Monday, January 21, 2013

Bosonization Part III, Co-Bosonization

A few days ago I sketched out some of the basic category theoretic ideas behind the theory of bosonization of quantum groups. Today I want to talk about the dual version, co-bosonization.

First note that this whole theory is trivial, as dualizing such a theory is a straightforward and obvious process, but I'm working through the proofs of bosonization in a dualized setting by hand, just so I can learn Quantum Groups.

So let $(H,\mathcal{R})$ be a dually quasitriangular Hopf algebra (I called this a co-braided Hopf algebra in an earlier post on braided categories) and let $\mathcal{C} = ^H\mathcal{M}$ by the braided category of $H$-comodules. If $(V, \beta_V: v \mapsto \sum h^{(1,v)}\otimes v^{(1)})$ and $(W, \beta_W: w \mapsto \sum h^{(1,w)}\otimes w^{(1)})$ are objects in the category, then the braiding is given by

$\displaystyle \Psi_{V,W}(v\otimes w) = \sum \mathcal{R}\left(h^{(1,v)} \otimes h^{(1,w)}\right) w^{(1)} \otimes v^{(1)}$

I won't give the proof here (I may not give it at all!). As before, a Hopf algebra $B$ in the category $^H\mathcal{M}$ means that $B$, $B\otimes B$ and all the structure maps are morphisms in the category, thus any twist involved is actually an invocation of the braiding $\Psi$.

Also as before, $^B \mathcal{C}$ is the category of $B$-comodules within the category $\mathcal{C}$, and the idea behind co-bosonization is to find a Hopf algebra $\text{cobos}(B)$ such that $^B\mathcal{C} \cong \, ^{\text{cobos}(B)}\mathcal{M}$.

We're going to propose what this new Hopf algebra $\text{cobos}(B)$ should be. I won't give any proofs in this post (because I haven't done them yet), but hopefully I've dualized things correctly and I'm not stating anything that's not true. We'll start with the tensor product $B \otimes H$ and we'll put an coalgebra structure on it by first noting that in the bosonization case, the coalgebra $\Delta B \otimes H \rightarrow (B\otimes H) \otimes (B \otimes H)$ is

$\displaystyle \left(\text{Id}_B \otimes \Psi_{B,H} \otimes \text{Id}_H\right) \circ \left( \Delta_B \otimes \Delta_H \right)$

This coproduct almost makes sense here, we just need to treat $H$ as a co-module of itself (so the braiding makes sense). But this is no difficult matter, because every Hopf algebra coacts on itself as a coalgebra by

$\displaystyle \beta_H (h) = \sum h_{(1)} S(h_{(3)}) \otimes h_{(2)}$

So we just use the dual-braiding $\Psi$ instead of the original one, and write

$\displaystyle \Delta (b \otimes h) = \sum \mathcal{R}\left( h^{(1,b_{(2)})} \otimes h^{(1,h_{(2)})} \right) b_{(1)} \otimes h_{(1)}^{(1)} \otimes b_{(2)}^{(1)} \otimes h_{(2)}$

Now on to the product for $\text{cobos}(B) = B \otimes H$: in the bosonization case (were $H$ acts, rather than coacts, on $B$), the product is given by

$\displaystyle (b \otimes h) (c\otimes g) = \sum b(h_{(1)} \triangleright c) \otimes h_{(2)}g$

This doesn't immediately make sense in our case, but we have the following proposition:

Proposition. There exists a monoidal functor (functor that respect the tensor product of objects) $^H \mathcal{M} \rightarrow ^H_H\mathcal{M}$, the crossed modules we mentioned for the quantum double, by $(V, \beta_V) \mapsto (V,\beta_V,\triangleright)$, where $h \triangleright v = \sum \mathcal{R}\left( h \otimes h^{(1,v)} \right)v^{(1)}$

Hence we do have an action of $H$ on $B$, and we can write the product

$\displaystyle (b \otimes h) (c \otimes g) = \sum \mathcal{R}\left( h_{(1)} \otimes h^{(1,c)}\right) bc^{(1)} \otimes h_{(2)}g$

