# Generalised homomorphisms, measuring coalgebras and extended symmetries

Marj Batchelor Cambridge University,, Department of Pure Mathematics and
Mathematical Statistics<EMAIL_ADDRESS>Will Boulton  Daren Chen currently
studying at Stanford University Jonathan Rawlinson Bristol University Mustafa
Warsi

###### Abstract

Three categories of algebras with morphisms generalising the usual set of
algebra homomorphisms are described. The Sweedler product provides a hom-
tensor equivalence relating these three categories, and a tool enabling the
universal measuring coalgebra to be calculated in small cases. A parallel
theory for modules is presented.

In comparing algebras $A$ and $B$, while homomorphisms remain the standard
morphisms between algebras, there are other linear maps - derivations, higher
order derivations, twisted derivations and other non (co) commutative variants
useful in non-commutative geometry - which record more subtle information than
can be monitored through algebra homomorphisms. The theory of measuring
coalgebras developed by Sweedler, in particular the universal measuring
coalgebra $P(A,B)$, the terminal object in a category of measuring coalgebras,
provides one framework for these more subtle comparisons. The principle
drawback to working with the universal measuring coalgebra has been the
absence of methods to describe and compute $P(A,B)$ even in simple cases.

The search for a method by which the universal measuring coalgebra may be
calculated has led to the study of an algebra, the Sweedler product,
$A\triangleleft K$ which is a tensor between an algebra $A$ and a coalgebra
$K$, creating a functorial action

$\triangleleft:Alg\times Coalg\to Alg,$

where $Alg,\ Coalg$ are the (usual) categories of algebras and coalgebras
respectively. The result is simple: the universal measuring coalgebra $P(A,B)$
is the dual coalgebra $(A\triangleleft B^{\circ})^{\circ}$.

This construction had also been developed by Joyal and Anel [1] for the
differential graded setting, and many of their results are duplicated here -
including the result stated above. This paper emphasises the computability of
$A\triangleleft K$, and hence $P(A,B)$, explores examples, and further
generalises the Sweedler product to categories of modules. We hope this paper
will encourage those working in fields beyond categorical algebra to apply the
concepts in their own fields. In that hope we include basic definitions,
results and simple examples.

The paper is organised as follows.

Section 1 introduces measuring coalgebras together with examples assuming no
prior knowledge. It presents two bicategories of algebras, $\widecheck{Alg}$
and $\widehat{Alg}$ with morphisms in the first being categories of measuring
coalgebras and in the second categories of extensions of algebras. These two
categories offer related means of exploring the more subtle transformations of
algebras.

Section 2 reviews the basic facts about universal measuring coalgebras and the
enriched category of algebras $\underline{Alg}$ whose hom-set
$\underline{Alg}(A,B)$ is the universal measuring coalgebra $P(A,B)$. It
includes the categorical framework for $A\triangleleft K$ and the principle
results establishing natural correspondences

$Alg(A\triangleleft K,B)\cong Alg(A,[K,B])\cong Coalg(K,P(A,B)),$

where $[K,B]$ denotes the convolution algebra of $K$ with $B$. The universal
measuring algebra $F(A,B)=A\triangleleft B^{\circ}$ is introduced, and the
properties which justify the name are summarised in Theorem 2.15.

Section 3 computes elementary examples of $F(A,B)$. Readers whose interest
lies in potential applications may prefer to work through the examples before
engaging with the categorical technicalities of section 2.

Section 4 follows the development of section 2 in the context of modules and
comodules, arriving at a corresponding sequence of natural equivalences. While
the interpretation and understanding of the corresponding objects in module
categories is challenging, the beauty of the categorical approach is that the
structure of the theory lies parallel with the more familiar theory in section
2. The module - comodule theory has one result without parallel in the algebra
- coalgebra theory: the global category of modules enriched over the universal
measuring comodule is fibred over a 2-category of modules with the category of
measuring comodules as morphisms.

The final potential application in Section 4 relates to internal symmetry
groups. Given an algebra $A$, (for example, the real algebra generated by
$\sqrt{-1}$ and a conjugation operator $J$), the theory provides a universal
algebra $F(A,A)$ and a canonical extension $A\to F(A,A)\otimes A$. Moreover,
for any $A$ module $W$ there is a universal module $D(W,W)$ for $F(A,A)$ and
an extension $W\to D(W,W)\otimes W$. If $W$ is a representation of a group
$G$, any $D(W,W)$ submodule $Z$ determines a group $G_{Z}$ of
$\operatorname{End}(Z)$ generated by invertible elements in the image of
$F(A,A)$. If $W$ is also a representation of a symmetry group $G$, any choice
of $F(A,A)$ module $Z$ determines an associated symmetry group $G_{Z}$ and
$G_{Z}\times G$ acts on $Z\otimes W$.

About the authorship of this paper. This work was done over a period of
several years with the help of summer research students, William Boulton,
Jonathan Rawlinson, Daren Chen and Mustafa Warsi who were undergraduates at
the time. All have moved on to work in other fields or other areas of
mathematics. William Boulton returned to the project to help with completion,
and others have contributed advice and corrections.

We would like to thank Queens’ College Cambridge, Trinity College Cambridge
and the Cambridge Summer Research in Mathematics programme for support for
these students during their summer projects. In addition Marj Batchelor would
like to thank Ed Miller for several very useful conversations in the early
development of the ideas. We are deeply indebted to Martin Hyland for his
encouragement and guidance in categorical matters.

##  1 Measuring coalgebras, notation and examples

###  1.1 Measuring coalgebras

The goal of this section is to introduce measuring coalgebras and provide
explicit examples. Here and throughout the paper vector spaces and algebras
and their tensor products are over an arbitrary field $k$ unless otherwise
specified.

A coalgebra $H$ is a vector space with a coassociative comultiplication.

$\triangle:H\rightarrow H\otimes H,\ \ \triangle h=\sum_{(h)}h_{(2)}\otimes
h_{(1)}$

together with a counit $\epsilon:H\to k$ satisfying

$h=\sum_{(h)}h_{(2)}\epsilon(h_{(1)})=\sum_{(h)}\epsilon(h_{(2)})h_{(1)}.$

where the sum $\sum_{(h)}h_{(2)}\otimes h_{(1)}$ is Sweedler notation
identifying the equivalence class rather than any particular representative of
the element in a tensor product. Coassociativity enables one to write powers
of the comultiplication map

$\triangle^{n}:H\to H\otimes...\otimes H,\ \
\triangle^{n}(h)=\sum_{(h)}h_{(n+1)}\otimes...\otimes h_{(1)}.$

######  Definition 1.1.

Let $A$ and $B$ be algebras. If $H$ is a coalgebra, a map

$\rho:H\rightarrow Hom_{k}(A,B)$

_measures_ if

$<\rho(h),aa^{\prime}>=\sum_{(h)}<\rho(h_{(2)}),a><\rho(h_{(1)}),a^{\prime}>,\
\ \ <\rho(h),1_{A}>=\epsilon(h)1_{B}$

for $a,a^{\prime}$ in $A$.

Here, and throughout, < _, _ > will denote evaluation.

The fundamental idea behind the paper springs from the following elementary
observation.

######  Proposition 1.2.

Let $H$ be a coalgebra, and let $A,\ B$ be algebras as above.

$\rho:H\rightarrow Hom_{k}(A,B)$

is a measuring map if and only if the corresponding map

$\rho:A\rightarrow Hom_{k}(H,B)$

is an algebra homomorphism.

The proof is immediate given the convolution product: for $\alpha,\ \beta$ in
$Hom_{k}(H,B)$, the product $\alpha*\beta$ is given by

$\alpha*\beta(h)=\sum_{(h)}<\alpha,h_{(2)}><\beta,h_{(1)}>.$

The space $Hom_{k}(H,B)$, equipped with this product, is called the
_convolution product of $H$ with $B$_, and is denoted by $[H,B]$.

###  1.2 Examples of measuring coalgebras

######  Example 1.3.

Homomorphisms.

Let $H$ be the coalgebra $kg$, where $\triangle g=g\otimes g$ and
$\epsilon(g)=1$. Observe that a linear map
$\varphi:H\to\operatorname{Hom}_{k}(A,B)$ measures if and only if $\varphi(g)$
is an algebra homomorphism.

######  Example 1.4.

: Derivations.

Let $H$ be the coalgebra with basis $\\{g,\gamma\\}$ with $\triangle g,\
\epsilon(g)$ as above, and

$\triangle\gamma=\gamma\otimes g+g\otimes\gamma,\ \ \ \epsilon(\gamma)=0.$

A linear map $\varphi:H\to\operatorname{Hom}_{k}(A,B)$ measures if and only if
$\varphi(g)$ is a homomorphism and $\varphi(\gamma)$ is a derivation with
respect to $\varphi(g)$.

Measuring coalgebras for a given pair $A,\ B$ of algebras forms a category
$\widecheck{Alg}(A,B)$ where maps between measuring coalgebras defined as
follows.

######  Definition 1.5.

Let $H_{1}$ and $H_{2}$ be coalgebras with measuring maps $\rho_{1},\
\rho_{2}:H_{1},H_{2}\to\operatorname{Hom}_{k}(A,B)$ respectively. A _map of
measuring coalgebras_ $\tau:H_{1}\to H_{2}$ is a coalgebra map such that the
following diagram commutes:

$\textstyle{H_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tau}$$\scriptstyle{\rho_{1}}$$\textstyle{\operatorname{Hom}_{k}(A,B)}$$\textstyle{H_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho_{2}}$

The identity $1_{H}:H\to H$ provides the unit in $\widecheck{Alg}(A,B)$.

Dual coalgebras provides a rich collection of less familiar examples.

######  Example 1.6.

Extensions and dual coalgebras as measuring coalgebras.

Let $S$ be a finite dimensional algebra, and suppose we have an algebra
homomorphism

$\sigma:A\to S\otimes B.$

Since $S$ is finite dimensional, the full linear dual $S^{*}$ is a coalgebra.
The map corresponding map

$\sigma:S^{*}\to\operatorname{Hom}(A,B)$

is a measuring map. Thus extensions $\sigma:A\to S\otimes B$ for finite
dimensional algebras $S$ are a source of examples of measuring coalgebras.

The finiteness restriction on $S$ can be dropped by introducing the dual
coalgebra.

######  Definition 1.7.

. Let $S$ be an algebra. The _dual coalgebra_ $S^{\circ}$ of an algebra $S$ is
the subspace of the full linear dual $S^{*}$ of elements which vanish on a
cofinite ideal of $S$.

See Sweedler [2] Chapter VI.

Another familiar example is that of $\mathbb{C}[z]^{\circ}$, which has a basis
$\\{\frac{d^{n}}{dz^{n}}|_{b}\\}$ of elements vanishing on the ideal generated
by $(z-b)^{n+1}$ with comultiplication familiar from the generalised product
rule:

$\triangle\frac{d^{n}}{dz^{n}}|_{b}=\sum_{k}{n\choose
k}\frac{d^{n-k}}{dz^{n-k}}|_{b}\otimes\frac{d^{k}}{dz^{k}}|_{b}.$

The notation $\frac{d^{n}}{dx^{n}}|_{b}$ is extended to the case $n=0$,
denoting evaluation at $b$.

Extensions $\sigma:A\to S\otimes B$ form a category in their own right, which
we will denote by $\widehat{Alg}(A,B)$ in a manner parallel with
$\widecheck{Alg}(A,B)$. Define morphisms of extensions as follows.

######  Definition 1.8.

Let $S_{1},\ S_{2}$ be algebras with homomorphisms $\sigma_{i}:A\to
S_{i}\otimes B$. A homomorphism $\theta:S_{1}\to S_{2}$ is a map of extensions
if the following diagram commutes.

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma_{1}}$$\scriptstyle{\sigma_{2}}$$\textstyle{S_{1}\otimes
B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\theta\otimes
1_{B}}$$\textstyle{S_{2}\otimes B}$

Again the identity map $1_{S}:S\to S$ provides the unit in
$\widehat{Alg}(A,B)$

###  1.3 The bicategories $\widecheck{Alg}$ and $\widehat{Alg}$

The purpose of the paper is to establish a framework for comparing algebras in
ways less restrictive than homomorphisms. Measuring coalgebras and extension
offer two means of generalising the concept of homomorphism, offering new
categories, but the increased generality needs additional structure enabling
the comparison of these maps: each can be used to define new categories of
algebras which carry the additional structure of a bicategory.

######  Proposition 1.9.

  1. 1.

There is a bicategory $\widecheck{Alg}$ whose objects are algebras, and whose
set of morphisms $\widecheck{Alg}(A,B)$ is the category of measuring
coalgebras $\rho:H\to\operatorname{Hom}_{k}(A,B)$.

  2. 2.

There is a bicategory $\widehat{Alg}$ whose objects are algebras, and whose
set of morphisms $\widehat{Alg}(A,B)$ is the category of extensions $\rho:A\to
S\otimes B$.

###### Proof.

The case for $\widecheck{Alg}$.

The task is to show that morphisms - measuring coalgebras - "compose", that a
suitable unit morphism exists, and that the composition of morphisms commutes
suitably with the "vertical" morphisms of measuring coalgebras.

_"Horizontal" morphisms $\odot$ \- composition of measuring coalgebras._

Suppose $H,\ K$ are measuring coalgebras with measuring maps
$\phi:H\to\operatorname{Hom}(A,B),\ \psi:K\to\operatorname{Hom}(B,C)$
respectively. Then $K\otimes H$ is a coalgebra and the composition

$\psi\otimes\phi:K\otimes
H\to\operatorname{Hom}(B,C)\otimes\operatorname{Hom}(A,B)\to\operatorname{Hom}(A,C)$

measures:

$\displaystyle<h\otimes k,aa^{\prime}>=$
$\displaystyle\sum_{(h)}<h,<k_{(2)},a><k_{(1)},a^{\prime}>>$ (1)
$\displaystyle=\sum_{(h),(k)}<h_{(2)},<k_{(2)},a>><h_{(1)},<k_{(1)},a^{\prime}>>$
(2)

as required, since $\triangle(h\otimes k)=\sum_{(h),(k)}(h_{(2)}\otimes
k_{(2)})\otimes(h_{(1)}\otimes k_{(1)})$

Define the composition of $H$ and $K$ in $\widecheck{Alg}$ to be the coalgebra
$H\otimes K$ with the measuring coalgebra structure given by
$\psi\otimes\phi$. This "horizontal" composition will be denoted by $H\odot
K$. Composition of "vertical" morphisms - maps between measuring coalgebras
will be denoted by $\circ$.

_The unit morphism in $\widecheck{Alg}$._ Let E be the one dimensional
coalgebra with the distinguished element $e$ with $\epsilon(e)=1$, and
$\triangle e=e\otimes e$. For an algebra $A$ define
$\eta_{A}:E\to\operatorname{Hom}(A,A)$ by $\eta(e)=1_{A}$ giving $E$ the
structure of a measuring coalgebra. Direct calculation confirms that the pair
$E,\eta_{A}$ has the properties required of a unit completing the requirements
that $\widecheck{Alg}$ is a category.

