---
abstract: 'Coalgebra-Galois extensions generalise Hopf-Galois extensions, which can be viewed as non-commutative torsors. In this paper it is analysed when a coalgebra-Galois extension is a separable, split, or strongly separable extension.'
address: 'Department of Mathematics, University of York, Heslington, York YO10 5DD, U.K.'
author:
- Tomasz Brzeziński
title: 'Coalgebra-Galois extensions from the extension theory point of view'
---

\#1 \#1 \#1 \#1 \#1

0.6in 23.6cm -1.5cm

\[section\] \[proposition\][Lemma]{} \[proposition\][Corollary]{} \[proposition\][Theorem]{}

\[proposition\][Definition]{} \[proposition\][Example]{}

\[proposition\][Remark]{}

[^1] [^2]

Introduction
============

Given a coalgebra $C$, an algebra $A$ and a right coaction $\rho^A: A\to A\otimes C$ one can define a fixed point subalgebra $B$ of $A$ as consisting of all those elements of $A$ over which the coaction is left-linear. In this way one obtains an extension $B\hookrightarrow A$, which is called a [*coalgebra-Galois extension*]{} if a certain canonical left $A$-module, right $C$-comodule map is bijective [@BrzMa:coa] [@BrzHaj:coa]. The aim of this article is to analyse such coalgebra-Galois extensions from the extension theory point of view. In particular we study the problem when such extensions are separable, split or strongly separable extensions. This problem is put in a broader context of [*entwining structures*]{} and [*entwined modules*]{} introduced in [@BrzMa:coa] [@Brz:mod], as a generalisation of a Doi-Hopf datum and Doi-Koppinen modules [@Doi:uni] [@Kop:var], respectively. We make use of the notion of a separability of a functor introduced in [@Nas:sep], and, as a byproduct, we generalise some of the results of [@CaeMil:sep] obtained recently for Doi-Koppinen modules.

The paper is organised as follows. In Section 2 we recall definitions and give examples of entwining structures and entwined modules. In Section 3 we analyse when certain functors between categories of entwined modules induced by morphisms of entwining structures are separable. In Section 4 we apply the results of Section 3 to prove that a sufficient and necessary condition for a coalgebra-Galois extension to be separable is the separability of a certain induction functor. This, in turn, is equivalent to the existence of a normalised integral in the canonical entwining structure. In Section 5 we analyse when a coalgebra-Galois extension is a split extension. This turns out to be related to the separability of the forgetful functor from the category of entwined modules to the category of right modules - another special case of the main theorem in Section 3. Finally, in Section 6 we study the problem when a coalgebra-Galois extension is a strongly separable extension in the sense of [@Kad:Jon].

We work over a commutative ring $k$ with identity 1. We assume that all the algebras are over $k$, associative and unital, and the coalgebras are over $k$, coassociative and counital. Unadorned tensor product is over $k$. For any $k$-modules $V, W$ the symbol ${\rm Hom}(V,W)$ denotes the $k$-module of $k$-linear maps $V\to W$ and the identity map $V\to V$ is denoted by $V$. The twist map between $k$-modules $V, W$ is denoted by $\twist: V\otimes W\to W\otimes V$, $v\otimes
w\mapsto w\otimes v$. We also implicitly identify $V$ with $V\otimes k$ and $k\otimes V$ via the canonical isomorphisms.

For a $k$-algebra $A$ we use $\mu_A$ to denote the product as a map and $1_A$ to denote the identity both as an element of $A$ and as a map $k\to
A$, $\alpha\mapsto \alpha 1_A$. $\M_A$ (resp. ${}_A\M$) denotes the category of right (resp. left) $A$-modules. The morphisms in this category are denoted by ${\rm Hom}_A(M,N)$ (resp. ${}_A{\rm Hom}(M,N)$). For any $M\in \M_A$ (resp. $M\in {}_A\M$), the symbol $\rho_M$ (resp.${}_M\rho$) denotes the action as a map (on elements the action is denoted by a dot). We often write $M_A$ (resp. ${}_AM$) to indicate in which context the $A$-module $M$ appears. For any $M\in \M^A$, $N\in {}^A\M$ we will write $\eq_{M_AN}:M\otimes
A\otimes N\to M\otimes N$ for the action equalising map defining tensor product $M\otimes_A N$, i.e., $\eq_{M_AN} = \rho_M\otimes
N-M\otimes{}_N\rho$, $M\otimes_A N = \coker(\eq_{M_AN})$.

For a $k$-coalgebra $C$ we use $\Delta_C$ to denote the coproduct and $\eps_C$ to denote the counit. Notation for comodules is similar to that for modules but with subscripts replaced by superscripts, i.e. $\M^C$ is the category of right $C$-comodules, $\rho^M$ is a right coaction etc. We use the Sweedler notation for coproducts and coactions, i.e. $\Delta_C(c)
= c\sw 1\otimes c\sw 2$, $\rho^M(m) = m\sw 0\otimes m\sw 1$ (summation understood). For any $V \in \M^C$, $W\in {}^C\M$, $V \square_{C}W$ denotes the cotensor product, which is defined by the exact sequence $$\begin{CD}
0 @>>> V\square_{C}W @>>> V\otimes W
@>{\eq^{V^CW}}>> V \otimes C\otimes W,  
\end{CD}$$ where $\eq^{V^CW}$ is the coaction equalising map, i.e., $\eq^{V^CW} = \rho^{V}\otimes W -
  V\otimes {}^{W}\!\rho$.

Preliminaries on entwining structures and coalgebra-Galois extensions
=====================================================================

An [*entwining structure*]{} (over $k$) is a triple $(A,C)_\psi$ consisting of a $k$-algebra $A$, a $k$-coalgebra $C$ and a $k$-linear map $\psi: C\tens A\to A\tens C$ satisfying $$\psi\circ(C\tens \mu_A) = (\mu_A\tens C)\circ (A\tens\psi)\circ(\psi\tens A),
\quad \psi\circ (C\tens 1_A) = 1_A\tens C,
\label{diag.A}$$ $$(A\tens\Delta_C)\circ\psi = (\psi\tens
C)\circ(C\tens\psi)\circ(\Delta_C\tens A), \quad (A\tens \eps_C)\circ\psi =
\eps_C\tens A.
\label{diag.B}$$ A morphism of entwining structures is a pair $(f,g):(A,C)_\psi\to
(\tilde{A}, \tilde{C})_{\tilde{\psi}}$, where $f:A\to \tilde{A}$ is an algebra map, $g: C\to \tilde{C}$ is a coalgebra map, and $(f\otimes g)\circ\psi =
\tilde{\psi}\circ(g\otimes f)$.

The category of entwining structures is a tensor category with tensor product $(A,C)_\psi \otimes (\tA, \tC)_\tpsi =
(A\otimes\tA, C\otimes\tC)_{(A\otimes\twist\otimes \tC)\circ
(\psi\otimes\tpsi)\circ (C\otimes\twist\otimes \tA)}$, and unit object $(k,k)_\twist$.

For $(A,C)_\psi$ we use the notation $\psi(c\otimes a) = a_\alpha \otimes c^\alpha$ (summation over a Greek index understood), for all $a\in
A$, $c\in C$. The notion of an entwining structure was introduced in [@BrzMa:coa Definition 2.1]. It is self-dual in the sense that conditions in Definition \[ent\] are invariant under the operation consisting of interchanging of $A$ with $C$, $\mu_A$ with $\Delta_C$, and $1_A$ with $\eps_C$, and reversing the order of maps. Below are two classes of examples of entwining structures coming from Galois-extensions.

Let $C$ be a coalgebra, $A$ an algebra and a right $C$-comodule. Let $B:= \{b\in A\; | \; \rho^A(ba) =
b\rho^A(a)\}$ and assume that the canonical left $A$-module, right $C$-comodule map $
\can:A\otimes _BA\to A\otimes C$, $a\otimes a'\mapsto a\rho^A(a')$, is bijective. Let $\psi:C\otimes A\to A\otimes C$ be a $k$-linear map given by $
\psi(c\otimes a) = \can(\can^{-1}(1_A\otimes c)a).
$ Then $(A,C)_\psi $ is an entwining structure. The extension $B\hookrightarrow A$ is called a [*coalgebra-Galois extension*]{} (or a [*$C$-Galois extension*]{}) and is denoted by $A(B)^C$. $(A,C)_\psi$ is the [*canonical entwining structure*]{} associated to $A(B)^C$. A coalgebra-Galois extension $A(B)^C$ is said to be [*copointed*]{} if there exists a group-like $e\in C$ such that $\rho^A(1_A)=1_A\otimes e$. \[can.ex\]

Dually we have

Let $A$ be an algebra, $C$ a coalgebra and a right $A$-module. Let $B:= C/I$, where $I$ is a coideal in $C$, $$I:={\rm span}\{(c\cdot a)\sw 1\xi((c\cdot a)\sw 2)-c\sw 1\xi(c\sw
2\cdot a)\; |\; a\in A,\; c\in C,\; \xi\in C^*\},$$ and assume that the canonical left $C$-comodule, right $A$-module map $
\cocan:C\otimes A\to C\square_B C$, $c\otimes a \mapsto c\sw 1\otimes c\sw
2\cdot a$, is bijective. Let $\psi:C\otimes A\to A\otimes C$ be a $k$-linear map given by $
\psi = (\eps_C\otimes A\otimes C)\circ (\cocan^{-1}\otimes C)\circ
(C\otimes \Delta_C)\circ \cocan.
$ Then $(A,C)_\psi $ is an entwining structure. The coextension $C\twoheadrightarrow B$ is called an [*algebra-Galois coextension*]{} (or an [*$A$-Galois coextension*]{}) and is denoted by $C(B)_A$. $(A,C)_\psi$ is the [*canonical entwining structure*]{} associated to $C(B)_A$. An algebra-Galois coextension $C(B)_A$ is said to be [*pointed*]{} if there exists an algebra map $\kappa:A\to k$ such that $\eps_C\circ \rho_C =
\eps_C\otimes\kappa$. \[cocan.ex\]

Associated to an entwining structure is the category of entwined modules.

