D((A^e)^{op})$ induces, by restriction, a duality
(see \cite[3.2]{G06})
\begin{equation}
  \ts{RHom}_{A^e}(-,A\otimes A)\colon \ts{per}(A^e)\to \ts{per}((A^e)^{op})\,.\notag
\end{equation}

For convenience, this duality will sometimes be denoted by $M\mapsto M^\vee$.

Let $X\to A$ be a cofibrant resolution in $\mathcal C(A^e)$. Then
$\ts{RHom}_{A^e}(A,A\otimes A)=\mathcal Hom_{A^e}(X,A\otimes A)$. Let
$\theta_A\to \ts{RHom}_{A^e}(A,A\otimes A)$ be a cofibrant resolution
in $\mathcal C((A^e)^{op})$. Then $\theta_A$ is unique up to
quasi-isomorphism and is called the \emph{inverse dualizing complex}
of $A$ (\cite{V97}).

Let $d$ be an integer. Following \cite[3.2.3]{G06}, the dg algebra $A$
is called \emph{Calabi-Yau of dimension $d$} if the two following
conditions hold:
\begin{enumerate}
\item $A$ is \emph{homologically smooth}, that is,
  $A\in\ts{per}(A^e)$,
\item there is an isomorphism $f\colon\ts{RHom}_{A^e}(A,A\otimes
  A)[d]\to A$ in $\mathcal D((A^e)^{op})$ such that
  $\ts{RHom}_{(A^e)^{op}}(f,A\otimes A)[d]=f$.
\end{enumerate}

In the last equality, left and right dg $A^e$-modules are identified
using the canonical isomorphism $A^e\to (A^e)^{op}$, and
$A^{\vee\vee}$ is identified with $A$ (because $A\in\ts{per}(A^e)$),
so that $\ts{RHom}_{(A^e)^{op}}(f,A\otimes A)$ is indeed a morphism
$A^\vee[d]\to A$ in $\mathcal D((A^e)^{op})$.

 \subsection{Crossed-products} Let $G$ be a finite group and $A$ be a
 dg algebra. As usual,
 conjugation of an element $g\in G$ by an element $h\in G$ is denoted
 by $^hg=hgh^{-1}$.
 An action of
 $G$ on $A$ (by dg automorphisms) is an action $(g,a)\mapsto \,^ga$
 of $G$ on $A$ by automorphisms of algebras such that:
\begin{enumerate}[-]
\item $\,^ga$ is homogeneous and $\ts{deg}(\,^ga)=\ts{deg}(a)$ if
  $a\in A$ is homogeneous,
\item the differential $d$ of $A$ verifies $d(\,^ga)=\,^gd(a)$,
\end{enumerate}
for $a\in A$ and $g\in G$. This defines the
 \emph{crossed-product} (or \emph{skew-group}) dg algebra $A\rtimes G$ as follows:
As
 a complex of vector spaces it equals $A\otimes \ts k G$ where $\ts k
 G$ is concentrated in degree $0$; the product is defined by
 $(a\otimes g)\cdot (b\otimes h)=a\,^gb\otimes gh$; and the
 differential is defined by $d(a\otimes g)=d(a)\otimes g$, for $a,b\in A$ and
 $g,h\in G$. As usual, $A$ and $\ts kG$ are considered as subalgebras
 of $A\rtimes G$ and $a\otimes g$ is written $ag$, for short. Hence,
 $ga=\,^gag$, for $a\in A$ and $g\in G$.

\section{The main theorem}
\label{sec:s2}

This section proves Theorem~\ref{thm}. The first paragraphs introduce
the needed material for the proof, it is inspired from the work of
Witherspoon in \cite{W04}.
\subsection{The dg algebra $\Delta$}
\label{sec:s2.1}

Let $g\in G$. Denote by $Ag=gA$ the subcomplex of vector
spaces of $\Lambda$ equal to $\{ag\ |\ g\in G\}=\{ga\ |\ g\in
G\}$. Then $Ag=gA$ is a dg $A^e$-submodule of $\Lambda$ and $A\simeq
Ag$ both in $\mathcal C(A)$ and $\mathcal C(A^{op})$ (but not in
$\mathcal C(A^e)$). Note that $\Lambda=\bigoplus\limits_{g\in G}Ag$
in $\mathcal C((A^e)^{op}\otimes A^e)$.

Let $g,h\in G$. Then, $Ag\otimes hA$ is a dg $A^e$-sub-bimodule of
$\Lambda\otimes \Lambda$. There are obvious isomorphisms $A\otimes
A\simeq Ag\otimes hA$ in $\mathcal C(A^e)$ and $\mathcal
C((A^e)^{op})$. Note that
$\Lambda\otimes\Lambda=\bigoplus\limits_{g,h\in G}Ag\otimes hA$ both
in $\mathcal C(A^e)$ and in $\mathcal C((A^e)^{op})$. Besides, the
product in $\Lambda^e$ induces a map $(Ag\otimes hA)\otimes
(Ag'\otimes h'A)\to Agg'\otimes h'hA$, for all $g,g',h,h'\in G$. In
particular, $\bigoplus\limits_{g\in G} Ag\otimes g^{-1}A$ is a dg
subalgebra of $\Lambda^e$. In the sequel, it will be denoted by
$\Delta$:
\begin{equation}
  \Delta=\bigoplus\limits_{g\in G}Ag\otimes Ag^{-1}.\notag
\end{equation}
Note that $A^e$ is a dg subalgebra of $\Delta$.

The dg algebra $\Delta$ has the following simple interpretation: The
group $G$ acts diagonally on the dg algebra $A^e$: $^g(a\otimes
b)=\,^ga\otimes \,^gb$, for $a,b\in A$ and $g\in G$. For this action,
the map $\Delta\to A^e\rtimes G,\,ag\otimes g^{-1}b\mapsto (a\otimes b).g$ is
an isomorphism of dg algebras. The dg $\Delta$-modules also have a
similar interpretation. By definition, \emph{an action of $G$ on
  $M\in\mathcal C(A^e)$ compatible with the action of $A^e$} is an
action $G\times M\to M,\,(g,m)\mapsto \,^gm$ of $G$ on $M$ by
automorphism of complexes of vector spaces such that $^g(m.(a\otimes
b))=\,^gm.(\,^ga\otimes\,^gb)$, for $a,b\in A$, $m\in, M$ and $g\in
G$. Then, the dg $\Delta$-modules are precisely the dg $A^e$-modules
$M$ endowed with a compatible action of $G$ (with $m.(g\otimes
g^{-1})=\,^{g^{-1}}m$, for $m\in M$ and $g\in G$). Similarly, one
defines the notion of \emph{compatible action of $G$ on left dg
  $A^e$-modules}, and left dg $\Delta$-modules are precisely left dg
$A^e$-modules endowed with a compatible action of $G$ (with $(g\otimes
g^{-1}).m=\,^gm$, for $m\in M$ and $g\in G$).

In the sequel, the following collection $\{M_g\}_{g\in G}$ of dg
$A^e-\Delta$-bimodules will be useful. Let $g\in G$. The complex of
vector spaces $\bigoplus\limits_{h\in G}Ah\otimes h^{-1}gA$ is a dg
$A^e-\Delta$-sub-bimodule of $\Lambda^e$. It will be denoted by $M_g$:
\begin{equation}
  M_g=\bigoplus\limits_{h\in G}Ah\otimes h^{-1}gA.\notag
\end{equation}
The following lemma collects some information on these
bimodules. \emph{Here and after, products are computed in $\Lambda^e$,
unless otherwise specified}.
\begin{lem}
  Let $t\in G$.
  \begin{enumerate}[(a)]
  \item There is an isomorphism in $\mathcal C(\Delta)$:
    \begin{equation}
      \begin{array}{rcll}
        \Delta & \longrightarrow & M_t\\
        u & \longmapsto & (1\otimes t).u&.
      \end{array}\notag
    \end{equation}
  \item $\Lambda^e=\bigoplus\limits_{g\in G}M_g$ in $\mathcal
    C((A^e)^{op}\otimes\Delta)$.
  \item Let $g\in G$. Multiplication by $g\otimes g^{-1}$ induces an
    isomorphism in $\mathcal C(\Delta)$:
    \begin{equation}
      \begin{array}{rcll}
        M_t & \longrightarrow & M_{gtg^{-1}}\\
        u & \longmapsto & (g\otimes g^{-1}).u&.
      \end{array}\notag
    \end{equation}
  \item There is an isomorphism in $\mathcal C((A^e)^{op}\otimes
    \Delta)$:
    \begin{equation}
      \begin{array}{rcll}
        \mathcal Hom_{A^e}(\Delta,A\otimes tA) & \longrightarrow &
        M_t\\
        \varphi & \longmapsto & \sum\limits_{g\in
          G}\varphi(g^{-1}\otimes g).(g\otimes g^{-1})&.
      \end{array}\notag
    \end{equation}
  \end{enumerate}
\end{lem}
\begin{proof}
  Only (d) needs details. The given map is a morphism in $\mathcal
  C((A^e)^{op}\otimes \Delta)$. One easily checks that the following
  map is its inverse:
  \begin{equation}
    \begin{array}{rcll}
      M_t & \longrightarrow & \mathcal Hom_{A^e}(\Delta,A\otimes tA)\\
      \bigoplus\limits_{g\in G}Ag\otimes g^{-1}tA\ni(m_g)_g & \longmapsto
      & \left(ag^{-1}\otimes gb\mapsto  m_g.(ag^{-1}\otimes gb)\right)&,
    \end{array}\notag
  \end{equation}
where, on the right-hand side, $a,b\in A$ and $g\in G$.
\end{proof}

\subsection{Extensions of scalars between $\mathcal C(A^e)$, $\mathcal C(\Delta)$ and $\mathcal C(\Lambda^e)$}
\label{sec:s2.2}

Recall (\cite[3.8]{K06}) the following adjunctions:
\begin{enumerate}[-]
\item $-\underset{A^e}{\otimes}\Lambda^e\colon \mathcal C(A^e)\to
  \mathcal C(\Lambda^e)$ is left adjoint to the restriction-of-scalars
  functor,
\item $-\underset{A^e}{\otimes}\Delta\colon\mathcal C(A^e)\to\mathcal
  C(\Delta)$ is left adjoint to the restriction-scalars-functor,
\item $-\underset{\Delta}{\otimes}\Lambda^e\colon\mathcal C(\Delta)\to
  \mathcal C(\Lambda^e)$ is left adjoint to the restriction-of-scalars
  functor,
\item $\mathcal Hom_{A^e}(\Delta,-)\colon\mathcal C(A^e)\to \mathcal
  C(\Delta)$ is right adjoint to the restriction-of-scalars functor.
\end{enumerate}
Of course, similar adjunctions hold true for left dg modules. In
particular, $\Lambda^e\underset{\Delta}{\otimes}-\colon\mathcal
C(\Delta^{op})\to\mathcal C((\Lambda^e)^{op})$ has the following
well-known interpretation that will be used later. Let $M$ be a left
dg $A^e$-module with a compatible action of $G$. Equivalently
(\ref{sec:s2.1}), it is naturally a left dg $\Delta$-module. Then, the
\emph{crossed-product $M\rtimes G$ of $M$ by $G$} is the left dg
$\Lambda^e$-module constructed as follows: As a complex of vector
spaces, it is equal to $M\otimes \ts kG$ (with $\ts{deg}(m\otimes
g)=\ts{deg}(m)$, for $g\in G$ and homogeneous $m$); and the action of $\Lambda^e$ on $M\rtimes G$ is
defined by $(a\otimes b).(m\otimes g)=((a\otimes\,^gb).m)\otimes g$ and
$(h\otimes k).(m\otimes g)=\,^hm\otimes hgk$, for $a,b\in A$, $m\in M$
and $g,h,k\in G$. Then, $M\rtimes G\simeq
\Lambda^e\underset{\Delta}{\otimes} M$ in $\mathcal
C((\Lambda^e)^{op})$. Actually, the two following maps are
isomorphisms in $\mathcal C((\Lambda^e)^{op})$ and inverse to each
other:
\begin{equation}
  \begin{array}{rcll}
    M\rtimes G & \longrightarrow &
    \Lambda^e\underset{\Delta}{\otimes}M\\
    m\otimes g & \longmapsto & (1\otimes g)\otimes m &,\ \text{and}\\
\Lambda^e\underset\Delta\otimes M & \longrightarrow & M\rtimes G\\
(ag\otimes hb)\otimes m & \longmapsto & ((a\otimes
\,^{gh}b).\,^gm)\otimes gh&,
  \end{array}\notag
\end{equation}
where $a,b\in A$, $m\in M$ and $g,h\in G$.

The adjunctions recalled at the beginning of the paragraph satisfy the
following properties that will be useful later. Notice that (d) was
first proved in \cite[Lem. 3.3]{B92} for group algebras and in
\cite[Lem. 3.5]{W04} for crossed-products of algebras. The proof for
 is  recalled
for convenience.
\begin{lem}
  Denote by $N$ the order of $G$.
  \begin{enumerate}[(a)]
  \item There are isomorphisms
    \begin{equation}
      \left\{
        \begin{array}{ll}
          \Delta\simeq (A\otimes A)^N & \text{in $\mathcal C(A^e)$,
            and}\\
          \Lambda\otimes\Lambda\simeq \Delta^N & \text{in $\mathcal C(\Delta)$.}
        \end{array}\right.\notag
    \end{equation}
\item The following functors preserve quasi-isomorphisms and
  componentwise surjective morphisms:
  \begin{enumerate}[-]
  \item $\mathcal Hom_{A^e}(\Delta,-)\colon\mathcal C(A^e)\to \mathcal
    C(\Delta)$,
  \item $-\underset{A^e}{\otimes}\Lambda^e\colon \mathcal C(A^e)\to
    \mathcal C(\Lambda^e)$,
  \item $-\underset{\Delta}{\otimes}\Lambda^e\colon\mathcal
    C(\Delta)\to \mathcal C(\Lambda^e)$.
  \end{enumerate}
\item The following functors preserve cofibrant objects and cofibrant
  resolutions:
  \begin{enumerate}[-]
  \item $-\underset{A^e}{\otimes}\Delta\colon\mathcal
    C(A^e)\to\mathcal C(\Delta)$,
  \item $-\underset{\Delta}{\otimes}\Lambda^e\colon\mathcal
    C(\Delta)\to\mathcal C(\Lambda^e)$,
  \item the restriction-of-scalars functor $\mathcal
    C(\Delta)\to\mathcal C(A^e)$.
  \end{enumerate}
\item The following map is an isomorphism in $\mathcal C(\Lambda^e)$:
  \begin{equation}
    \begin{array}{rcl}
      A\underset{\Delta}{\otimes}\Lambda^e & \longrightarrow &
      \Lambda\\
      a\otimes (x\otimes y) & \longmapsto & a.(x\otimes y)=(-1)^{\ts{deg}(y).(\ts{deg}(x)+\ts{deg}(a))}yax
    \end{array}\notag
  \end{equation}
(defined on homogeneous $a\in A$ and $x,y\in \Lambda$).
\item Let $M\in\mathcal C(\Delta)$, then:
  \begin{enumerate}[-]
  \item $M$ is a direct summand of
    $M\underset{\Delta}{\otimes}\Lambda^e$ in $\mathcal C(\Delta)$,
  \item $M$ is a direct summand of $M\underset{A^e}{\otimes}\Delta$ in
    $\mathcal C(\Delta)$ if, moreover, $N\in\ts k^\times$,
  \item if $M$ is cofibrant in $\mathcal C(\Delta)$, then so is it in
    $\mathcal C(A^e)$. The converse holds true if $N\in\ts k^\times$.
  \end{enumerate}
  \end{enumerate}
\end{lem}
\begin{proof}
  (a) follows directly from the discussion made in \ref{sec:s2.1}.