_Compatibility of "vertical" and "horizontal" maps_ Let $H_{i},\ i=1,\ 2,\ 3$
be in $\widecheck{Alg}(A,B)$ and $K_{i}$ in $\widecheck{Alg}(B,C)$, and let
$\varphi_{i}:H_{i}\to H_{i+1}$, ($\psi_{i}:K_{i}\to K_{i+1}$) for $i=1,\ 2$ be
maps of measuring coalgebras. The requirement is to show that there are
measuring maps

$\varphi_{i}\circ\psi_{i}:H_{i}\odot K_{i}\to H_{i+1}\odot K_{i+1},\ \ i=1,\
2$

and that these maps satisfy

$(\psi_{1}\circ\psi_{2})\odot(\varphi_{1}\circ\varphi_{2})\cong(\psi_{2}\odot\varphi_{2})\circ(\psi_{1}\odot\varphi_{1})$

The notation $\cong$ denotes a natural isomorphism guaranteed by the
uniqueness of the tensor product rather than an equality. That this
compatibility requirement is fulfilled is follows from the fact that the
horizontal composition $\odot$ is exactly the tensor product $\otimes$ for
vector spaces, and the statement is true for tensor products of vector spaces.

The case for $\widehat{Alg}$.

Again, the task is to define "horizontal" composition $\odot$ of extensions,
verify that $\widehat{Alg}$ forms a category with objects being algebras, and
morphisms being the categories $\widehat{Alg}(A,B)$, and confirm appropriate
compatibility of "horizontal" and "vertical" morphisms.

_"Horizontal morphisms - composition of extensions"_

If $(S,\sigma)$ and $(T,\tau)$ are extensions from algebras $A$ to $B$ and $B$
to $C$ respectively, the composition $(T,\tau)\circ(S,\sigma)$ is defined via

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho}$$\textstyle{S\otimes
B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1\otimes\tau}$$\textstyle{S\otimes(T\otimes
C)\cong(S\otimes T)\otimes C}$

making use of the natural isomorphism inherited from vector spaces.

The pair $(k,\mu)$ where $\mu:k\to k\otimes A=A$ is the unit in $A$ provides
an identity map in the set of extensions from $A$ to $A$, so that
$\widehat{Alg}(-,-)$ has the properties required of morphisms.

_Compatibility of "vertical" and "horizontal" maps._ The compatibility
condition is the same as required for $\widecheck{Alg}$. The condition is
satisfied for exactly the same reason: the "horizontal" morphisms again are
simply the tensor products of vector spaces.

∎

Let $\widecheck{Alg}(A,B)_{f}$ denote the full subcategory of
$\widecheck{Alg}(A,B)$ whose objects are all finite dimensional measuring
coalgebras. Let $\widecheck{Alg}_{f}$ be the sub-bicategory whose morphisms
are the restricted subcategories $\widecheck{Alg}(A,B)_{f}$. The full linear
dual $H^{*}$ of a coalgebra $H$ is (always) an algebra. The content of
proposition 1.2 is precisely that in the case where $H$ is finite dimensional,
the dual of a measuring $H\to\operatorname{Hom}_{k}(A,B)$ is precisely an
extension $A\to H^{*}\otimes B$. Thus, proposition 1.2 can be recast in
categorical terms as follows.

######  Theorem 1.10.

For $A$, $B$ algebras, taking the dual establishes an equivalence of
categories

$*:\widecheck{Alg}(A,B)_{f}^{op}\cong\widehat{Alg}(A,B)_{f}:*$

This establishes an equivalence of bicategories:

$*:\widecheck{Alg}_{f}^{co}\cong\widehat{Alg}_{f}:*.$

When the finiteness restriction is dropped the subtle differences between
coalgebras and algebras becomes significant. The next section introduces
universal measuring coalgebra $P(A,B)$ which is the terminal object in
$\widecheck{Alg}(A,B)$. The search for a corresponding initial object $F(A,B)$
in $\widehat{Alg}(A,B)$ generated this work.

##  2 Universal measuring coalgebras

This section further develops the theory of measuring coalgebras through the
introduction of the universal measuring coalgebra and the Sweedler product
which can be combined to introduce an algebra, the universal measuring
algebra, which serves as a pre-dual for the universal measuring coalgebra.

###  2.1 The universal measuring coalgebra

The contents of this subsection are covered in Sweedler for example. This
section is included to provide sufficient background to enable those
unfamiliar with coalgebras to use the results.

Coalgebras have a strong finiteness property which is fundamental for the
constructions in this paper. One consequence is that the universal measuring
coalgebra $P(A,B)$, the terminal element in the category
$\widecheck{Alg}(A,B)$, exists.

######  Proposition 2.1.

Given a coalgebra $C(\Delta,\epsilon)$ and $x\in C$, there exists a finite
dimensional subcoalgebra $D\subset C$ such that $x\in D$.

See Sweedler page 46. Essentially, consider $\triangle^{2}x=\sum
x_{(3)}\otimes x_{(2)}\otimes x_{(1)}$ and take $D$ to be the span of the
elements $x_{(2)}$ in a minimal expression for $\triangle^{2}$.

######  Definition 2.2.

The _universal measuring coalgebra_ $\pi:P(A,B)\to\operatorname{Hom}_{k}(A,B)$
is the unique measuring coalgebra with the property that there is a unique
coalgebra map $P(\rho)$ such that the diagram

$\textstyle{P(A,B)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{\operatorname{Hom}_{k}(A,B)}$$\textstyle{H\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{P(\rho)}$$\scriptstyle{\rho}$

commutes for any measuring coalgebra $\rho:H\to Hom_{k}(A,B)$.

The existence of the universal measuring coalgebra is guaranteed because the
finiteness property ensures that the coproduct over all _finite dimensional_
coalgebras measuring $A$ to $B$ is sufficiently large to capture all measuring
coalgebras. The fact that the category of coalgebras has both colimits and
coequalisers, then allows $P(A,B)$ to be constructed as follows.

######  Proposition 2.3.

$P(A,B)$ exists, and is a terminal object in $\widecheck{Alg}(A,B)$.

###### Proof.

See Sweedler [2] p143, or CV [3] p18. $P(A,B)$ can be constructed as a
coequaliser as follows. For algebras $A$ and $B$, construct the coproduct.

$K=\coprod_{\kappa}H_{\kappa}$

where the coproduct is taken over all isomorphism classes of _finite
dimensional_ coalgebras $(H_{\kappa},\rho_{\kappa})$ that measure A to B. The
finite dimensionality is critical for the coproduct to be well defined. Define
$L$ to be the coproduct

$L=\coprod_{\lambda:H_{\kappa}\to H_{\kappa^{\prime}}}H_{\kappa,\lambda}$

where $H_{\kappa,\lambda}$ is a copy of $H_{\kappa}$ for all
$\lambda\in\widecheck{Alg}(A,B)$. Define two maps $f,g:L\to H$ by the image of
each factor $H_{\kappa,\lambda}$. Define $f$ to be the map that sends
$H_{\kappa,\lambda}$ identically to $H_{\kappa}$ followed by the inclusion of
$H_{\kappa}$ in $K$. Let $g$ be the map that sends $H_{\kappa,\lambda}$ to
$H_{\kappa^{\prime}}$ followed by the inclusion of $H_{\kappa^{\prime}}$ in
$K$.

$P(A,B)$ to be the coequaliser of this diagram; effectively

$P(A,B)=K/im(f-g).$

That the image of $f-g$ is a coideal of the coalgebra $K$, and that it is sent
to 0 in $\operatorname{Hom}(A,B)$ need to be verified.

That $P(A,B)$ is a terminal object in $\widecheck{Alg}(A,B)$ depends again
upon the finiteness property of coalgebras: $P(A,B)$ is the ascending union of
finite dimensional coalgebras. For an arbitrary measuring coalgebra $H$, the
map of measuring coalgebras from $H$ to $P(A,B)$ can be built up as the union
of the inclusions of finite dimensional subcoalgebras of $H$ in $K$. Being a
terminal object in this category immediately shows uniqueness. ∎

######  Corollary 2.4.

For any algebra $S$, $S^{\circ}=P(S,k)$.

###### Proof.

Evidently $S^{\circ}\to\operatorname{Hom}(S,k)$ measures so that there is a
coalgebra map $S^{\circ}\to P(S,k)$. This map is injective since composition
with the measuring map $\pi$ is just the inclusion
$S^{\circ}\to\operatorname{Hom}(S,k)$. Conversely, if $p$ is any element of
$P(S,k)$, it is an element of a finite dimensional subcoalgebra $H$ of
$P(S,k)$. The measuring map $\pi:H\to\operatorname{Hom}(S,k)=S^{*}$
corresponds to an algebra map $\pi:S\to H^{*}$, thus $\pi(p)$ vanishes on the
cofinite kernel of that homomorphism, and hence $\pi(p)$ is in $S^{\circ}$. ∎

The correspondence between coalgebra maps $H\to P(A,B)$ and algebra
homomorphisms $A\to[H,B]$ is made formal in the following corollary.

######  Proposition 2.5.

There is a natural equivalence

$Coalg(-,P(A,B))\cong Alg(A,[-,B]).$

###### Proof.

Again, see Sweedler, p 143. The symbol $\cong$ indicates that this is a
natural equivalence, meaning that the correspondence between the two functors
$Coalg(-,P(A,B))$ and $Alg(A,[-,B])$ observed in 1.2 commutes appropriately
with coalgebra maps $\varphi:H\to K$: $Coalg(\varphi,P(A,B))$ corresponds to
$Alg(A,[\varphi,B])$. ∎

###  2.2 The Sweedler product

The principle tool in this paper, the Sweedler product, is an algebra,
$A\triangleleft H$, a tensor of an algebra A with a coalgebra H which serves
as an adjoint to $\operatorname{Hom}$ so that

$\operatorname{Hom}(A\triangleleft-,B)\cong\operatorname{Hom}(A,[-B])$

The idea was introduced in Anel and Joyal [1] for use in the differential
graded setting, and the main theorems in this section are found in their paper
on pages 122 to 124. This paper studies the construction as a practical tool
for extending the scope of "morphism" in other settings, and generalises the
construction to the setting of modules and comodules.

######  Definition 2.6.

Define $A\triangleleft H$, the _Sweedler product_ to be the quotient of the
tensor algebra $T(A\otimes H)/J$ where $J$ is the minimal ideal which makes
the following diagrams commute.

$\textstyle{A\otimes A\otimes
H\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mu\otimes
1}$$\scriptstyle{1\otimes\triangle}$$\textstyle{A\otimes A\otimes H\otimes
H\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(A\otimes
H)\otimes(A\otimes
H)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{T(A\otimes
H)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{s}$$\textstyle{A\otimes
H\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r}$$\textstyle{T(A\otimes
H)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\triangleleft
H}$ $\textstyle{k\otimes
H\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1_{A}\otimes
1}$$\scriptstyle{1\otimes\epsilon}$$\textstyle{A\otimes
H\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{T(A\otimes
H)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{k\otimes
k\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{k\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{T(A\otimes
H)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\triangleleft
H}$

######  Theorem 2.7.

There are functorial actions

$\triangleleft:Alg\times Coalg\to Alg$

$[-,-]:Coalg^{op}\times Alg\to Alg$

providing a tensor hom adjunction

$Alg(A,[H,B])\cong Alg(A\triangleleft H,B).$

###### Proof.

_Functoriality_ The functoriality of the action $[-,-]$ is familiar. For the
construction of the Sweedler product, the outline of the argument is as
follows.

Let $A_{i}$, $i=1,\ 2,\ 3$ be algebras and $H_{i}$ be coalgebras, let
$f_{i}:A_{i}\to A_{i+1},\ g_{i}:H_{i}\to H_{i+1}$, $i=1,\ 2$ be algebra and
coalgebra homomorphisms respectively. The first requirement is that the linear
maps $f_{i}\otimes g_{i}:A_{i}\otimes A_{i+1}\otimes K_{i+1}$ should pass to
algebra homomorphisms $f_{i}\triangleleft g_{I}:A_{i}\triangleleft K_{i}\to
A_{i+1}\triangleleft K_{i+1}$, the second is that the composition should
behave in a functorial manner:

$(f_{2}\circ f_{1})\triangleleft(g_{2}\circ g_{1})=(f_{2}\triangleleft
g_{2})\circ(f_{1}\triangleleft g_{1})$

For $f_{i}\otimes g_{i}$ to pass to a homomorphism $f_{i}\triangleleft g_{i}$
it is necessary that the ideals $J_{i}$ defining the algebras
$A_{i}\triangleleft H_{i}$ satisfy $T(f_{i}\otimes g_{i})J_{i}\subset
J_{i+1}$. Consider the diagrams

$\textstyle{A_{i}\otimes A_{i}\otimes
H_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{m}$$\scriptstyle{a}$$\textstyle{A_{i}\otimes
H_{i}\otimes A_{i}\otimes
H_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{c}$$\scriptstyle{f}$$\textstyle{A_{i}\otimes
H_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{j}$$\textstyle{T(A_{i}\otimes
H_{i})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{k}$$\textstyle{A_{i+1}\otimes
H_{i+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j}$$\textstyle{T(A_{i+1}\otimes
H_{i+1})}$$\textstyle{A_{i+1}\otimes A_{i+1}\otimes
H_{i+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a}$$\scriptstyle{m}$$\textstyle{A_{i+1}\otimes
H_{i+1}\otimes A_{i+1}\otimes
H_{i+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{c}$

Here maps labelled $a$ have to do with comultiplication on $H$, maps labelled
$m$ involve multiplication in $A$, maps labelled $f$ have to do with maps
$A_{i}\to A_{i+1}$ and $H_{i}\to H_{i+1}$, are the inclusion of the
appropriate subspaces in the tensor algebra, and $k$ is the map induced by
$f_{i}\otimes g_{i}$ on the tensor algebra.

$\textstyle{k\otimes
H_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{e}$$\scriptstyle{u}$$\textstyle{A_{i}\otimes
H_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a}$$\scriptstyle{f}$$\textstyle{k\otimes
k\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{Id}$$\scriptstyle{b}$$\textstyle{T(A_{i}\otimes
H_{i})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{k}$$\textstyle{k\otimes
k\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{b}$$\textstyle{T(A_{i+1}\otimes
H_{i+1})}$$\textstyle{k\otimes
H_{i+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{u}$$\scriptstyle{e}$$\textstyle{A_{I+1}\otimes
H_{i+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g}$

Here the maps labelled $u$ are the inclusions of $k$ as units in the $A_{i}$,
the maps $e$ have to do with the counits in the $A_{i}$, $b$ is the inclusion
of $k\otimes k\to k$ in the tensor algebra. $k$ again is the map on the tensor
algebra induced by $f_{i}\otimes g_{i}$.

The requirement is that $k(J_{i})\subset J_{i+1}$. Observe that the ideal
$J_{i}$ is generated by the images of $j\circ m-c\circ a$ in the first diagram
and $a\circ u-e\circ f$ in the second. The commutativity of the two diagrams
ensures that the generators of $J_{i}$ are sent by $k$ into the generators of
$J_{i+1}$, hence $k(J_{i})\subset J_{i+1}$.