Let $(A,C)_\psi$ be an entwining structure. An (entwined) [*$(A,C)_\psi$-module*]{} is a right $A$-module, right $C$-comodule $M$ such that $$\rho^M\circ\rho_M = (\rho_M\tens C)\circ(M\tens\psi)\circ(\rho ^M\otimes
A),$$ (explicitly: $
\rho^M(m\cdot a) = m\sw 0\cdot a_\alpha\tens m\sw 1^\alpha$, $ \forall
a\in A, m\in M$). A morphism of $(A,C)_\psi$-modules is a right $A$-module map which is also a right $C$-comodule map. The category of $(A,C)_\psi$-modules is denoted by $\M_A^C(\psi)$. \[def.mpsi\]

The category $\M_A^C(\psi)$ was introduced and studied in [@Brz:mod]. An example of such modules are Doi-Koppinen modules introduced in [@Doi:uni], [@Kop:var]. In this paper we will be concerned with two covariant functors between categories of entwined modules, which are special cases of the construction in [@Brz:mod Section 3][^3] (see also [@CaeRai:ind] for the Doi-Koppinen case). These functors are induced by certain morphisms of entwining structures.

Let $(f,g) :(A,C)_\psi \to (\tA,\tC)_\tpsi$ be a morphism of entwining structures. View $C$ as a left $\tC$-comodule via ${}^C\!\rho = (g\otimes C)\circ\Delta_C$ and $C\otimes\tA$ as a right $\tC$-comodule via $\rho^{C\otimes\tA} = (C\otimes\tpsi)\circ(C\otimes
g\otimes\tA)\circ(\Delta_C\otimes\tA)$. Then $(f,g)$ is said to be an [*admissible morphism*]{} iff:\
(i) for all $\tM\in\M_\tA^\tC(\tpsi)$, $\tM\square_\tC (C\otimes C) = (\tM\square_\tC C)\otimes C$,\
(ii) for all $M\in \M_A$, $(M\otimes
(C\otimes\tA))\square_\tC C = M\otimes
((C\otimes\tA)\square_\tC C)$. \[admissible\]

For example, if $C,\tC$ are $k$-flat then $(f,g)$ is an admissible morphism provided that ${}^\tC \!C$ is coflat. On the other hand if $k$ is a regular ring or a field every morphism is admissible. Also, it can be easily checked that the following morphisms $(A,\eps_C): (A,C)_\psi\to (A,k)_\twist$ and $(1_A,
C): (k,C)_\twist\to (A,C)_\psi$ are admissible.

Let $(f,g) :(A,C)_\psi \to (\tA,\tC)_\tpsi$ be an admissible morphism of entwining structures. View $\tA$ as a right $A$-module via $\rho_\tA = \tmu\circ(\tA\otimes
f)$, and $C$ as a right $\tC$-comodule via $\rho^C = (C\otimes g)\circ\Delta_C$. Then:

\(1) For any $\tM\in\M_\tA^\tC(\tpsi)$, $\tM\square_\tC C$ is an $(A,C)_\psi$-module with structure maps $\rho^{\tM\square_\tC C} =
\tM\otimes \Delta_C$ and $$\rho_{\tM\square_\tC C} :\tM\square_\tC C\otimes A\to \tM\square_\tC C, \qquad
\sum_i \tm_i\otimes  c_i\otimes a =
\sum_i\tm_if(a_\alpha)\otimes c_i^\alpha.$$

\(2) For any $M\in\M_A^C(\psi)$, $M\otimes_A\tA$ is an $(\tA,\tC)_\tpsi$-module with structure maps $\rho_{M\otimes_A\tA} =
M\otimes_A\tmu$ and $$\rho^{M\otimes_A\tA}:M\otimes_A\tA\to M\otimes_A\tA\otimes\tC, \qquad
m\otimes\ta\mapsto m\sw0\otimes\ta_\alpha\otimes g(m\sw 1)^\alpha.$$

\(3) The covariant functor $-\square_\tC C:\M_\tA^\tC(\tpsi) \to
\M_A^C(\psi)$ is the right adjoint of $-\otimes_A\tA:\M_A^C(\psi)\to
\M_\tA^\tC(\tpsi)$. The adjunctions are: $$\forall M\in\M_A^C(\psi), \qquad \Phi_M: M\to (M\otimes_A\tA)\square_\tC C,
\qquad m\mapsto m\sw 0\otimes 1_\tA\otimes m\sw 1,$$ $$\forall \tM\in \M_\tA^\tC(\tpsi), \quad \Psi_\tM : (\tM\square_\tC
C)\otimes _A\tA\to \tM, \quad \sum_i \tm_i\otimes c_i\otimes \ta
\mapsto \sum_i\tm_i\cdot\ta\eps_C(c_i).$$ \[ex.functors\]

Applying Example \[ex.functors\] to morphisms $(A,\eps_C): (A,C)_\psi\to (A,k)_\twist$ and $(1_A,
C): (k,C)_\twist\to (A,C)_\psi$ one obtains

Let $(A,C)_\psi$ be an entwining structure. Then

\(1) If $M$ is a right $A$-module then $M\otimes C$ is an $(A,C)_\psi$-module with the coaction $M\otimes \Delta_C$ and the action $(m\otimes c)\cdot a =
m\cdot\psi(c\otimes a)$, for all $a\in A, c\in C$ and $m\in M$. In particular $A\otimes C\in \M^C_A(\psi)$. The operation $M\mapsto M\otimes C$ defines a covariant functor $-\otimes C:\M_A\to \M_A^C(\psi)$ which is the right adjoint of the forgetful functor $\M_A^C(\psi)\to \M_A$.

\(2) If $V$ is a right $C$-comodule then $V\otimes A\in \M^C_A(\psi)$ with the action $V\otimes\mu_A$ and the coaction $v\otimes a\mapsto v\sw
0\otimes \psi(v\sw 1\otimes a)$ for any $a\in A$ and $v\in V$. In particular $C\otimes A\in \M^C_A(\psi)$. The operation $V\mapsto V\otimes A$ defines a covariant functor $-\otimes C:\M^C\to \M_A^C(\psi)$, which is the left adjoint of the forgetful functor $\M_A^C(\psi)\to \M^C$. \[ex.psi.modules\]

Another class of examples of entwined modules comes from (co)algebra-Galois (co)extensions [@BrzHaj:coa]

 \
(1) Let $(A,C)_\psi$ be the canonical entwining structure associated to a coalgebra-Galois extension $A(B)^C$. Then $A$ is an $(A,C)_\psi$-module via $\rho^A$ and $\mu_A$.

\(2) Let $(A,C)_\psi$ be the canonical entwining structure associated to an algebra-Galois coextension $C(B)_A$. Then $C$ is an $(A,C)_\psi$-module via $\rho_C$ and $\Delta_C$. \[ex.psi.galois\]

Separable functors of entwined modules
======================================

In this section we analyse when functors described in Example \[ex.functors\] are separable. Recall from [@Nas:sep] that a covariant functor $F :{\cal C}\to {\cal D}$ is [*separable*]{} if the natural transformation ${\rm Hom}_{\cal C}(-,-)\to {\rm Hom}_{\cal
D}(F(-), F(-))$ splits. In this paper we are dealing with the pairs of adjoint functors, so that the following characterisation of separable functors, obtained in [@Raf:sep] [@Rio:cat], is of great importance

Let $G:{\cal D}\to {\cal C}$ be the right adjoint of $F :{\cal C}\to {\cal
D}$ with adjunctions $\Phi: 1_{\cal C}\to GF$ and $\Psi:FG\to 1_{\cal
D}$. Then

\(1) $F$ is separable if and only if $\Phi$ splits, i.e., for all objects $C\in{\cal C}$ there exists a morphism $\nu_C \in{\rm Mor}_{\cal C}
(GF(C), C)$ such that $\nu_C\circ \Phi_C = C$ and for all $f\in {\rm
Mor}_{\cal C} (C, \tC)$, $\nu_\tC\circ GF(f) = f\circ\nu_C$.

\(2) $G$ is separable if and only if $\Psi$ cosplits, i.e., for all objects $D\in {\cal D}$ there exists a morphism $\nu_D \in{\rm Mor}_{\cal D}
(D, FG(D))$ such that $\Psi_D\circ \nu_D = D$ and for all $f\in {\rm
Mor}_{\cal D} (D, \tD)$, $\nu_\tD\circ f = FG(f)\circ\nu_D$. \[thm.sep\]

An admissible morphism $(f,g): (A,C)_\psi\to (\tA,\tC)_\tpsi$ of entwining structures is said to be:

\(1) [*integrable*]{} if there exists $\lambda \in {\rm Hom}_A((C\otimes
\tA)\square_\tC C, A)$ such that the following diagrams commute $$\begin{CD}
(C\otimes A\otimes \tA)\square_\tC C @>{\psi\otimes\tA\otimes C}>>
A\otimes (C\otimes \tA)\square_\tC C @>{A\otimes\lambda}>> A\otimes A\\
@VV{C\otimes f\otimes\tA\otimes C}V     @.     @VV{\mu_A}V\\
(C\otimes \tA\otimes \tA)\square_\tC C @>{C\otimes\tmu\otimes C}>>
(C\otimes \tA)\square_\tC C @>{\lambda}>> A,
\end{CD}
\label{int.a}$$ $$\begin{CD}
(C\otimes\tA)\square_\tC C @>{C\otimes\tA\otimes \Delta_C}>> 
(C\otimes \tA)\square_\tC C\otimes C @>{\lambda\otimes C}>> A\otimes C\\
@VV{\Delta_C\otimes\tA\otimes C}V   @.   @|\\
C\otimes (C\otimes \tA)\square_\tC C @>{C\otimes\lambda}>> C\otimes A
@>{\psi}>> A\otimes C.
\end{CD}
\label{int.b}$$ The right $A$-module structure of $(C\otimes\tA)\square_\tC C$ is as in Example \[ex.functors\](1), explicitly $\rho_{(C\otimes\tA)\square_\tC C}:
c'\otimes \ta\otimes c\otimes a\mapsto c'\otimes \ta f(a_\alpha)\otimes
c^\alpha$.