(b) follows from (a). Notice that all restriction-of-scalars functors
preserve quasi-isomorphisms and componentwise surjective morphisms.

(c) Recall the following basic fact on adjunctions: If $B$ and $C$ are
dg algebras and $F_1\colon\mathcal C(B)\to
\mathcal C(C)$ and $F_2\colon\mathcal C(C)\to\mathcal C(B)$ are
additive functors such that $(F_1,F_2)$ is adjoint and $F_2$ preserves
morphisms which are both quasi-isomorphisms and componentwise
surjective, then $F_1$ preserves cofibrant objects. This and (b) show
that (c) holds true.

(d) The given map is a morphism in $\mathcal C(\Lambda^e)$. It is
straightforward  to check that the following map is its inverse:
\begin{equation}
  \begin{array}{rcll}
    \Lambda & \longrightarrow & A\underset{\Delta}{\otimes}\Lambda^e\\
ag & \longmapsto &a\otimes (g\otimes 1)&.
  \end{array}\notag
\end{equation}

(e) Recall that $M_t=\bigoplus\limits_{g\in G}Ag\otimes
g^{-1}tA\in\mathcal C((A^e)^{op}\otimes \Delta)$ was defined in
\ref{sec:s2.1}. Then, both $M_1$ and $M_{\neq 1}:=\bigoplus\limits_{t\neq 1}M_t$ are dg
$\Delta$-sub-bimodules of $\Lambda\otimes \Lambda$. Therefore,
$\Lambda\otimes \Lambda=M_1\oplus M_{\neq 1}$ and $M_1=\Delta$ in
$\mathcal C(\Delta)$. Thus, $M=M\underset{\Delta}{\otimes}M_1$ is a
direct summand of $M\underset{\Delta}{\otimes}\Lambda^e$ in $\mathcal
C(\Delta)$.

Next, assume that $N\in\ts k^\times$. The functor
$-\underset{A^e}{\otimes}\Delta\colon\mathcal C(A^e)\to \mathcal
C(\Delta)$ is left adjoint to the restriction-of-scalars functor and
the counit of this adjunction evaluates at $M$ as:
\begin{equation}
  \begin{array}{crcl}
    \eta_M \colon & M\underset{A^e}{\otimes}\Delta & \longrightarrow
    & M \\
    &m\otimes u & \longmapsto & mu
  \end{array}\ \ \ \ \ \ \text{in $\mathcal C(\Delta)$.}\notag
\end{equation}
On the other hand, there is a well-defined morphism in $\mathcal
C(\Delta)$:
\begin{equation}
  \begin{array}{rcll}
    M & \longrightarrow & M\underset{A^e}{\otimes}\Delta\\
m & \longmapsto & \frac{1}{\sharp G}\sum\limits_{g\in G}
\left(m.(g\otimes g^{-1})\right)\otimes (g^{-1}\otimes g)&.
  \end{array}\notag
\end{equation}
A simple verification shows that it is right inverse to $\eta_M$. Thus,
$M$ is a direct summand of $M\underset{A^e}{\otimes}\Delta$ in
$\mathcal C(\Delta)$.

Finally, if $M$ is cofibrant in $\mathcal C(\Delta)$, then so is it in
$\mathcal C(A^e)$, because of (c). Conversely, if $M$ is cofibrant in
$\mathcal C(A^e)$ and $\sharp G\in\ts k^\times$, then
$M\underset{A^e}{\otimes}\Delta$ is cofibrant in $\mathcal C(\Delta)$,
because of (c), and it admits $M$ as a direct summand in $\mathcal
C(\Delta)$, as proved before. Thus, $M$ is cofibrant in $\mathcal C(\Delta)$.
\end{proof}

The last assertion in (e) has the following practical consequence,
where $A$ is a dg algebra acted on by a finite group $G$ whose order
is invertible in $\ts k$: Cofibrant objects in $\mathcal C(\Delta)$
are precisely cofibrant objects in $\mathcal C(A^e)$ endowed with a
compatible action of $G$; in particular, given $M\in\mathcal
C(\Delta)$, the cofibrant resolutions of $M$ in $\mathcal C(\Delta)$
are precisely the cofibrant resolutions $P\to M$ in $\mathcal C(A^e)$
which are $G$-equivariant, where $P$ is endowed with a compatible
action of $G$. Of course, if $\sharp G=0$ in $\ts k$, then a cofibrant
object in $\mathcal C(A^e)$ endowed with a compatible action of $G$
needs not be cofibrant in $\mathcal C(\Delta)$.

\subsection{Interaction between duality and extension of scalars}
\label{sec:s2.3}

This paragraph is a key-step towards a comparison of the inverse
dualizing complexes of $A$ and $\Lambda$. For this purpose, it is
convenient to introduce a functor $\mathcal C(\Delta)\to\mathcal
C(\Delta^{op})$ induced by $\mathcal Hom_{A^e}(-,A\otimes
A)\colon\mathcal C(A^e)\to \mathcal C((A^e)^{op})$.

Let $M\in\mathcal C(\Delta)$. Then, the left dg $A^e$-module structure
on $\mathcal Hom_{A^e}(M,A\otimes A)$ arises from the action of $A^e$
on $A\otimes A$ on the left. Let $g\in G$ and $f\in\mathcal
Hom_{A^e}(M,A\otimes A)$. Let $^gf\colon M\to A\otimes A$ be the map
defined by:
\begin{equation}
  \left(\forall m\in M\right)\ \ \ (\,^gf)(m)=(g\otimes
  g^{-1}).f(m.(g\otimes g^{-1})).(g^{-1}\otimes g).\notag
\end{equation}
This defines an action of $G$ on $\mathcal Hom_{A^e}(M,A\otimes A)$
compatible with the structure of left dg $A^e$-module. In other words,
the structure of left dg $A^e$-module on $\mathcal Hom_{A^e}(M,A\otimes
A)$ extends to a structure of left dg $\Delta$-module by setting
$(g\otimes g^{-1}).f=\,^gf$ for $g\in G$ and $f\in \mathcal
Hom_{A^e}(M,A\otimes A)$. This defines an additive functor $\mathcal
Hom_{A^e}(-,A\otimes A)\colon\mathcal C(\Delta)\to \mathcal
C(\Delta^{op})$ such that the following diagram commutes:
\begin{equation}
\label{d1}
  \xymatrix{
\mathcal C(\Delta) \ar[rrr]^{\mathcal Hom_{A^e}(-,A\otimes A)} \ar[d] &&&
\mathcal C(\Delta^{op}) \ar[d]\\
\mathcal C(A^e) \ar[rrr]_{\mathcal Hom_{A^e}(-,A\otimes A)} &&& \mathcal C((A^e)^{op})
}\tag{$D_1$}
\end{equation}
where the vertical arrows are the restriction-of-scalars functors,
they preserve quasi-isomorphisms and cofibrant resolutions
(\ref{sec:s2.2}).
Therefore, the preceding diagram yields a commutative diagram (up to
isomorphism of functors), by taking right total derived functors:
\begin{equation}
\label{d2}
  \xymatrix{
\mathcal D(\Delta) \ar[rrr]^{\ts{RHom}_{A^e}(-,A\otimes A)} \ar[d] &&& \mathcal D(\Delta^{op})\ar[d]\\
\mathcal D(A^e)  \ar[rrr]_{\ts{RHom}_{A^e}(-,A\otimes A)} &&& \mathcal D((A^e)^{op})
}\tag{$D_2$}
\end{equation}
where the notation ${\ts{RHom}_{A^e}(-,A\otimes A)}\colon\mathcal
D(\Delta)\to \mathcal D(\Delta^{op})$ is adopted for the right total
derived functor of $\mathcal Hom_{A^e}(-,A\otimes A)\colon\mathcal
C(\Delta)\to\mathcal C(\Delta^{op})$. This notation is justified by
(\ref{d1}).

The following result makes a comparison between
$\ts{RHom}_{A^e}(-,A\otimes A)\colon\mathcal D(\Delta)\to\mathcal
D(\Delta^{op})$ and $\ts{RHom}_{\Lambda^e}(-,\Lambda\otimes
\Lambda)\colon\mathcal D(\Lambda^e)\to\mathcal D((\Lambda^e)^{op})$,
using extensions of scalars.
\begin{lem}
  The following diagram commutes up to isomorphism of functors:
  \begin{equation}
    \label{d3}
\xymatrix{
\mathcal D(\Delta) \ar[rrr]^{\ts{RHom}_{A^e}(-,A\otimes A)}
\ar[d]_{-\underset{\Delta}{\otimes}\Lambda^e} &&& \mathcal
D(\Delta^{op}) \ar[d]^{\Lambda^e\underset{\Delta}{\otimes}-} \\
\mathcal D(\Lambda^e) \ar[rrr]_{\ts{RHom}_{\Lambda^e}(-,\Lambda\otimes
  \Lambda)} &&& \mathcal D((\Lambda^e)^{op})&.
}\tag{$D_3$}
  \end{equation}
\end{lem}
\begin{proof}
  Recall (\ref{sec:s2.2}) that
  $-\underset{\Delta}{\otimes}\Lambda^e\colon\mathcal
  C(\Delta)\to\mathcal C(\Lambda^e)$ and
  $\Lambda^e\underset{\Delta}{\otimes}-\colon \mathcal
  C(\Delta^{op})\to \mathcal C((\Lambda^e)^{op})$ preserve
  quasi-isomorphisms and cofibrant resolutions. Therefore, the
  vertical arrows of the diagram make sense and, in order to prove the
  lemma, it suffices to prove that the following diagram commutes up
  to an isomorphism of functors:
  \begin{equation}
\label{d4}
    \xymatrix{
\mathcal C(\Delta) \ar[rrr]^{\mathcal Hom_{A^e}(-,A\otimes A)}
\ar[d]_{-\underset{\Delta}{\otimes}\Lambda^e} &&& \mathcal
C(\Delta^{op}) \ar[d]^{\Lambda^e\underset{\Delta}{\otimes}-} \\
\mathcal C(\Lambda^e) \ar[rrr]_{\mathcal Hom_{\Lambda^e}(-,\Lambda\otimes
  \Lambda)} &&& \mathcal C((\Lambda^e)^{op})&.
}\tag{$D_4$}
  \end{equation}
Let $M\in\mathcal C(\Delta)$. The construction of a functorial
isomorphism $\mathcal
Hom_{\Lambda^e}(M\underset{\Delta}{\otimes}\Lambda^e,\Lambda\otimes
\Lambda)\to \Lambda^e\underset{\Delta}{\otimes}\mathcal
Hom_{A^e}(M,A\otimes A)$ will be made in three steps:
\begin{enumerate}[(a)]
\item First, construct an isomorphism $\mathcal
  Hom_{\Lambda^e}(M\underset{\Delta}{\otimes}\Lambda^e,\Lambda\otimes
  \Lambda)\to \bigoplus\limits_{g\in G}\mathcal Hom_{A^e}(M,A\otimes
  gA)$ in $\mathcal C((A^e)^{op})$, using the tools introduced in
  \ref{sec:s2.1} and \ref{sec:s2.2}.
\item Then, endow $\bigoplus\limits_{g\in G}\mathcal Hom_{A^e}(M,A\otimes
  gA)$ with a structure of left dg $\Lambda^e$-module such that the
  isomorphism of the first step actually lies in  $\mathcal C((\Lambda^e)^{op})$.
\item Finally, construct an isomorphism $\bigoplus\limits_{g\in G}\mathcal Hom_{A^e}(M,A\otimes
  gA)\to \Lambda^e\underset{\Delta}{\otimes}\mathcal Hom_{A^e}(M,A\otimes
  A)$ in $\mathcal C((\Lambda^e)^{op})$.
\end{enumerate}

(a) By adjunction, the following map is an isomorphism in $\mathcal
C((\Lambda^e)^{op})$ and functorial in $M$:
\begin{equation}
  \label{eq1}
  \begin{array}{rcll}
    \mathcal Hom_{\Lambda^e}(M\underset{\Delta}{\otimes}\Lambda^e,
    \Lambda\otimes \Lambda) & \longrightarrow & \mathcal
    Hom_{\Delta}(M,\Lambda\otimes \Lambda)\\
    f&\longmapsto&  \left(m\mapsto f(m\otimes (1\otimes 1))\right)&.
  \end{array}
\end{equation}
Moreover, $\Lambda\otimes\Lambda=\bigoplus\limits_{g\in G}M_g$ in
$\mathcal C((A^e)^{op}\otimes\Delta)$ (\ref{sec:s2.1}, (b)),
therefore:
\begin{equation}
  \label{eq2}
\mathcal Hom_{\Delta}(M,\Lambda\otimes \Lambda)=\bigoplus\limits_{g\in
  G} \mathcal Hom_{\Delta}(M,M_g)\ \ \text{in $\mathcal C((A^e)^{op})$.}
\end{equation}
On the other hand, \ref{sec:s2.1}, (d), yields an isomorphism in
$\mathcal C((A^e)^{op})$, for every $g\in G$:
\begin{equation}
  \label{eq3}
  \begin{array}{rcll}
    \mathcal Hom_{\Delta}(M,M_g) & \longrightarrow & \mathcal
    Hom_{\Delta}(M,\mathcal Hom_{A^e}(\Delta,A\otimes gA))\\
    f &\longmapsto & \left( m\mapsto \left( at^{-1}\otimes tb\mapsto\
        \text{(projection of $f(m)$ on $At\otimes
          t^{-1}gA).at^{-1}\otimes tb$}\right)\right)&,
  \end{array}
\end{equation}
where, on the right-hand side, $a,b\in A$ and $t\in G$. Finally, by
adjunction, the following map is an isomorphism in $\mathcal
C((A^e)^{op})$ and functorial in $M$, for every $g\in G$:
\begin{equation}
  \label{eq4}
  \begin{array}{rcll}
    \mathcal Hom_{\Delta}(M,\mathcal Hom_{A^e}(\Delta,A\otimes gA)) &
    \longrightarrow & \mathcal Hom_{A^e}(M,A\otimes gA)\\
    f & \longmapsto & \left(m\mapsto (f(m))(1\otimes 1)\right)&.
  \end{array}
\end{equation}
Note that (\ref{eq2}) and (\ref{eq3}) are functorial in $M$. Putting
together (\ref{eq1}), (\ref{eq2}), (\ref{eq3}) and (\ref{eq4}) yields
the following functorial isomorphism in $\mathcal C((A^e)^{op})$:
\begin{equation}
  \label{eq5}
  \begin{array}{rcll}
    \mathcal
    Hom_{\Lambda^e}(M\underset{\Delta}{\otimes}\Lambda^e,\Lambda\otimes
    \Lambda) & \longrightarrow  & \bigoplus\limits_{g\in G}\mathcal
    Hom_{A^e}(M,A\otimes gA)\\
    f & \longmapsto & \sum\limits_{g\in G} \left(m\mapsto \
      \text{projection of $f(m\otimes (1\otimes 1))$ on $A\otimes gA$}\right)&.
  \end{array}
\end{equation}
This achieves the first step.