Functoriality follows as a consequence of the fact that the statement is true
for the respective tensor algebras:

$T(f_{2}\circ f_{1}\otimes g_{2}\circ g_{1})=T(f_{2}\otimes g_{2})\circ
T(f_{1}\otimes g_{1}).$

To see that $A\triangleleft H$ has the desired adjunction properties, observe
that the following three sets are naturally identified as vector spaces:

$\operatorname{Hom}_{k}(A,[H,B])\cong\operatorname{Hom}_{k}(A\otimes
H,B)\cong\operatorname{Hom}_{k}(H,\operatorname{Hom}_{k}(A,B)).$

An element $\varphi$ in $\operatorname{Hom}_{k}(A,[C,B])$ is an algebra
homomorphism exactly when the corresponding element in
$\operatorname{Hom}_{k}(C,\operatorname{Hom}_{k}(A,B))$ is a measuring map.
But the condition that $\varphi$ measures is exactly the statement that the
algebra map

$T\phi:T(A\otimes H)\to B$

factors through $A\triangleleft H$. As with proposition 2.5 naturality is a
consequence of the underlying fact for vector spaces. ∎

###  2.3 Measuring coalgebras for coalgebras

In the previous section, the bicategories $\widecheck{Alg}$ and
$\widehat{Alg}$ were introduced to provide a context which includes more
general morphisms between algebras. The existence of the universal measuring
coalgebra provides another approach, that of enriched categories, to be
introduced in the next section, replacing the morphism set $Alg(A,B)$ with the
measuring coalgebra $P(A,B)$.

Coalgebras carry a similar structure. There is a parallel concept of measuring
coalgebra for coalgebras which introduces the possibility of generalised maps
between coalgebras, and a final object in the category of measuring coalgebras
for coalgebras, which will enable us to establish a full equivalence between
these three approaches to the generalisation of homomorphisms.

######  Definition 2.8.

Let $H,\ K,\ L$ be coalgebras. A linear map
$\lambda:L\to\operatorname{Hom}_{k}(H,K)$ is said to _measure_ if the map
$\lambda$ regarded as an element of $\operatorname{Hom}_{k}(L\otimes H,K)$ is
a coalgebra map. As with measuring coalgebras for algebras, maps between
measuring coalgebras $L_{1},\ L_{2}$ for coalgebras are those for which the
diagram

$\textstyle{L_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\lambda_{1}}$$\textstyle{\operatorname{Hom}_{k}(H,K)}$$\textstyle{L_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\lambda_{2}}$

commutes.

######  Definition 2.9.

For coalgebras $H,\ K$, the _universal measuring coalgebra_ $P(H,K)$ is a
coalgebra with a map $\pi:P(H,K)\to\operatorname{Hom}_{k}(H,K)$ with the
property that given any measuring coalgebra
$\lambda:L\to\operatorname{Hom}_{k}(H,K)$ there is a unique map of coalgebras
making the following diagram commute.

$\textstyle{L\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\lambda}$$\textstyle{\operatorname{Hom}_{k}(H,K)}$$\textstyle{P(H,K)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$

As for algebras, there is no obstruction to forming the required coequalisers,
so that there is a terminal object in the category of measuring coalgebras for
any pair of coalgebras $(H,K)$; i.e. $P(H,K)$ exists.

Proposition 2.5 for the universal measuring coalgebra for algebras has its
parallel for the universal measuring coalgebra for coalgebras.

######  Proposition 2.10.

There is a natural equivalence

$Coalg(-,P(H,K))\cong Coalg(-\otimes H,K)$

###  2.4 Enriched categories of algebras and coalgebras

The concept of enriched categories, in which morphisms and compositions of
morphisms may be objects and morphisms in a category more general than sets,
provides an alternative to the bicategory $\widecheck{Alg}$ as a means of
including maps between algebras respecting the multiplicative structure in
more sensitive ways.

######  Proposition 2.11.

  1. 1.

There is a category $\underline{Alg}$ whose objects are algebras and whose
hom-sets are the universal measuring coalgebras $P(A,B)$. $\underline{Alg}$
has the structure of an enriched category.

  2. 2.

There is a category $\underline{Coalg}$ whose objects are coalgebras and whose
hom-sets are the universal measuring coalgebras $P(H,K)$. $P(H,K)$ has the
structure of an enriched category.

###### Proof.

The universal property of the universal measuring coalgebra guarantees the
composition of morphisms. The uniqueness of the universal measuring map
guarantees associativity of composition and also guarantees that the identity
morphism satisfies its requirements. ∎

Proposition 2.5 extends to the enriched setting as follows.

######  Theorem 2.12.

There are natural equivalences

$\underline{Alg}(A\triangleleft-,B)\cong\underline{Alg}(A,[-,B])\cong\underline{Coalg}(-,\underline{Alg}(A,B))$

###### Proof.

Let $L$ be a coalgebra. The aim is to establish the existence of natural
isomorphisms

$P(A\triangleleft L,B)\cong P(A,[L,B])\cong P(L,P(A,B))$

By Yoneda’s lemma, it is sufficient to show that for any coalgebra $C$ there
are natural isomorphisms of sets

$Coalg(C,P(A\triangleleft L,B))\cong Coalg(C,P(A,[L,B]))$
$Coalg(C,P(A,[L,B]))\cong Coalg(C,P(L,P(A,B)))$

The work has either already been done or is familiar from vector spaces. For
the first equivalence:

$\displaystyle Coalg(C,P(A\triangleleft L,B))$ $\displaystyle\cong^{1}$
$\displaystyle Alg(A\triangleleft L,[C,B])$ $\displaystyle\cong^{2}$
$\displaystyle Alg(A,[L,[C,B]])$ $\displaystyle\cong^{3}$ $\displaystyle
Alg(A,[L\otimes C,B])$ $\displaystyle\cong^{4}$ $\displaystyle Alg(A,[C\otimes
L,B])$ $\displaystyle\cong^{5}$ $\displaystyle Alg(A,[C,[L,B]])$
$\displaystyle\cong^{6}$ $\displaystyle Coalg(C,P(A,[L,B]).$

Here, equivalence 1 is the consequence of theorem 1.2 and the universal
property of universal measuring coalgebras. Equivalence 2 is the adjunction of
theorem 2.7. The final equivalence again follows from proposition 1.2 and the
property of the universal measuring coalgebra. For the second:

$\displaystyle Coalg(C,P(L,P(A,B)))$ $\displaystyle\cong^{7}$ $\displaystyle
Coalg(C\otimes L,P(A,B))$ $\displaystyle\cong^{8}$ $\displaystyle
Alg(A,[C\otimes L,B])$ $\displaystyle\cong^{9}$ $\displaystyle
Alg(A,[C,[L,B]])$ $\displaystyle\cong^{10}$ $\displaystyle
Coalg(C,P(A,[L,B])).$

Here, equivalence 7 is proposition 2.10, the universal property for the
universal measuring coalgebra for coalgebras. Equivalence 8 makes use of
proposition 2.5, as does 10. ∎

###  2.5  The universal measuring algebra $F(A,B):=A\triangleleft B^{\circ}$
and the calculation of $P(A,B)$

The motivation behind our development of the Sweedler algebra was the
observation that it provided a pre-dual which enabled explicit computation of
$P(A,B)$ in the case where $B$ is finite dimensional. The fundamental
observation is the following specific case of theorem 2.12

######  Corollary 2.13.

$P(A,[B^{\circ},k])\cong P(A\triangleleft B^{\circ},k)\cong(A\triangleleft
B^{\circ})^{\circ}.$

In the case where $B$ is finite dimensional, $P(A,B)\cong P(A\triangleleft
B^{\circ},k).$

######  Definition 2.14.

Define the _universal measuring algebra_ $F(A,B)$ to be the Sweedler product

$F(A,B):=A\triangleleft B^{\circ}.$

The name is chosen to suggest its role linking the algebra structures of $A$
and $B$. The properties of $F(A,B)$ are summarised in the following theorem.

######  Theorem 2.15.

The properties of $F(A,B)$ are as follows.

  1. 1.

For any finite dimensional algebra $B$ and $A$ any algebra there is an algebra
homomorphism

$\eta(A,B):A\to F(A,B)\otimes B$

such that for any extension $\sigma:A\to S\otimes B$ there is a unique
homomorphism $F(\sigma):F(A,B)\to S$ such that the diagram

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma}$$\scriptstyle{\eta(A,B)}$$\textstyle{F(A,B)\otimes
B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{F(\sigma)\otimes
1_{B}}$$\textstyle{S\otimes B}$

commutes. Moreover, $\eta(-,B)$ is a natural transformation.

  2. 2.

For arbitrary B the algebra homomorphism $F(\sigma):F(A,B)\to S$ still exists:
the assignment

$F:\widehat{Alg}(A,B)\to Alg(F(A,B),-),\ \ \ F(S,\sigma)=F(\sigma):F(A,B)\to
S$

is a functor.

  3. 3.

If $B$ is finite dimensional, $F(A,B)$ is an initial object in
$\widehat{Alg}(A,B)$.

  4. 4.

For $A$ finite dimensional, $F(A,A)$ is a bialgebra. More generally for finite
dimensional algebras $A,\ B,\ C$ there is a factorisation

$\triangle_{B}:F(A,C)\to F(A,B)\otimes F(B,C)$

which satisfies a coassociativity identity:

$(1_{F(AB)}\otimes\triangle_{C})\circ\triangle_{B}=(\triangle_{B}\otimes
1_{F(C,D)})\circ\triangle_{C}:F(A,D)\to F(A,B)\otimes F(B,C)\otimes F(C,D).$

  5. 5.

The evaluation map

$A\to[P(A,B),B]$

is an algebra homomorphism.

  6. 6.

$P(A,B)\cong F(A,B)^{\circ}$

###### Proof.

  1. 1.

Define $\eta(A,B):A\to F(A,B)\otimes B$ using the identity element $e$ in
$B^{\circ}\otimes B$ so that $\eta_{A}(a)$ is the image of $a\otimes e$ in
$A\otimes B^{\circ}\otimes B\subset T(A\otimes B^{\circ})\otimes B$ in the
quotient $F(A,B)\otimes B$.

Check that $\eta(A,B):A\to F(A,B)\otimes B$ is an algebra homomorphism.
Consider the diagram:

$\textstyle{A\otimes
A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\otimes
A\otimes B^{\circ}\otimes B\otimes B^{\circ}\otimes
B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\otimes
B^{\circ}\otimes B\otimes A\otimes B^{\circ}\otimes
B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\otimes
A\otimes B^{\circ}\otimes
B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\otimes
A\otimes B^{\circ}\otimes B^{\circ}\otimes
B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(A\otimes
B^{\circ})\otimes(A\otimes B^{\circ})\otimes
B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\otimes
B^{\circ}\otimes
B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{F(A,B)\otimes
B}$$\textstyle{F(A,B)\otimes B\otimes F(A,B)\otimes
B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

Going anticlockwise from top left, the diagram represents the product of
elements in $A$, followed by their inclusion into $F(A,B)\otimes B$. Going
clockwise, the morphisms map $A$ to $F(A,B)\otimes B$ by $\eta$ first, and
then multiply elements (first multiplying elements in $B$, then in $F(A,B)$).
Thus the perimeter arrows are the required identity to verify that
$\eta(A,B):A\to F(A,B)\otimes B$ is an algebra homomorphism.

To see that this diagram commutes; the bottom square is just the diagram
defining the quotient ideal of $F(A,B)$, tensored with $B$. For the top
square, consider the diagram relating multiplication in $B$ with
comultiplication in $B^{\circ}$:

$\textstyle{B\otimes B\otimes
B^{\circ}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B\otimes
B\otimes B^{\circ}\otimes
B^{\circ}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B\otimes
B^{\circ}\otimes B\otimes
B^{\circ}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B\otimes
B^{\circ}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{k}$$\textstyle{k\otimes
k\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

Since $B$ is finite dimensional the dual algebra of $B^{\circ}$ is just $B$,
this diagram dualises with reversed arrows; tensored with $A\otimes A$ this is
essentially the top square above - other than some shuffling, the factors of
$A$ are otherwise untouched.

The statement that $\eta(-,B)$ is natural means that If $A^{\prime}$ is
another algebra with a homomorphism $\tau:A\to A^{\prime}$ the diagram

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tau}$$\scriptstyle{\eta(A,B)}$$\textstyle{F(A,B)\otimes
B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{F(\tau,1_{B})}$$\textstyle{A^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\eta(A^{\prime}B)}$$\textstyle{F(A^{\prime},B)\otimes
B}$

commutes. (A more correct statement would be that $\eta:\mathbb{Id}\to
F(-,B)\otimes B$ is a natural transformation.) The proof is a matter of
checking that the map

$\tau\otimes e:A\otimes B^{\circ}\otimes B\to A^{\prime}\otimes
B^{\circ}\otimes B$

induces a homomorphism

$\tau\otimes e:T(A\otimes B^{\circ})\otimes B\to T(A^{\prime}\otimes
B^{\circ})\otimes B$

that respects the ideal defining $F(A,B)$. The dependence on the second
variable, $\eta(A,-)$ is not so simple.

  2. 2.

A homomorphism $\sigma:A\to S\otimes B$ can be regarded as an algebra
homomorphism $\sigma:A\to[B^{\circ},S]$ by mapping $s\otimes b\mapsto
s\otimes\psi_{b}$; $\psi_{b}$ denoting evaluation at $b$. The equivalence in
2.7 $Alg(A,[B^{\circ},S])\cong Alg(A\triangleleft B^{\circ},S)$ provides the
required map $A\triangleleft B^{\circ}=F(A,B)\to S$.

The category $Alg(F(A,B),-)$ is the category with pairs $(\rho,T)$ where $T$
is an algebra and $\rho:F(A,B)\to T$ is an algebra homomorphism. Morphisms
$\tau:(\rho,T)\to(\rho^{\prime},T^{\prime})$ are homomorphisms $\tau:T\to
T^{\prime}$ such that $\rho^{\prime}=\tau\circ\rho$. Functoriality follows
from the functoriality of $\triangleleft$, theorem 2.7.

  3. 3.

From part 1 above, $\eta(A,B):A\to F(A,B)\otimes B$ is an extension. Check
that the map $\sigma$ of part 2 above is a morphism of extensions.

  4. 4.

This follows from uniqueness of $\eta$ For algebras $A,\ B,C$ consider the
composition

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\eta(A,B)}$$\textstyle{F(A,B)\otimes
B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1\otimes\eta(B,C)}$$\textstyle{F(A,B)\otimes
F(B,C)\otimes C.}$

This provides a factorisation $\triangle_{B}:F(A,C)\to F(A,B)\otimes F(B,C)$.
The coassociativity follows from the universal property of $\eta$. In the case
$A=B=C$, this provides F(A, A) with a coassociative comultiplication. $F(A,\
A)$ has a counit determined on generators by the contraction $A\otimes
A^{\circ}\to k$.

  5. 5.

This is 1.2.

  6. 6.

This is a restatement of proposition 2.13.

∎

Some elementary examples of $F(A,B)$ are collected in the following
proposition.

######  Proposition 2.16.

  1. 1.

$F(A,k)=A$

  2. 2.

$F(k,A)=k$

_Remark_ Part 2 of 2.15 justifies the naming of this construction as the
universal measuring algebra. For $A=B$ it seems to play a role reminiscent of
universal enveloping algebras, providing an object that maps to all
representations of $A$.