The unit and counit are obvious in this case ($1 \otimes 1$ and $\epsilon(b \otimes h) = \epsilon(b)\epsilon(h)$), but we still need an antipode. Again, we refer to bosonization and try to find the correct dual notion. In bosonization, the antipode $S: B \otimes H \rightarrow B \otimes H$ is given by:

$\displaystyle B\otimes H \xrightarrow{S_B \otimes S_H} B \otimes H \xrightarrow{\Psi_{B,H}} H \otimes B \xrightarrow{\Delta_H \otimes \text{Id}_B} H \otimes H \otimes B \xrightarrow{\text{Id}_H \otimes \Psi_{H,B}} H \otimes B \otimes H \xrightarrow{\triangleright \otimes \text{Id}_H} B \otimes H$

This makes perfect sense for co-bosonization without modification. To summarize, we need proofs for the claims

1. That $\Psi$ is the braiding for the category $^H \mathcal{M}$. (I have this.)
2. That the product and coproduct as defined above give us a bialgebra $\text{cos}(B)$ for $B$ a Hopf algebra in the category $^H \mathcal{M}$. (Not yet proved.)
3. That the above antipode gives us a Hopf algebra $\text{cos}(B)$. (Not yet proved.)
4. That $^B\mathcal{C} = ^{\text{cos}(B)}\mathcal{M}$. (Not yet proved).

Saturday, January 19, 2013

Category Theoretic Methods and Drinfeld's Quantum Double

Earlier this month I discussed braided Hopf algebras, but I neglected to give an example of one. Today I'll describe how we can find a large collection of braided Hopf algebras via a construction by Drinfeld called the Quantum Double.

Let $H$ be a finite dimensional Hopf algebra with invertible antipode. Then $H^* = \text{Hom}(H, k)$ is also a Hopf algebra: it's a basic result to show that the dual vector space of a coalgebra is an algebra with the product map $\Delta^*$, the transpose of the original coproduct, moreover, in the finite dimensional case, $m^*$ becomes a coproduct, making the dual of a finite dimensional algebra a coalgebra. Since $H$ is finite dimensional by assumption, and the product and coproduct maps are [co]algebra morphisms, we see we have a Hopf algebra structure on $H^*$, too. That the transpose of the antipode is an antipode and that we have the necessary co/units is a more straightfoward result. The finite dimensional requirement is needed because $H^* \otimes H^*$ is not necessarily isomorphic to $(H\otimes H)^*$ without it.

To be explicit, we have the product

$\displaystyle \phi \psi (h) = \sum \phi(h_1) \psi(h_2)$

and the coproduct

$\displaystyle \Delta \phi (h \otimes g) = \phi(hg)$

along with the unit $1(h) = \epsilon(h)$, counit $\epsilon(\phi) = \phi(1)$ and antipode $S(\phi)(h) = \phi(h)(Sh)$. Of course, $H^*$ and $H$ are dually paired by the evaluation map

$\displaystyle \langle \phi, h \rangle = \text{ev}_\phi(h) = \phi(h)$

we also have the reverse dual pairing $\langle , \rangle : H\otimes H^{\text{op}*}$ for $H$ and $H^*$ with the opposite multiplication. From this, we can define the Quantum Double $D(H)$ as the vector space $H^* \otimes H$ with the product:

$\displaystyle (\phi \otimes h) (\psi \otimes g) = \sum \left( \langle Sh_1, \psi_2 \rangle \langle h_3, \psi_3\rangle \right) \psi_2 \phi \otimes h_2 g$

the coproduct:

$\displaystyle \Delta(\phi, h) = \sum (\phi_1 \otimes h_1) \otimes (\phi_2 \otimes h_2)$

the antipode:

$\displaystyle S(\phi\otimes h) = (1\otimes Sh) (S^{-1}\phi \otimes 1)$

and finally with the unit $1\otimes1$ and counit $\epsilon(\phi \otimes h) = \epsilon(\phi)\epsilon(h)$.