(b) The second step consists in extending the action of $A^e$ on
$\bigoplus\limits_{g\in G}\mathcal Hom_{A^e}(M,A\otimes gA)$ to an
action of $\Lambda^e$, such that (\ref{eq4}) is an isomorphism in
$\mathcal C((\Lambda^e)^{op})$. For this purpose, note that the
compatible action of $G$ on $\mathcal Hom_{A^e}(M,A\otimes
A)\in\mathcal C((A^e)^{op})$ may be extended to $\mathcal
Hom_{A^e}(M,\Lambda\otimes \Lambda)$ in a similar way: If $g\in G$ and
$\varphi\in\mathcal Hom_{A^e}(M,\Lambda\otimes\Lambda)$, then let
$^g\varphi$ be the map $M\to \Lambda\otimes\Lambda,\,m\mapsto
(g\otimes g^{-1}).\varphi(m.(g\otimes g^{-1})).(g^{-1}\otimes g)$. It
is easy to check that this defines an action $(g,\varphi)\mapsto
\,^g\varphi$ on $\mathcal Hom_{A^e}(M,\Lambda\otimes \Lambda)$
compatible with the action of $A^e$ (but not with that of $\Lambda^e$)
on the left. Therefore, the subspace $\mathcal Hom_{A^e}(M,A\otimes
A)$ of $\mathcal Hom_{A^e}(M,\Lambda\otimes \Lambda)$ is stabilized by
this action; and the action of $g\in G$ maps the
subspace $\mathcal Hom_{A^e}(M,Ah\otimes kA)$ to the subspace
$\mathcal Hom_{A^e}(M,A\,^gh\otimes \,^gkA)$, for $h,k\in G$.
 In order to extend the left action of $A^e$ on
 $\bigoplus\limits_{g\in G}\mathcal Hom_{A^e}(M,A\otimes gA)$ to a
 left action of $\Lambda^e$, proceed as follows. Let $t,g\in G$ and
 $\varphi\in\mathcal Hom_{A^e}(M,A\otimes gA)$. Denote by $(1\otimes
 t)*\varphi$ the following map:
 \begin{equation}
   \begin{array}{crcll}
     (1\otimes t)*\varphi\colon & M &\longrightarrow & A\otimes gtA\\
     &m & \longmapsto & (1\otimes t).\varphi(m)&.
   \end{array}\notag
 \end{equation}
It obviously lies in $\mathcal Hom_{A^e}(M,A\otimes gtA)$. Denote by
$(t\otimes 1)*\varphi$ the following map:
 \begin{equation}
   \begin{array}{crcll}
     (t\otimes 1)*\varphi\colon & M &\longrightarrow & A\otimes gtA\\
     &m & \longmapsto & (1\otimes t).(\,^t\varphi)(m)&.
   \end{array}\notag
 \end{equation}
It obviously lies in $\mathcal Hom_{A^e}(M,A\otimes tgA)$. Finally,
let $(a\otimes b)*\varphi$ be $(a\otimes b).\varphi$, for $a,b\in
A$. It is then easy to check that the assignment:
\begin{equation}
  (at\otimes sb)*\varphi:=(a\otimes b)*\left[(t\otimes
    1)*\left((1\otimes s)*\varphi\right)\right],\ a,b\in A,\,s,t\in G\notag
\end{equation}
extends the action of $A^e$ on $\bigoplus\limits_{g\in G}\mathcal
Hom_{A^e}(M,A\otimes gA)$ on the left to an action of
$\Lambda^e$. Also easy is the verification that (\ref{eq4})  then
becomes an isomorphism in $\mathcal C((\Lambda^e)^{op})$. This finishes the second step.

(c) Now, proceed the third step. The following map is certainly an isomorphism in
$\mathcal C((\Lambda^e)^{op})$ and functorial in $M$:
\begin{equation}
  \label{eq6}
  \begin{array}{rcll}
    \bigoplus\limits_{g\in G}\mathcal Hom_{A^e}(M,A\otimes gA) &
    \longrightarrow & \Lambda^e\underset{\Delta}{\otimes}\mathcal
    Hom_{A^e}(M,A\otimes A)\\
\sum\limits_{g\in G}\varphi_g & \longmapsto & \sum\limits_{g\in G}
(1\otimes g)\otimes \left((1\otimes g^{-1})*\varphi_g\right)&.
  \end{array}
\end{equation}
Putting together (\ref{eq4}) and (\ref{eq6}) shows that (\ref{d4})
commutes up to an isomorphism of functors. This proves the lemma.
\end{proof}

\subsection{Main theorem}
\label{sec:s2.4}

All the necessary tools have been introduced in order to prove the
main result of the text. Recall that crossed-products of left dg
$A^e$-modules acted on compatibly by $G$ were defined in
\ref{sec:s2.2} and that this operation is precisely the extension of
scalars from $\Delta$ to $\Lambda^e$. The following implies Theorem~\ref{thm}
\begin{thm}
\label{thm1}
  Let $A$ be a dg algebra over the field $\ts k$. Assume that a finite
  group $G$ acts on $A$ by dg automorphisms. Let $\Lambda=A\rtimes G$
  be the crossed-product and $\Delta=\bigoplus\limits_{g\in
    G}Ag\otimes g^{-1}A$.
  \begin{enumerate}[(a)]
  \item If $\Lambda\in\ts{per}(\Lambda^e)$, then
    $A\in\ts{per}(A^e)$. If $\sharp G\in\ts k^\times$, then the
    converse holds true.
  \item The dg algebra $A$ admits an inverse dualizing complex
    $\theta_A\in\mathcal C((A^e)^{op})$ endowed with a compatible
    action of $G$. For this action, $\theta_A\rtimes G$ is the inverse
    dualizing complex of $\Lambda$.
  \item Assume that $A$ is Calabi-Yau of dimension $d$ and
    $\Lambda\in\ts{per}(\Lambda^e)$ (see (a)). If an isomorphism
    $f\colon \ts{RHom}_{A^e}(A,A\otimes A)[d]\to A$ exists in
    $\mathcal D(\Delta^{op})$ and is such that
    $\ts{RHom}_{A^e}(f,A\otimes A)[d]=f$ in $\mathcal D(\Delta)$, then
    $\Lambda$ is Calabi-Yau of dimension $d$.
  \end{enumerate}
\end{thm}
\begin{proof}
  Recall that the functor $\ts{RHom}_{A^e}(-,A\otimes A)\colon\mathcal
  D(\Delta)\to \mathcal D(\Delta^{op})$ was defined in \ref{sec:s2.3},
  (\ref{d2}).

Fix a cofibrant resolution $p\colon X\to A$ in $\mathcal
C(\Delta)$. Following \ref{sec:s2.2}, (c), it is also a cofibrant
resolution in $\mathcal C(A^e)$ and so is
$p\underset{\Delta}{\otimes}\Lambda^e\colon
X\underset{\Delta}{\otimes}\Lambda^e\to
A\underset{\Delta}{\otimes}\Lambda^e$. Composing with the
isomorphism $A\underset{\Delta}{\otimes}\Lambda^e\to \Lambda$ in
$\mathcal C(\Lambda^e)$ (\ref{sec:s2.2}, (d)) therefore gives a
cofibrant resolution $X\underset{\Delta}{\otimes}\Lambda^e\to \Lambda$
in $\mathcal C(\Lambda^e)$.

(a) Assume first that $\Lambda\in\ts{per}(\Lambda^e)$. Therefore,
$A\underset{\Delta}{\otimes}\Lambda^e\simeq\Lambda\in
\ts{per}(\Lambda^e)$. The restriction-of-scalars functor $\mathcal
C(\Lambda^e)\to\mathcal C(\Delta)$ preserves quasi-isomorphisms and
maps $\Lambda^e$ to $\Lambda^e\simeq \Delta^{\sharp G}$ in $\mathcal
C(\Delta)$ (\ref{sec:s2.2}, (a)). Therefore, it induces a functor
$\mathcal D(\Lambda^e)\to \mathcal D(\Delta)$ which maps
$\ts{per}(\Lambda^e)$ into $\ts{per}(\Delta)$. In particular,
$A\underset{\Delta}{\otimes}\Lambda^e\in\ts{per}(\Delta)$. Since
$A$ is a direct summand of $A\underset{\Delta}{\otimes}\Lambda^e$ in
$\mathcal C(\Delta)$ (\ref{sec:s2.2}, (e)), and therefore in $\mathcal
D(\Delta)$, this implies that $A\in\ts{per}(\Delta)$. On the other hand the
restriction-of-scalars functor $\mathcal C(\Delta)\to \mathcal C(A^e)$
preserves quasi-isomorphisms and maps $\Delta$ to $\Delta\simeq
(A^e)^{\sharp G}$ in $\mathcal C(A^e)$
(\ref{sec:s2.2}, (a)). Therefore, it induces a functor $\mathcal
D(\Delta)\to \mathcal D(A^e)$ which maps $\ts{per}(\Delta)$ into
$\ts{per}(A^e)$. It also maps $A$ to $A$ Thus, $A\in\ts{per}(A^e)$.

Now, assume that $A\in\ts{per}(A^e)$ and $\sharp G\in\ts
k^\times$. The two functors $-\underset{A^e}{\otimes}\Delta\colon
\mathcal C(A^e)\to \mathcal C(\Delta)$ and
$-\underset{\Delta}{\otimes}\Lambda^e\colon\mathcal C(\Delta)\to
\mathcal C(\Lambda^e)$ preserve quasi-isomorphisms (\ref{sec:s2.2},
(b)), the former maps $A^e$ to
$A^e\underset{A^e}{\otimes}\Delta=\Delta$ and the latter maps
$\Delta$ to
$\Delta\underset{\Delta}{\otimes}\Lambda^e=\Lambda^e$. Therefore,
these functors induce functors $-\underset{A^e}{\otimes}\Delta\colon
\mathcal D(A^e)\to \mathcal D(\Delta)$ and
$-\underset{\Delta}{\otimes}\Lambda^e\colon\mathcal D(\Delta)\to
\mathcal D(\Lambda^e)$, the former maps $\ts{per}(A^e)$ into
$\ts{per}(\Delta)$, the latter maps $\ts{per}(\Delta)$ into
$\ts{per}(\Lambda^e)$. The former maps $A$ to
$A\underset{A^e}{\otimes}\Delta$ and $A$ is a direct summand of
$A\underset{A^e}{\otimes}\Delta$ in $\mathcal C(\Delta)$, thus
$A\in\ts{per}(\Delta)$. The latter maps $A$ to
$A\underset{\Delta}{\otimes} \Lambda^e\simeq\Lambda$, thus $\Lambda\in\ts{per}(\Lambda^e)$.

(b) By construction, $\mathcal Hom_{A^e}(X,A\otimes A)\in\mathcal
C(\Delta^{op})$. Let $Y\to \mathcal Hom_{A^e}(X,A\otimes A)$ be a
cofibrant resolution in $\mathcal C(\Delta^{op})$. Therefore, it is
also a cofibrant resolution in $\mathcal C((A^e)^{op})$
(\ref{sec:s2.2}, (c)). Thus, $\theta_A$ may be chosen equal to $Y$
and, as a left dg $\Delta$-module, it has an action of $G$ compatible
with the left action of $A^e$. Of course, not any cofibrant resolution of
$\mathcal Hom_{A^e}(X,A\otimes A)$ has this property.

By definition of the inverse dualizing complex of $\Lambda$, there is a
quasi-isomorphism $\theta_{\Lambda}\colon \mathcal
Hom_{A^e}(X\underset{\Delta}{\otimes}\Lambda^e,\Lambda\otimes\Lambda)$
in $\mathcal C((\Lambda^e)^{op})$. On the other hand, the proof in
\ref{sec:s2.3} gave an isomorphism $\mathcal
Hom_{A^e}(X\underset{\Delta}{\otimes}\Lambda^e,\Lambda\otimes\Lambda)\to
\Lambda^e\underset{\Delta}{\otimes} \mathcal Hom_{A^e}(X,A\otimes A)$
in $\mathcal C((\Lambda^e)^{op})$ and
$\Lambda^e\underset{\Delta}{\otimes}\mathcal Hom_{A^e}(M,A\otimes
A)\simeq \mathcal Hom_{A^e}(X,A\otimes A)\rtimes G$ in $\mathcal
C((\Lambda^e)^{op})$. Thus, there is a quasi-isomorphism
$\theta_{\Lambda}\to\theta_A\rtimes G$ in $\mathcal
C((\Lambda^e)^{op})$. For this reason, $\theta_{\Lambda}$ may be
chosen equal to $\theta_A\rtimes G$.