\(2) [*totally integrable*]{}, if there exists $\lambda\in {\rm Hom}_A((C\otimes
\tA)\square_\tC C, A)$ making it an integrable morphism and such that the following diagram $$\begin{CD}
C @>{\Delta_C}>> C\otimes C\\
@VV{1_A\circ\eps_C}V   @VV{C\otimes 1_\tA\otimes C}V\\
A @<{\lambda}<<  (C\otimes \tA)\square_\tC C
\end{CD}
\label{int.c}$$ commutes. \[def.int.morphism\]

Notice that the condition (\[int.a\]) makes sense because $\psi$ is a morphism in $\M_A^C(\psi)$, $(f,g)$ is admissible and $(C\otimes\tmu)\circ(C\otimes f\otimes
\tA): C\otimes A\otimes\tA\to C\otimes\tA$ is a left $\tC$-comodule map, where the $k$-modules involved are left $\tC$-comodules via $(g\otimes C)\circ\Delta_C \otimes A\otimes\tA$ and $(g\otimes C)\circ\Delta_C \otimes
\tA$, respectively. Similarly, condition (\[int.b\]) makes sense because $\Delta_C$ is a left $\tC$-comodule map and $\Delta_C\otimes\tA$ is a morphism in $\M_\tA^\tC(\tpsi)$. Dually to Definition \[def.int.morphism\] one considers

An admissible morphism $(f,g): (A,C)_\psi\to (\tA,\tC)_\tpsi$ of entwining structures is said to be:

\(1) [*cointegrable*]{} if there exists $\z \in {\rm Hom}^\tC(\tC, 
(\tA\otimes
C)\otimes_A \tA)$ such that the following diagrams commute $$\begin{CD}
\tC @>{\z}>> (\tA\otimes C)\otimes_A\tA
@>{(\tA\otimes\Delta_C)\otimes_A\tA}>> (\tA\otimes C\otimes C)\otimes_A\tA
\\ 
@VV{\tDelta}V    @.     @VV{(\tA\otimes g\otimes C)\otimes_A\tA}V\\
\tC\otimes \tC @>{\tC\otimes \z}>> \tC\otimes (\tA\otimes C)\otimes_A\tA
@>{\tpsi\otimes\tC\otimes\tA}>> (\tA\otimes\tC\otimes C)\otimes_A\tA
\end{CD}
\label{coint.a}$$ $$\begin{CD}
\tC\otimes\tA @>{\tpsi}>> \tA\otimes\tC @>{\tA\otimes \z}>> (\tA\otimes
\tA\otimes C)\otimes_A\tA\\
@|   @.   @VV{(\tmu\otimes C)\otimes_A\tA}V\\
\tC\otimes\tA @>{\z\otimes\tA}>> (\tA\otimes C)\otimes_A\tA\otimes\tA
@>{(\tA\otimes C)_A\otimes\tmu}>> (\tA\otimes C)\otimes_A\tA.
\end{CD}
\label{coint.b}$$ The right $\tC$-comodule structure of $(\tA\otimes C)\otimes_A\tA$ is as in Example \[ex.functors\](2), explicitly $\rho^{(\tA\otimes C)\otimes_A\tA}:
\ta\otimes c\otimes\ta' \mapsto \ta\otimes c\sw 1\otimes \ta'_\alpha
\otimes g(c\sw 2)^\alpha$.

\(2) [*totally cointegrable*]{}, if there exists $\z \in {\rm Hom}^\tC(\tC, 
(\tA\otimes C)\otimes_A \tA)$ making it a cointegrable morphism and such that the following diagram $$\begin{CD}
\tC @>{\z}>> (\tA\otimes C)\otimes_A\tA\\
@VV{1_\tA\circ\teps}V   @VV{(\tA\otimes\eps_C)\otimes_A\tA}V\\
\tA @<{\mu_{\tA A}}<<  \tA\otimes_A\tA
\end{CD}
\label{coint.c}$$ commutes. Here $\mu_{\tA, A}:\tA\otimes_A\tA\to \tA$ is the natural map induced by $\mu_\tA$. \[def.coint.morphism\]

The right actions of $A$ on the $k$-modules involved in the above definition are as follows. For any $a\in A$, $\ta,\ta'\in\tA$, $c,c'\in C$, $\tc\in
\tC$: $(\ta\otimes c)\cdot a = \ta f(a_\alpha)\otimes c^\alpha$, $(\ta\otimes c\otimes c') = \ta f(a_{\alpha\beta})\otimes c^\beta\otimes
c^{\prime\alpha}$, $(\ta\otimes \tc\otimes c)\cdot a = \ta
f(a_\alpha)_\beta\otimes \tc^\beta\otimes c^\alpha$, $(\tc\otimes
\ta\otimes c)\cdot a = \tc\otimes \ta f(a_\alpha)\otimes c^\alpha$, $(\ta\otimes\ta'\otimes c)\cdot a  = \ta\otimes \ta' f(a_\alpha)\otimes
c^\alpha$. Using properties of entwining structures and the fact that $(f,g)$ is a morphism of entwining structures one can easily convince oneself that all the maps featuring in Definition \[def.coint.morphism\] are well-defined.

With these definitions at hand we can now state the main result of this section.

Let $(f,g): (A,C)_\psi\to (\tA,\tC)_\tpsi$ be an admissible morphism of entwining structures.

\(1) If for all $M\in \M_A^C(\psi)$, $(M\otimes_A\tA)\square_\tC C \subseteq
\coker(\eq_{M_A\tA}\square_\tC C)$, then the functor $-\otimes_A\tA:\M_A^C(\psi)\to \M_\tA^\tC(\tpsi)$ is separable if and only if $(f,g)$ is totally integrable.

\(2) If for all $\tM\in\M_\tA^\tC(\tpsi)$, $\ker(\eq^{\tM^\tC C}\otimes_A\tA)
\subseteq 
(\tM\square_\tC C)\otimes_A\tA$, then the functor $-\square_\tC C:\M_\tA^\tC(\tpsi)\to \M_A^C(\psi)$ is separable if and only if $(f,g)$ is totally cointegrable. \[thm.main\]

\(1) Let $(f,g)$ be totally integrable and assume that $\lambda$ is as in Definition \[def.int.morphism\]. For all $M\in \M_A^C(\psi)$ define $$\tnu_M : (M\otimes\tA)\square_\tC C\to M, \qquad
\sum_im_i\otimes\ta_i\otimes c_i \mapsto \sum_i m_i\sw
0\cdot\lambda(m_i\sw 1\otimes\ta_i\otimes c_i).$$ Notice that the map $\tnu_M$ is well-defined since the fact that $(f,g)$ is admissible implies that for any $x\in (M\otimes\tA)\square_\tC C$, one has $(\rho^M\otimes \tA\otimes
C)(x)\in (M\otimes (C\otimes\tA))\square_\tC C) =  M\otimes 
((C\otimes\tA)\square_\tC C)$. Take any $x= \sum_im_i\cdot
a_i\otimes \ta_i\otimes c_i\in (M\otimes\tA)\square_\tC C$. Then $$\begin{aligned}
\tnu_M(x) & = & \sum_i(m_i\cdot a_i)\sw 0\cdot\lambda((m_i\cdot a_i)\sw
1\otimes \ta_i\otimes c_i)\\
& = &  \sum_im_i\sw 0\cdot a_{i\alpha}\lambda(m_i\sw
1^\alpha \otimes \ta_i\otimes c_i)\qquad \qquad \mbox{\rm
($M\in\M_A^C(\psi)$)} \\
& = &  \sum_im_i\sw 0\cdot \lambda(m_i\sw
1 \otimes f(a_i)\ta_i\otimes c_i)\qquad \qquad \mbox{\rm
(by (\ref{int.a}))} \\
& = & \tnu_M(\sum_i m_i\otimes f(a_i)\ta_i\otimes c_i).\end{aligned}$$ The above calculation means that ${\rm Im}(\eq_{M_A\tA}\square_\tC
C)\subseteq \ker\tnu_M$, and together with the assumption that $-\square_\tC C$ preserves the cokernel of the action equalising map $\eq_{M_A\tA}$ imply that one can define the map $\nu_M:(M\otimes_A\tA)\square_\tC C\to
M$ by the diagram $$\begin{CD}
(M\otimes_A \tA)\square_\tC C \!@>>> \!\coker(\eq_{M_A\tA}\square_\tC
C)\!  @>>>
\!((M\otimes \tA)\square_\tC C)/\ker\tnu_M \! @>>>\!  M\\
@. @. @AAA @|\\
@. @. (M\otimes \tA)\square_\tC C @>{\tnu_M}>> M
\end{CD}$$ Slightly abusing the notation we will still write $\nu_M: 
\sum_im_i\otimes\ta_i\otimes c_i \mapsto \sum_i m_i\sw
0\cdot\lambda(m_i\sw 1\otimes\ta_i\otimes c_i)$.