We can also describe the quantum double as the bicrossed product of $H$ with $H^{\text{op}*}$, but we aren't focusing on that here. Rather, let me point out that $D(H)$ is a braided Hopf algebra with

$\displaystyle \mathcal{R} = \sum_{k=1}^n (f^k \otimes 1) \otimes (1 \otimes e_k)$

Where $\langle e_1, \ldots e_n \rangle$ is a basis for $H$ and $\langle f^1,\ldots f^n \rangle$ is a dual basis.

We can also define the Quantum Double for an infinite dimensional Hopf algebra so long as we can find another dually paired Hopf algebra $H'$, which means finding the map $\langle, \rangle: H \otimes H' \rightarrow k$. But I want to describe it in a better (in my opinion) way. First, I need to quote some results:

Let $_H^H \mathcal{M}$ denote the category of crossed $H$-modules, that is, vector spaces $V$ that are compatibly both $H$-modules and $H$-comodules in the following sense: $\triangleright: H \otimes V \rightarrow V$ is the module action and $v \mapsto \sum h^{(1)} \otimes v^{(1)}$ is the coaction, then

$\displaystyle \sum h_1 h^{(1)} \otimes h_2 \triangleright v^{(1)} = \sum (h_1 \triangleright v)^{(1,H)} h_2 \otimes (h_1 \triangleright v)^{(1,V)}$

for all $h \in H$ and $v \in V$ (the pair $(1,V)$, et al, is to denote which is the $H$ component and which is the $V$ component of the coaction). The morphisms in this category are linear maps that commute with both the action and the coaction. This category is braided with

$\displaystyle \Psi_{V,W}(v \otimes w) = \sum h^{(1,v)} \triangleright w \otimes v^{(1)}.$

Here the pair $(1,v)$ denotes that this is the coaction on $V$ and not on $W$. Now, we have the following result:

Theorem Let $H$ be a finite dimensional Hopf algebra with invertible antipode. Then $_H^H \mathcal{M} = _{D(H)}\mathcal{M}$, that is, the category of crossed $H$-modules is category of modules of the quantum double of $H$.

I won't (and can't yet) prove this, but you can see an argument for it by noting that $D(H)$ contains both $H$ and $H^{\text{op}*}$ as Hopf sub-algebras; a $D(H)$ module is then an $H$ module and an $H^{\text{op}*}$ left module, but that's the same as a $H^*$ right module, which is the same as a left $H$-comodule.

Now, if $H$ is not finite dimensional, then can we find a Hopf algebra $\mathcal{H}$ such that $_H^H\mathcal{M} = _\mathcal{H}\mathcal{M}$? We again invoke those Tannakian reconstruction theorems and thus have a suitable definition for $D(H) = \mathcal{H}$ for $H$ of any dimension, and without having to specify a dually paired Hopf algebra $H'$.

The obvious question is, what does $D(H)$ look like for an infinite dimensional $H$? Perhaps I'll try answering this later.

Monday, January 14, 2013

Bosonisation of Quantum Groups, part II (Braided Categories)

Let's talk about the category of modules ${_H\mathcal{M}}$ for a braided Hopf algebra ${H}$ for a minute. If ${V}$ and ${W}$ are ${H}$-modules, then we have an easy action of ${H}$ on ${V \otimes W}$ as well, by:

$\displaystyle h \triangleright(v\otimes w) = \sum h_1\triangleright v \otimes h_2\triangleright w$

Hence we can think of ${\otimes}$ as a functor ${_H\mathcal{M} \times _H\mathcal{M} \rightarrow _H\mathcal{M}}$. This sort of functor is what turns ${_H\mathcal{M}}$ into a monoidal category. We also have the opposite function ${\otimes^\text{op}}$, which does what one would expect: ${V \otimes^\text{op} W = W \otimes V}$.