(c) Assume that $A\in\ts{per}(A^e)$, $\Lambda\in\ts{per}(\Lambda^e)$
and that there is an isomorphism $f\colon\ts{RHom}_{A^e}(A,A\otimes
A)[d]\to A$ in $\mathcal D(\Delta^{op})$ such that
$\ts{RHom}_{A^e}(f,A\otimes A)[d]=f$ in $\mathcal D(\Delta)$. Applying
the functor $\Lambda^e\underset{\Delta}{\otimes} -\colon \mathcal
D(\Delta^{op})\to \mathcal D((\Lambda^e)^{op})$ yields an isomorphism
in $\mathcal D((\Lambda^e)^{op})$:
\begin{equation}
  \Lambda^e\underset{\Delta}{\otimes} f\colon
  \Lambda^e\underset{\Delta}{\otimes} \ts{RHom}_{A^e}(A,A\otimes
  A)[d]\xrightarrow{\sim} \Lambda^e\underset{\Delta}{\otimes}A\simeq
  \Lambda.\notag
\end{equation}
As explained above,
$\Lambda^e\underset{\Delta}{\otimes}\ts{Rhom}_{A^e}(A,A\otimes A)$
identifies with $\ts{RHom}_{\Lambda^e}(\Lambda,\Lambda\otimes
\Lambda)$. Thus,
in order to conclude, it only remains to prove that
$\ts{RHom}_{(\Lambda^e)^{op}}(\Lambda^e\underset{\Delta}{\otimes}
f,\Lambda\otimes \Lambda)[d]=\Lambda^e\underset{\Delta}{\otimes}
f$. But this follows from the equality
$\ts{RHom}_{A^e}(f,A\otimes A)[d]=f$ in $\mathcal D(\Delta)$ and from
the commutativity of the following diagram obtained by applying
\ref{sec:s2.3} (and interchanging the rôles played by left and right
modules):
\begin{equation}
  \xymatrix{
\mathcal D(\Delta^{op}) \ar[rrr]^{\ts{RHom}_{(A^e)^{op}}(-,A\otimes
  A)} \ar[d]_{\Lambda^e\underset{\Delta}{\otimes}-} &&& \mathcal
D(\Delta) \ar[d]^{-\underset{\Delta}{\otimes}\Lambda^e} \\
\mathcal D((\Lambda^e)^{op})
\ar[rrr]_{\ts{RHom}_{(\Lambda^e)^{op}}(-,\Lambda\otimes \Lambda)} &&&
\mathcal D(\Lambda^e) &.
}\notag
\end{equation}
Thus, $\Lambda$ is Calabi-Yau of dimension $d$.
\end{proof}

Here are some technical comments on the hypotheses made in part (c) of
the above theorem. Let $X\to A$ be a cofibrant resolution in $\mathcal
C(\Delta)$ and $\theta_A\to \mathcal Hom_{A^e}(X,A\otimes A)$ be a
cofibrant resolution in $\mathcal C(\Delta^{op})$. The data of an
isomorphism $\ts{RHom}_{A^e}(A,A\otimes A)[d]\to
A$ in $\mathcal D(\Delta^{op})$ is equivalent to that of a
quasi-isomorphism $\theta_A[d]\to A$ in $\mathcal C(\Delta^{op})$ that
is, a quasi-isomorphism in $\mathcal C((A^e)^{op})$ which is
compatible with the actions of $G$ on $\theta_A$ and on $A$.
Given an isomorphism $f\colon \ts{RHom}_{A^e}(A,A\otimes
A)[d]\to A$ in $\mathcal D(\Delta^{op})$, the equality
$\ts{RHom}_{(A^e)^{op}}(f,A\otimes A)[d]=f$ holds true in $\mathcal
D(A^e)$ if it does so in $\mathcal D(\Delta)$, because of (\ref{d2})
(after interchanging the rôles of left and right modules). The
converse needs not hold true because the restriction-of-scalars
functor $\mathcal D(\Delta)\to \mathcal D(A^e)$ is not necessarily
faithful. However, $\mathcal D(\Delta)\to \mathcal D(A^e)$ is faithful
if $\sharp G\in\ts k^\times$: Indeed, it is right adjoint to
$-\underset{A^e}{\otimes}\Delta\colon \mathcal D(A^e)\to\mathcal
D(\Delta)$ and the counit of this adjunction splits if $\sharp G\in\ts
k^\times$ (see the proof of \ref{sec:s2.1}, (e)).

These considerations give Theorem~\ref{thm} as a simplification of
Theorem~\ref{thm1} in case $\sharp G\in\ts k^{\times}$:
\setcounter{thm}{0}
\begin{thm}
  \label{thm2}
Let $A$ be a dg algebra acted on by a finite group $G$ such that
$\sharp G\in\ts k^\times$. Fix the following data:
\begin{enumerate}[-]
\item $X\to A$ is a cofibrant resolution in
$\mathcal C(A^e)$ such that $X$ is endowed with a compatible action of
$G$ and the resolution if $G$-equivariant,
\item  $\theta_A\to \mathcal
Hom_{A^e}(X,A\otimes A)=\ts{RHom}_{A^e}(A,A\otimes A)$ is a cofibrant
resolution in $\mathcal C((A^e)^{op})$ such that $\theta_A$ is endowed
with a compatible action of $G$ and the resolution is $G$-equivariant.
\end{enumerate}
Then:
\begin{enumerate}
\item $\theta_A\rtimes G$ is the inverse dualizing complex of
  $A\rtimes G$.
\item Assume that $A$ is homologically smooth and there exists a
$G$-equivariant quasi-isomorphism $f\colon \theta_A[d]\to A$ in
$\mathcal C((A^e)^{op})$ such that $\ts{RHom}_{(A^e)^{op}}(f,A\otimes
A)[d]=f$ in $\mathcal D(A^e)$ (in particular, $A$ is Calabi-Yau of
dimension $d$). Then $A\rtimes G$ is Calabi-Yau of dimension $d$.
\end{enumerate}
\end{thm}

Deleting the equivariance condition on $f$ in Theorem~\ref{thm2} gives
exactly the definition of the Calabi-Yau property for $A$. Later, this
equivariance condition will be discussed on and referred to as the
\emph{compatibility condition}, for short.

\section{Crossed-products of Calabi-Yau completions}

This section investigates when the crossed-product of a dg Calabi-Yau algebra
$A$ by a finite group $G$ is still Calabi-Yau when $A$ arises from a
particular construction introduced in \cite{K09} and called
\emph{deformed Calabi-Yau completion}.

\subsection{A reminder on deformed Calabi-Yau completions}
\label{sec:s3.1}

The reader is referred to \cite{K09} for a detailed account on deformed
Calabi-Yau completions. This paragraph recalls this construction briefly.
Let $A$ be a dg algebra, $d$ be an integer and $\theta_A\in\mathcal
C((A^e)^{op})$ be the inverse dualizing complex of $A$. The
\emph{$d$-Calabi-Yau completion} of $A$ is defined as the dg tensor
algebra
\begin{equation}
  \Pi_d(A)=T_A(\theta_A[d-1])=A\oplus
  \theta_A[d-1]\oplus(\theta_A[d-1]\underset{A}{\otimes}\theta_A[d-1])\oplus\cdots \,.\notag
\end{equation}
It does not depend on the choice of $\theta_A$ up to quasi-isomorphism
of dg algebras. If $A$ is homologically smooth, then it is Calabi-Yau
of dimension $d$ (\cite[Thm. 4.8]{K09}).

The definition of the deformed Calabi-Yau completion uses the
following canonical isomorphisms where $A$ is assumed to be
homologically smooth:
\begin{equation}
  \label{eq7}
  \begin{array}{rcl}
    \ts{HH}_{d-2}(A) = \ts{Tor}_{d-2}^{A^e}(A,A) & = & \ts
    H^{d-2}(A\overset{\mathbb L}{\underset{A^e}{\otimes}} A) \simeq
    \ts H^{d-2}(A\overset{\mathbb{L}}{\underset{A^e}{\otimes}}
    A^{\vee\vee})\\
&\simeq & \ts{Hom}_{\mathcal D((A^e)^{op})}(A^\vee,A[d-2])\\
&\simeq & \ts{Hom}_{\mathcal D((A^e)^{op})}(\theta_A[d-1],A[1])
  \end{array}
\end{equation}
where, for short, $-^\vee$ stands for $\ts{RHom}_{A^e}(-,A\otimes A)$
(hence $\theta_A$ is any cofibrant resolution of $A^\vee$ in $\mathcal
C((A^e)^{op})$, and $A\simeq A^{\vee\vee}$ in $\mathcal D(A^e)$
because $A$ is homologically smooth). Also $\ts{HH}$ denotes the
Hochschild homology. If $X\to A$ and $p_A\colon \theta_A\to \mathcal
Hom_{A^e}(X,A\otimes A)$ are cofibrant resolutions in $\mathcal
C(A^e)$ and $\mathcal C((A^e)^{op})$, respectively, then
$\ts{HH}_{d-2}(A)=\ts H^{d-2}(X\underset{A^e}{\otimes} A)$. Also
$\ts{Hom}_{D((A^e)^{op})}(\theta_A[d-1],A[1])=\ts H^0\left(\mathcal
  Hom_{(A^e)^{op}}(\theta_A[d-1],A[1])\right)$, and the isomorphism in
(\ref{eq7}) is induced by the following morphism of complexes (which
is therefore a quasi-isomorphism):
\begin{equation}
  \label{eq8}
  \begin{array}{rcll}
    X\underset{A^e}{\otimes}A[2-d] & \longrightarrow & \mathcal
    Hom_{(A^e)^{op}}(\theta_A[d-1],A[1])\\
x\otimes a &\longmapsto& \left(\alpha\mapsto (p_A(\alpha))(x).a\right)&.
  \end{array}
\end{equation}
Let $c\in\ts{HH}_{d-2}(A)$. The associated morphism in
$\ts{Hom}_{D((A^e)^{op})}(\theta_A[d-1],A[1])$ (see (\ref{eq7}))
admits, say, $D\colon \theta_A[d-1]\to A[1]$ as a representative. Then
$D$ is a morphism of graded left $A^e$-modules. Therefore, it induces
a unique $A^e$-linear (skew-)derivation of degree $1$:
\begin{equation}
  \overline D\colon T_A(\theta_A[d-1])\to T_A(\theta_A[d-1])\,.\notag
\end{equation}
Writing $\delta\colon T_A(\theta_A[d-1])\to T_A(\theta_A[d-1])$ for the
differential of the dg algebra $\Pi_d(A)$, the \emph{deformed
  Calabi-Yau completion} $\Pi_d(A,c)$ is defined as the dg algebra
\begin{equation}
  \Pi_d(A,c)=(T_A(\theta_A), \delta+\overline D)\,.\notag
\end{equation}
Here, $\delta+\overline D$ is indeed a differential because
$c\in\ts{HH}_{d-2}(A)$. It is proved in \cite[Thm. 5.2]{K09} that
$\Pi_d(A,c)$ is Calabi-Yau of dimension $d$, for every $c$ (under the
assumption that $A$ is homologically smooth).

Throughout this section, $A$ will denote a dg algebra, acted on by a
finite group $G$. The crossed-product $A\rtimes G$ will be denoted by
$\Lambda$ and the dg subalgebra $\bigoplus\limits_{g\in G}Ag\otimes
g^{-1}A$ of $\Lambda$ will be denoted by $\Delta$ (see
\ref{sec:s2.1}). Also, a cofibrant resolution $X\to A$ in $\mathcal
C(\Delta)$ will be fixed, recall that this is also a cofibrant
resolution in $\mathcal C(A^e)$ and it induces a cofibrant resolution
$X\underset{\Delta}{\otimes}\Lambda^e\to
A\underset{\Delta}{\otimes}\Lambda^e\xrightarrow{\sim}\Lambda$ in
$\mathcal C(\Lambda^e)$ (\ref{sec:s2.2}, (c) and (d)). Finally, a
cofibrant resolution $p_A\colon \theta_A\to \mathcal
Hom_{A^e}(X,A\otimes A)$ in $\mathcal C(\Delta^{op})$ is fixed (so it
is a cofibrant resolution in $\mathcal C((A^e)^{op})$).

Following \ref{sec:s2.3}, left dg $\Delta$-modules will be considered
as left dg $A^e$-modules endowed with a compatible action of $G$. In
particular, given $M\in\mathcal C(\Delta)$, the compatible action of
$G$ on $\mathcal Hom_{A^e}(M,A\otimes A)\in\mathcal C(\Delta^{op})$ is
such that $^gf=(g\otimes g^{-1}).f$ maps $m\in M$ to $(g\otimes
g^{-1}).f(m.(g\otimes g^{-1})).(g^{-1}\otimes g)$, for $g\in G$ and
$f\in\mathcal Hom_{A^e}(M,A\otimes A)$.

\subsection{Group actions on deformed Calabi-Yau completions}
\label{sec:s3.2}

Since $\theta_A\in\mathcal C(\Delta^{op})$, there is an action of $G$ on $\theta_A$
which is compatible with the action of $A^e$ on the left. This induces
an action of $G$ on $\Pi_d(A)=T_A(\theta_a[d-1])$ by dg automorphisms
($g\in G$ acts diagonally on
$\left(\theta_A[d-1]\right)^{\underset{A}{\otimes}^n}$ for all
  $n\geqslant 1$).

On the other hand, the actions of $G$ on $X\in\mathcal C(\Delta)$ and
on $A$ induce a diagonal action of $G$ on $X\underset{A^e}{\otimes}A$
and, therefore, on $\ts{HH}_*(A)$. To be more precise: $^g(x\otimes
a)=\,^gx\otimes \,^ga=\left(x.(g^{-1}\otimes g)\right)\otimes
\left((g\otimes g^{-1}).a\right)$, for $x\in X$, $a\in A$ and $g\in G$
(see \ref{sec:s2.1}). As usual, $\ts{HH}_{d-2}(A)^G$ denotes the set
of fixed points in $\ts{HH}_{d-2}(A)$ under this action. The following
lemma gives a necessary and sufficient condition on
$c\in\ts{HH}_{d-2}(A)$ for the action of $G$ on $\Pi_d(A)$ to be an
action on $\Pi_d(A,c)$ by dg automorphisms (of course, $G$ already
acts on $\Pi_d(A,c)$ by graded automorphisms, since
$\Pi_d(A,c)=\Pi_d(A)$ as graded algebras).
\begin{lem}
  Let $A$ be a homologically smooth dg algebra acted on by a finite
  group $G$ such that $\sharp G\in\ts k^\times$. The following
  conditions are equivalent:
  \begin{enumerate}[(i)]
  \item $c\in\ts{HH}_{d-2}(A)^G$,
  \item there exists a cocycle $D\in\mathcal
    Hom_{(A^e)^{op}}(\theta_A[n-1],A[1])$ which corresponds to $c$
    under (\ref{eq7}) (see also (\ref{eq8})) and such that the action
    of $G$ on $\Pi_d(A)$ is also an action on
    $\Pi_d(A,c)=(T_A(\theta_A[d-1]),\delta+\overline D)$ by dg automorphisms.
  \end{enumerate}
\end{lem}
\begin{proof}
  Let $u\in X\underset{A^e}{\otimes}A$  be any representative of
  $c\in\ts{HH}_{d-2}(A)=\ts H^0(X\underset{A^e}{\otimes}A)$. Let
  $D_u\in\mathcal Hom_{(A^e)^{op}}(\theta_A[d-1],A[1])$ be the image
  of $u$ under the isomorphism (\ref{eq8}). Note that (\ref{eq8}) is
  $G$-equivariant. On the other hand, $c$ lies in $\ts{HH}_{d-2}(A)^G$
  if and only if there is a choice of the representative $u$ such that
  $^gu=u$ for all $g\in G$, because $\sharp G\in\ts
  k^\times$. Therefore, $c$ lies in $\ts{HH}_{d-2}(A)^G$ if and only
  if there is a representative $u$ of $c$ such that $^gD_u=D_u$ for
  all $g\in G$.