To show that $\nu_M$ is a right $A$-module map, take any $a\in A$ and $x=\sum_i
m_i\otimes \ta_i\otimes c_i \in (M\otimes_A\tA)\square_\tC C$ and compute $$\begin{aligned}
\nu_M(x\cdot a) & = & \nu_M(\sum_im_i\otimes \ta_if(a_\alpha)\otimes
c_i^\alpha) \\
& = & \sum_i m_i\sw 0\cdot \lambda(m_i\sw 1\otimes
\ta_if(a_\alpha)\otimes c_i^\alpha)\\
& = & \sum_i m_i\sw 0\cdot \lambda(m_i\sw 1\otimes
\ta_i\otimes c_i)a \qquad \mbox{\rm ($f\in {\rm Hom}_A((C\otimes
\tA)\square_\tC C, A)$)}\\
& = & \nu_M(x)\cdot a.\end{aligned}$$ Furthermore we have $$\begin{aligned}
\nu_M(x\sw 0)\otimes x\sw 1 & = & \sum_i m_i\sw 0\cdot 
\lambda(m_i\sw 1\otimes
\ta_i\otimes c_i\sw 1)\otimes c_i\sw 2\\
& = &\sum_i m_i\sw 0\cdot \lambda(m_i\sw 2\otimes
\ta_i\otimes c_i)_\alpha\otimes m_i\sw 1^\alpha \qquad \mbox{\rm (by
(\ref{int.b}))} \\
& = &  \sum_i \rho^M(m_i\sw 0\cdot \lambda(m_i\sw 1\otimes
\ta_i\otimes c_i)) \qquad \qquad \mbox{\rm ($M\in \M_A^C(\psi)$)} \\
& = &  \rho^M\circ\nu_M(x),\end{aligned}$$ which proves that $\nu_M$ is a right $C$-comodule map. Using (\[int.c\]) one easily finds that the adjunction $\Phi_M$ is splitted by $\nu_M$. It remains to be shown that $\nu_M$ is natural in $\M_A^C(\psi)$. Take any $M,N\in
\M_A^C(\psi)$ and $\phi \in {\rm Hom}_A^C(M,N)$. Then $$\begin{aligned}
\nu_N(\sum_i\phi(m_i)\otimes \ta_i\otimes c_i) &=& \sum_i \phi(m_i)\sw
0\cdot\lambda(\phi(m_i)\sw 1\otimes\ta_i\otimes c_i)\\
&=& \sum_i \phi(m_i\sw 0)\cdot\lambda(m_i\sw 1\otimes\ta_i\otimes c_i)
\\
&=& \sum_i \phi(m_i\sw 0\cdot\lambda(m_i\sw 1\otimes\ta_i\otimes c_i))
\\
&=& \phi\circ\nu_M(x),\end{aligned}$$ where we used that $\phi$ is a right $C$-comodule and right $A$-module map to derive the second and the third equalities respectively. This completes the proof that the functor $-\square_\tC C$ is separable.

Conversely, assume that $-\square_\tC C$ is separable and let $\nu_M$ be the corresponding splitting of $\Phi_M$. Define $$\lambda : (C\otimes\tA)\square_\tC C\to A, \qquad \lambda =  
(A\otimes\eps_C)\circ\nu_{A\otimes C}(1_A\otimes (C\otimes\tA)\square_\tC
C).$$ Since $\nu_{A\otimes C}$ is a right $A$-linear map, so is $\lambda$. We first show that $\nu_M$ can be expressed in terms of $\lambda$. For any $M\in \M_A$ and $m\in M$ consider a morphism $\ell_m: A\otimes C\to M\otimes C$ in $\M_A^C(\psi)$ given by $a\otimes c\mapsto m\cdot
a\otimes c$. Since the splitting of the adjunction $\Phi$ is natural in $\M_A^C(\psi)$ we have $$\ell_m\circ \nu_{A\otimes C} = \nu_{M\otimes C}\circ
((\ell_m\otimes_A \tA)\square_\tC C).
\label{ell}$$ In particular, choosing $M=A$ one easily finds that (\[ell\]) implies that $\nu_{A\otimes
C}$ is a left $A$-module map. Now, if $M\in \M_A^C(\psi)$ one can take the morphism $\rho^M\in {\rm
Hom}_A^C(M, M\otimes C)$, and thus using the naturality of $\nu$, obtain $
\rho^M\circ\nu_M = \nu_{M\otimes C}\circ (\rho^M\otimes C)$. In view of (\[ell\]) this reads for all $\sum_i m_i\otimes\ta_i\otimes c_i\in
(M\otimes \tA)\square_\tC C$ projected down to $(M\otimes_A \tA)\square_\tC C$ $$\rho^M\circ\nu_M(\sum_i m_i\otimes\ta_i\otimes c_i) = \sum_i\ell_{m_i\sw 0}\circ\nu_{A\otimes
C}(1_A\otimes m_i\sw 1\otimes \ta_i \otimes c_i).$$ Applying $M\otimes \eps_C$ to this last equality and using assumption that $(f,g)$ is admissible one obtains $$\nu_M(\sum_i m_i\otimes \ta_i\otimes c_i) = \sum_i m_i\sw
0\cdot\lambda(m_i \sw 1\otimes\ta_i\otimes c_i).$$ In particular, the choice $M=A\otimes C$ gives for all $a\in A$, $\sum_ic_i\otimes\ta_i\otimes c'_i \in (C\otimes\tA)\square_\tC C$ $$\nu_{A\otimes C}(\sum_i a\otimes c_i\otimes\ta_i \otimes c'_i) = 
a\sum_i\lambda( c_i\sw 2\otimes \ta_i \otimes c'_i)_\alpha\otimes 
c_i\sw 1^\alpha .
\label{nuac}$$ We are now ready to show that $\lambda$ satisfies all the conditions of Definition \[def.int.morphism\]. Take any $x= \sum_ic_i\otimes\ta_i\otimes c'_i\in (C\otimes \tA)\square_\tC
C$, then $$\begin{aligned}
\sum_i \lambda(c_i\otimes\ta_i\otimes c_i'\sw 1) \otimes c_i'\sw 2 
&=& \sum_i (A\otimes \eps_C)\circ
\nu_{A\otimes C} (1_A\otimes c_i\otimes\ta_i \otimes c_i'\sw 1)\otimes
c'_i \sw 2\\
& = & (A\otimes \eps_C\otimes
C)\circ(A\otimes\Delta_C)\circ\nu_{A\otimes C}(1_A\otimes x)\\
& = & \nu_{A\otimes C}(1_A\otimes x)\\
& = & \sum_i\lambda(c_i\sw 1\otimes\ta_i\otimes c_i')_\alpha\otimes c_i
\sw 2^\alpha 
\qquad \mbox{\rm (by (\ref{nuac}))},\end{aligned}$$ where we used that $\nu_{A\otimes C}$ is a right $C$-comodule map to derive the second equality. This proves that $\lambda$ satisfies (\[int.b\]). Furthermore, for all $\sum_i c_i\otimes a_i\otimes\ta_i
\otimes c'_i\in (C\otimes A\otimes \tA)\square_\tC C$ we have $$\begin{aligned}
\lambda(\sum_ic_i\otimes f(a_i)\ta_i\otimes c'_i) & = &
(A\otimes\eps_C)\circ\nu_{A\otimes C} (\sum_i 1_A\otimes c_i\otimes 
f(a_i)\ta_i\otimes c'_i) \\
&=& (A\otimes\eps_C)\circ\nu_{A\otimes C} (\sum_i (1_A\otimes c_i)\cdot a_i
\otimes \ta_i\otimes c'_i) \\
&=& (A\otimes\eps_C)\circ\nu_{A\otimes C} (\sum_i a_{i\alpha} \otimes
c_i^\alpha \otimes \ta_i\otimes c'_i) \\
&=& \sum_i a_{i\alpha}\lambda(c_i^\alpha \otimes \ta_i\otimes c'_i),\end{aligned}$$ where we used the properties of the domain of $\nu_{A\otimes C}$ and the assumption that $-\square_\tC C$ preserves cokernel of $\eq_{M_A\tA}$ to derive the second equality. This proves that $\lambda$ satisfies (\[int.a\]). Finally, for all $c\in C$, $\Phi_{A\otimes C}(1_A\otimes c) = 1_A\otimes c\sw 1\otimes 1_\tA\otimes
c\sw 2$. Since $\nu_{A\otimes C}$ splits $\Phi_{A\otimes C}$ we have $1\otimes c = \nu_{A\otimes C}( 1_A\otimes c\sw 1\otimes 1_\tA\otimes
c\sw 2)$. Applying $A\otimes\eps_C$ to this equality one immediately deduces that $\lambda$ satisfies (\[int.c\]). Therefore the morphism $(f,g)$ is totally integrable. This completes the proof of the first statement of the theorem.

\(2) Given $\z$ as in Definition \[def.coint.morphism\] define for all $\tM\in \M_A^C(\tpsi)$, $\nu_\tM: \tM\to (\tM\square_\tC C)\otimes_A\tA$, $\nu_\tM = (\rho_\tM\otimes C\otimes_A
\tA)\circ(\tM\otimes\z)\circ\rho^\tM$. The proof that $\nu_\tM$ is the required cosplitting is dual to the proof of the corresponding part of assertion (1). Conversely, given a cosplitting $\nu_\tM$ define $\zeta =
(\eps_\tC\otimes\tA\otimes C\otimes_A \tA)\circ\nu_{\tC\otimes\tA}\circ 
(\tC\otimes 1_\tA)$.