######  Corollary 2.17.

If $\theta:A\to\operatorname{End}(W)$ is a representation of $A$, $F(\theta)$
is an algebra homomorphism $F(\theta):F(A,A)\to End(W)$

###### Proof.

This is immediate from part 2 of 2.15 above, as $\theta$ can be considered as
the extension

$A\to\operatorname{End}(V)\otimes A,\ \ \ a\mapsto\theta(a)\otimes 1.$

∎

##  3 Computations

The categorical structure behind the constructions have the consequence that
they are easy to work with. They are also computable in small cases. This
section begins with the basic example that initiated the project, the
computation of the universal measuring coalgebra $P(\mathbb{C},\mathbb{C})$ as
$F(\mathbb{C},\mathbb{C})^{\circ}$, where $\mathbb{C}$ is considered as an
algebra over $\mathbb{R}$. A second set of computable examples are the
algebras $F(\mathbb{Q}[x]/p(x),\mathbb{Q}[x]/p(x))$ for a polynomial $p(x)$.
Finally for the particular case of $p(x)=x^{2}$,
$F(\mathbb{R}[x]/x^{2},\mathbb{R}[x]/x^{2})$ coincides with the Pareigis Hopf
algebra, connecting comodules over this coalgebra with chain complexes.

###  3.1 The algebra $F(\mathbb{C},\mathbb{C})$

Throughout this subsection let $A=\mathbb{R}[x]/(x^{2}+1)$, and let $F$ be the
(real) algebra $F(A,A)$. The generators of the ideal defining $F(A,A)$ can be
got directly from the definitions, but it is easier to compute postulate an
algebra $F$ together with a homomorphism $\eta:A\to F\otimes A$ with the
required universal property.

Write $\eta(x)=f_{0}\otimes 1+f_{1}\otimes x$. Since $\eta$ is assumed to be
an algebra homomorphism, it must be the case that $\eta(x)^{2}=-1$ or

$\displaystyle f_{0}^{2}-f_{1}^{2}$ $\displaystyle=-1$ (3) $\displaystyle
f_{0}f_{1}+f_{1}f_{0}$ $\displaystyle=0$ (4)

Let $F$ be the algebra over $\mathbb{R}$ generated by $\\{f_{0},\ f_{1}\\}$
subject to the above identities. If $S$ is another algebra with a homomorphism
$\sigma:A\to S\otimes A$, write $\sigma(x)$ as $\sigma(x)=s_{0}\otimes
1+s_{1}\otimes x$ with $s_{i}$ in $S$.. Evidently $s_{0},\ s_{1}$ must satisfy
the same identities, and hence there is an algebra homomorphism
$F(\sigma):F\to S$, satisfying the universal property required of
$F(\mathbb{C},\mathbb{C})$.

$F(\mathbb{C},\mathbb{C})$ can then be given explicitly in terms of a basis:

$F=<f_{0}^{\epsilon}f_{1}^{k}:\epsilon\in\\{0,1\\},k\in\mathbb{Z}^{+}>.$

####  3.1.1 The coalgebra structure on $F(\mathbb{C},\mathbb{C})$

The coalgebra structure on $F$ can be given by considering the composition

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\eta}$$\textstyle{F\otimes
A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1\otimes\eta}$$\scriptstyle{\triangle\otimes
1}$$\textstyle{F\otimes F\otimes A}$

The comultiplication $\triangle$ is defined so that both compositions are
equal.

$\displaystyle(1\otimes\eta)\circ\eta(x)$
$\displaystyle=(1\otimes\eta)(f_{0}\otimes 1+f_{1}\otimes x)$ (5)
$\displaystyle=f_{0}\otimes 1\otimes 1+f_{0}\otimes f_{0}\otimes
1+f_{1}\otimes f_{1},$ (6) $\displaystyle(\triangle\otimes 1)\circ\eta$
$\displaystyle=\triangle f_{0}\otimes 1+\triangle f_{1}\otimes x$ (7)

By comparing coefficients of $A$

$\triangle f_{1}=f_{1}\otimes f_{1},\ \ \triangle f_{0}=f_{0}\otimes
1+f_{1}\otimes f_{0}.$

####  3.1.2 Finite dimensional modules of $F$ and $P(\mathbb{C},\mathbb{C})$

The quest that generated the development of the theory contained in this paper
was the quest to calculate $P(\mathbb{C},\mathbb{C})$ considering $\mathbb{C}$
as an $\mathbb{R}$ algebra.

######  Proposition 3.1.

Irreducible subcoalgebras of $P(\mathbb{C},\mathbb{C})$ are of the form
$J_{b}^{\perp}$ where $b\in\mathbb{C}$ and $J_{b}$ is the ideal of $F$
generated by $f_{1}^{2}-b^{2}$ in the case that $b^{2}$ is real, or
$(f_{1}-b)(f_{1}-\bar{b})$ otherwise, and $J_{b}^{\perp}$ is the subcoalgebra
of $F$ which vanishes on $J_{b}$.

###### Proof.

To calculate $P(\mathbb{C},\mathbb{C})=F^{\circ}$, the strategy is to explore
finite dimensional complex representations $\theta:F\to\operatorname{End}(V)$.
The real dual coalgebra $F^{\circ}$ is then the union of the real dual
coalgebras $\theta^{*}(\operatorname{End}(V)^{*})$.

Suppose $V$ is such a finite dimensional complex representation of $F$. Let
$V_{b}$ be an $f_{1}$ eigenspace with eigenvalue $b$, and suppose that $v\in
V_{b}$ is an eigenvector. Then $f_{0}v$ is also an eigenvector of $f_{1}$ with
eigenvalue $-b$. The subspace spanned by $\\{v,f_{0}v\\}$ is then an
irreducible $F$ submodule, since $f_{0}^{2}v=(f_{1}^{2}-1)v$.

Thus finite dimensional irreducible modules of $\mathbb{C}F$ are indexed by
pairs of complex numbers $\\{b,a\\}$ with $a^{2}-b^{2}+1=0$. The image of $F$
is isomorphic to $M_{2}(\mathbb{C})$. The kernel $J_{b}$ of
$\mathbb{C}F\to\operatorname{End}(V)$ is generated by $(f_{1}^{2}-b^{2})$. If
$b^{2}$ is real, this generates a maximal ideal of $F$. If $b^{2}$ is not
real, $(f_{1}-b)(f_{1}-\bar{b})$ is in $F$ and generates an ideal. ∎

###  3.2 Representations of $F$ in $M_{2}(\mathbb{R}(\lambda))$

The algebra $F$ can be represented as a subalgebra of
$M_{2}(\mathbb{R}(\lambda))$ by assigning

$f_{1}\to\begin{bmatrix}a(\lambda)&0\\\ 0&-a(\lambda)\end{bmatrix},\ \ \ \ \
f_{0}=\begin{bmatrix}0&b(\lambda)\\\ c(\lambda)&0\end{bmatrix}$

where $a(\lambda),\ b(\lambda),\ c(\lambda)$ are such that
$b(\lambda)c(\lambda)-a(\lambda)^{2}+1=0$. The dependence on an indeterminate
$\lambda$ is needed to ensure that powers of $f_{0}$ and $f_{1}$ are
algebraically independent from one another (excluding the two defining
relations $f_{0}^{2}-f_{1}^{2}+1=f_{0}f_{1}+f_{1}f_{0}=0$). Examples
suggesting a interpretation as deformations of complex numbers are

$f_{1}\to\begin{bmatrix}\lambda&0\\\ 0&-\lambda\end{bmatrix},\ \ \ \ \
f_{0}=\begin{bmatrix}0&\lambda+1\\\ \lambda-1&0\end{bmatrix}$

and

$f_{1}\to\begin{bmatrix}\sin(\lambda)&0\\\ 0&-\sin(\lambda)\end{bmatrix},\ \ \
\ \ f_{0}=\begin{bmatrix}0&\cos(\lambda)\\\ -\cos(\lambda)&0\end{bmatrix}$

(allowing more general functions of $\lambda$). These can also be viewed as
curves in $\mathfrak{sl}_{2}(\mathbb{R})$.

Such representations of $F$ enables us to give a more concrete description of
$F^{\circ}$. Since $F\subset M_{2}(\mathbb{R}[\lambda])$, there is a coalgebra
map

$M_{2}(\mathbb{R}[\lambda])^{\circ}=\mathbb{R}[\lambda]^{\circ}\otimes
M_{2}(\mathbb{R})^{\circ}\to F^{\circ}$

The coalgebras $\mathbb{R}[\lambda]^{\circ}$ and $M_{2}(\mathbb{R})^{\circ}\to
F^{\circ}$ were described in Section 1.2, although we need to note the slight
subtlety that in section 1.2 we considered $\mathbb{C}[\lambda]^{\circ}$
rather than $\mathbb{R}[\lambda]^{\circ}$. The necessary modification for the
real case is that $\mathbb{R}[\lambda]^{\circ}$ has a real basis
$\\{\mathfrak{Re}\frac{d^{n}}{d\lambda^{n}}|_{b},\mathfrak{Im}\frac{d^{n}}{d\lambda^{n}}|_{b}\\}$
for complex $b$. Thus $M_{2}(\mathbb{R}[\lambda])^{\circ}$ has a basis
$\\{\mathfrak{Re}\frac{d^{n}}{d\lambda^{n}}|_{b}\xi_{ij},\mathfrak{Im}\frac{d^{n}}{d\lambda^{n}}|_{b}\xi_{ij}\\}$.

$F$ is a subalgebra of $M_{2}(\mathbb{R}[\lambda])$, so the map
$M_{2}(\mathbb{R}[\lambda])^{\circ}\to F^{\circ}$ has a kernel. A general
element $M(\lambda)$ in $F$ can be written as

$M(\lambda)=\begin{bmatrix}p(\lambda)&q(\lambda)(\lambda+1)\\\
-q(-\lambda)(-\lambda+1)&p(-\lambda)\end{bmatrix},$

for poilynomials $p(\lambda),\ q(\lambda)$. Thus

$\displaystyle\frac{d^{n}}{d\lambda^{n}}|_{b}<\xi_{22},M(\lambda)>=$
$\displaystyle(-1)^{n}\frac{d^{n}}{d\lambda^{n}}|_{-b}<\xi_{22},M(\lambda)>$
(8) $\displaystyle\frac{d^{n}}{d\lambda^{n}}|_{b}<\xi_{12},M(\lambda)>=$
$\displaystyle(-1)^{n+1}\frac{d^{n}}{d\lambda^{n}}|_{-b}<\xi_{21},M(\lambda)>.$
(9)

Thus $F^{\circ}$ has a basis
$\\{\mathfrak{Re}\frac{d^{n}}{d\lambda^{n}}|_{b}\xi_{11},\mathfrak{Im}\frac{d^{n}}{d\lambda^{n}}|_{b}\xi_{11},\mathfrak{Re}\frac{d^{n}}{d\lambda^{n}}|_{b}\xi_{12},\mathfrak{Im}\frac{d^{n}}{d\lambda^{n}}|_{b}\xi_{12}\\}$.

and comultiplication given by

$\displaystyle\triangle\mathfrak{Re}\frac{d^{n}}{d\lambda^{n}}|_{b}\xi_{ij}=$
$\displaystyle\sum_{m,k}\binom{n}{m}\mathfrak{Re}\frac{d^{m}}{d\lambda^{m}}|_{b}\xi_{ik}\otimes\mathfrak{Re}\frac{d^{n-m}}{d\lambda^{n-m}}|_{b}\xi_{kj}-\sum_{m,k}\binom{n}{m}\mathfrak{Im}\frac{d^{m}}{d\lambda^{m}}|_{b}\xi_{ik}\otimes\mathfrak{Im}\frac{d^{n-m}}{d\lambda^{n-m}}|_{b}\xi_{kj}$
(10)
$\displaystyle\triangle\mathfrak{Im}\frac{d^{n}}{d\lambda^{n}}|_{b}\xi_{ij}=$
$\displaystyle\sum_{m,k}\binom{n}{m}\mathfrak{Re}\frac{d^{m}}{d\lambda^{m}}|_{b}\xi_{ik}\otimes\mathfrak{Im}\frac{d^{n-m}}{d\lambda^{n-m}}|_{b}\xi_{kj}+\sum_{m,k}\binom{n}{m}\mathfrak{Im}\frac{d^{m}}{d\lambda^{m}}|_{b}\xi_{ik}\otimes\mathfrak{Re}\frac{d^{n-m}}{d\lambda^{n-m}}|_{b}\xi_{kj}$
(11)

This presentation of $F$ and $F^{\circ}$ provides a hands-on approach to using
$F$. The finite dimensional subcoalgebras

$F^{\circ}_{b}=<\frac{d^{n}}{d\lambda^{n}}|_{b}\xi_{ij}|n\geq 0,i,\
j,\in\\{1,2\\}>$

for $n\leq N$ provide finite dimensional representations of $F$ which could be
induced to provide representations of
$M_{2}(\mathbb{R}[\lambda,\lambda^{-1}])$ and related algebras.

###  3.3 Matrix methods

The method used to calculate $F(\mathbb{C},\mathbb{C})$ extends to cases where
$A$ and $B$ are finite dimensional with dim$B=n$ using matrices. Choose a
basis $\\{a_{i}\\}$ (resp. $\\{b_{r}\\}$) for $A$ (resp. $B$) and let $B_{r}$
be the matrices representing $b_{r}$ in the left regular representation of
$B$. Let $\\{b_{r}^{*}\\}$ denote the dual basis. Let $\\{c_{ij}\\}$ and
$\\{d_{rs}\\}$ be the structure constants

$a_{i}a_{j}=\sum_{k}c_{ij}^{k}a_{k},\ \ \ b_{r}b_{s}=\sum_{t}d_{rs}^{t}b_{t}$

Writing $f_{ir}=a_{i}\otimes b^{*}_{r}$, the Write $\eta:A\to F(A,B)\otimes B$
in terms of these bases so that $\eta(a_{i})$ is an $n\times n$ matrix

$\eta(a_{i})=\sum_{r}f_{ir}B_{r}$

with coefficients in $F(A,B)$. The identities required of $F(A,B)$ are given
by comparing the matrix entries:

$\displaystyle(\sum_{r}f_{ir}B_{r})(\sum_{s}f_{js}B_{s})$
$\displaystyle=\eta(a_{i})\eta(a_{j})$ (12)
$\displaystyle=\sum_{k}c_{ij}^{k}\eta(a_{k})$ (13) $\displaystyle=\sum_{k\
r}c_{ij}^{k}f_{kr}B_{r}$ (14)

This can be computed by matrix multiplication, taking care not to assume
commutativity of the $f_{ij}$. This method is particularly useful in
calculations where $A=B=\mathbb{Q}[x]/p$. It will also be helpful in
understanding and computing with the module version of this construction.

###  3.4 Calculating $F(\mathbb{Q}[x]/p,\mathbb{Q}[x]/p)$ over $\mathbb{Q}$

Here $k=\mathbb{Q}$ and $p=p(x)=\sum p_{i}x^{i}$ is a polynomial. The problem
of computing $F(\mathbb{Q}[x]/p,\mathbb{Q}[x]/p)$ can be broken into three
stages as follows.

######  Proposition 3.2.

  1. 1.

$F(\mathbb{Q}[x],\mathbb{Q}[x])=T(\mathbb{Q}[x]^{\circ})$.

  2. 2.

$F(\mathbb{Q}[x],\mathbb{Q}[x]/p)=T((\mathbb{Q}[x]/p)^{\circ})$.

  3. 3.