Next, let $D\in\mathcal Hom_{(A^e)^{op}}(\theta_A[d-1],A[1])$ be a
$0$-cocycle. Then, the action of $G$ on $\Pi_d(A)$ is an action by dg
automorphisms on $(T_A(\theta_A[d-1]),\delta+\overline D)$ if and only
if $\delta+\overline D$ is a $G$-equivariant differential. Since
$\delta$ is itself $G$-equivariant, this is equivalent to say that
$\overline D$ is $G$-equivariant. Clearly, this is equivalent to $D$
being $G$-equivariant. Note that if this is the case and if the
cohomology class of $D$ in $\ts{Hom}_{\mathcal D((A^e)^{op})}(\theta_A[d-1],A[1])$
corresponds to $c$ under (\ref{eq7}), then $c$ lies in
$\ts{HH}_{d-2}(A)^G$, because (\ref{eq7}) is $G$-equivariant.

These considerations show the equivalence between ($i$) and ($ii$).
\end{proof}

Note that not all the actions of $G$ on $\Pi_d(A,c)$ arise from
actions on $A$ such that $c\in\ts{HH}_{d-2}(A)^G$.

\subsection{Crossed-products of Calabi-Yau completions}
\label{sec:s3.3}
The following lemma shows that taking crossed-products commutes with
Calabi-Yau completion.
\begin{lem}
  Let $A$ be a dg algebra acted on  by a finite group $G$.
Let $d$ be an integer. Then, the dg algebras $\Pi_d(A\rtimes G)$ and
$\Pi_d(A)\rtimes G$ are quasi-isomorphic.
\end{lem}
\begin{proof}
  Recall that $\theta_A\rtimes G$ is the inverse dualizing complex of
  $\Lambda=A\rtimes G$ (\ref{sec:s2.4}, (b)). Therefore,
  $\Pi_d(A\rtimes G)=T_{A\rtimes G}(\theta_A\rtimes G[d-1])$. The
  following map is therefore a morphism of dg algebras:
  \begin{equation}
    \begin{array}{rcll}
      \Pi_d(A)\rtimes G & \longrightarrow & \Pi_d(A\rtimes G)\\
      a\otimes g & \longmapsto & a\otimes g\\
      (f_1\otimes \cdots\otimes f_n)\otimes g & \longmapsto &
      f_1\otimes \cdots \otimes f_{n-1}\otimes (f_n\otimes g) &,
    \end{array}\notag
  \end{equation}
where $a\in A$, $f_1,\ldots,f_n\in\theta_A[d-1]$ and $g\in G$. It is
straightforward to check that the following map is its inverse:
\begin{equation}
  \begin{array}{rcll}
    \Pi_d(A\rtimes G) & \longrightarrow & \Pi_d(A)\rtimes G\\
    a\otimes g &\longmapsto & a\otimes g\\
    (f_1\otimes g_1)\otimes\cdots \otimes (f_n\otimes g_n) & \longmapsto & \left(f_1\otimes
    \,^{g_1}f_2\otimes\cdots \otimes \,^{g_1\cdots
      g_l}f_{l+1}\otimes\cdots\otimes \,^{g_1\cdots
      g_{n-2}}f_{n-1}\otimes \,^{g_1\cdots
      g_{n-1}}f_{n}\right)\otimes g_1\cdots
    g_n& ,
  \end{array}\notag
\end{equation}
where $a\in A$, $g,g_1,\ldots,g_n\in G$ and $f_1,\ldots,f_n\in\theta_A[d-1]$.
\end{proof}

\subsection{Compatibility between crossed-products and deformed Calabi-Yau completions}
\label{sec:s3.4}
The following proposition is the main result of the section. It shows
that under compatibility conditions, the crossed-product of a deformed
Calabi-Yau completion is still a deformed Calabi-Yau completion and,
therefore, is Calabi-Yau.
\begin{prop}
  Let $A$ be a homologically smooth algebra. Let $G$ be a finite group
  acting on $A$ by dg automorphisms and such that $\sharp G\in\ts
  k^\times$. Finally let $c\in\ts{HH}_{d-2}(A)^G$. Then, up to a
  choice of the inverse dualizing complex of $A$, the action
  of $G$ on $A$ induces an action on $\Pi_d(A,c)$ by dg
  automorphisms. For this action, there is an isomorphism of dg
  algebras:
  \begin{equation}
    \Pi_d(A,c)\rtimes G\simeq \Pi_d(A\rtimes G, \overline c)\,,\notag
  \end{equation}
where $\overline c$ is the image of $c$ under the canonical map
$\ts{HH}_{d-2}(A)\to \ts{HH}_{d-2}(A\rtimes G)$. In particular,
$\Pi_d(A,c)\rtimes G$ is Calabi-Yau of dimension $d$.
\end{prop}
\begin{proof}
  For the purpose of the proposition, it is necessary to first
  introduce an adequate cofibrant resolution $p_{\Lambda}\colon
  \theta_{\Lambda}\to \ts{RHom}_{\Lambda^e}(\Lambda,\Lambda\otimes
  \Lambda)$ in $\mathcal C((\Lambda^e)^{op})$. Following
  \ref{sec:s2.4}, the inverse dualizing complex of $\Lambda$ is
  $\theta_{\Lambda}:=\Lambda^e\underset{\Delta}{\otimes}\theta_A$. Moreover,
  $X\underset{\Delta}{\otimes}\Lambda^e$ is a cofibrant resolution of
  $\Lambda$ in $\mathcal C(\Lambda^e)$, because $X$ is a cofibrant
  resolution of $A$ in $\mathcal C(\Delta)$ (see \ref{sec:s2.2}, (c)
  and (d)). Therefore, an explicit quasi-isomorphism
  $p_{\Lambda}\colon \theta_{\Lambda}\to \mathcal
  Hom_{\Lambda^e}(X\underset{\Delta}{\otimes}\Lambda^e,\Lambda\otimes
  \Lambda)=\ts{RHom}_{\Lambda^e}(\Lambda,\Lambda\otimes\Lambda)$ in
  $\mathcal C((\Lambda^e)^{op})$ is obtained as follows. On the one
  hand, $\Lambda^e\underset{\Delta}{\otimes}p_A\colon
  \theta_{\Lambda}\to \Lambda^e\underset{\Delta}{\otimes} \mathcal
  Hom_{A^e}(X,A\otimes A)$ is a quasi-isomorphism in $\mathcal
  C((\Lambda^e)^{op})$, because $p_A\colon \theta_A\to \mathcal
  Hom_{A^e}(X,A\otimes A)$ is a quasi-isomorphism in
  $\mathcal C(\Delta^{op})$ (see \ref{sec:s2.2}, (c)). On the other
  hand, the following map is an isomorphism in $\mathcal
  C(\Delta^{op})$, inverse to the composition of (\ref{eq5}) with
  (\ref{eq6}):
  \begin{equation}
    \begin{array}{rcll}
      \Lambda^e\underset{\Delta}{\otimes}\mathcal Hom_{A^e}(X,A\otimes
      A) & \longrightarrow & \mathcal
      Hom_{A^e}(X\underset{\Delta}{\otimes} \Lambda^e,
      \Lambda\otimes\Lambda)\\
      (ag\otimes hb)\otimes f & \longmapsto & \left( x\otimes u
        \mapsto (a\otimes ghb).\left((\,^gf)(x)\right).u\right) &,
    \end{array}\notag
  \end{equation}
where $a,b\in A$, $g,h\in G$, $f\in\mathcal Hom_{A^e}(X,A\otimes A)$
and $u\in\Lambda^e$. Thus, the following map is an isomorphism in
$\mathcal C((\Lambda^e)^{op})$:
\begin{equation}
  \begin{array}{crcll}
    p_{\Lambda}\colon & \theta_{\Lambda} & \longrightarrow & \mathcal
    Hom_{\Lambda^e}(X\underset{\Delta}{\otimes}
    \Lambda^e,\Lambda\otimes \Lambda)\\
   & (ag\otimes hb)\otimes f & \longmapsto & \left(x\otimes u \mapsto
      (a\otimes ghb).(\,^g(p_A(f)))(x).u\right) &.
  \end{array}\notag
\end{equation}

Following \ref{sec:s3.3}, there exists $D\in\mathcal
Hom_{(A^e)^{op}}(\theta_A[d-1],A[1])$ which is $G$-invariant (that is,
$^gD=D$ for every $g\in G$) and such that $\Pi_d(A,c)$ is the graded
algebra $T_A(\theta_A[d-1])$ endowed with the differential
$\delta_A+\overline D$, where $\delta_A$ denotes the differential of
$\Pi_d(A)$. In particular, the action of $G$ on $A$ induces an action
on $\Pi_d(A,c)$. Therefore, $\Pi_d(A,c)\rtimes G$ is the graded
algebra $T_A(\theta_A[d-1])\rtimes G$ endowed with the differential
whose restriction to $\theta_A[d-1]$ equals to $\delta_A+\overline D$
and which vanishes on $G$. It then follows from \ref{sec:s3.2} that
$\Pi_d(A,c)\rtimes G$ is isomorphic to the graded algebra $T_{A\rtimes
  G}(\theta_{\Lambda}[d-1])$ endowed with the differential
$\delta_{\Lambda}+\overline{\Lambda^e\underset{\Delta}{\otimes}D}$,
where $\Lambda^e\underset{\Delta}{\otimes}D\in\mathcal
Hom_{(\Lambda^e)^{op}}(\theta_{\Lambda}[d-1],\Lambda[1])$ (recall that
$\theta_{\Lambda}=\Lambda^e\underset{\Delta}{\otimes} \theta_A$ and
$\Lambda^e\underset{\Delta}{\otimes}A$ is identified to $\Lambda$ in
$\mathcal C((\Lambda^e)^{op})$, via $u\otimes a\mapsto u.a$). In other
words, $\Pi_d(A,c)\rtimes G$ is the deformed Calabi-Yau completion
$\Pi_d(\Lambda, c')$ where $c'\in\ts{HH}_{d-2}(\Lambda)$ corresponds
to $\Lambda^e\underset{\Delta}{\otimes}D\in\mathcal
Hom_{(\Lambda^e)^{op}}(\theta_{\Lambda}[d-1], \Lambda[1])$ under the
map (\ref{eq8}) corresponding to $\Lambda$.