Notice that the assumption of Theorem \[thm.main\](1) is satisfied if ${}^\tC C$ is coflat. Dually, the assumption of Theorem \[thm.main\](2) is satisfied if ${}_A\tA$ is flat. The remainder of the paper is devoted to the analysis of special cases of Theorem \[thm.main\].

Separable coalgebra-Galois extensions
=====================================

The following notion was introduced in [@Brz:Fro]. It generalises the notion of an $H$-integral for a Doi-Hopf datum [@CaeMil:Doi Definition 2.1].

Let $(A,C)_\psi$ be an entwining structure. An [*integral*]{} in $(A,C)_\psi$ is an element $\z=\sum_i a_i\otimes c_i\in A\otimes C$ such that for all $a\in A$, $a\cdot \z = \z\cdot a$. Explicitly, we require $\sum_i
aa_i\otimes c_i = \sum_ia_i\psi(c_i\otimes a)$. An integral $\z=\sum_i a_i\otimes c_i$ is said to be [*normalised*]{} if $\sum_i a_i\eps_C(c_i)=1$. \[def.int\]

Let $A$ be a Hopf algebra and $B\subset A$ be a left $A$-comodule subalgebra, i.e., a subalgebra of $A$ such that $\Delta_A(B) \subset
A\otimes B$. Consider the coalgebra $C/B^+A$. $C$ is a right $A$-module in the natural way and there is an entwining structure $(A,C)_\psi$ with $\psi: c\otimes a\mapsto a\sw 1\otimes c\cdot a\sw 2$. Let $\Lambda \in
C$ be such that for all $a\in A$, $\Lambda\cdot a = \eps_A(a)\Lambda$ and $\eps_C(\Lambda) = 1$. Then $\z = 1\otimes \Lambda$ is an integral in $(A,C)_\psi$. \[ex.homog.int\]

Clearly, $1_A\eps(\Lambda) = 1_A$. Take any $a\in A$, then $
(1\otimes\Lambda)\cdot a = a\sw 1\otimes \Lambda\cdot a\sw 2 =
a\otimes\Lambda = a\cdot(1\otimes \Lambda).
$

In [@Brz:Fro] it has been shown that the existence of an integral in $(A,C)_\psi$ is closely related to the fact that the functor $-\otimes
C:\M_A\to \M_A^C(\psi)$ of Example \[ex.psi.modules\](1) is both left and right adjoint of the forgetful functor $\M_A^C(\psi)\to \M_A$. The following theorem, which is an entwining structure version of [@CaeMil:sep Theorem 2.14], shows that integrals are closely related to the separability of $-\otimes C$.

Let $(A,C)_\psi$ be an entwining structure. The functor $-\otimes C
:\M_A\to \M_A^C(\psi)$ is separable if and only if there exists a normalised integral in $(A,C)_\psi$. \[thm.int\]

Consider an admissible morphism $(A,\eps_C) : (A,C)_\psi\to
(A,k)_\twist$. Then $-\otimes C = -\square_kC:\M_A\to \M_A^C(\psi)$. Since ${}_AA$ is flat, Theorem \[thm.main\](2) can be applied and thus $-\otimes C$ is separable if and only if $(A,\eps_C)$ is totally cointegrable, i.e. there exists $\z\in {\rm
Hom}(k, (A\otimes C)\otimes_AA) \cong A\otimes C$ such that conditions (\[coint.a\])–(\[coint.c\]) are satisfied. In this case condition (\[coint.a\]) is empty, while condition (\[coint.b\]) means that $\z$ is an integral in $(A,C)_\psi$. Finally, condition (\[coint.c\]) states that $\z$ is normalised.

The existence of normalised integrals in the canonical entwining structure associated to a coalgebra-Galois extensions turns out to be equivalent to the separability of such an extension. First, recall from [@HirSug:sem]

An extension of algebras $B\hookrightarrow A$ is [*separable*]{} if there exists $u\in A\otimes_B A$ such that for all $a\in A$, $au=ua$ and $\mu_{A,B}(u)=1_A$, where $\mu_{A,B}:A\otimes_B A\to A$ is the natural map induced by $\mu_A$. The element $u$ is called a [*separability idempotent*]{}. \[def.sep\]

A coalgebra-Galois extension $A(B)^C$ is separable if and only if there exists a normalised integral in the canonical entwining structure. \[prop.galois.sep\]

We first show that $\can^{-1}: A\otimes C\to A\otimes_B A$ is an $(A,A)$-bimodule map, where the $(A,A)$-bimodule structure on $A\otimes
C$ is as in Definition \[def.int\]. By construction, $\can^{-1}$ is a left $A$-module map. For all $\z=\sum_i a_i\otimes c_i\in A\otimes C$, $a\in A$ $$\begin{aligned}
\can^{-1}(\z\cdot a) &=& \can^{-1}(\sum_i a_ia\sb\alpha\otimes
c_i\sp\alpha) = \sum_i a_i\can^{-1}(a\sb\alpha\otimes
c_i\sp\alpha)\\
& = & \sum_ia_i\can^{-1}(\can(\can^{-1}(1_A\otimes c_i)a))
\qquad\qquad 
\mbox{\rm (def.\ of canonical $\psi$)}\\
& = & \sum_ia_i\can^{-1}(1_A\otimes c_i)a = \can^{-1}(\z)a.\end{aligned}$$ Therefore $\z$ is an integral in $(A,C)_\psi$ if and only if for all $a\in A$, $au=ua$, where $u=\can^{-1}(\z)$. Furthermore, directly from the definition of the canonical map $\can$, one finds that $(A\otimes \eps_C)\circ \can =
\mu_{A,B}$. Therefore $\z$ is normalised if and only if $\mu_{A,B}(u) = 1_A$.

In the setting of Example \[ex.homog.int\], view $A$ as a right $C$-comodule via $\rho^A = (A\otimes\pi)\circ\Delta_A$, where $\pi: A\to
C = A/B^+A$ is the canonical surjection, and assume that $B = \{b\in B
\; | \; \forall a\in A, \; \rho^A(ba) = b\rho^A(a)\}$ (for example, this holds if either ${}_BA$ or $A_B$ is faithfully flat). Then $A(B)^C$ is a coalgebra-Galois extension, and if there is $\Lambda \in
C$ such that for all $a\in A$, $\Lambda\cdot a = \eps_A(a)\Lambda$ and $\eps_C(\Lambda) = 1$, then $B\hookrightarrow A$ is separable. \[ex.homog.sep\]

The introduction of separable extensions in [@HirSug:sem] was motivated by the Hochschild relative homological algebra [@Hoc:rel]. In the case of a coalgebra-Galois extension the relationship between cohomology and separable extensions can be expressed in terms of integrals in the canonical entwining structure. Recall from [@Hoc:rel] that if $B$ is a subalgebra of $A$ then for every $(A,A)$-bimodule $M$ the [*relative Hochschild cohomology groups*]{} $H^n(A,B, M)$ are defined as cohomology groups of the complex $(\bigoplus
_{n=0}^\infty C^n(A,B,M), \delta)$, where $
C^0(A,B,M) = \{m\in M \; |\; \forall b\in B,\; b\cdot m = m \cdot b\}$, $$C^n(A,B,M) = {}_B{\rm
Hom}_B(\underbrace{A\otimes_B A\otimes_B \cdots \otimes_B A}_{\mbox{\rm
$n$-times}},M),  \qquad n>0,$$ and the coboundary $ \delta :C^{n}(A,B,M)\to C^{n+1}(A,B,M)$ is given by $$\begin{aligned}
\delta(f)(a_1,\ldots 
,a_{n+1}) &=& a_1\cdot f(a_2,\ldots a_{n+1}) + \sum_{i=1}^{n} (-1)^if(a_1,\ldots
,a_ia_{i+1}, \ldots ,a_{n+1})\\
&& +(-1)^{n+1} f(a_1,\ldots , a_n)\cdot
a_{n+1}. \end{aligned}$$

Let $A(B)^C$ be a coalgebra-Galois extension. Then a normalised integral in the associated canonical entwining structure exists if and only if for all $(A,A)$-bimodules $M$, $H^1(A,B,M) = 0$. \[cor.sep1\]

By an argument similar to [@DeMIng:sep p. 76], one shows, that the first relative Hochschild cohomology group is trivial for all $(A,A)$-bimodules if and only if the extension $B\hookrightarrow A$ is separable. Then the assertion follows from Proposition \[prop.galois.sep\].

Let $A(B)^C$ be a coalgebra-Galois extension with a normalised integral in the canonical entwining structure. Then any $(A,A)$-bimodule which is semisimple as a $(B,B)$-bimodule is semisimple as an $(A,A)$-bimodule. \[cor.sep2\]

By Corollary \[cor.sep1\], for all $(A,A)$-bimodules $M$, $H^1(A,B,M)= 0$. Then [@Hoc:rel Theorem 1] implies the assertion.