We have another operation on this category, the flip map ${\tau}$, which carries the action of one functor ${\otimes}$ to another ${\otimes^\text{op}}$. We'd like it to be a natural transformation, actually a natural isomorphism (one can see already that it's invertible) from ${\otimes \rightarrow \otimes^\text{op}}$. For this to work we must have, for each ${f: V \rightarrow V'}$ and ${g: W \rightarrow W'}$ in the category, a morphism ${\tau_{V \otimes W}: V \otimes W \rightarrow W \otimes V}$ such that the square

commutes. But before we check that, let's first verify that ${\tau_{V \otimes W}}$ is even a morphism in the category (to take a step back from category-theoretic language, this means that we need to check that ${\tau_{V \otimes W}(v\otimes w) = w \otimes v}$ is an ${H}$-linear map). But it's not necessarily true that ${h \triangleright \tau_{V \otimes W}(v\otimes w) = \tau_{V \otimes W}(\sum h_{(1)}\triangleright v \otimes h_{(2)}\triangleright w)}$. In particular, this is only true when ${H}$ is co-commutative.

But ${H}$ is a braided Hopf algebra, meaning that its failure to be co-commutative is controlled by an invertible element ${\mathcal{R} \in H \otimes H}$. Can we use this to construct an natural transformation ${\otimes \rightarrow \otimes^{op}}$? Yes, we can, and this is exactly the structure that gives us a braided category.

To be brief, we can define a braided category to be a monoidal category together with a natural isomorphism ${\Psi : \otimes \rightarrow \otimes^\text{op}}$ that obeys the following conditions: ${ \Psi_{V \otimes W, Z} = \Psi_{V,Z} \circ \Psi_{W,Z}}$ and ${\Psi_{V,W\otimes Z} = \Psi_{V,Z} \circ \Psi_{V,W}}$. (These are the hexagon conditions''. I'm not writing the structure maps of the monodial category, but when you include them, you express these as two hexagonal commutative diagrams.)

In the case of a braided, or quasitriangular Hopf algebra, our ${\Psi_{V,W} = \tau(\mathcal{R} \triangleright v\otimes w) = \sum R_2 \triangleright w \otimes R_1 \triangleright v}$. The reader can verify that this meets the axioms of a braided category.

There's more we can say about the braiding ${\Psi}$ and it's connection to braid groups, et cetera, but I don't want to go down that route this evening. Rather, I want to discuss the dual version of this theory. First, I need find the dual notion of a braided Hopf algebra. We'll call this a co-braided Hopf algebra. By this I mean that we want to control how far ${H}$ is from being commutative, instead of co-commutative.

A co-braided Hopf algebra is a Hopf algebra ${H}$ together with a linear functional ${R: H\otimes H \rightarrow k}$. We require ${R}$ to be convolution invertible, id est, that there exists another linear functional ${R^{-1}}$ such that ${\sum R^{-1}(h_1 \otimes g_1) R(h_2 \otimes g_2) = \epsilon(h)\epsilon(g) = \sum R(h_1 \otimes g_1) R^{-1}(h_2\otimes g_2)}$. We also require it to obey three other relations:

$\displaystyle \sum g_1h_1 R(h_2\otimes g_2) = \sum R(h_1 \otimes g_1) h_2g_2$

$\displaystyle R (hf \otimes f) = \sum R(h\otimes f_1)R(g\otimes f_2)$

$\displaystyle R(h\otimes gf) = \sum R(h_1 \otimes f)R(h_2 \otimes g)$

for ${g,h,f \in H}$.

Let ${V}$ and ${W}$ be ${H}$-co-modules with structure maps ${\beta_V: V \rightarrow H \otimes V}$ by ${v \mapsto \sum g_1 \otimes v_1}$ and ${\beta_W: W \rightarrow H \otimes W}$ by ${w \mapsto \sum h_1 \otimes w_1}$ The tensor functor works like such:

$\displaystyle \beta_{V\otimes W}(v\otimes w) = \sum g_1 h_1 \otimes v_1 \otimes w_1$

What does the braiding on the category of co-modules ${^H\mathcal{M}}$ look like? I propose that the braiding ${\Psi}$ is then:

$\displaystyle \Psi_{V,W}(v\otimes w) = \left(\sum R(g_1 \otimes h_1) \right) w_1 \otimes v_1$

I've not proved it yet.