$F(\mathbb{Q}[x]/p,\mathbb{Q}[x]/p)=T((\mathbb{Q}[x]/p^{\circ}))/J$ where $J$
is the ideal generated by

$p_{0}\epsilon(\alpha)+\sum p_{i}\triangle^{i-1}\alpha$

for $\alpha\in(\mathbb{Q}[x]/p)^{\circ}$

###### Proof.

Yoneda’s lemma provides a strategy for proving all three parts. For algebras
$A,\ B,C$, from 2.7 we have $Alg(F(A,B),C)\cong Alg(A,[B^{\circ},C])$.
Yoneda’s lemma says that $Alg(F(A,B),C)\cong Alg(A,[B^{\circ},C])\cong
Alg(F^{\prime},C)$ (with the isomorphisms being natural) for all algebras $C$
if and only if $F^{\prime}=F(A,B)$. Therefore it is sufficient to show that

  * •

Step 1. $Alg(\mathbb{Q}[x],[\mathbb{Q}[x]^{\circ},C])\cong
Alg(T(\mathbb{Q}[x]^{\circ}),C)$

  * •

Step 2. $Alg(\mathbb{Q}[x],[(\mathbb{Q}[x]/p)^{\circ},C])\cong
Alg(T((\mathbb{Q}[x]/p)^{\circ}),C)$

  * •

Step 3. $Alg(\mathbb{Q}[x]/p,[(\mathbb{Q}[x]/p)^{\circ},C])\cong
Alg((T(\mathbb{Q}[x]/p)^{\circ})/J,C)$.

_Step 1_. An algebra homomorphism
$\theta:\mathbb{Q}[x]\to[\mathbb{Q}[x]^{\circ},C]$ is determined by the image
of $\theta(x):\mathbb{Q}[x]^{\circ}\to C$ with no restrictions.

Similarly an algebra homomorphism $\tilde{\theta}:T(\mathbb{Q}[x]^{\circ})\to
C$ is determined by its restriction to
$T^{1}(\mathbb{Q}[x]^{\circ})=\mathbb{Q}[x]^{\circ}$. There are no
restrictions; any linear map $\tilde{\theta}$ gives rise to an algebra
homomorphism $\tilde{\theta}:T(\mathbb{Q}[x]^{\circ})\to C$.

The correspondence $\theta\leftrightarrow\tilde{\theta}$ thus establishes a
(natural) correspondence $Alg(\mathbb{Q}[x],[\mathbb{Q}[x]^{\circ},C])\cong
Alg(T(\mathbb{Q}[x]^{\circ},C)$ and the first part of the proposition.

_Step 2_ The argument is identical, replacing $\mathbb{Q}[x]^{\circ}$ with
$\mathbb{(Q}[x]/p)^{\circ}$.

_Step 3_ Observe that
$\theta:\mathbb{Q}[x]/p\to[(\mathbb{Q}[x]^{/}p)^{\circ},C]$ is an algebra
homomorphism if $\theta(f(x)p)=0$ for any product $f(x)p$ in the ideal
generated by $p$. That is, $\theta(f(x)p)=0$ if and only if for any $\alpha$
in $(\mathbb{Q}[x]^{/}p)^{\circ}$

$<\alpha,f(x)p>=\sum_{(\alpha)}<\alpha_{(2)},f(x)><\alpha_{(1)},p>=0.$

Therefore, $\theta(f(x)p)=0$ if and only if for any $\alpha$,

$\displaystyle<\alpha,\theta(p)>$
$\displaystyle=\sum^{n-1}_{i=0}p_{i}<\alpha,\theta(x^{i})>$ (15)
$\displaystyle=p_{0}\epsilon(\alpha)+p_{1}<\alpha,\theta(x)>+\sum_{k=2}^{n-1}\sum_{(\alpha)}<\alpha_{(k)},\theta(x)>...<\alpha_{(1)},\theta(x)>$
(16)

On the other hand, $\tilde{\theta}:T(\mathbb{Q}[x]/p)^{\circ})\to C$ vanishes
on $J$ if and only if

$\displaystyle 0$
$\displaystyle=\tilde{\theta}(p_{0}\epsilon(\alpha)+p_{1}\alpha+\sum_{i=2}^{n-1}\sum_{(\alpha)}\alpha_{(i)}\otimes...\otimes\alpha_{(1)})$
(17)
$\displaystyle=p_{0}\epsilon(\alpha)+p_{1}\tilde{\theta}(\alpha)+\sum_{i=2}^{n-1}p_{i}\sum_{(\alpha)}\tilde{\theta}(\alpha)_{(i)}...\tilde{\theta}(\alpha)_{(1)}$
(18)
$\displaystyle=p_{0}\epsilon(\alpha)+p_{1}<\alpha,\theta(x)>+\sum_{k=2}^{n-1}\sum_{(\alpha)}<\alpha_{(k)},\theta(x)>...<\alpha_{(1)},\theta(x)>$
(19)

as required. ∎

###  3.5 Extensions of $\mathbb{Q}[x]/p$

Let $A=\mathbb{Q}[x]/p(x)$. While it is inconvenient to calculate $F(A,A)$ in
its entirety, extensions $\theta:A\to k\otimes A$ for a field $k$ correspond
to algebra homorphisms $F(\theta):F(A,A)\to k$. These can be calculated as
follows.

The algebra $\operatorname{End}(A)$ can be identified with $M_{n}(\mathbb{Q})$
via the basis of powers of $x$, $\\{x^{i}\\}$. Left multiplication defines a
representation of $L:A\to M_{n}(\mathbb{Q})$ in which $x$ is represented by
its companion matrix $C$. Denote by $<C>$ the subalgebra of
$M_{n}(\mathbb{Q})$ generated by $C$. We can further identify $k\otimes
A=k[x]$ with its image in $M_{n}(k)$ for $k$ a field containing $\mathbb{Q}$,
so that if $Y$ is any matrix in $<C>$, the first column of $Y$ gives the
coefficients $Y=\sum y_{i}X^{i}$.

Now assume the roots of $p$ are all distinct and are contained in the field
$k$, and choose an ordering $\\{\mu_{1},...,\mu_{n}\\}$. The Vandermonde
matrix $V$ expresses roots in terms of the basis $\\{x^{i}\\}$, so that if
$V=(v_{ij})$, ${\mu_{j}=\sum v_{ij}x^{i}}$. Conjugation by $V$ diagonalises
$k\otimes A$ considered as a subalgebra of $M_{n}(k)$, identifying $k\otimes
A$ with the subalgebra $D_{n}(k)$ of diagonal matrices.

Now let $[n]$ denote the set of integers from 1 to n, and let
$\sigma:[n]\to[n]$ be any function. Then $\sigma$ induces a homomorphism
$\tilde{\sigma}:D_{n}(k)\to D_{n}(k)$ simply by repositioning the entries.
These homomorphisms form a monoid which includes all of the symmetric group
$S_{n}$ as its invertible elements.

Conjugating $\tilde{\sigma}$ by Vandermonde matrices provides a homomorphism

$W_{\sigma}=V^{-1}\tilde{\sigma}V:k\otimes A\to k\otimes A$

or, by restriction, an extension $W_{\sigma}:A\to k\otimes A.$

Consider the matrix representing $W_{\sigma}$ in terms of the basis
$\\{1,C,C^{2},...,C^{n-1}\\}$. With respect to this basis writing
$W_{\sigma}=\sum w_{i}C^{i}$, the coefficients $w_{i}$ are exactly the first
column of $W_{\sigma}$.

Write $\eta=\eta(A,A):A\to F(A,A)$ as
$\eta(x)=f_{0}+f_{1}x+f_{2}x^{2}+...+f_{n-1}s^{n-1}$. The homomorphism
$F(W_{\sigma}):F(A,A)\to k$ is given by $F(f_{i})=w_{i}$

One can ask if the homomorphisms $W_{\sigma}$ correspond to Galois
transformations of $k$. For each root $\mu$ there is a homomorphism
$\pi_{\mu}:\mathbb{Q}[x]/p(x)\to k$ sending $x\mapsto\mu$. The requirement is
that $\pi_{\bar{\sigma}(\mu)}(x)=\pi_{\mu}W_{\sigma}(x)$ for all roots $\mu$.

###  3.6 Representations of $F(\mathbb{Q}[x]/p,\mathbb{Q}[x]/p)$ in
$M_{n}(k[\lambda])$

. The loop algebra representations of $F(\mathbb{C},\mathbb{C})$ have
analogues for $\mathbb{Q}[x]/p$. With notation as in the previous subsection,
let $Z$ be any matrix in $M_{n}$, and let $g(u)=exp(uZ)$. Then $g(u)Cg(-u)$
also satisfies the polynomial $p$.

Expand $g(u)Cg(-u)$ as the exponential of $ad(Z)$:

$\displaystyle g(u)Cg(-u)$ $\displaystyle=exp(ad(uZ))C$ (20)
$\displaystyle=C+u[Z,C]+\frac{1}{2}u^{2}[Z,[Z,C]]+...$ (21)

This expression satisfies the polynomial $p$ _for any indeterminate t_. Now
let $u=\lambda x$. Then

$\displaystyle g(u)Cg(-u)$ $\displaystyle=exp(x.ad(\lambda Z))C$ (22)
$\displaystyle=C+[\lambda Z,C]x+\frac{1}{2}[\lambda Z,[\lambda Z,C]]x^{2}+...$
(23)

which can be considered as an element in
$M_{n}(\mathbb{Q}[\lambda])\otimes\mathbb{Q}[x]/p$.

The assignment $x\mapsto g(u)C(g-u)$ written out this way then provides an
extension homomorphism $\sigma_{Z}:\mathbb{Q}[x]/p\to
M_{n}(\mathbb{Q}[\lambda])\otimes\mathbb{Q}[x]/p$ as promised. The
corresponding map $F(\sigma)$ represents $F(\mathbb{Q}[x]/p,\mathbb{Q}[x]/p)$
in $M_{n}(k[\lambda])$ in $M_{n}(\mathbb{Q}[\lambda])$.

###  3.7 $F(B,B)$ for $B=k[d]/d^{2}$, and relation to the Pareigis Hopf
algebra

By analogy with $F(\mathbb{C},\mathbb{C})$, the algebra $F(B,B)$ for
$B=k[d]/d^{2}$, The algebra $F(B,B)$ is generated by two elements, $g_{0},\
g_{1}$, such that

$\eta:B\to F(B,B)\otimes B,\ \ \ \eta(d)=g_{0}\otimes 1+g_{1}\otimes d.$

The same reasoning gives explicit defining identities,

$g_{0}g_{1}=g_{1}g_{0}=0=g_{0}^{2}$

and an explicit basis

$F(B,B)=<g_{1}^{i},g_{0}g_{1}^{i}:i\geq 0>.$

Comultiplication is determined: since $1\otimes\eta\circ\eta=\triangle\otimes
1\circ\eta$,

$\triangle g_{1}=g_{1}\otimes g_{1},\ \ \ \triangle g_{0}=g_{1}\otimes
g_{0}+g_{0}\otimes 1.$

The counit $\epsilon:F(B,B)\to k$ is determined by the evaluation map
$B\otimes B^{*}\to k$, so that

$\epsilon(g_{0})=0,\epsilon(g_{1})=1.$

This bialgebra is closely related to a well understood Hopf algebra, first
presented by Pareigis [4].

######  Definition 3.3.

There is a non-commutative, non-cocommutative Hopf algebra $H$, the Pareigis
Hopf Algebra, defined over any field $k$, which has presentation given by
generators and relations:

  1. 1.

$H=k[x,y,1/y]/(xy+yx,x^{2})$ as an algebra (1/y is a left and right
multiplicative inverse for y).

  2. 2.

The antipode $s:H\to H$ is defined by $y\mapsto 1/y,x\mapsto xy$

  3. 3.

$\triangle:H\to H\otimes H,y\mapsto y\otimes y,x\mapsto x\otimes
1+(1/y)\otimes x$

  4. 4.

$\epsilon:H\to k,x\mapsto 0,y\mapsto 1$

The relationship of $H$ with $F(B,B)$ is as follows. Let $H^{-}$ be the
subalgebra of $H$ generated by $x$ and $1/y$. This is a bialgebra, but not a
Hopf algebra, as it is not closed under the antipode.

######  Proposition 3.4.

There is a homomorphism

$p:F(B,B)\to H$

establishing an isomorphism of $F(B,B)$ with $H^{-}$ as bialgebras.

###### Proof.

Set $p(g_{0})=x,\ p(g_{1})=1/y.$ and observe that the generators and relations
coincide. ∎

##  4 Comodules

A (left) _comodule_ $X$ over a coalgebra $H$ is a vector space with a
comultiplication

$\triangle_{X}:X\to H\otimes X,\ \ \ \triangle_{X}(d)=\sum_{(d)}d_{(1)}\otimes
d_{(0)}$

satisfying coassociativity:

$(1_{C}\otimes\triangle_{X})\circ\triangle_{X}=(\triangle_{C}\otimes
1_{X})\circ\triangle_{X}:X\to C\otimes C\otimes X$

and counit

$(\epsilon\otimes 1_{X})\circ\triangle_{X}=1_{X}:X\to k\otimes X=X.$

When it is helpful, a left comodule $X$ may be denoted ${}_{H}X$ to emphasise
that $X$ is an $H$ comodule. Similarly, left modules $M$ over $A$ may be
denoted ${}_{A}M$.

If ${}_{H}D_{1},\ _{H}D_{2}$ are two comodules, a linear map $\zeta:D_{1}\to
D_{2}$ is a map of $H$-comodules if

$\triangle_{2}\circ\zeta=1_{H}\otimes\zeta\circ\triangle_{1}:D_{1}\to H\otimes
D_{2}.$

Denote by ${}_{H}Comod$ the category of (left) $H$-comodules.

#### Example: Duals of modules

The dual coalgebra model provides examples of comodules. Let $V$ be a (finite
dimensional) vector space on which a finite dimensional algebra $S$ acts on
the right. Then dualising the action gives rise to a comultiplication

$\triangle:V^{*}\to(V\otimes S)^{*}=S^{*}\otimes V^{*}$

#### Example: Comodules of $F(B,B)$ for $B=k[d]/d^{2}$

Let $M=\oplus M_{i}$ be a chain complex with $M_{i}=0$ for $i\geq 1$ with
boundary map $d:M_{i}\to M_{i+1}$. Define a comodule structure on $M$ by
setting

$\rho:M\to F(B,B)\otimes M,\ \ \ m_{i}\mapsto g_{1}^{i}\otimes
m_{i}+g_{0}g_{1}^{i+1}\otimes dm_{i}.$

That $M_{i}$ is a comodule can be verified directly.

Conversely, given an $F(B,B)$ comodule $Z$, set $\widetilde{Z}_{k}$ to be the
subspace of $Z$ spanned by elements $z_{(1,i)},\ z_{(0,i)}$ in the expression

$\triangle z=\sum_{i}g_{i}^{-i}\otimes z_{(1,i)}+g_{0}g_{1}^{-i+1}\otimes
z_{(0,i+1)}$

This connection provides homological links with extensions more generally. Let
$K$ be a coalgebra and $Z$ a $K$ comodule. A coalgebra homomorphism
$\kappa:F(B,B)\to K$ gives $Z$ the structure of an $F(B,B)$ comodule, and
hence defines a chain complex $\widetilde{Z}$. Suppose $A$ is a finite
dimensional algebra and let $V$ be a finite dimensional $A$ module. The theme
of the present section is the construction of a universal $F(A,A)$ comodule
$D(V,V)$. In this way, coalgebra homomorphisms $\rho:F(B,B)\to F(A,A)$ provide
a source of chain complexes. Thus the coalgebra structure of $F(A,A)$ is
closely linked to the homological information pertaining to $A$.