Therefore, in order to prove the proposition, it suffices to prove
that the following diagram commutes:
\begin{equation}
  \label{d5}
\xymatrix{
\ts{HH}_{d-2}(A)^G \ar@{^(->}[r] \ar[d]& \ts{HH}_{d-2}(A)
\ar[rrr]^{(\ref{eq7})} &&& \ts H^0\left(\mathcal
  Hom_{(A^e)^{op}}(\theta_A[d-1],A[1])\right) \ar[d]^{\ts
  H^0(\Lambda^e\underset{\Delta}{\otimes}-)
}\\
\ts{HH}_{d-2}(\Lambda) \ar[rrrr] &&&& \ts H^0\left(\mathcal
  Hom_{(\Lambda^e)^{op}}(\theta_\Lambda[d-1],\Lambda[1])\right)
}\tag{$D_5$}
\end{equation}
where:
\begin{enumerate}[-]
\item the vertical arrow on the left is the composition
  $\ts{HH}_{d-2}(A)^G\hookrightarrow \ts{HH}_{d-2}(A)\to
  \ts{HH}_{d-2}(\Lambda)$,
\item the bottom horizontal arrow is the map (\ref{eq7}) corresponding
  to $\Lambda$,
\end{enumerate}
For this purpose, it suffices to prove that the following diagram
commutes (recall that $\ts{HH}_{d-2}(A)^G=\ts
H^{d-2}((X\underset{A^e}{\otimes}A)^G)$ because $\sharp G\in\ts
k^{\times}$):
\begin{equation}
  \label{d6}
\xymatrix{
(X\underset{A^e}{\otimes}A)^G \ar[d] \ar@{^(->}[r] &
X\underset{A^e}{\otimes} A \ar[rrr]^{(\ref{eq8})} &&& \mathcal
Hom_{(A^e)^{op}}(\theta_A[d-1],A[1])
\ar[d]^{\Lambda^e\underset{\Delta}{\otimes}-}\\
(X\underset{\Delta}{\otimes}\Lambda^e)\underset{\Lambda^e}{\otimes}\Lambda
= X\underset{\delta}{\otimes}\Lambda \ar[rrrr] &&&& \mathcal
Hom_{(\Lambda^e)^{op}}(\theta_{\Lambda}[d-1],\Lambda[1])
}\tag{$D_6$}
\end{equation}
where:
\begin{enumerate}[-]
\item the vertical arrow on the left is the composition
  $(X\underset{A^e}{\otimes} A)^G\hookrightarrow
  X\underset{A^e}{\otimes} A\to X\underset{\Delta}{\otimes} A$,
\item the bottom horizontal arrow is the map (\ref{eq8}) corresponding
  to $\Lambda$.
\end{enumerate}
Let $\sum\limits_ix_i\otimes a_i\in(X\underset{A^e}{\otimes}A )^G$,
that is, $\sum\limits_i\,^gx_i\otimes
\,^ga_i=\sum\limits_ix_i\otimes a_i$ for every $g\in G$. Then, the
  image of $\sum\limits_ix_i\otimes a_i$ under the composition of the
  top horizontal arrows in (\ref{d6}) is the map (see (\ref{eq8})):
  \begin{equation}
    \begin{array}{rcll}
      \mathcal Hom_{A^e}(X,A\otimes A)[d-1] &\longrightarrow & A[1]\\
      f &\longmapsto & \sum\limits_i(p_A(f))(x_i).a_i&.
    \end{array}\notag
  \end{equation}
Applying the right-hand side vertical arrow in (\ref{d6}) yields the
map:
\begin{equation}
  \label{eq9}
  \begin{array}{rcll}
    \Lambda^e\underset{\Delta}{\otimes}\mathcal Hom_{A^e}(X,A\otimes
    A)[d-1] & \longrightarrow & \Lambda[1]\\
    u\otimes f & \longmapsto & \sum\limits_iu.(p_A(f))(x_i).a_i&.
  \end{array}
\end{equation}
On the other hand, the image of $\sum\limits_ix_i\otimes a_i$ under
the left-hand side vertical arrow of (\ref{d6}) is equal to
$\sum\limits_ix_i\otimes a_i\in
X\underset{\Delta}{\otimes}\Lambda$. Applying the bottom horizontal
arrow of (\ref{d6}) to $\sum\limits_ix_i\otimes a_i$ yields the
following map (see (\ref{eq8})):
\begin{equation}
  \label{eq10}
  \begin{array}{rcll}
    \Lambda^e\underset{\Delta}{\otimes}\mathcal Hom_{A^e}(X,A\otimes
    A)[d-1] & \longrightarrow & \Lambda[1]\\
    (ag\otimes hb)\otimes f & \longmapsto &
    \sum\limits_i\left(p_{\Lambda}((ag\otimes hb)\otimes g)\right)(x_i).a_i&.
  \end{array}
\end{equation}
The proof of the commutativity of (\ref{d6}) therefore reduces to
proving that the two maps (\ref{eq9}) and (\ref{eq10}) are
equal. Let $a,b\in A$, $g\in G$ and $f\in\mathcal Hom_{A^e}(X,A\otimes
A)$. Using the expression of $p_{\Lambda}$ that was computed at the
beginning of the proof, gives:
\begin{equation}
  \sum\limits_i\left(p_{\Lambda}((ag\otimes hb)\otimes
    f)\right)(x_i).a_i=\sum\limits_i(a\otimes ghb).\left(\,^g(p_A(f))\right)(x_i).a_i\,.\notag
\end{equation}
Recall that $^g(p_A(f))(x_i)=(g\otimes g^{-1}).\left((p_A(f))(x_i.(g\otimes
g^{-1}))\right).(g^{-1}\otimes g)$, therefore:
\begin{equation}
  \sum\limits_ip_{\Lambda}((ag\otimes hb)\otimes
  f)(x_i).a_i=\sum\limits_i (a\otimes ghb).(g\otimes
  g^{-1}).\left(p_A(f)(x_i.(g\otimes g^{-1}))\right).(g^{-1}\otimes g).a_i\,.\notag
\end{equation}
Note that $(x_i.(g\otimes g^{-1}))\otimes ((g^{-1}\otimes
g).a_i)=\,^{g^{-1}}a_i\otimes \,^{g^{-1}}a_i$, by definition of the
action of $G$ on $X\underset{A^e}{\otimes} A$. Therefore:
\begin{equation}
\begin{array}{rcl}
  \sum\limits_ip_{\Lambda}((ag\otimes hb)\otimes g)(x_i).a_i&=&
  \sum\limits_i(ag\otimes hb).p_A(f)(\,^{g^{-1}}x_i).\,^{g^{-1}}a_i\\
  &=&\sum\limits_i(ag\otimes hb)p_A(f)(x_i).a_i
\end{array}\notag
\end{equation}
where the last equality holds true because $\sum\limits_ix_i\otimes
a_i\in (X\underset{A^e}{\otimes}A)^G$. This proves that the maps
(\ref{eq9}) and (\ref{eq10}) are equal. Thus, (\ref{d6}) and
(\ref{d5}) commute, which proves that $\Pi_d(A,c)\rtimes G\simeq
\Pi_d(A\rtimes G,\overline c)$ where $\overline c$ is the image of $c$
under the map $\ts{HH}_{d-2}(A)\to \ts{HH}_{d-2}(A\rtimes G)$. The last assertion of the proposition
holds true because $\Pi_d(A\rtimes G,\overline c)$ itself is
Calabi-Yau of dimension $d$.
\end{proof}

\section{Application to Ginzburg algebras}

This section applies the previous study on crossed-products of
deformed Calabi-Yau completions to the particular case of Ginzburg
algebras.

\subsection{A reminder on Ginzburg algebras}
\label{sec:s4.1}

Ginzburg algebras were introduced by Ginzburg in \cite[4.2]{G06}. The
definition used here it the more general one of \cite[Sect. 6.3]{K09}.

Let $Q$ be a finite \emph{graded $\ts k$-quiver}, that is, the set
$Q_0$ of vertices is finite and the $\ts kQ_0$-bimodule $\ts kQ_1$
spanned by the (finite) set $Q_1$ of arrows has a
$\mathbb{Z}$-grading. Thus, the path algebra $\ts kQ$ is a graded
algebra, such that the idempotents associated to trivial paths have
degree $0$. As usual, the idempotent associated to a vertex $x$ is
denoted by $e_x$. A \emph{potential} on $Q$ is an element $W$ of the
quotient vector space $\left(\ts kQ\underset{(\ts kQ_0)^e}{\otimes}\ts
  kQ_0\right)/\left\langle uv\otimes
  e_x-(-1)^{\ts{deg}(u).\ts{deg}(v)}vu\otimes e_y\ |\
  u,v\right\rangle$ of $\ts kQ\underset{(\ts kQ_0)^e}{\otimes}\ts kQ_0$ by the
subspace generated by all elements of the form $uv\otimes
e_x-(-1)^{\ts{deg}(u).\ts{deg}(v)}vu\otimes e_y$ for homogeneous
 $u\in e_x\ts kQ$ and $v\in e_y\ts kQ$. Hence, $W$ is a
linear combination of oriented cycles in $Q$, each of which is
considered up to cyclic permutation with Koszul-type signs. The
grading on $\ts kQ$ defines a grading on the space of potentials.

Let $d\geqslant 3$ be an integer and $W$  a homogeneous of degree
$3-d$ potential on $Q$. The Ginzburg algebra $\mathcal A(Q,W)$
associated to $(Q,W)$ is defined as follows. As a graded algebra, it
is freely generated over $\ts kQ_0$ by the direct sum of the following
graded $\ts kQ_0$-bimodules:
\begin{enumerate}[-]
\item $\ts kQ_1$, hence $\ts kQ$ is a graded subalgebra of $\mathcal
  A(Q,W)$,
\item $\ts{Hom}_{(\ts kQ_0)^e}(\ts kQ_1,\ts kQ_0\otimes \ts
  kQ_0)[d-2]$, hence, every arrow $a\colon x\to y$ in $Q$ defines a
  new generator $a^*\colon y\to x$ and the linear combination
  $\sum\limits_t\lambda_ta_t^*\colon y\to x$ has degree $2-d-n$ if
  $\sum\limits_t\lambda_ta_t\colon x\to y$ has degree $n$ (with
  $\lambda_t\in\ts k$ and $a_t\in Q_1$),
\item $\ts kQ_0[d-1]$, hence, every vertex $i\in Q_0$ defines a new
  generator $c_i\colon i\to i$ of degree $1-d$.
\end{enumerate}
The differential on $\mathcal A(Q,W)$ (of degree $+1$) is uniquely
determined by the following rules:
\begin{enumerate}[-]
\item it vanishes on $\ts kQ$,
\item it maps $a^*\colon y\to x$ to the partial derivative
  $\partial_aW$,
\item it maps $c_i\colon i\to i$ to $\sum\limits_{a\colon i\to
    \cdot}aa^*-\sum\limits_{a\colon \cdot\to i}a^*a$, where the first
  sum runs over all the arrows in $Q$ starting in $i$ and the second
  sum runs over all the arrows in $Q$ arriving in $i$.
\end{enumerate}
Recall that the partial derivation $\partial_a$ is the linear map
which takes a path $p$ to:
\begin{equation}
  \partial_ap=\sum\limits_{p=p_1ap_2}p_2p_1\notag
\end{equation}
where the sum runs over all decompositions $p=p_1ap_2$ with paths
$p_1$ and $p_2$. Of course, this extends linearly to define the
partial derivation $\partial_u$ with respect to any $u\in\ts kQ_1$.

As said above, this definition is taken from
\cite[Sect. 6.2]{K09}. The original definition of Ginzburg corresponds
to the case where $\ts kQ$ has the trivial grading and $d=3$. It is
proved in \cite[Thm. 6.3]{K09} (see also \cite[Thm. A.12]{K09}) that
$\mathcal A(Q,W)$ is quasi-isomorphic to the deformed Calabi-Yau
completion $\Pi_d(\ts kQ,c)$, where the graded algebra $\ts kQ$ is
endowed with the zero differential and $c\in\ts{HH}_{d-2}(\ts kQ)$ is
as follows (see \cite[6.1 and 6.2]{K09} for more details): As a
homogeneous of degree $3-d$ potential, $W$ may be considered as lying
in the cyclic homology space $\ts{HC}_{d-3}(\ts kQ)$ (and any
element of $\ts{HC}_{d-3}(\ts kQ)$ arises in this way if $\ts{car}(\ts
k)=0$ or if $n=3$), then $c\in\ts{HH}_{d-2}(\ts kQ)$ is defined as the
image of $W$ under Connes' map $\ts{HC}_{d-3}(\ts
kQ)\xrightarrow{B}\ts{HH}_{d-2}(\ts kQ)$.

\subsection{The Morita reduced version of $\ts kQ\rtimes G $}
\label{sec:s4.2}

Let $Q$ be a graded $\ts k$-quiver and $G$ a finite group such that
$\ts{car}(\ts k)$ does not divide $\sharp G$. Assume that $G$ acts
on the
set $Q_0$ (with action denoted by $(g,i)\mapsto g.i$) and also
on the $\ts kQ_0$-bimodule $\ts kQ_1$ (with action denoted by
$(g,u)\mapsto \,^gu$ ), the
two
actions are supposed to be compatible to each other, that is, $^ga\in
e_{g.i}\ts kQ_1e_{g.j}$ if $a\in e_i\ts kQ_1e_j$ and $g\in G$. This defines an action of $G$
on $\ts kQ$ by graded automorphisms and any action of $G$ on $\ts kQ$
by graded automorphisms stabilising $Q_0$ and $\ts kQ_1$ arises in
this way.

It is proved in \cite{D09a} that if $\ts kQ$ has the trivial grading,
then there exists a quiver
$Q'$ such that $\mathcal D(\ts kQ')$ and $\mathcal D(\ts kQ\rtimes G)$
are equivalent. It is not difficult to check that the proof given
there works for the graded situation and actually shows the
existence of  a graded $\ts
k$-quiver $Q'$ and of an injective homomorphism of graded algebras $\ts
kQ'\to \ts kQ\rtimes G$ which, by restriction, induces an equivalence
$\mathcal D(\ts kQ\rtimes G)\to \mathcal D(\ts kQ')$. Here is a
description of $Q'$, the reader is referred to \cite{D09a} for a
detailed proof.

If a group $H$ acts on a set $E$ on the left, $\left[H\backslash
  E\right]$ will denote an arbitrarily fixed set of representatives
of the orbit set $H\backslash E$. For every $i\in Q_0$, let
$\kappa_i\in G$ be such that $\kappa_i.i\in\left[G\backslash
  Q_0\right]$. The stabilizer of $i$ is denoted by $G_i$.

The construction of $\ts kQ'$ is done in two steps (see
\cite[3.1]{D09a} for more details). Let $S$ be the semi-simple algebra
\begin{equation}
  S=\prod\limits_{i\in\left[G\backslash Q_0\right]}\ts kG_i\,.\notag
\end{equation}
For $i,j\in\left[G\backslash Q_0\right]$, consider the following
graded $\ts kG_i-\ts kG_j$-sub-bimodule of $\ts kQ\rtimes G$:
\begin{equation}
  \bigoplus\limits_{(i',j')\in\left[G\backslash G.i\times G.j\right]}
  G_i\kappa_{i'}^{-1} e_{i'}\ts kQ_1e_{j' }\kappa_{j'}G_j\subseteq\ts
  kQ\rtimes G\,.\notag
\end{equation}
Here, the action of $G$ on $G.i\times G.j$ is diagonal. Taking the
direct sum of all these modules for $(i,j)\in\left[G\backslash
  Q_0\right]\times \left[G\backslash Q_0\right]$ yields a graded
$S$-subbimodule of $\ts kQ\rtimes G$. Denote it by $M$. Then, the
inclusion $M\hookrightarrow \ts kG\rtimes G$ induces a
homomorphism of graded algebras $T_S(M)\to \ts kQ\rtimes G$. It is injective and the
associated restriction-of-scalars functor $\mathcal D(\ts kQ\rtimes
G)\to \mathcal D(T_S(M))$ is an equivalence. The second step is as
follows. For every $i\in\left[G\backslash Q_0\right]$, let
$e_1^i,\ldots,e_{n_i}^i$ be primitive (orthogonal) idempotents in $\ts
kG_i$ such that $e_1^i\ts kG_i,\ldots,e_{n_i}^i\ts kG_i$ form a
complete set of representatives of the irreducible representations of
$G_i$ (recall that $\sharp G\in \ts k^\times$). Let
$e=\sum\limits_{i,j}e_i^j\in S$. This is an idempotent and $eSe$ is
the algebra:
\begin{equation}
  eSe=\prod\limits_{i,j}\ts ke_j^i\,.\notag
\end{equation}
Moreover, $eMe$ is a graded $eSe$-sub-bimodule of $M$. Then, any
choice of a basis of $eMe$ yields a quiver $Q'$ and an isomorphism of graded algebras
$\ts kQ'\simeq T_{eSe}(eMe)$. Besides, the inclusion
$eMe\hookrightarrow M$ induces a  homomorphism of graded
algebras $\ts kQ'\simeq T_{eSe}(eMe)\to T_S(M)$. It is injective and the
associated restriction-of-scalars functor $\mathcal D(T_S(M))\to
\mathcal D(T_{eSe}(eMe))\simeq\mathcal D(\ts kQ')$ is an
equivalence. The composition $\ts kQ'\simeq T_{eSe}(eMe)\to T_S(M)\to
\ts kQ\rtimes G$ is the desired homomorphism of graded algebras whose
associated restriction-of-scalars functor $\mathcal D(\ts kQ\rtimes
G)\to \mathcal D(\ts kQ')$ is an equivalence. By construction, if $\ts
kQ$ has the trivial grading, then so does $\ts kQ'$.