Sunday, January 13, 2013

Bosonisation of Quantum Groups, an Introduction

Let $H$ be a quasitriangular (or braided) Hopf algebra and let $\mathcal{C} = _H\mathcal{M}$ be it's category of modules (that is, a category where the objects are $H$-modules and the morphisms are $H$-linear maps between them.) Let $B \in \mathcal{C}^0$ be a Hopf algebra such that it's structure maps $m_B : B\otimes B \rightarrow B$, $\Delta_B : B \rightarrow B \otimes B$, et cetera, and all their respective objects, are all morphisms and objects in the category.

Naturally, $B$ will have it's own modules. But what if we restrict the $B$-modules we want to look at? In fact, let's only look at $B$-modules in $\mathcal{C}$, we'll call these the braided $B$-modules (the reason for this language will become clear later on). More precisely, let $_B\mathcal{C}$ be the sub-category of $B$-modules in $\mathcal{C}$ such that the action of $B$ on the module $V$, $B \otimes V \rightarrow V$, is a morphism in the category.

Does there exist a Hopf algebra $\mathcal{H}$ such that it's category of modules is exactly $_B\mathcal{C}$ ? The answer is yes. From a general abstract nonsense point of view, the affirmation comes from the Tannakian reconstruction theorems which allows us to construct certain algebraic objects from its category of modules. This is a sort of duality between the object and it's category of modules. I don't know enough about this sort of thing to comment any more.

But for us, we'll answer this question by means of Majid's theory of bosonisation of quantum groups. We won't use higher category theory, rather, we will construct the Hopf algebra $\mathcal{H}$ explicitly from $H$ and $B$, and call this construction the bosonisation of $B$. To do this, I need to study a few things:

1. Braided Categories. When $H$ is quasitriangular, it's module category has an additional structure, a braiding, that will color all calculations in $_H\mathcal{M}$ and $_B\mathcal{C}$. I barely know what a braided category is at the moment, so I'll have to cover this in a blog post or two.
2. Hopf algebras in braided categories. (aka braided groups'' in Majid's lexicon) A more concise way of describing my choice of $B$ is to say that $B$ is a braided group in $_M\mathcal{H}$. All the structure maps of $B$ will have to contend with the braiding from $H$, hence why we called $_B\mathcal{C}$ the category of braided modules. I'll study these objects in a subsequent blog post, too.
3. Constructing the bosonisation $\mathcal{H} = B \rtimes H$ as a semidirect product of $B$ and $H$, with its structure maps built from the various morphisms in $\mathcal{C}$

The next step is to repeat this construction for a dual quasitriangular Hopf algebra to arrive at a theory of co-bosonisation, and then to repeat the process for double bosonisation.

Saturday, January 5, 2013

What I learned today: Braided (or quasi-triangular) Hopf Algebras

This is a post in my psuedo-daily'' series What I learned Today''

A Hopf algebra $H$ is endowed with both a multiplication map $m: H \otimes H \rightarrow H$ and a comultiplication map $\Delta: H \rightarrow H \otimes H$. Many of our quantum'' objects lack commutivity or co-commutitivity, but not completely. In cases of co-commutitivity, the entities that lack it often have the lack of co-commutivity controlled by an invertible element $\mathcal{R} \in H \otimes H$ in the following manner:

$\displaystyle \tau \circ \Delta h = \mathcal{R}(\Delta h)\mathcal{R}^{-1}$

Where $\tau(x\otimes y) = y \otimes x$ is the flip operator. If we also put some restrictions on what element $\mathcal{R}$ we can chose, we'll arrive at Drinfeld's theory of Quasi-tranigular, or Braided, Hopf Algebras. But first, let's look at an example: Let $G$ be the Klein four-group, id est, $G = \langle x, y : x^2 = y^2 = (xy)^2 = 1 \rangle$ and let $kG$ be the group Hopf algebra. For $q \in k$ define

$\displaystyle \mathcal{R}_q = \frac{1}{2} (1\otimes 1 + 1\otimes x + x\otimes 1 - x\otimes x) + \frac{q}{2}(y \otimes y + \otimes xy +xy \otimes xy - xy \otimes y)$