This relationship between the Pareigis Hopf algebra and homologies are
developed extensively (and correctly in the differential graded setting) in
[1].

###  4.1 Measuring comodules

######  Definition 4.1.

Let $A$, $B$ be algebras, let ${}_{A}M,\ _{B}N$ be $A$ and $B$ modules
respectively. Let $\rho:H\to\operatorname{Hom}_{k}(A,B)$ be a measuring
coalgebra, and let ${}_{H}X$ be an $H$ comodule. Say that
$\gamma:X\to\operatorname{Hom}_{k}(M,N)$ _measures_ if

$\gamma(x)(am)=\sum_{(x)}<\rho(x_{(1)}),a><\gamma(x_{(0)}),m>$

for all $x$ in $X$, $a$ in $A$, and $m$ in $M$.

As with measuring coalgebras (proposition 1.2), there is an equivalent
condition.

######  Proposition 4.2.

$\gamma:_{H}X\to\operatorname{Hom}_{k}(_{A}M,_{B}N)$ measures if and only if
the corresponding map

$\gamma:_{A}M\to\operatorname{Hom}_{k}(_{H}X,_{B}N)$

is a map of $A$-modules.

As for algebras and coalgebras, use the notation

$\operatorname{Hom}_{k}(_{H}X,_{B}N):=[X,N]=_{[H,B]}[X,N]$

to emphasise the module structure.

###### Proof.

$\operatorname{Hom}_{k}(_{H}X,_{B}N)$ is an $[H,B]$ module with the action
given by

$<\alpha*\zeta,x>=\sum_{(x)}<\alpha,x_{(1)}><\zeta,x_{(0)}>$

for $\alpha$ in $[H,B]$, $\zeta$ in $[X,N]$ and $x$ in $X$. The measuring map
$H\to\operatorname{Hom}_{k}(A,B)$ corresponds to an algebra homomorphism
$A\to[H,B]$. With respect to this map, $[X,N]$ becomes an $A$-module. ∎

.

###  4.2 Examples of measuring comodules

#### Example: Module maps

Let $kf$ be the one dimensional comodule over the one dimensional coalgebra
$ke$ and let $\triangle f=e\otimes f$. Suppose
$\rho:ke\to\operatorname{Hom}(A,B)$ measures (so that $\rho(e)$ is a
homomorphism). A linear map $\theta:kf\to\operatorname{Hom}(_{A}M,_{B}N)$ is a
measuring map if and only if $\theta(f)$ is a module map with respect to the
homomorphism $\rho(e)$.

#### Example: Modules of extensions

Let $\psi:A\to S\otimes B$ is an extension, Let $M,\ N,\ V$ be modules over
$A,\ B,S$ respectiviely and suppose that $V$ is finite dimensional. Suppose
also that there is a map $\theta:M\to V\otimes N$ which is an $A$ module map
with the $A$ module structure on $V\otimes N$ that conferred by $\psi$. That
$V^{*}$ is an $S^{\circ}$ module can be seen by dualising the maps defining
the action of $S$ on $V$ so that for $v^{*}$ in $V^{*}$, $s$ in S and $w$ in
$V$,

$<v^{*},sw>=\sum_{(v^{*})}<v^{*}_{(1)},s><v^{*}_{(0)},w>.$

It is helpful to extend the application of Sweedler notation to express
elements in tensor products. Set

$\displaystyle\psi(a)=$
$\displaystyle\sum_{(\psi(a))}\psi(a)_{(1)}\otimes\psi(a)_{(0)}\in S\otimes B$
(24) $\displaystyle\theta(m)=$
$\displaystyle\sum_{\theta(m)}\theta(m)_{(1)}\otimes\theta(m)_{(0)}\in
V\otimes N.$ (25)

The claim is that map $\theta$ corresponds to a map

$\theta^{*}:V^{*}\to\operatorname{Hom}(M,N),\ \
<\theta^{*}v^{*},m>:=<v^{*},\theta(m)_{(1)}>\theta(m)_{(0)}.$

is a measuring map, just as the map $\psi$ corresponds to a measuring map (of
algebras)

$\psi^{*}:S^{\circ}\to\operatorname{Hom}(A,B),\ \
<\psi^{*}s^{*},a>=\sum_{(\psi(a))}<s^{*},\psi(a)_{(1)}>\psi(a)_{(0)}$

The statement that $\theta$ is a module map written out in Sweedler notation
is

$\displaystyle\theta(am)=$ $\displaystyle\psi(a)\theta(m)$ (26)
$\displaystyle=$
$\displaystyle\sum_{(\psi(a))(\theta(m))}\psi(a)_{(1)}\theta(m)_{(1)}\otimes\psi(a)_{(0)}\theta(m)_{(0)}.$
(27)

Then $\theta^{*}:V^{*}\to\operatorname{Hom}(M,N)$ measures if

$<\theta^{*}v^{*},am>=\sum_{(v^{*})}<\psi^{*}(v^{*}_{(1)}),a><\theta^{*}(v^{*}_{(0)}),m>$

With the notation in place, this is a matter of direct calculation:

$\displaystyle<\theta^{*}v^{*},am>=$
$\displaystyle\sum_{(\theta(am))}<v^{*},\theta(am)_{(1)}>\theta(am)_{(0)}$
(28) $\displaystyle=$
$\displaystyle\sum_{(\psi(a))(\theta(m))}<v^{*},\psi(a)_{(1)}\theta(m)_{(1)}>\psi(a)_{(0)}\theta(m)_{(0)}$
(29) $\displaystyle=$
$\displaystyle\sum_{(v^{*})(\psi(a))(\theta(m))}<v^{*}_{(1)},\psi(a)_{(1)}><v^{*}_{(0)},\theta(m)_{(1)}>\psi(a)_{(0)}\theta(m)_{(0)}$
(30) $\displaystyle=$
$\displaystyle\sum_{(v^{*})}<\psi^{*}v^{*}_{(1)},a><\theta^{*}v^{*}_{(0)},m>$
(31)

as required.

#### Example: Connections

A familiar example of measuring comodules arises in differential geometry as
connections. Let $A$ be an algebra of (differentiable) functions on a manifold
$Y$, and let $Der$ be the Lie algebra of derivations of $A$, that is, the
vector fields on $Y$. Then define $H$ to be $Der\oplus\mathbb{R}e$, where
$\mathbb{R}e$ is a vector space, and endow $H$ with a coalgebra structure by
setting

$\triangle e=e\otimes e,\ \ \triangle\eta=\eta\otimes e+e\otimes\eta,\ \
\epsilon(e)=1,\ \ \epsilon(\eta)=0.$

for $X$ in $Der$. The linear map $\rho:H\to\operatorname{Hom}_{k}(A,A)$
defined by

$\rho(e)=1_{A},\ \ \rho(X)=\eta\in Der\subset\operatorname{Hom}_{k}(A,A)$

for $X$ in $Der$ measures.

Sections $S$ of a bundle $E$ over $Y$ is just an $A$ module. $H$ can be
considered a comodule over itself. If $\gamma$ is a connection, $\gamma$ is a
linear map

$\gamma:Der\to\operatorname{Hom}(_{A}S,_{A}S).$

satisfying

$\gamma(X)(as)=(Xa)s+a\gamma(X)s$

for $a\in A$, $s\in S$, Extend $\gamma$ to all of $H$ by setting
$\gamma(e)=1_{S}$. Thus

$\gamma:H\to\operatorname{Hom}(S,S)$

The condition for $\gamma$ to be a connection can be rewritten

$<\gamma(X),as>=<\rho(X),a><\gamma(e),s>+<\rho(e),a><\gamma(X),s>$

which is just the measuring condition.

###  4.3 $\widecheck{Mod}$ and $\widehat{Mod}$

Just as measuring coalgebras for a fixed pair of algebras $A,\ B$ can be
compared by maps of measuring coalgebras, measuring comodules for a pair
${}_{A}M,\ _{B}N$ can be compared.

######  Definition 4.3.

A map of comodules $\zeta:X\to X^{\prime}$ is a _map of measuring comodules_
if $\zeta$ respects the measuring maps:

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\zeta}$$\textstyle{\operatorname{Hom}_{k}(_{A}M,_{B}N)}$$\textstyle{X^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

The concept of extensions of algebras has a corresponding concept for modules
over algebras.

######  Definition 4.4.

If ${}_{A}M,\ _{B}N,\ _{S}T$ are modules over $A,\ B,S$ respectively, an
_extension of modules_ is a map

$\zeta:M\to T\otimes N$

which is a module map with respect to an extension $\psi:A\to S\otimes B$. If
${}_{S^{\prime}}T^{\prime}$ is an $S^{\prime}$ module, and $\zeta:M\to
T^{\prime}\otimes N$ another extension with respect to an algebra extension
$\psi^{\prime}:A\to S^{\prime}\otimes B$, a map of extensions
$\zeta\to\zeta^{\prime}$ consists of a map of (algebra) extensions
$\theta:S\to S^{\prime}$ and a map of $S$ modules $\upsilon:T\to T^{\prime}$
so that the following diagram commutes.

$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\zeta}$$\scriptstyle{\zeta^{\prime}}$$\textstyle{T\otimes
N\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\upsilon\otimes
1_{N}}$$\textstyle{T^{\prime}\otimes N}$

The categories $\widecheck{Alg}$ and $\widehat{Alg}$ have analogues in
modules.

######  Proposition 4.5.

Let $M,\ N$ be modules over algebras $A,\ B$ respectively.

  1. 1.

There is a category $\widecheck{Mod}(M,N)$ whose objects are measuring
comodules $X\to\operatorname{Hom}_{k}(M,N)$ and whose morphisms are maps
$\kappa:X\to X^{\prime}$ of measuring comodules.

  2. 2.

There is a category $\widehat{Mod}(M,N)$ whose objects are module maps
$M\to_{S}T\otimes N$ for some $\sigma:A\to S\otimes B$ in $\widehat{Alg}(A,B)$
and whose morphisms are maps $\nu:T\to T^{\prime}$ of extensions.

###### Proof.

The procedure to prove this is identical to that of the corresponding proof
for algebras. Ultimately, as before, the proof rests on the fact that for both
$\widecheck{Mod}$ and $\widehat{Mod}$, composition of horizontal maps in both
categories is simply the tensor product of comodules and modules respectively.
∎

######  Theorem 4.6.

$\widecheck{Mod}$ and $\widehat{Mod}$ are bicategories

Again, the proof is parallel to the proof for the corresponding statement for
algebras.

The example 4.2 enables enables a comparison of the subcategories
$\widecheck{Mod}_{f}$ and $\widehat{Mod}_{f}$ where the morphisms are
restricted to finite dimensional measuring comodules and finite dimensional
modules ${}_{S}U$ respectively. This comparison is made formal in the
following equivalence.

######  Proposition 4.7.

There is a natural equivalence of categories

$*:\widecheck{Mod}_{f}\cong\widehat{Mod}_{f}^{co}:*$

The proof follows the same steps as 1.10.

###  4.4 Universal measuring comodules

Comodules share the same profound finiteness property that coalgebras have:
_every element of a comodule is an element of a finite dimensional
subcomodule_ (Sweedler, Chapter II [2]). As with coalgebras, a consequence is
that the category of comodules over a given coalgebra has colimits and
coequalisers, ensuring the existence of universal measuring comodules.

######  Definition 4.8.

Let ${}_{A}M,\ _{B}N$ be algebras. The _universal measuring comodule_ $Q(M,N)$
is the $P(A,B)$ comodule with a measuring map
$\psi:Q(M,N)\to\operatorname{Hom}_{k}(M,N)$ with the property that if
$\gamma:X\to\operatorname{Hom}_{k}(M,N)$ is any measuring map over any
measuring coalgebra $\rho:H\to\operatorname{Hom}_{k}(A,B)$, then there is a
unique map of measuring comodules $Q(\gamma):X\to Q(M,N)$ such that the
following diagram commutes.

$\textstyle{Q(M,N)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi}$$\textstyle{\operatorname{Hom}_{k}(M,N)}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\gamma}$$\scriptstyle{Q(\gamma)}$

Notice that by the universal property for the universal measuring coalgebra,
any measuring comodule over a measuring coalgebra $H$ can be considered as a
measuring comodule over $P(A,B)$.

$Q(M,N)$ exists. The finiteness properties of comodules enable $Q(M,N)$ to be
constructed as a coequaliser in the same manner as the proof outlined in 2.3.
A more detailed study of these structures can be found in [3].

The fundamental property of the universal measuring comodule is the analogue
of proposition 2.5.

######  Proposition 4.9.

Let $M,\ N$ be modules over algebras $A,\ B$ respectively.

${}_{P(A,B)}Comod(-,Q(M,N))\cong_{A}Mod(M,[-,N]).$

###### Proof.

Let $X$ be a $P(A,B)$ comodule. First observe that $[X,N]$ is a $[P(A,B),B]$
module. For $\alpha\in[P(A,B),B]$ and $\nu\in[X,N]$, define the action by
setting

$<\alpha*\nu,x>=\sum_{x}<\alpha,x_{(1)}><\nu,x_{(0)}>.$

Then $[X,N]$ becomes an $A$ module in the usual way via the homomorphism
$A\to[P(A,B),B]$ of theorem 2.15. As with the corresponding statement for
algebras (proposition 2.5), naturality and equivalence must be checked, but
present no exceptional problems.

∎

The universal measuring comodule enables the definition of a global enriched
category of modules.

######  Proposition 4.10.

There is an enriched category $\underline{Mod}$ whose objects are modules and
whose morphisms are the universal enveloping comodules $Q(M,N)$.

###### Proof.

The slightly uncomfortable fact that requires verification is that an element
$\alpha$ in $Q(_{A}L,\ _{B}M)$ can compose with an element $\beta$ in
$Q(_{B}M,\ _{C}N)$ to give an element in $Q(_{A}L,\ _{C}N)$, but that is a
consequence of the universal property, as follows. Tensoring followed by
composition provides a map

$Q(M,N)\otimes
Q(L,M)\to\operatorname{Hom}(M,N)\otimes\operatorname{Hom}(L,M)\to\operatorname{Hom}(L,N)$

By the universal map $P(B,C)\otimes P(A,B)\to P(A,C)$, $Q(M,N)\otimes Q(L,M)$
can be considered a $P(A,C)$ comodule. That the composition of maps above is a
measuring map can be verified directly. ∎

Thus $\underline{Mod}$ is a global category of modules: a map between an $A$
module $M$ and a $B$ module $N$ is simply an element of the universal
measuring comodule $Q(M,N)$; there is no need to specify an algebra
homomorphism relating $A$ and $B$.

This global category is fibred over a 2-category $\widecheck{Alg1}$ of
algebras adapted from the bicategory $\widecheck{Alg}$ as follows.