Dually one can consider

Let $(A,C,\psi)$ be an entwining structure. A $k$-module map $\y:
C\otimes A\to k$, such that for all $a\in A$, $c\in C$, $c\sw
1\y(c\sw 2\otimes a) = \y(c\sw 1\otimes a_\alpha)c\sw 2^\alpha$ is called a [*cointegral*]{} in $(A,C)_\psi$. A cointegral $\y$ is said to be [*normalised*]{} if $\y\circ(C\otimes 1_A) = \eps_C$. \[def.coint\]

Let $C$ be a Hopf algebra and let $A$ be a right $C$-comodule algebra. Then $(A,C)_\psi$ is an entwining structure with $\psi: c\otimes a\mapsto a\sw 0\otimes ca\sw
1$. Let $\kappa\in A^*$ be such that $\kappa(1_A) = 1$ and for all $a\in
A$, $1_C\kappa(a) =
\kappa(a\sw 0)a\sw 1$. Then $\y = \eps_C\otimes \kappa$ is a normalised cointegral in $(A,C)_\psi$. \[ex.homog.coint\]

Let $(A,C)_\psi$ be an entwining structure. The functor $-\otimes A
:\M^C\to \M_A^C(\psi)$ is separable if and only if there exists a normalised cointegral in $(A,C)_\psi$. \[thm.coint\]

Consider the morphism $(1_A, C): (k,C)_\twist\to (A,C)_\psi$ and apply Theorem \[thm.main\](1).

A coextension of coalgebras $C\twoheadrightarrow B$ is said to be a [ *separable coextension*]{} if there exists a $k$-module map $\upsilon : C\square_B C\to k$ such that $(C\otimes\upsilon)\circ (\Delta_C\otimes C) = (\upsilon\otimes
C)\circ(C\otimes \Delta_C)$ on $C\square_B C$, and $\upsilon\circ\Delta_C = 
\eps_C$. \[def.coseparable\]

An algebra-Galois coextension $C(B)_A$ is separable if and only if there exists a normalised cointegral in the associated canonical entwining structure. \[prop.coseparable\]

Let $C$ be a Hopf algebra and $A\subset C$ a right comodule subalgebra of $C$, i.e., $\Delta_C(A)\subset A\otimes C$, so that we are in the setting of Example \[ex.homog.coint\]. Consider the coalgebra $B = C/CA^+$, and assume that $A= \{a\in C\; | \; \pi(a\sw 1)\otimes
a\sw 2 = \pi(1_C)\otimes a\}$, where $\pi:C\to B$ is the canonical surjection (this assumption is satisfied if either ${}_AC$ or $C_A$ is faithfully flat). Then $C\twoheadrightarrow B$ is an $A$-Galois coextension and if there exists $\kappa\in A^*$ such that for all $a\in A$, $\kappa(a\sw 1)a\sw 2 =
\kappa(a)\eps_C$ and $\kappa(1_A) = 1$, then this coextension is separable. \[ex.homog.sep\*\]

When $k = \bf C$, a rich source of separable coalgebra-Galois extensions is provided by quantum homogeneous spaces of compact quantum groups [@Wor:com]. In this case we are in the setting of Example \[ex.homog.sep\*\], with $C$ a compact quantum group and $A$ a right $C$-homogeneous quantum space. In many cases $C$ is a faithfully flat right or left $A$-module (see [@MulSch:hom] for examples). The map $\kappa$ is the Haar measure on $C$ restricted to $A$. Perhaps the simplest example of this situation is when $C$ is the quantum $SU(2)$ group and $A$ is any of the quantum 2-spheres of Podleś [@Pod:sph].

Split coalgebra-Galois extensions
=================================

The following definition is a slightly modified version of [@Brz:Fro Definition 4.1]; both definitions describe the same object if a coalgebra $C$ is a finitely-generated projective $k$-module.

Let $(A,C)_\psi$ be an entwining structure. Any $\gamma\in {\rm
Hom}(C\otimes C, A)$ such that the following diagrams $$\begin{CD}
C\otimes C @>{C\otimes \Delta_C}>> C\otimes C\otimes C @>{
\gamma\otimes C}>> A \otimes C \\
@VV{\Delta_C\otimes C}V @.       @|\\
C\otimes C\otimes C @>{C\otimes \gamma}>> C\otimes A @>{\psi}>> A\otimes C
\end{CD}
\label{int.i}$$ $$\begin{CD}
C\otimes C\otimes A @>{\gamma\otimes A}>> A\otimes A @>{\mu_A}>> A\\
@VV{C\otimes\psi}V  @.  @AA{\mu_A}A\\
C\otimes A\otimes C @>{\psi\otimes C}>> A\otimes C\otimes C
@>{A\otimes\gamma}>> A\otimes A
\end{CD}
\label{int.ii}$$ commute is called an [*integral map*]{} in $(A,C)_\psi$. An integral map $\gamma$ is said to be [*normalised*]{}, if for all $c\in C$, $\gamma(c\sw 1\otimes c\sw 2) = \eps_C(c)1_A$. \[def.integral.map\]

The following theorem is an entwining structure version of [@CaeMil:sep Theorem 2.3].

The forgetful functor $\M_A^C(\psi)\to \M_A$ is separable if and only if there exists a normalised integral map in $(A,C)_\psi$. \[thm.separable\]

Consider an admissible morphism $(A,\eps_C):(A,C)_\psi\to (A,k)_\twist$. Then $-\otimes_AA:\M_A^C(\psi)
\to \M_A$ is the forgetful functor. In this case $(C\otimes A)\square_k C =
C\otimes A\otimes C$, and for all $M\in\M_A$, $\eq_{M_AA} = M$ so that the assumption of Theorem \[thm.main\](1) holds. Therefore the forgetful functor is separable if and only if $(A,\eps_C)$ is totally integrable, i.e., iff there exists $\lambda\in {\rm Hom}_A(C\otimes
A\otimes C,A)$ satisfying all the conditions of Definition \[def.int.morphism\]. Assume that such a $\lambda$ exists and define $\gamma = \lambda\circ(C\otimes 1_A\otimes C):C\otimes C\to
A$. Then for all $a\in A$, $c,c'\in C$ we have $$\begin{aligned}
a_{\alpha\beta}\gamma(c^\beta\otimes c^{\prime\alpha}) & = &
a_{\alpha\beta}\lambda(c^\beta\otimes 1_A\otimes c^{\prime\alpha}) =
\lambda(c\otimes a_\alpha\otimes c^{\prime\alpha}) \qquad\mbox{\rm (by
(\ref{int.a}))}\\ 
& = & \lambda(c\otimes 1_A\otimes c')a = \gamma(c\otimes c')a,\end{aligned}$$ where we used that $\lambda$ is a right $A$-module map to derive the penultimate equality. Hence the diagram (\[int.i\]) commutes. Also, (\[int.b\]) implies that the diagram (\[int.ii\]) commutes, while the normalisation of $\gamma$ follows immediately from (\[int.c\]). Thus we conclude that $\gamma$ is a normalised integral map as required.

Conversely, assume that $\gamma$ is a normalised integral map and define $\lambda :C\otimes A\otimes C\to A$, $c\otimes a\otimes c' \mapsto
a_\alpha\gamma(c^\alpha\otimes c')$. For all $a,a'\in A$, $c,c'\in C$ we have $$a_\alpha\lambda(c^\alpha\otimes a'\otimes c') = a_\alpha
a'_\beta\gamma(c^{\alpha\beta}\otimes c') =
(aa')_\alpha\gamma(c^\alpha\otimes c') = \lambda(c\otimes aa'\otimes
c'),$$ where (\[diag.A\]) was used to obtain the third equality. This proves that the diagram (\[int.a\]) commutes. Furthermore $$\begin{aligned}
\lambda(c\sw 2\otimes a\otimes c')_\alpha\otimes c\sw 1^\alpha & = &
(a_\delta\gamma(c\sw 2^\delta\otimes c'))_\alpha\otimes c\sw 1^\alpha\\
& = &
a_{\delta\alpha}\gamma(c\sw 2^\delta\otimes c')_\beta\otimes c\sw
1^{\alpha\beta} \qquad \qquad \mbox{\rm (by (\ref{diag.A}))}\\
& = & a_{\alpha}\gamma(c^\alpha\sw 2\otimes c')_\beta\otimes c^\alpha \sw
1^{\beta} \qquad\qquad  \mbox{\rm (by (\ref{diag.B}))}\\ 
& = & a_{\alpha}\gamma(c^\alpha\otimes c'\sw 1)_\beta\otimes c'\sw
2 \qquad \qquad\mbox{\rm (by (\ref{int.ii}))}\\
& = & \lambda(c\otimes a\otimes c'\sw 1)\otimes c'\sw 2.\end{aligned}$$ This proves that diagram (\[int.b\]) commutes. Also, $$\begin{aligned}
\lambda(c\otimes aa'_\alpha\otimes c^{\prime\alpha}) & = &
(aa'_\alpha)_\beta \gamma (c^\beta\otimes c^{\prime\alpha}) = a_\beta
a'_{\alpha\delta}  \gamma (c^{\beta\delta}\otimes c^{\prime\alpha})
\qquad \mbox{\rm (by (\ref{diag.A}))}\\
& = & a_\beta\gamma(c^\beta\otimes c')a' = \lambda(c\otimes a\otimes a')
\qquad\qquad \quad  \mbox{\rm (by (\ref{int.i}))}.\end{aligned}$$ Therefore $\lambda$ is a right $A$-module map, and, consequently the morphism $(A,\eps_C)$ is integrable. The fact that it is totally integrable follows immediately from the normalisation of $\gamma$.