\subsection{Crossed-products of Ginzburg algebras by finite groups}
\label{sec:s4.3}

Let $d\geqslant 3$ be an integer, $Q$ a graded $\ts k$-quiver and $W$
a homogeneous of degree $3-d$ potential on $Q$. Let $G$ be a finite
group such that $\sharp G\in\ts k^\times$ and assume that $G$ acts on the
set $Q_0$ and on the $\ts kQ_0$-bimodule $\ts kQ_1$ (with the two
actions compatible to each other), thus defining an action on $\ts kQ$
by graded automorphisms. This action extends to an action on $\mathcal
A(Q,W)$ by graded automorphisms as follows:
\begin{enumerate}[-]
\item given $g\in G$ and an arrow $a\colon x\to y$, if
  $^ga=\sum\limits_t\lambda_ta_t$ as linear combination of arrows in
  $Q$, then $^g(a^*)$ is defined as $^g a=\sum\limits_t\lambda_ta_t^*$,
\item given $g\in G$ and $i\in Q_0$, then $^gc_i=c_{g.i}$.
\end{enumerate}
Recall (\ref{sec:s4.2}) that
there exists a graded $\ts k$-quiver $Q$ and a morphism of graded
algebras $\ts kQ'\to \ts kQ\rtimes G$ whose associated
restriction-of-scalars functor $\mathcal D(\ts kQ\rtimes G)\to\mathcal
D(\ts kQ')$ is an equivalence.
\begin{prop}
  Keep the above setting. If $W\in\ts{HC}_{d-3}(\ts kQ)^G$, then the action of $G$ on
  $\mathcal A(Q,W)$ is an action by dg automorphisms. Moreover, there
  is a quasi-isomorphism of dg algebras
  \begin{equation}
  \mathcal A(Q,W)\rtimes G  \underset{\ts{qis}}{\simeq}\Pi_d(\ts kQ\rtimes G,c_1)\notag
  \end{equation}
   where
  $c_1\in\ts{HH}_{d-2}(\ts kQ\rtimes G)$ is the image of
  $W\in\ts{HH}_{d-3}(\ts kQ)$ under the composite map
  $\ts{HC}_{d-3}(\ts kQ)\to \ts{HC}_{d-3}(\ts kQ\rtimes
  G)\xrightarrow{B}\ts{HH}_{d-2}(\ts kQ\rtimes G)$. In particular,
  $\mathcal A(Q,W)\rtimes G$ is Calabi-Yau of dimension $d$.

  If, moreover, $\ts{car}(\ts k)=0$ or $d=3$, then the image of $W$
  under the composite map $\ts{HC}_{d-3}(\ts kQ)\to \ts{HC}_{d-3}(\ts kQ\rtimes
  G)\simeq \ts{HC}_{d-3}(\ts kQ')$ is a homogeneous of degree $d-3$
  potential $W'$ on $Q'$ and there exists a homomorphism of dg
  algebras
  \begin{equation}
    \mathcal A(Q',W')\to \mathcal A(Q,W)\rtimes G\notag
  \end{equation}
whose associated restriction-of-scalars functor $\mathcal D(\mathcal
A(Q,W)\rtimes G)\to \mathcal D(\mathcal A(Q',W))$ is an equivalence.
\end{prop}
\begin{proof}
  Since the restriction-of-scalars functor $\mathcal D(\ts kQ\rtimes
  G)\to \mathcal D(\ts kQ')$ is an equivalence, the canonical maps
  $\ts{HC}_{d-3}(\ts kQ')\to \ts{HC}_{d-3}(\ts kQ\rtimes G)$ and
  $\ts{HH}_{d-2}(\ts kQ')\to \ts{HH}_{d-2}(\ts kQ\rtimes G)$ are
  isomorphisms (\cite{K04}). These embed into the diagram:
  \begin{equation}
    \xymatrix{
\ts{HC}_{d-3}(\ts kQ)^G \ar@{^(->}[r] \ar[d] & \ts{HC}_{d-3}(\ts kQ)
\ar[r] \ar[d]_B & \ts{HC}_{d-3}(\ts kQ\rtimes G)  \ar[d]_B &
\ts{HC}_{d-3}(\ts kQ') \ar[l]_{\sim} \ar[d]^B\\
\ts{HH}_{d-2}(\ts kQ)^G \ar@{^(->}[r] & \ts{HH}_{d-2}(\ts kQ) \ar[r] &
\ts{HH}_{d-2}(\ts kQ\rtimes G) & \ts{HH}_{d-2}(\ts kQ') \ar[l]^{\sim}
}\notag
  \end{equation}
where $B$ stands for the Connes' map, the leftmost vertical arrow is
induced by $B$ and the horizontal arrows are the canonical
ones. Denote by $c\in\ts{HH}_{d-2}(\ts kQ)$, $c_1\in\ts{HH}_{d-2}(\ts
kQ\rtimes G)$, $c'\in\ts{HH}_{d-2}(\ts kQ')$ and
$W'\in\ts{HC}_{d-3}(\ts kQ')$ the corresponding images of
$W\in\ts{HC}_{d-3}(\ts kQ)$. Note that, if $\ts{car}\,\ts k=0$ or if
$d=3$, then $W'$ is a homogeneous of degree $3-d$ potential on
$Q'$.

Note that
$\partial_{^ga}(p)=\,^g\left(\partial_a(\,^{g^{-1}}p)\right)$, for
$g\in G$, $p\in \ts kQ$ and $a\in \ts kQ_1$. Therefore, if
$W\in\ts{HC}_{d-3}(\ts kQ)^G$, then:
\begin{enumerate}[-]
\item the action of $G$ on $\mathcal A(Q,W)$ is an action by dg
  automorphisms, and
\item $c\in\ts{HH}_{d-2}(\ts kQ)^G$, so that $G$ acts on $\Pi_d(\ts
  kQ,c)$ by dg automorphisms, following \ref{sec:s3.4}.
\end{enumerate}

From now on, suppose that $W\in\ts{HC}_{d-3}(\ts kQ)^G$. In
\cite[Thm. 6.3]{K09} is constructed a quasi-isomorphism of dg algebras
$\mathcal A(Q,W)\simeq \Pi_d(\ts kQ,c)$ which is easily checked to be
$G$-equivariant for the corresponding actions of $G$ on $\mathcal
A(Q,W)$ and $\Pi_d(\ts kQ,c)$, respectively. Whence a
quasi-isomorphism of dg algebras
\begin{equation}
  \label{eq15}
\mathcal A(Q,W)\rtimes G\underset{\ts{qis}}{\simeq}\Pi_d(\ts
kQ,c)\rtimes G\,.
\end{equation}
In particular, $\mathcal A(Q,W)\rtimes G$ is Calabi-Yau of dimension
$d$
because $c\in\ts{HH}_{d-2}(\ts kQ)^G$ and because of
\ref{sec:s3.4}. Next, according to \ref{sec:s3.4}, there is a
quasi-isomorphism of dg algebras
\begin{equation}
  \label{eq16}
\Pi_d(\ts kQ,c)\rtimes G\underset{\ts{qis}}{\simeq} \Pi_d(\ts
kQ\rtimes G,c_1)\,.
\end{equation}
Combining (\ref{eq15}) and (\ref{eq16}) shows that $\mathcal
A(Q,W)\rtimes G$ and $\Pi_d(\ts kQ\rtimes G,c_1)$ are
quasi-isomorphic.

Now assume that $\ts{car}(\ts k)=0$ or $d=3$. Then $W'$ is a
homogeneous of degree $3-d$ potential on $W$, as recalled
earlier. Applying  \cite[Thm. 5.5]{K09} to $\ts kQ'\to \ts kQ\rtimes G$
gives a morphism of dg algebras
\begin{equation}
  \label{eq17}
\Pi_d(\ts kQ',c')\to \Pi_d(\ts kQ\rtimes G,c_1)
\end{equation}
whose associated restriction-of-scalars functor $\mathcal D(\Pi_d(\ts
kQ\rtimes G,c_1))\to \mathcal D(\Pi_d(\ts kQ',c'))$ is an
equivalence. Finally, \cite[Thm. 6.3]{K09} gives a quasi-isomorphism
of dg algebras
\begin{equation}
  \label{eq18}
\Pi_d(\ts kQ',c')\underset{\ts{qis}}{\simeq} \mathcal A(Q',W')\
\text{if $\ts{car}(\ts k)=0$ or $d=3$}.
\end{equation}
Combining (\ref{eq15}), (\ref{eq16}), (\ref{eq17}) and (\ref{eq18})
shows that $\mathcal A(Q,W)\rtimes G$ and $\mathcal A(Q',W')$ have
equivalent derived categories.
\end{proof}

\section{Geometric interpretation of the compatibility condition}

Recall the following examples of Calabi-Yau algebras: The
algebra $\mathbb{C}[x_1,\ldots,x_n]$ of polynomials in $n$
variables; the coordinate ring $\mathbb{C}[X]$ of a smooth affine
variety with trivial canonical sheaf; and the
Weyl algebra. The crossed-product of any of these algebras by
a finite group $G$ has the Van den Bergh duality
(\cite[Prop. 7.1]{AFLS02}). Actually, it can be checked that the
compatibility condition of Theorem~\ref{thm2} is satisfied (and,
therefore, the crossed-product is Calabi-Yau, in each of these three examples) under the following
geometric conditions:
\begin{enumerate}[-]
\item $G<\ts{SL}_n(\mathbb{C})$, in the case of the algebra
  $\mathbb{C}[x_1,\ldots,x_n]$,
\item the action of $G$ on $\mathbb{C}[X]$ arises from an action of
  $G$ on $X$ such that the canonical sheaf is $G$-invariant,
\item $G<\ts{Sp}_{2n}(\mathbb{C})$, in the case of the Weyl algebra.
\end{enumerate}
Here are some details on this fact in the case of the Weyl algebra $A_n$ (the three
cases are very similar).

Denote by $V$ the vector space
$\mathbb{C}^{2n}=\bigoplus\limits_{i=1}^n\mathbb{C}.x_i\oplus\bigoplus\limits_{i=1}^n\mathbb{C}.\frac{\partial}{\partial
x_i}$
and let $G$ be a subgroup of
$\ts{Sp}(V)=\ts{Sp}_{2n}(\mathbb{C})$. Then $G$ acts on $V$ and this
action extends naturally to an action on $A_n$ by algebra
automorphisms. Recall that $A_n$ admits the following projective
resolution (that is, a cofibrant resolution) in $\mathcal C(A_n^e)$
(\cite[3.2]{S02a}):
\begin{equation}
  \label{eq11}
0\to \Lambda^{2n}V\otimes A_n^e\to \cdots\to\Lambda^dV\otimes
A_n^e\to\cdots\to V\otimes A_n^e\to A_n^e\to A_n\to 0
\end{equation}
where the differential $\Lambda^{d+1}V\otimes A_n^e\to
\Lambda^dV\otimes A_n^e$ is defined by
\begin{equation}
 (v_1\wedge\cdots\wedge v_{d+1})\otimes
u\mapsto
\sum\limits_{i=1}^{d+1}(-1)^{d-i+1}(v_1\wedge\cdots\wedge\widehat{v_i}\wedge\cdots\wedge
v_{d+1})\otimes (v_i\otimes 1-1\otimes v_i).u\,.\notag
\end{equation}
 This differential is
$G$-equivariant for the diagonal action of $G$ on $\Lambda^\bullet V\otimes
A_n^e$. As earlier, let $\Delta=\bigoplus\limits_{g\in G}A_ng\otimes
g^{-1}A_n$. Therefore, (\ref{eq11}) is a cofibrant resolution of $A_n$ in
$\mathcal C(\Delta)$ (\ref{sec:s2.2}, (e)). This implies that
$\ts{RHom}_{A_n^e}(A_n,A_n\otimes A_n)=\mathcal
Hom_{A_n^e}(\Lambda^\bullet V\otimes A_n^e,A_n\otimes A_n)$ in $\mathcal
C(\Delta^{op})$, which is isomorphic in $\mathcal C(\Delta^{op})$ to:
\begin{equation}
  \label{eq12}
0\to A_n^e\otimes \Lambda^0V^*\to\cdots\to A_n^e\otimes
\Lambda^dV^*\to A_n^e\otimes \Lambda^{d+1}V^*\to\cdots\to
A_n^e\otimes \Lambda^{2n}V^*\to 0
\end{equation}
where $A_n^e\otimes \Lambda^dV^*$ is in degree $d$; the action of $G$
on $A_n^e\otimes \Lambda^dV^*$ is diagonal (and given by the adjoint
action of $G$ on $V^*$); and the differential is defined by:
\begin{equation}
  \begin{array}{rcll}
    A_n^e\otimes \Lambda^dV^* & \longrightarrow & A_n^e\otimes
    \Lambda^{d+1}V^*\\
u\otimes e_{i_1}^*\wedge\cdots\wedge e_{i_d}^* & \longmapsto &
\sum\limits_{j=1}^{2n} u.(e_j\otimes 1-1\otimes e_j)\otimes
e_{i_1}^*\wedge\cdots\wedge e_{i_d}^*\wedge e_j^*&,
  \end{array}\notag
\end{equation}
where $(e_1^*,\ldots,e_{2n}^*)$ is the basis of $V^*$ dual to
$(e_1,\ldots,e_{2n})=(p_1,\ldots,p_n,q_1,\ldots,q_n)$. Since
$A_n^e\otimes \Lambda^dV^*$ is a projective left $A_n^e$-module for
every $d$, the complex (\ref{eq12}) is cofibrant in $\mathcal
C(\Delta^{op})$ and is therefore equal to $\theta_{A_n}$. Consider the
morphism of complexes $\theta_{A_n}[2n]\to A_n$ given by the
multiplication $A_n^e\otimes \Lambda^{2n}V^*=A_n^e\to
A_n$ in degree $0$ (and the zero map in non zero degrees). Since $A_n$
is Calabi-Yau of dimension $2n$, the cohomology of
$\theta_{A_n}[2n]$ vanishes in non-zero degrees. Besides,
$\theta_{A_n}[2n]\to A_n$ is clearly a quasi-isomorphism in degree
$0$. Finally, $\theta_{A_n}[2n]\to A_n$ is obviously a morphism in
$\mathcal C(\Delta^{op})$, that is, it is $G$-equivariant. This shows
that the compatibility condition
in Theorem~\ref{thm2} holds true if $G<\ts{Sp}_{2n}(\mathbb{C})$ acts
on $A_n$. As a consequence, the crossed-product $A_n\rtimes G$ is
Calabi-Yau of dimension $2n$.