First, for algebras $A,\ B$ define a category $\mathcal{P}(A,B)$ whose objects
are are subcoalgebras of the measuring coalgebra $P(A,B)$ and whose morphisms
are inclusions, that is, if $H_{1},\ H_{2}$ are subcoalgebras of $P(A,B)$ the
morphism set $\mathcal{P}(A,B)(H_{1},H_{2})$ contains a single element,
$i_{12}$ if $H_{1}\subset H_{2}$ or is the empty set. Composition of
morphisms, the associativity and unit properties are immediate.

For the bicategory $\widecheck{Alg1}$, objects are algebras and the set of
morphisms $\widecheck{Alg1}(A,B)$ is $\mathcal{P}(A,B)^{op}$,

Composition $H_{2}\circ H_{1}$ in $\widecheck{Alg1}$ is defined by setting

$H_{2}\circ H_{1}=P(i_{2}\otimes i_{1})(H_{2}\otimes H_{1})$

the image of $H_{2}\otimes H_{1}$ in $P(A_{1},A_{3})$.

######  Proposition 4.11.

$\widehat{Alg1}$ defined above is a 2 category.

###### Proof.

Associativity can be checked by inspecting the following diagram. Let
$H_{i}:A_{i}\to A_{I+1},i=1,\ 2,\ 3$ be morphisms (subcoalgebras of
$P(A_{i},A_{i+1})$). Consider the diagram

$\textstyle{H_{3}\otimes H_{2}\otimes
H_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H_{3}\otimes
H_{2}\circ
H_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H_{3}\circ
H_{2}\otimes
H_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P(A_{1},A_{3})}$

The critical observation is that any element in $H_{3}\circ H_{2}\subset
P(A_{2},A_{4})$ is in the image of an element in $H_{3}\otimes H_{2}$.
Therefore for $k\in H_{3}\circ H_{2}\otimes H_{1}$, there is an element $h\in
H_{3}\otimes H_{2}\otimes H_{1}$ with $q(h)=k$. By the uniqueness of the
universal map, it must be that $s(k)=r(p(h))$. Thus $s(H_{3}\circ H_{2}\times
H_{1})\subset r(H_{3}\otimes H_{2}\circ H_{1})$. Similarly $r(H_{3}\otimes
H_{2}\circ H_{1})\subset s(H_{3}\circ H_{2}\times H_{1})$.

The interchange law requires more unravelling than proof. It is satisfied
since if $H_{i},\subset J_{i}\subset K_{i}$ are subcoalgebras of
$P(A_{i},A_{i+1})$ the composition of inclusions $H_{2}\circ H_{1}\subset
J_{2}\circ J_{1}\subset K_{2}\circ K_{1}$ is just the composition $H_{2}\circ
H_{1}\subset K_{2}\circ K_{1}=(H_{2}\subset J_{2}\subset
K_{2})\circ(H_{1}\subset J_{1}\subset k_{1})$. ∎

The fibred structure of $\underline{Mod}$ can now be given.

######  Proposition 4.12.

There is a functor $\Pi:\underline{Mod}\to\widecheck{Alg1}$ giving
$\underline{Mod}$ the structure of a fibred category.

###### Proof.

The steps in the proof are as follows.

  1. 1.

Construct the projection functor $\Pi:\underline{Mod}\to\widecheck{Alg1}$, and
verify that it is a functor.

  2. 2.

For an element $d\in Q(M,N)$ construct the "pullback" $d^{*}(N)$ of $d$.

  3. 3.

Construct an $A$ module map $j:M\to d^{*}(N)$

  4. 4.

Construct an element $e:d^{*}(N)\to N$ with $\Pi(e)=\Pi(d)$ and $e\circ j=d$.

1\. _The functor $\Pi$._ For an element $d$ in $Q(M,N)$ let $D$ be the minimal
measuring subcomodule containing $d$, (which will be finite dimensional) and
let $\Pi(d)$ be the minimal subcoalgebra of $P(A,B)$ over which $D$ is a
subcomodule. $\Pi(d)$ is also finite dimensional.

$\Pi$ is functorial in the sense that that there is a natural transformation,

$\kappa:\Pi(-)\circ\Pi(-)\to\Pi(-\circ-).$

(It is at this point that we need the morphism to be $\mathcal{P}(A,B)^{op}$
rather than $\mathcal{P}(A,B)$.) This is a consequence of uniqueness in the
universal property of $Q(-,-)$ as follows. By their respective universal
properties there are unique maps

$Q(M_{2},M_{3})\otimes Q(M_{1},M_{2})\to Q(M_{1},M_{3}),\ \
P(A_{2},A_{3})\otimes P(A_{1},A_{2})\to P(A_{1},A_{3})$

The image of $\Pi(d_{2})\otimes\Pi(d_{1})$ is a measuring subcoalgebra of
$P(A_{1},A_{3})$ and

$\triangle(d_{2}\circ d_{1})=\triangle(d_{2})\otimes\triangle(d_{1}).$

Thus $\Pi(d_{2})\otimes\Pi(d_{1})\to\operatorname{Hom}(A_{1},A_{3})$ measures
and its imagine in $P(A_{1},A_{3})$ is $\Pi(d_{2})\circ\Pi(d_{1})$. But
computing $\triangle(d_{2}\circ d_{1})$ as above, it must be that
$\Pi(d_{2}\circ d_{1})\subset P(\Pi(d_{2})\otimes\Pi(d_{1})$ since
$\Pi(d_{2}\circ d_{1})$ is minimal, or otherwise put, there is a natural
transformation

$\Pi(d_{2})\circ\Pi(d_{1})\to\Pi(d_{2}\circ d_{1})$

as required.

2\. _The pullback of $d$_. Define $d^{*}(N)=:\Pi(d)\otimes N$. The $A$ action
on $\Pi(d)\otimes N$ is given by

$\textstyle{A\otimes\Pi(d)\otimes
N\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1\otimes\triangle\otimes
1}$$\textstyle{A\otimes\Pi(d)\otimes\Pi(d)\otimes
N\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{z}$$\textstyle{\Pi(d)\otimes
B\otimes
N\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Pi(d)\otimes
N}$

where $z$ is evaluation on the second factor:

$z(a\otimes h_{2}\otimes h_{1}\otimes n)=h_{2}\otimes<a,h_{1}>\otimes N.$

3\. _The module map $j:M\to d^{*}(N)$_. Define $j:M\to d^{*}(N)$ be the
composition

$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{d\otimes
M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Pi(d)\otimes
Q(M,N)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Pi(d)\otimes
N}$

That this is a map of $A$ modules can be verified directly.

4\. For any finite dimensional coalgebra $K$, the dual algebra $K^{*}$ acts on
$K$:

$\textstyle{K^{*}\otimes
K\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1\otimes\triangle}$$\textstyle{K^{*}\otimes
K\otimes
K\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tau\otimes
1}$$\textstyle{K\otimes K^{*}\otimes
K\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\ \ \ \
1\otimes<-,->}$$\textstyle{K}$

where second map twists the first two factors, and the third map evaluates the
second and third factor.

Reversing these arrows gives $K^{*}$ the structure of a coalgebra:

$\textstyle{K^{*}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{K^{*}\otimes
K\otimes
K^{*}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{K\otimes
K^{*}\otimes
K^{*}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{K\otimes
K^{*}}$

or with respect to a basis $\\{k_{i}\\}$, its dual basis $\\{k_{i}^{*}\\}$ and
structure constants $\triangle k_{i}=\sum c_{i}^{pq}k_{p}\otimes k_{q}$, so
that $k^{*}_{m}k^{*}_{n}=\sum_{t}c_{t}^{mn}k*_{t}$

$\textstyle{k_{i}^{*}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\sum_{m}k_{i}^{*}\otimes
k_{m}\otimes
k^{m}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\sum_{m}k_{m}\otimes
k^{*}_{i}\otimes
k^{*}_{m}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\sum_{m}c_{t}^{im}k_{m}\otimes
k^{*}_{t}}$

Now apply this to $K=\Pi(d)$, so that $\Pi(d)^{*}$ is a comodule over
$\Pi(d)$. Then

$\Pi(d)^{*}\to\operatorname{Hom}(d^{*}(N),N),\ \ \beta\mapsto<\beta,->\otimes
1_{N}:\Pi(d)*(N)=\Pi(d)\otimes N\to N$

is a measuring map. Thus $\epsilon\otimes 1_{N}\in\Pi(d)^{*}$ represents an
element in $Q(\Pi^{*}(d),N)$ and $\Pi(\epsilon\otimes 1_{N})=\Pi(d)$.

The observation that the composition $e\circ j$ is exactly $d$ completes the
proof. ∎

###  4.5 The Sweedler product for modules and comodules

As $A\triangleleft H$ fills the role of tensor enabling a tensor - hom
adjunction for the enriched category of algebras, so there is a similar
construction $M\triangleleft X$ enabling a corresponding tensor - hom
adjunction for the enriched category of modules. More precisely, the aim of
this section is to define an $A\triangleleft H$ module $M\triangleleft_{H}X$
such that there is a natural isomorphism

$Q(M\triangleleft X,N)\cong Q(M,[X,N])$

At first glance, it is not even immediately apparent that the two sides are
comodules with respect to the same coalgebra; the left hand side being a
comodule over $P(A\triangleleft H,B)$ and the right hand side being a comodule
over $P(A,[H,B])$. However, the natural isomorphism between these two is the
first part of theorem 2.12, and the arguments to establish this result will
follow along the same lines as for the corresponding result for algebras.

####  4.5.1 Definition

Let $H\to\operatorname{Hom}_{k}(A,B)$ measure. For an $H$-comodule $X$ and a
$A$-module $M$ define $M\triangleleft X$ to be the coequaliser of the
following diagram:

$\textstyle{A\otimes M\otimes
X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\otimes
M\otimes H\otimes
X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\otimes
H\otimes M\otimes
X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M\otimes
X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\triangleleft
H\otimes M\otimes
X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M\triangleleft
X.}$

That is, set

$M\triangleleft X=(A\triangleleft H\otimes M\otimes X)/R$

where R is the minimal $A\triangleleft H$ submodule such that the diagram
commutes.

This construction has the same properties for modules as 2.7 for algebras.

######  Theorem 4.13.

$\triangleleft:_{A}Mod\times_{P(A,B)}Comod\to_{B}Mod$ $[-,-]:\
_{P(A,B)}Comod^{op}\times_{B}Mod\to_{A}Mod$

providing a tensor - hom adjunction

$Mod(M,[-,N])\cong Mod(M\triangleleft-,N)$

where $M,\ N$ are modules over $A,\ B$ respectively.

###### Proof.

In essence, as with theorem 2.7, this is true because the statements are true
for the underlying vector spaces. The steps required for verification are the
same, the work required again reduces to brute computation.

∎

###  4.6 Measuring comodules for comodules

The concept of measuring comodules extends to measuring comodules for
comodules, in parallel with measuring coalgebras for coalgebras. Again, the
reason for introducing this concept is to complete a chain of three functorial
equivalences parallel to theorem 2.12. Let $X,\ Y$ be comodules over $H,\ K$
respectively, and let $Z$ be a $P(H,K)$ comodule. A map
$\theta:Z\to\operatorname{Hom}_{k}(X,Y)$ measures if

$\triangle<\theta(z),x>=\sum<\theta(z_{(1)}),x_{(1)}>\otimes<\theta(z_{(0)}),x_{(0)}>.$

Observe that the evaluation map

$P(H,K)\otimes H\to K$

is a coalgebra map. As with the modules, this coalgebra homomorphism provides
comodules over $P(H,K)\otimes H$ with a $K$ comodule structure. This in turn
provides an alternative description of measuring comodules.

######  Proposition 4.14.

With $X,\ Y,\ Z$ comodules over $H,\ K,\ P(H,K)$ respectively, a map
$\theta:Z\to\operatorname{Hom}_{k}(X,Y)$ is a measuring map if and only if the
corresponding map

$\theta:Z\otimes H\to K$

is a map of comodules.

.

As for measuring coalgebras, the strong finiteness property holds equally for
comodules and ensures that colimits and coequalisers exist in the category of
comodules (over a given coalgebra). As for coalgebras, there is a terminal
element for measuring comodules for comodules. The procedure for proving

######  Proposition 4.15.

For comodules $X,\ Y$ over a coalgebras $H,\ K$, there is an $P(H,K)$ comodule
$Q(X,Y)$ and a measuring map $\ Q(X,Y)\to\operatorname{Hom}_{k}(X,Y)$ such
that if $Z$ is another $P(H,K)$ comodule and
$\zeta:Z\to\operatorname{Hom}_{k}(X,Y)$ measures, there is a unique map
$Q(\zeta):Z\to Q(X,Y)$ such that the following diagram commutes.

$\textstyle{Q(X,Y)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Hom}_{k}(X,Y)}$$\textstyle{Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{Q(\zeta)}$$\scriptstyle{\zeta}$

This establishes a natural equivalence

$Comod(-,Q(X,Y))\cong Comod(-\otimes X,Y)$

###### Proof.

The proof of the existence of $Q(M,N)$ follows the same arguments as the
corresponding proposition 2.3 for algebras. The natural equivalence can be
shown applying the Yoneda and using proposition 4.15. ∎

###  4.7 Enriched categories of modules and comodules

As for the case of algebras, the existence of the universal measuring
comodules enables a categories of modules enriched over comodules and of
comodules enriched over comodules. The principle results about the universal
measuring comodule are collected in the following theorem.

######  Theorem 4.16.

  1. 1.

There is a category $\underline{Mod}$ enriched over comodules whose objects
are modules and whose morphisms $\underline{Mod}(_{A}M,_{B}N)$ is the
measuring comodule $Q(M,N)$

  2. 2.

There is a category $\underline{Comod}$ enriched over comodules whose objects
are comodules and whose morphisms $\underline{Comod}(_{H}X,_{K}Y)$ is the
universal measuring comodule $Q(X,Y)$

  3. 3.

Let $M,\ N$ be modules over $A,\ B$ respectively. There are natural
equivalences

$\underline{Comod}(-,Q(M,N))\cong\underline{Mod}(M,[-,N])\cong\underline{Mod}(M\triangleleft-,N)$

###### Proof.

1\. Composition of morphisms is a consequence of the universal property of
measuring comodules. For modules $M_{i}$ over $A_{i}$, $i\in\\{1,\ 2,\ 3\\}$

$\textstyle{Q(M_{2},M_{3})\otimes
Q(M_{1},M_{2})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Hom}(M_{2},M_{3})\otimes\operatorname{Hom}(M_{1},M_{2})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Hom}(M_{1},M_{3})}$

is a measuring map.

2\. This is identical to 1. in its logic.

3\. The great simplification of working within the categorical framework is
that the parallels with the enriched categories of algebras and coalgebras are
exact; the proof of this statement is formally identical to its parallel
theorem 2.12. However, the added subtlety is that care must be taken to
identify which algebras or coalgebras over which the modules or comodules are
acting or coacting.

Let $X$ be a comodule over a coalgebra $H$. By Yoneda’s lemma, it is
sufficient to show that for a comodule $Y$ over $P(A,[H,B])\cong
P(A\triangleleft H,B)\cong P(H,P(A,B))$, there are natural isomorphisms of
sets

$Comod(Y,Q(M\triangleleft X,N))\cong Comod(Y,Q(M,[X,N]))$
$Comod(Y,Q(M,[X,N]))\cong Comod(Y,Q(X,Q(M,N)))$

The challenging aspect of the proof is to keep track of the underlying algebra
or coalgebra homomorpshisms.