Let $(A,C)_\psi$ be the canonical entwining structure associated to a pointed algebra-Galois coextension $C(k)_A$ of $k$. Then the forgetful functor $\M_A^C(\psi)\to \M_A$ is separable. \[ex.separable\]

Since $B=k$, $C\square_B C = C\otimes C$, and we define $\gamma = (\eps_C\otimes A)\circ \cocan^{-1}:C\otimes C\to A$. We show that $\gamma$ is a normalised integral map. First notice that since $\cocan^{-1}$ is a left $C$-comodule map, one has $\cocan^{-1} =
(C\otimes\gamma)\circ(\Delta_C\otimes C)$. Applying the definition of the canonical entwining map in Example \[cocan.ex\] to $\cocan^{-1}$ one immediately obtains $\psi\circ \cocan^{-1} = 
(\gamma\otimes C)\circ(C\otimes\Delta_C)$, i.e. $\psi\circ(C\otimes\gamma)\circ(\Delta_C\otimes C) = (\gamma\otimes
C)\circ(C\otimes\Delta_C)$. Thus we conclude that $\gamma$ satisfies condition (\[int.i\]).

Let $\kappa:A\to k$ be the algebra map making $C(k)_A$ a pointed algebra-Galois coextension. One easily finds that $\rho_C =
(\kappa\otimes C)\circ \psi$ and $C\otimes \kappa = (C\otimes\eps_C)\circ
\cocan$. The map $\gamma$ is the [*cotranslation map*]{}, so, as explained in [@BrzHaj:coa Theorem 3.5], it has the following properties $$\mu_A\circ(\gamma\otimes A) = \gamma\circ(C\otimes\rho_C),
\label{cot.2}$$ $$\mu_A\circ(\gamma\otimes\gamma)\circ(C\otimes \Delta_C\otimes C) =
\gamma\circ(C\otimes\eps_C\otimes C).
\label{cot.3}$$ Using all these properties we obtain $$\begin{aligned}
\mu_A\circ(\gamma\otimes A) & = &
\gamma\circ(C\otimes\rho_C)\qquad\qquad\qquad\qquad\qquad
\qquad\qquad\qquad\qquad  \mbox{\rm (by
(\ref{cot.2}))}\\ 
& = & \gamma\circ(C\otimes\kappa\otimes C)\circ(C\otimes\psi) \\
& = & \gamma\circ(C\otimes\eps_C\otimes  C)\circ(\cocan\otimes
C)\circ(C\otimes\psi)\\ 
& = & \mu_A\circ(\gamma\otimes\gamma)\circ(C\otimes \Delta_C\otimes
C)\circ(\cocan\otimes C)\circ (C\otimes\psi) \quad \mbox{\rm (by
(\ref{cot.3}))}\\ 
& = & \mu_A\circ(A\otimes\gamma)\circ(\psi\otimes C)\circ (C\otimes\psi)
\qquad\qquad\qquad \qquad \mbox{\rm (def. of $\psi$)}.\end{aligned}$$ This proves that $\gamma$ is an integral map. Finally, $\gamma$ is normalised by the normalisation property of the cotranslation map (cf. [@BrzHaj:coa Theorem 3.5]).

As explained in [@CaeMil:sep] the separability of the forgetful functor implies various Maschke-type theorems. Thus, similarly as in [@Brz:Fro] we have

If there is a normalised integral map in $(A,C)_\psi$, then

\(1) Every object in $\M_A^C(\psi)$ which is semisimple as an object in $\M_A$ is semisimple as an object in $\M_A^C(\psi)$.

\(2) Every object in $\M_A^C(\psi)$ which is projective (resp. injective) as a right $A$-module is a projective (resp. injective) object in $\M_A^C(\psi)$.

\(3) $M\in \M_A^C(\psi)$ is projective as a right $A$-module if and only if there exists $V\in \M^C$ such that $M$ is a direct summand of $V\otimes A$ in $\M_A^C(\psi)$ ($V\otimes A$ is an entwined module by Example \[ex.psi.modules\](2)).

In the case of a coalgebra-Galois extension, the existence of normalised integral maps in the canonical entwining structure is closely related to the coalgebra-Galois extension being a split extension. Recall from [@Pie:ass][@Kad:Jon]

An extension of algebras $B\hookrightarrow A$ is called a [*split extension*]{} if there exists a unital $(B,B)$-bimodule map $E:A\to B$. The map $E$ is called a [*conditional expectation*]{}. \[def.split\]

A coalgebra-Galois extension $A(B)^C$ is a split extension if and only if there exists $\phi\in{\rm Hom}(C,A)$ such that

\(i) $\forall c\in C, \qquad \psi(c\sw 1\otimes \phi(c\sw 2)) =
\phi(c)\rho^A(1_A)$,

\(ii) $\sum_i a^i\phi(c_i) = 1_A$, where $\sum_i a^i\otimes c_i =
\rho^A(1_A)$.

\(iii) $\forall b\in B,\; c\in C,\qquad  b_\alpha\phi(c^\alpha) = 
\phi(c)b$. \[proposition.split\]

As explained in the proof of [@Brz:mod Proposition 4.4], given a unital $(B,B)$-bimodule map $E:A\to B$ there exists $\phi\in{\rm
Hom}(C,A)$ satisfying conditions (i)–(iii). Explicitly, $\phi =
(A\otimes_B E)\circ \can^{-1}\circ (1_A\otimes C)$. Conversely, given $\phi\in{\rm Hom}(C,A)$ satisfying (i), [@Brz:mod Theorem 4.3] implies that $E:A\to B$, $a\mapsto a\sw 0\phi(a\sw 1)$ is a left $B$-module map. Clearly, condition (ii) implies $E$ is unital. Furthermore, for all $a\in A$, $b\in B$ $$E(ab) = (ab)\sw 0\phi((ab)\sw 1) = a\sw 0b_\alpha\phi(a\sw 1^\alpha) =
a\sw 0\phi(a\sw 1) b = E(a)b,$$ where we used that $A\in \M_A^C(\psi)$ and the assumption (iii) to derive the second and third equalities respectively. This completes the proof.

Let $(A,C)_\psi$ be an entwining structure and assume that $A\in\M_A^C(\psi)$. Define $B$ as in Example \[can.ex\]. Then one can consider a covariant functor $(-)_0: \M_A^C(\psi)
\to \M_{B}$ $$M\mapsto M_0 := \{m\in M\; | \; \forall a\in A, \;\; \rho^M(m\cdot a) =
m\rho^A(a)\}.$$ Notice, in particular, that $B=A_0$. As explained in [@Brz:mod] the functor $(-)_0$ is the right adjoint of the functor $-\otimes_BA:\M_B\to \M_A^C(\psi)$.

If a coalgebra-Galois extension $A(B)^C$ is a split extension then ${}_BA$ is a faithfully flat module. Consequently, the functors $-\otimes_B A:\M_B\to
\M_A^C(\psi)$ and $(-)_0 :\M_A^C(\psi)\to \M_B$ are inverse equivalences. \[cor.split\]

The first assertion follows from [@Brz:mod Proposition 4.4], while the second is the consequence of [@Brz:mod Corollary 3.11].

Let $(A,C)_\psi$ be the canonical entwining structure associated to a coalgebra-Galois extension $A(B)^C$. If there is a normalised integral map in $(A,C)_\psi$ then $B\hookrightarrow A$ is a split extension. \[prop.split\]

Let $\gamma: C\otimes C\to A$ be a normalised integral map in $(A,C)_\psi$, and take $\phi:C \to A$, $c\mapsto \sum_i
a^i_{\alpha}\gamma(c^\alpha\otimes c_i)$, where $\sum_ia^i\otimes c_i = \rho^A(1_A)$. Notice that the fact that $A$ is an $(A,C)_\psi$ module implies that for all $a\in A$, $\rho^A(a) =
\sum_ia^ia_\alpha\otimes c_i^\alpha$. Furthermore, since $\rho^A$ is a coaction we have $$\sum_{i,j}a^ja^i_\alpha\otimes c_j^\alpha \otimes c_i = \sum_i a^i\otimes c_i\sw
1\otimes c_i\sw 2.
\label{star}$$ We now show that $\phi$ satisfies all the conditions of Proposition \[proposition.split\]. For all $c\in C$ $$\begin{aligned}
\psi(c\sw 1\otimes \phi(c\sw 2)) & = &  (\sum_i
a^i_{\alpha}\gamma(c\sw 2^\alpha\otimes c_i))_\beta\otimes c\sw
1^\beta \\
& = &\sum_ia^i_{\alpha\delta}\gamma(c\sw 2^\alpha\otimes c_i)_\beta\otimes c\sw
1^{\delta\beta} \qquad\qquad \mbox{\rm (by (\ref{diag.A}))}\\
& = & \sum_ia^i_{\alpha}\gamma(c^\alpha \sw 2\otimes c_i)_\beta\otimes 
c^{\alpha}\sw 1^\beta  \qquad\qquad \mbox{\rm (by (\ref{diag.B}))}\\ 
& = & \sum_ia^i_{\alpha}\gamma(c^\alpha \otimes c_i\sw 1)\otimes 
c_i\sw 2 \qquad\qquad \mbox{\rm (by (\ref{int.i}))}\\ 
& = &    \sum_{i,j}(a^ja^i_{\beta})_{\alpha}\gamma(c^\alpha \otimes 
c_j^\beta)\otimes 
c_i \qquad\qquad \mbox{\rm (by (\ref{star}))}\\ 
& = &    \sum_{i,j}a^j_\alpha a^i_{\beta\delta}\gamma(c^{\alpha\delta}
\otimes 
c_j^\beta)\otimes 
c_i \qquad\qquad \mbox{\rm (by (\ref{diag.A}))}\\ 
& = &  \sum_{i,j}a^j_\alpha\gamma(c^\alpha\otimes c_j)a^i\otimes c_i 
= \phi(c)\rho^A(1_A) 
\qquad \mbox{\rm (by (\ref{int.ii}))}  \end{aligned}$$ Using normalisation of $\gamma$ as well as (\[star\]) one easily finds that $\sum_i a^i\phi(c_i) = 1_A$. Finally, take any $b\in B$, $c\in C$ and compute $$\begin{aligned}
b_\alpha\phi(c^\alpha) & = & \sum_i b_\alpha 
a^i_{\beta}\gamma(c^{\alpha\beta}\otimes c_i)
=  \sum_i (b 
a^i)_\alpha\gamma(c^\alpha\otimes c_i) \qquad \mbox{\rm
(by (\ref{diag.A}))} \\
& = &  \sum_i ( 
a^ib_\beta )_\alpha\gamma(c^\alpha\otimes c_i^\beta ) =   \sum_i  
a^i_\alpha b_{\beta\delta}\gamma(c^{\alpha\delta}\otimes c_i^\beta )\qquad \mbox{\rm
($b\in B$,  (\ref{diag.A}))} \\
& = &   \sum_i  
a^i_\alpha \gamma(c^{\alpha}\otimes c_i )b = \phi(c)b \qquad \qquad 
\qquad\qquad \mbox{\rm
(by (\ref{int.ii}))} \end{aligned}$$ Therefore $\phi$ satisfies all the conditions of Proposition \[proposition.split\] and, consequently, $B\hookrightarrow A$ is a split extension.