\section*{Acknowledgements}

The author acknowledges Jacques Alev for stimulating comments
and questions

\bibliographystyle{plain}
\bibliography{biblio}

\end{document}


TITLE: Derive formula for the area of triangle provided that you know 2 angles and side between them.
QUESTION [1 upvotes]: I am trying to derive the formula for finding the area of the triangle that uses 2 angles and and 1 side between them.
Formula provided by the book:

If we know two angles (call them $A$ and $B$) and the side between
  them call it $b$, then Area = $\frac{b^2 \sin B \sin C}{2 \sin A}$

However, I keep getting different result.
My attempt as follows:
First we draw following picture:

$\angle YPX = 90^{\circ}$ and $\angle YPZ = 90^{\circ}$
We know that  $\angle YXZ = A$ and $\angle YZX = C$ and XZ = $b$
Let $XP = d$ and $PZ = b-d$
$\tan A = \frac{YP}{XP} = \frac{YP}{d} \implies YP = \tan A \cdot d$
$\tan C = \frac{YP}{PZ} = \frac{YP}{b-d} \implies YP = \tan C \cdot (b-d)$
It follows that $$\tan A \cdot d = \tan C \cdot (b-d)$$
$$d = \frac{\tan C \cdot (b-d)}{\tan A}$$
$$\frac{d}{b-d} = \frac{\tan C}{\tan A}$$
$$\frac{b-d}{d} = \frac{\tan A}{\tan C}$$
$$\frac{b}{d} - 1 = \frac{\tan A}{\tan C}$$
$$\frac{b}{d} = \frac{\tan A}{\tan C} + 1$$
$$\frac{1}{d} = \frac{\tan A}{\tan C \cdot b} + \frac{1}{b}$$
$$\frac{1}{d} = \frac{\tan A}{\tan C \cdot b} + \frac{\tan C}{\tan C \cdot b}$$
$$\frac{1}{d} = \frac{\tan A + \tan C}{\tan C \cdot b} $$
$$ d= \frac{\tan C \cdot b}{\tan A + \tan C} $$
$$ d= \frac{\frac{\sin C}{\cos C} \cdot b}{\frac{\sin A}{\cos A} + \frac{\sin C}{\cos C}} $$
$$ d= \frac{\frac{\sin C}{\cos C} \cdot b}{\frac{\sin A \cos C + \sin C \cos A}{\cos C \cos A} }$$
$$ d = \frac{\frac{\sin C}{\cos C} \cdot b \cdot \cos C \cos A}{\sin A \cos C + \sin C \cos A}$$
$$ d = \frac{\sin C \cos A\cdot b}{\sin A \cos C + \sin C \cos A}$$
We know that $$YP = \tan A \cdot d$$
$$YP = \frac{\sin A}{\cos A} \frac{\sin C \cos A\cdot b}{\sin A \cos C + \sin C \cos A}$$
$$YP = \frac{\sin A}{\cos A} \frac{\sin C \cos A\cdot b}{\sin(A+C)}$$
$$YP =  \frac{\sin A \sin C \cdot b}{\sin(A+C)}$$
Because
$$\text{Area} =\frac{1}{2} \cdot YP \cdot b$$
Then
$$\text{Area} =\frac{1}{2} \cdot \frac{\sin A \sin C \cdot b}{\sin(A+C)} \cdot b$$
$$\text{Area} = \frac{\sin A \sin C \cdot b^2}{2 \sin(A+C)} $$
Which, as can be seen, is different from the one provided by the textbook. I've tried to find areas of several triangles using formula above and this calculator, both give the same results. However, I'm still unsure whether it is correct. Thus I would like to ask you, is the the derivation (and the formula) above correct?

REPLY [1 votes]: It is simply use are of triangle is ∆=(1/2)cbSinA=(1/2)abSinB
Therefore ∆^2 = (1/4) acb^2 SinA.SinB
Also ∆=(1/2)acSinC
Dividing both we get Area ∆ =( c^2 SinA.SinB)/2 Sin C


TITLE: Given that $n=\prod^r_{i=1}p_i^{k_i}$, find $|\{x\in \mathbb{Z}/n\mathbb{Z}: x^a\}|$.
QUESTION [0 upvotes]: Given that $$n=\prod^r_{i=1}p_i^{k_i}$$
where $k_i\in \mathbb Z_+$ and $p_i$ is prime $\forall i\in\mathbb Z_+<r$, find $$|\{x\in \mathbb{Z}/n\mathbb{Z}: x^a\}|$$
(Note: This is a proof verification question. See my proof below. Feel free to comment on it, or provide your own proof if you want.)

REPLY [0 votes]: First, a useful lemma: $ad\equiv bd\pmod{dc}\Rightarrow a\equiv b\pmod c$. The proof goes as follows: Let $ad=ecd+f$ and $bd=gcd+f$. We note $f=ad-ecd\Rightarrow f=hd$ for an integer $h$. Dividing $d$ from both equations, we have $a=ec+h$ and $b=gc+h$. So, $a\equiv h\equiv b\pmod c$
$b\in\{x\in \mathbb{Z}/n\mathbb{Z}: x^a\}$ iff there exists $x$ such that $x^a\equiv b\pmod n$. By Chinese Remainder Theorem, we can split $x^a\equiv b\pmod n$ into $x^a\equiv b\pmod {p_1^{k_1}}$, $x^a\equiv b\pmod {p_2^{k_2}}$, $x^a\equiv b\pmod {p_3^{k_3}}$, e.t.c.
Consider the equation $x^a\equiv b\pmod{p_i^{k_i}}$ where $0\leq b<p_i^{k_i}$, and suppose $p_i\neq2$ and $p_i\not\mid b$. Then there exists a primitive root $g$ modulo $p_i^{k_i}$. Suppose $x\equiv g^k\pmod{p_i^{k_i}}$ and $b\equiv g^m\pmod{p_i^{k_i}}$. The equation becomes $g^{ka}\equiv g^m\pmod{p_i^{k_i}}$. As the multiplicative order of $g$ is $\varphi\left(p_i^{k_i}\right)=p_i^{k_i-1}(p_i-1)$ we need $ka\equiv m\pmod{p_i^{k_i-1}(p_i-1)}$. Let $a_i=\gcd(a,p_i^{k_i-1}(p_i-1))$. We note that $\gcd(m,p_i^{k_i-1}(p_i-1))=\gcd(ka,p_i^{k_i-1}(p_i-1))=ca_i$ for some integer $c$, hence $\gcd(m,p_i^{k_i-1}(p_i-1))=ca_i$. This implies $a_i|m$.
Now let $a=a_ib_i$, $m=m_ia_i$, and $p_i^{k_i-1}(p_i-1)=a_iq_i$, then $ka_ib_i\equiv m_ia_i\pmod{a_iq_i}\Rightarrow kb_i\equiv m_i\pmod{q_i}$. By definition $b_i$ and $q_i$ are coprime, hence by Beztout's Theorem there exists an inverse of $b_i$, and we can write $k\equiv m_ib^{-1}\pmod{q_i}$. Hence  $x^a\equiv b\pmod{p_i},p_i\not\mid b$  has a solution iff $b\equiv g^m\pmod{p_i}$ where $g$ is a primitive root and $\gcd(a,p_i-1)|k$. There are $\frac{p_i^{k_i-1}(p_i-1)}{\gcd\left(a,p_i^{k_i-1}(p_i-1)\right)}$ possible values of $k$ mod $p_i^{k_i-1}(p_i-1)$ and each one corresponds to a different value of $b\pmod{p_i}$. Hence we have $\frac{p_i^{k_i-1}(p_i-1)}{\gcd\left(a,p_i^{k_i-1}(p_i-1)\right)}$ possible values of $b\pmod{p_i^{k_i}}$.
When $k_i<3$ and $p_i=2$ and $2\not\mid b$, there exists a primitive root $g$. We can use the argument above to conclude there are $\frac{p_i^{k_i-1}(p_i-1)}{\gcd\left(a,p_i^{k_i-1}(p_i-1)\right)}$ possible values of $b$.
Now suppose $p_i=2$, $a_i\neq2$, $k_i\geq3$ and $2\not\mid b$. By Hensel's lifting lemma, for each solution to $y^a\equiv b\pmod{p_i}$ where $0\leq b<p_i$, there exists $x\pmod{p_i^{k_i}}$ such that $x\equiv y\pmod{p_i}$ and $x^a\equiv b\pmod{p_i^{k_i}}$, provided that $\frac d{dx}[x^a-b]=ax^a\not\equiv0\pmod{p_i}$. The above requirements prevent $ax^a\equiv0\pmod{p_i}$ so Hensel's lemma applies. Excluding the case $b\equiv 0\pmod{p_i}$ (as that means $p|b$), there are $\frac{p_i-1}{\gcd(a,p_i-1)}$ possible values of $b<p_i$ such that $y$ exists. We note that if $y^a\equiv b\pmod{p_i}$, then $y^a\equiv b+ck\pmod{p_i}$. As $0\leq k<p_i^{k_i-1}$ (we need $0\leq b<p_i^{k_i}$), so there are $\frac{p_i^{k_i}(p_i-1)}{\gcd(a,p_i-1)}=\frac{p_i^{k_i}(p_i-1)}{\gcd\left(a,p_i^{k_i}(p_i-1)\right)}$ values of $b\pmod{p_i^{k_i}}$ such that $b\not\mid p$ and $x$ exists.
Now suppose $a=2$, $p_i=2$, $2\not\mid b$. We cannot apply Hensel's lemma as $\frac d{dx} x^2-b=2x\equiv 0\pmod 2$. Luckily, Gauss has shown for us that the only odd quadratic residues mod $2^k$ are of the form $8n+1$. Thus, there are $2^{k_i-3}$ possible values of $b\pmod{p_i^{k_i}}$.
In conclusion, if $p_i\not\mid b$, then there are $\frac{p_i^{k_i}(p_i-1)}{\gcd\left(a,p_i^{k_i}(p_i-1)\right)}$ ways of selecting $b$ such that there exists an $x$ such that $x^a\equiv b\pmod{p_i}$ if $a\neq2$ or $p_i\neq2$ and $k_i<3$, and $2^{k_i-3}$ otherwise. Let us denote this function as $f(p_i,k_i,a)$.
Now suppose $p|b$. If $k_i\geq\nu_p(b)$, then $\nu_p\left(x^a\right)=a\nu_p(x)=\nu_p(b)$, so $a|\nu_p(b)$. Let $b\equiv c\cdot p^{ka}\pmod{p_i^{k_i}}$ where $ka<k_i$ and $0<c<p^{ka-k_i}$ (i.e. $0<b<p^{k_i}$) and $p_i\not\mid c$. We have $\nu_p\left(x^a\right)=a\nu_p(x)=ka$, hence $\nu_p(x)=k$. Let $x=y\cdot p^k$, so $x^a\equiv b\pmod{p_i^{k_i}}\Rightarrow y^ap^{ak}\equiv c\cdot p^{ka}\pmod{p_i^{k_i}}\Rightarrow y^a\equiv c\pmod{p_i^{k_i-ka}}$. There are $f(p_i,k_i-ka,a)$ possible values of $c$, and hence, $f(p_i,k_i-ka,a)$ possible of values of $b$ where $b=c\cdot p^{ka}$ for a given $k$ where $ka<k_i$. This means, when $b\equiv c\cdot p^{ka}\pmod{p_i^{k_i}}$ where $k=0$ to $k=\left\lceil\frac{k_i}a\right\rceil-1$, we have $\sum^{\left\lceil\frac{k_i}a\right\rceil-1}_{k=0}f(p_i,k_i-ka,a)$.
Finally, consider when $b\equiv 0\pmod{p_i^{k_i}}$ Obviously there is a solution to when $x^a\equiv0\pmod{p_i^{k_i}}$ ($x\equiv0\pmod{p_i^{k_i}}$ is a solution). Hence $b\equiv0\pmod{p_i^{k_i}}$ is a possible value.
Thus, there are $1+\sum^{\left\lceil\frac{k_i}a\right\rceil-1}_{k=0}f(p_i,k_i-ka,a)$ possible values of $b$ given that there exists $x$ such that $x^a\equiv b\pmod{p_i^{k_i}}$.
Now there is a unique $b\pmod n$ satisfying the linear congurences $b\equiv b_1\pmod{p_1^{k_1}}$, $b\equiv b_2\pmod{p_2^{k_2}}$, $b\equiv b_3\pmod{p_3^{k_3}}$, \cdots, $b\equiv b_r\pmod{p_r^{k_r}}$ by Chinese Remainder Theorem. There are $1+\sum^{\left\lceil\frac{k_1}a\right\rceil-1}_{k=0}f(p_1,k_1-ka,a)$ ways of selecting $b_1$, $1+\sum^{\left\lceil\frac{k_2}a\right\rceil-1}_{k=0}f(p_2,k_2-ka,a)$ ways of selecting $b_2$, ... $1+\sum^{\left\lceil\frac{k_r}a\right\rceil-1}_{k=0}f(p_r,k_r-ka,a)$ ways of selecting $b_r$. This gives $\prod_{i=1}^r1+\sum^{\left\lceil\frac{k_i}a\right\rceil-1}_{k=0}f(p_i,k_i-ka,a)$ ways of selecting $b_1,b_2,\cdots,b_r$, and hence $b$.
In conclusion, the number of solutions for $b$ is
$$\prod_{i=1}^r1+\sum^{\left\lceil\frac{k_i}a\right\rceil-1}_{k=0}f(p_i,k_i-ka,a)$$
where
$$f(x)=\left\{
\begin{matrix}
2^{k_i-3} \text{ if }p_i=a_i=2\text{ and }k_i\geq3 \\
\frac{p_i^{k_i}(p_i-1)}{\gcd\left(a,p_i^{k_i}(p_i-1)\right)}\text{ otherwise}
\end{matrix}\right.
$$


TITLE: Find the limit of $\lim\limits_{x\rightarrow0}\frac{x}{\tan x}$.
QUESTION [6 upvotes]: Find the limit of $\lim\limits_{x\rightarrow0}\frac{x}{\tan x}$

Clearly, since the limit takes the form of $\frac{0}{0}$, one should try L'Hopital's Rule. If we apply L'Hopital's Rule, the problem is that $\frac{d(\tan x)}{dx}=\cot x$, and $\cot(0)$ is undefined. Thus, we cannot find a limit using L'Hopital's Rule.
My problem set suggests that this question can be answered using little more than L'Hopital and Cauchy's Mean Value Theorem. Not sure how to proceed. Please help.

REPLY [1 votes]: Here are the steps
$$ \lim\limits_{x\to 0} \frac{x}{\tan x} = \lim\limits_{x\to 0} \frac{x}{\frac{\sin x}{\cos x}} = \lim\limits_{x\to 0} \frac{x\cos x}{\sin x}$$
$$ = \left(\lim\limits_{x\to 0} \frac{x}{\sin x}\right)\left(\lim\limits_{x\to 0} \cos x\right) = 1\cdot 1 = 1$$
Note that we can split the original limit into two convergent limits, as shown above.