For the first, the sequence of equivalences is the following.

$\displaystyle Comod(Y,Q(M\triangleleft X,N))$ $\displaystyle\cong^{a}$
$\displaystyle Mod(M\triangleleft X,[Y,N])$ $\displaystyle\cong^{b}$
$\displaystyle Mod(M,[X,[Y,N]])$ $\displaystyle\cong^{c}$ $\displaystyle
Mod(M,[X\otimes Y,N])$ $\displaystyle\cong^{d}$ $\displaystyle Mod(M,[Y\otimes
X,N])$ $\displaystyle\cong^{e}$ $\displaystyle Mod(M,[Y,[X,N]])$
$\displaystyle\cong^{f}$ $\displaystyle Comod(Y,Q(M,[X,N])).$

The equivalence $a$ is the defining property of the universal measuring
comodule, and proposition 4.2. Equivalence $b$ is theorem 4.13, equivalences
$c,\ d,\ e$ are all familiar rearrangements. Equivalence $f$ is again theorem
4.13.

For the second the sequence of equivalences is the following.

$\displaystyle Comod(Y,Q(X,Q(M,N))$ $\displaystyle\cong^{g}$ $\displaystyle
Comod(Y\otimes X,Q(M,N))$ $\displaystyle\cong^{h}$ $\displaystyle
Mod(M,[Y\otimes X,N])$ $\displaystyle\cong^{k}$ $\displaystyle
Mod(M,[Y,[X,N]])$ $\displaystyle\cong^{l}$ $\displaystyle Comod(Y,Q(M,[X,N]).$

Equivalence $g$ is the fundamental property of the universal measuring
comodule, proposition 4.15. Equivalence $h$ is propositions 4.9. Equivalence
$k$ is a standard rearrangement, and $l$ is again proposition 4.9. ∎

The enriched categories $\underline{Mod}$ and $\underline{Comod}$ provide
global category of modules: _no reference need be made to the algebra or
coalgebra acting or coacting_. Either this enriched category structure, or the
bicategory structure of $\widecheck{Mod}$ seems to be required to recover the
more sensitive comparisons between modules.

$Q(M,N)$ is a distinguished element (the terminal element) of
$\widecheck{Mod}(M,N)$. There is no such distinguished element of
$\widehat{Mod}(M,N)$; the following section describes the closest thing to it,
the module analogue for the Sweedler product.

###  4.8 $D(M,N)=(M\triangleleft N^{\circ})$ and its properties

As with algebras and the universal measuring coalgebras, the Sweedler product
enables a pre-dual concept, which in turn provides methods to understand and
compute the universal measuring comodule. We will need the analogue of the
dual coalgebra $-^{\circ}$ for modules.

######  Proposition 4.17.

Let $N$ be an module over an algebra $B$. Define the _dual comodule_
$N^{\circ}$ to the the subspace of the full dual $N^{*}$.

$N^{\circ}=\\{\nu\in N^{*}:\emph{ker}\nu\supset W,\ W\emph{ is a submodule},\
dimN/W<\infty\\}$

Then $N^{\circ}$ is a $B^{\circ}$ comodule.

###### Proof.

Suppose $W$ is a submodule of $B$ such that $N/W$ is a finite dimensional $B$
module. The action $B\otimes N\to N$ then factors through $(B/Ann(N/W))\otimes
N/W\to(N/W)$. Thus any element $\nu$ in $n^{\circ}$ is an element of
$W^{\perp}$ which is an $Ann(N/W)^{\perp}$ comodule. But since
$Ann(N/W)^{\perp}\subset B^{\circ}$ $W^{\perp}$ is equally a $B^{\circ}$
module. ∎

######  Definition 4.18.

The _universal measuring module_ $D(M,N)$ is defined to be the Sweedler
product $M\triangleleft N^{\circ}$.

The categorical structure behind $D(M,N)$ is identical to that behind
$F(A,B)$, enabling us to lift properties and proofs from algebras to modules.

######  Proposition 4.19.

Properties of D(M,N).

  1. 1.

Let $B$ be a finite dimensional algebra. For any finite dimensional $B$ module
$N$ and (arbitrary) $A$ module $M$, there is a module map $\tau(M,N)$ relative
to $\eta:A\to F(A,B)\otimes B$

$\tau(M,N):M\to D(M,N)\otimes N$

such that for any extension $\rho:M\to W\otimes N$ relative to an extension
$\sigma:A\to S\otimes B$ there is a module map $D(\rho):D(M,N)\to W$ relative
to $F(\sigma)$ such that the diagram

$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho}$$\scriptstyle{\tau(M,N)}$$\textstyle{D(M,N)\otimes
N\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{D(\rho)\otimes
1_{B}}$$\textstyle{W\otimes N}$

commutes. Moreover, $\tau(\\_,N)$ is a natural transformation.

  2. 2.

For arbitrary $B$ and arbitrary $N$ the module map $D(\rho):D(M,N)\to W$ still
exists. The assignment

$D:\widehat{Mod(M,N)}\to Mod(D(M,N),-),\ \ D(W,\rho):=D(\rho):D(M,N)\to S$

is a functor.

  3. 3.

If $N$ is finite dimensional $D(M,N)$ is an initial object in the category
$\widehat{Mod}(M,N)$

  4. 4.

The evaluation map $M\to[D(M,N),N]$ is a module map relative to the algebra
homomorphism $A\to[P(A,B),B]$.

  5. 5.

$D(M,N)^{\circ}\cong Q(M,N).$

The great advantage of the categorical approach is that the proofs of these
statements are essentially the same as the corresponding statements for
$F(A,B)$ in Theorem 2.15. The module $D(M,N)$ is difficult to conceptualise,
but we have some familiar examples.

######  Corollary 4.20.

  1. 1.

Considering $A$ as a left $A$ module, $D(A,A)=F(A,A)\otimes A^{\circ}$ as a
left $F(A,A)$ module.

  2. 2.

For $A=k$ and $M,\ N$ finite dimensional vector vector spaces,
$D(M,N)=M\otimes N^{*}$.

  3. 3.

For $A$ modules $M_{1},\ M_{2}$ and $N$ a $B$ module, $D(M_{1}\oplus
M_{2},N)=D(M_{1},N)\oplus D(M_{2},N)$.

  4. 4.

For an $A$ module $M$ and $B$ modules $N_{1},\ N_{2}$, $D(M,N_{1}\oplus
N_{2})=D(M,N_{1})\oplus D(M,N_{2})$

###### Proof.

  1. 1.

Observe that $a\otimes\alpha=\sum a\otimes\alpha_{(1)}\otimes
1_{A}\otimes\alpha_{(0)}$ for $a$ in $A$, $\alpha$ in $A^{\circ}$. Identifying
$A^{\circ}$ with $1_{F(A,A)}\otimes(1_{A}\otimes A^{\circ})$, $D(A,A)$ is thus
seen to be generated by $A^{\circ}$ as an $F(A,A)$ module. To see that
$D(A,A)=F(A,A)\otimes A^{\circ}$, observe that he ideal $R$ defining the
module is generated by the set $U$ of elements of the form
$aa^{\prime}\otimes\alpha-\sum(a\otimes\alpha_{(1)})\otimes(a^{\prime}\otimes\alpha_{(0)})$.
But

$\displaystyle\sum_{(\alpha)}(aa^{\prime}\otimes\alpha_{(1)})\otimes(1_{A}\otimes\alpha_{(0)})=$
$\displaystyle aa^{\prime}\otimes\alpha$ (32) $\displaystyle=$
$\displaystyle\sum_{(\alpha)}(a\otimes\alpha_{(2)})\otimes(a^{\prime}\otimes\alpha_{(1)})\otimes(1_{A}\otimes\alpha_{(0)})$
(33)

But then, $\triangle^{2}=(\triangle\otimes 1)\circ\triangle$, for any
expression $\triangle\alpha=\sum_{\alpha}\alpha_{(1)}\otimes\alpha_{(0)}$, and
$U$ can be replaced by the set $U^{\prime}$ of elements of the form

$\sum_{(\alpha)}(aa^{\prime}\otimes\alpha_{(1)})\otimes(1_{A}\otimes\alpha_{(0)})-\sum_{(\alpha)}(a\otimes\alpha_{(2)})\otimes(a^{\prime}\otimes\alpha_{(1)})\otimes(1_{A}\otimes\alpha_{(0)})$

that is, elements

$\sum_{(\alpha)}(aa^{\prime}\otimes\alpha_{(1)})-\sum_{(\alpha)}(a\otimes\alpha_{(2)})\otimes(a^{\prime}\otimes\alpha_{(1)})\otimes(1_{A}\otimes\alpha_{(0)})$

but each term

$\sum_{(\alpha_{(0)})}(aa^{\prime}\otimes\alpha_{(1)})-\sum_{(\alpha_{(0)})}(a\otimes\alpha_{(2)})\otimes(a^{\prime}\otimes\alpha_{(1)}))$

is 0 in $F(A,A)$.

  2. 2.

This is immediate since $F(k,k)=k$.

  3. 3.

$F(A,B)\otimes((M_{1}\oplus M_{2})\otimes
N^{\circ})=F(A,B)\otimes(M_{1}\otimes N^{\circ})\oplus
F(A,B)\otimes(M_{2}\otimes N^{\circ}).$ The subspace $U$ generating the
submodule $R$ defining $D(M_{1}\oplus M_{2},N)$ similarly splits as a direct
sum $U_{1}\otimes U_{2}\subset F(A,B)\otimes(M_{1}\otimes N^{\circ})\oplus
F(A,B)\otimes(M_{2}\otimes N^{\circ})$

  4. 4.

This is similar to part 3.

∎

###  4.9 Coda: Internal symmetries?

It is our hope that the material in this paper will excite the imagination of
mathematicians working in fields beyond algebra and category theory. One
potential application that has motivated the research has been the possibility
of finding a context for internal symmetries in physics: not a conjecture of
what the "right" group should be (although the model would suggest various
groups) but a reason why internal symmetries should be expected and a
framework that would govern their operation.

The first observation is that for physical systems with a symmetry group $G$
acting, the representations of interest in quantum physics are complex
representations in which $G$ is represented as unitary matrices.

Let $A$ be the real algebra by two elements, $i$, complex multiplication, and
$J$, complex conjugation.

The theory of section 2 provides a canonical universal extension:

$\eta:A\to F(A,A)\otimes A$

The theory of section 4 provides, for _any_ representation $W$ of $A$ a
canonical universal extension

$\tau:W\to D(W,W)\otimes W$

where $D(W,W)$ is an $F(A,A)$ module, so that $D(W,W)\otimes W$ is an $A$
module via the homomorphism $\eta$.

The theory thus provides a universal symmetry algebra, $F(A,A)$; its
homomorphic images contain groups which could play the role of internal
symmetry groups. If $W$ is a (complex) module for the group $G$,
$D(W,W)\otimes W$ is a module for $F(A,A)\otimes\mathbb{C}G$. If $Z$ is any
module on which $F(A,A)$ acts, $F(A,A)\otimes\mathbb{C}G$ acts on $Z\otimes
W$. If $H$ is then a group generated by a set of invertible elements in the
image of $F(A,A)\in\operatorname{End}(Z)$, then $H\times G$ acts on $Z\otimes
W$. In this way, groups $H$ would appear as inevitable extra symmetries given
any complex representation of $G$.

Furthermore, $F(A,A)$ can be calculated explicitly, and its homomorphic images
contain algebras and groups which have been used to model internal symmetries.
The procedure used to calculate $F(\mathbb{C},\mathbb{C})$ in section 3.1
provides generators and relations for $F(A,A)$.

The canonical map $\eta:A\to F(A,A)\otimes A$ is a homomorphism thus writing
$\eta(x)(resp\eta(J))$

$\displaystyle\eta(x)=$ $\displaystyle f_{1}1+f_{x}X+f_{J}J+f_{xJ}xJ$ (34)
$\displaystyle\eta(J)=$ $\displaystyle g_{1}1+g_{x}X+g_{J}J+g_{xJ}xJ$ (35)

we need

$\eta(x)^{2}=-1,\ \ \ \eta(J)^{2}=1,\ \ \ \\{\eta(x),\eta(J)\\}=0.$

Direct computation of $\eta(x)^{2}$ gives

$\displaystyle-1=$ $\displaystyle f_{1}^{2}-f_{x}^{2}+f_{J}^{2}+f_{xJ}^{2}$
(36) $\displaystyle 0=$
$\displaystyle\\{f_{1},f_{x}\\}+[f_{xJ},f_{J}]=\\{f_{1},f_{J}\\}+[f_{xJ},f_{x}]=\\{f_{1},f_{xJ\\}}\\}+[f_{x},f_{J}]$
(37)

and similarly

$\displaystyle 1=$ $\displaystyle g_{1}^{2}-g_{x}^{2}+g_{J}^{2}+g_{xJ}^{2}$
(38) $\displaystyle 0=$
$\displaystyle\\{g_{1},g_{x}\\}+[g_{xJ},g_{J}]=\\{g_{1},g_{J}\\}+[g_{xJ},g_{x}]=\\{g_{1},g_{xJ\\}}\\}+[g_{x},g_{J}].$
(39)

The identity $\\{\eta(x),\eta(J)\\}=0$ gives

$\displaystyle 0=$
$\displaystyle\\{f_{1},g_{1}\\}-\\{f_{x},g_{x}\\}+\\{f_{j},g_{J}\\}+\\{f_{xJ},g_{xJ}\\}$
(40) $\displaystyle=$
$\displaystyle\\{f_{1},g_{x}\\}+\\{g_{1},f_{x}\\}+[f_{xJ},g_{J}]+[g_{xJ},f_{J}]$
(41) $\displaystyle=$
$\displaystyle\\{f_{1},g_{J}\\}+\\{g_{1},f_{J}\\}+[f_{xJ},g_{x}]+[g_{xJ},f_{x}]$
(42) $\displaystyle=$
$\displaystyle\\{f_{1},g_{xJ}\\}+\\{g_{1},f_{xJ}\\}+[f_{x},g_{J}]+[g_{x},f_{J}].$
(43)

Finite dimensional images of this algebra and $F(A,A)\otimes A$ contain many
familiar algebras. For example, setting

$f_{1}^{2}=f_{x}^{2}=f_{J}^{2}=f_{xJ}^{2}=1/2,\ \ g_{1}=g_{x}=g_{xJ}=0,\ \
g_{j}=1$

and

$\\{f_{1},f_{x}\\}=\\{f_{1},f_{j}\\}=\\{f_{1},f_{xJ}\\}=[f_{x},f_{J}]=[f_{J},f_{xJ}]=[f_{xJ},f_{x}]=0$

contains a Clifford algebra with an inner product $\\{1,-1,1,1\\}$ generated
by $\\{f_{1},f_{x}x,f_{J}J,f_{xJ}xJ\\}$. Additionally requiring
$f_{J}=f_{xJ}=0$ defines an algebra isomorphic to $F(\mathbb{C},\mathbb{C})$
whose complexification $F(\mathbb{C},\mathbb{C})\otimes\mathbb{C}$ has
$M_{2}(\mathbb{C})$ as simple homomorphic images.

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