Dually to Definition \[def.integral.map\] we can consider

Let $(A,C)_\psi$ be an entwining structure. Any $\zeta\in {\rm
Hom}(C, A\otimes A)$ such that the following diagrams $$\begin{CD}
C\otimes A@>{\zeta\otimes A}>> A\otimes A \otimes A@>{A\otimes
\mu_A}>> A\otimes A \\
@| @.       @AA{\mu_A\otimes A}A\\
C\otimes A @>{\psi}>> A\otimes C @>{A\otimes\zeta}>> A\otimes A\otimes A
\end{CD}
\label{coint.i}$$ $$\begin{CD}
C @>{\Delta_C}>> C\otimes C @>{\zeta\otimes C}>> A\otimes A\otimes C\\
@VV{\Delta_C}V  @.  @AA{A\otimes\psi}A\\
C\otimes C @>{C\otimes\zeta}>> C\otimes A\otimes A
@>{\psi\otimes A}>> A\otimes C\otimes A
\end{CD}
\label{coint.ii}$$ commute is called a [*cointegral map*]{} in $(A,C)_\psi$. A cointegral map $\zeta$ is said to be [*normalised*]{}, if $\mu_A\circ \zeta= 1_A\circ \eps_C$. \[def.cointegral.map\]

The forgetful functor $\M_A^C(\psi)\to \M^C$ is separable if and only if there exists a normalised cointegral map in $(A,C)_\psi$. \[thm.separable\*\]

Consider an admissible morphism $(1_A,C): (k,C)_\sigma\to
(A,C)_\psi$ and apply Theorem \[thm.main\](2).

Let $(A,C)_\psi$ be a canonical entwining structure associated to a copointed coalgebra-Galois extension $A(k)^B$ of $k$. Then the forgetful functor $\M_A^C(\psi)\to \M^C$ is separable. \[ex.separable\*\]

In this case a normalised cointegral map is $\zeta = \can^{-1}\circ (1_A\otimes C)$.

Strongly separable coalgebra-Galois extensions
==============================================

In this section we combine the results of previous two sections to determine when a coalgebra-Galois extension is a strongly separable extension. Such an extension was introduced in [@Kad:Jon] in order to describe algebraic aspects of the Jones knot polynomial.

An extension of algebras $B\hookrightarrow A$ is called a [*strongly separable extension*]{} if it is a separable and split extension, and there exist a separation idempotent $u = \sum_iu_i\otimes u^i$, a conditional expectation $E:A\to B$ and a unit $\tau\in k$ such that for all $a\in
A$,

\(i) $\sum_i E(au_i)u^i = a\tau$

\(ii) $\sum_i u_iE(u^ia) = a\tau$. \[def.str\]

Let $A(B)^C$ be a coalgebra-Galois extension. If there exist a normalised integral $\z=\sum_ia_i\otimes c_i$ and a normalised integral map $\gamma\in {\rm Hom}(C\otimes C,A)$ in the canonical entwining structure $(A,C)_\psi$, and a unit $\tau\in k$ such that\
(i) $
\sum_i a_i1_A\sw 0_\alpha\gamma(c_i^\alpha\otimes 1_A\sw 1) = \tau,
$\
(ii) $\sum_ia_i\sw 0\gamma(a_i\sw 1\otimes c_i) = \tau ,$\
then $B\hookrightarrow A$ is a strongly separable extension. \[prop.str\]

By Proposition \[prop.galois.sep\], $B\hookrightarrow A$ is separable with $u=\sum_i u_i\otimes u^i = \can^{-1}(\z)$, while by Proposition \[prop.split\], $B\hookrightarrow A$ is split with a conditional expectation $E:A\mapsto a\sw 01_A\sw 0_\alpha\gamma(a\sw 1^\alpha\otimes 1_A\sw 1) =
(a1_A\sw 0)\sw 0\gamma((a1_A\sw 0)\sw 1\otimes 1_A\sw 1)$. Take any $a\in A$ and compute: $$\begin{aligned}
\sum_i E(au_i)u^i & = & \sum_i E(u_i)u^i a \qquad \qquad \qquad \mbox{\rm ($u$
is an integral)}\\
& = & \sum_i u_i\sw 0 1_A\sw 0_\alpha \gamma(u_i\sw 1^\alpha\otimes
1_A\sw 1)u^ia\\
& = & \sum_i u_i\sw 0 1_A\sw 0_\alpha u^i_{\beta\delta}
\gamma(u_i\sw 1^{\alpha\delta}\otimes
1_A\sw 1^\beta)a \qquad \mbox{\rm (by (\ref{int.ii}))}\\
& = & \sum_i u_i\sw 0 (1_A\sw 0  u^i_{\beta})_\alpha
\gamma(u_i\sw 1^{\alpha}\otimes
1_A\sw 1^\beta)a \qquad \mbox{\rm (by (\ref{diag.A}))}\\
& = & \sum_i u_i\sw 0 u^i\sw 0_\alpha
\gamma(u_i\sw 1^{\alpha}\otimes
u^i\sw 1)a \qquad \mbox{\rm ($A\in \M_A^C(\psi)$)}\\
& = & \sum_i (u_i u^i\sw 0)\sw 0
\gamma((u_i u^i\sw 0)\sw 1\otimes
u^i\sw 1)a \qquad \mbox{\rm ($A\in \M_A^C(\psi)$)}\\
& = & \sum_i a_i\sw 0\gamma(a_i\sw 1\otimes c_i)a =\tau a \qquad \mbox{ \rm
($\z=\can(u)$)} \end{aligned}$$ Therefore the condition Definition \[def.str\](i) is satisfied. Furthermore $$\begin{aligned}
\sum_i u_iE(u^ia) & = & \sum_i au_iE(u^i) \qquad \qquad
\qquad \mbox{\rm ($u$
is an integral)}\\
& = & \sum_i au_i u^i\sw 01_A\sw 0 _\alpha\gamma(u^i\sw 1^\alpha
\otimes 1_A\sw 1)\\
& = & \sum_i aa_i1_A\sw 0_\alpha\gamma(c_i^\alpha\otimes 1_A\sw 1) 
= \tau a \qquad \mbox{ \rm ($\z=\can(u)$)}\end{aligned}$$ This proves Definition \[def.str\](ii) and thus completes the proof of the proposition.

Let $k$ be a field and let $A(B)^C$ be a coalgebra-Galois extension with both $A$ and $B$ finite dimensional. Suppose that $A_B$ is free. Then $B\hookrightarrow A$ is a strongly separable extension if and only if there exists a normalised integral $\z = \sum_i a_i\otimes
c_i$ in the canonical entwining structure $(A,C)_\psi$, a map $\phi:C\to
A$ satisfying conditions (i)–(iii) in Proposition \[proposition.split\], and a non-zero $\tau\in k$ such that $$\sum_ia_i\phi(c_i) = \tau.
\label{f.dim}$$ \[prop.f.dim\]

By Proposition \[prop.galois.sep\], $B\hookrightarrow A$ is separable with $u=\sum_i u_i\otimes u^i = \can^{-1}(\z)$, while by Proposition \[proposition.split\], $B\hookrightarrow A$ is split with a conditional expectation $E:a\mapsto a\sw 0\phi(a\sw 1)$. By [@FisMon:Fro Remark 1.4(d)], Definition \[def.str\](i) holds provided that condition Definition \[def.str\](ii) holds. Thus it suffices to prove that (\[f.dim\]) is a sufficient and necessary condition for Definition \[def.str\](ii). Take any $a\in A$ and compute: $$\begin{aligned}
\sum_i u_iE(u^ia) & = & \sum_i au_iE(u^i) \qquad \qquad
\qquad \mbox{\rm ($u$
is an integral)}\\
& = & \sum_i au_iu^i\sw 0\phi(u^i\sw 1) = \sum_i
aa_i\phi(c_i).\end{aligned}$$ Therefore $\sum_i u_iE(u^ia) = \tau a$ if and only if (\[f.dim\]) holds.

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[^1]: Lloyd’s of London Tercentenary Fellow.

[^2]: On leave from: Department of Theoretical Physics, University of Łódź, Pomorska 149/153, 90-236, Łódź, Poland

[^3]: Although the paper [@Brz:mod] is restricted to $k$ being a field, all the results of [@Brz:mod] quoted in the present paper can easily be seen to hold for a general $k$.