Let's change the coproduct on $kG$ to be $\Delta x = x\otimes x$ and $\Delta y = 1 \otimes y + y\otimes x$. Also change the antipode to $Sx = x$ and $Sy = xy$ and the counit $\epsilon (x) = 1$ and $\epsilon (y) = 0$. This new Hopf algebra indeed obeys the $\tau \circ \Delta h = \mathcal{R}_q (\Delta h) \mathcal{R}_q^{-1}$. It also obeys the additional relations

$\displaystyle (\Delta \otimes I) \mathcal{R} = \mathcal{R}_{13}\mathcal{R}_{23}$

$\displaystyle (I \otimes \Delta) \mathcal{R} = \mathcal{R}_{13}\mathcal{R}_{12}$

Where $\mathcal{R}_{23} = 1 \otimes \mathcal{R}$, for instance. These three conditions are what defines a Braided Hopf algebra (we call them braided because their category of modules generates a braided category. More on that in a later post.) These Hopf algebras generate solutions to the \href{http://en.wikipedia.org/wiki/Yang

$\displaystyle (c \otimes I)(I \otimes c)(c \otimes I) = (I \otimes c)(c \otimes I)(I \otimes c)$

for an automorphism $c$ of $V \otimes V$, where $V$ is an $H$-module. To see this, let $V$ and $W$ be $H$ modules and define an isomorphism of $V \otimes W$ and $W \otimes V$ by

$\displaystyle c_{V,W} (v\otimes w) = \tau_{V,W} (\mathcal{R} (v\otimes w))$

From here, one can show that for $H$-modules $U$, $V$, and $W$ we have

$\displaystyle c_{U\otimes V, W} = (c_{U,W} \otimes I_V)(I_U \otimes c_{V,W})$

$\displaystyle c_{U, V\otimes W} = (I_V \otimes c_{U,W})(c_{U,V} \otimes I_W)$

and finally

$\displaystyle (c_{V,W} \otimes I_U) (I_V \otimes c_{U,W})(c_{U,V} \otimes I_W) = (I_W \otimes c_{U,V})(c_{U,W} \otimes I_V)(I_U \otimes c_{V,W})$

Letting $V = W = U$, we see we have a solution of the Yang-Baxter equation. It's also worth noting that the element $\mathcal{R}$ itself satisfies a version of the Yang-Baxter equation: $\mathcal{R}_{12}\mathcal{R}_{13}\mathcal{R}_{23} = \mathcal{R}_{23}\mathcal{R}_{13}\mathcal{R}_{12}$.

Friday, January 4, 2013

Galois Quantum Groups of Finite Fields

In Chater 4 of Majid's A Quantum Groups Primer, he introduces an object he calls the automorphism quantum group'' $M(A)$ associated to an algebra $A$. At the end of the chapter he makes a comment the role of such objects in number theory, however, is totally unexplored at the moment'', which was my queue to start searching for references. After I couldn't find any, I thought I'd take a stab at computing this object for some simple number-theory flavored objects. I haven't got anywhere yet, but let's take a look at what these objects are:

Let $A$ be an algebra over a field $k$. We call $B$ a comeasuring of $A$ if it comes with an algebra map $\beta: A \rightarrow A \otimes B$, this is very similar to a coaction, except that we forget about a coalgebra structure. A morphism of comeasurings $B_1$ and $B_2$ is an algebra morphism $\phi: B_1 \rightarrow B_2$ such that $\beta_2 = (I \otimes \phi) \circ \beta_1$.

Hence we have a category where the objects are comeasurings of $A$ and the arrows are comeasuring morphisms. When $A$ is finite dimensional, we can prove that this category has an initial object, that is, an object $M(A)$ such that for every other object $B$ in the category, there exists one and only one unique morphism $M(A) \rightarrow B$.

The prove of the existence of $M(A)$ for finite dimensional $A$ is constructive, we won't go through the details here (see Majid's book), but we will state and use the result:

Let $e_1,\ldots,e_n$ be a basis for $A$ and let $c_{ij}^k$ be the structure constants of $A$, id est, defined by

$\displaystyle e_i e_j = \sum_{k=1}^n c_{ij}^k e_k$

and define $M'(A)$ as the free associative algebra $k\langle t^1_1, \ldots t^1_n, \ldots t^n_1, \ldots t^n_n \rangle$ modulo the relations

$\displaystyle \sum_{r=1}^n c_{ij}^r t^k_r = \sum_{r,s=1}^n c_{rs}^k t^r_i t^s_j$

for $i,j,k=1$ to $n$ The initial object $M(A)$ is $M'(A)$ modulo the additional relations $t^1_1 =1$, $t^i_1 =0$ for $1 < i \leq n$. The initial map $\beta_{M(A)}: A \rightarrow A \otimes M(A)$ is:

$\displaystyle \beta_{M(A)}(e_i) = \sum_{r=1}^n e_r \otimes t^r_i$

The reader can verify that this $\beta_{M(A)}$ is an algebra map and thus verify that this is indeed a comeasuring. To show that it's initial, we need to show that for any other comeasuring $B$ with map $\beta: A \rightarrow A \otimes B$, there exists a unique algebra morphism $\pi: M(A) \rightarrow B$. Let $b^i_j = \sum_r e_r \otimes t^r_i$ and observe that $\sum_r c_{ij}^a b^k_r = \sum_{r,s} c_{rs}^k b^r_i b^s_j$, so we can let $\pi(t^i_j) = b^i_j$ and extend this as multiplicative map. Uniqueness follows from having finite dimension and the by noting that another morphism $\pi'$ must obey the same relations above on the same image of $\beta$.

The algebra $M(A)$ is what Majid calls the automorphism quantum group. It is a bialgebra. I won't prove it, but you can see that it has a coproduct, et cetera by noting that $M(A) \otimes M(A)$ is a comeasuring with the map $(\beta_{M(A)} \otimes I) \circ \beta_{M(A)}$, hence we have a unique morphism $\Delta: M(A) \rightarrow M(A) \otimes M(A)$ by the initial object condition. The counit pops out in a similar way. The coproduct on the generators is

$\displaystyle \Delta t^i_j = \sum_{r=1}^n t^i_r \otimes t^r_j$

As an example of this nonsense, let $m(x) \in \mathbb{F}_p[x]$ be an irreducible polynomial of degree $n$ over the field with $p$ elements, and let $F_m$ by the corresponding field extension (which is an algebra over $\mathbb{F}_p$). For each other irreducible polynomial $g$ of degree $n$, we have an isomorphic field $F_g$. Each $F_g$ is also an object in the category of comeasurings of $F_m$, as the isomorphism $F_m \rightarrow F_g$ will give rise to an algebra map $F_m \rightarrow F_m \otimes F_g$. $M(F_m)$ is a $n^2 -n$ dimensional algebra in this category that has a unique isomorphism to each $F_g$, hence $M(F_m)$ and its maps to the $F_g$'s would capture the arithmetic of each other field extension of the same degree. I hope that I can use this object in my program for the quantum de rham cohomology over finite fields. One obvious task to describe how the various $M(F_g)$ relate? The relations will be governed by a rearrangement of the structure constants. Of course they should all be isomorphic, and each $M(F_g)$ exists in the other's $M(F_m)$ category of comeasurings, too...(I suspect my ideas are a bit vacuous.)

In the case of $\mathbb{F}_2$, consider the extension $\mathbb{F}_4 \cong \mathbb{F}_2[x]/(x^2 +x + 1)$ as a finite dimensional algebra. The structure constants are easy to compute, and we can easily see that the dimension of $M(\mathbb{F}_4)$ is 2. Let's call its generators $\alpha$ and $\beta$, then using the above formulas we have that $\beta_{M(\mathbb{F}_4)}(x) = x \otimes \alpha + x \otimes \beta$ and the relations are $1 + \alpha = \alpha^2 + \beta^2$ and $[\alpha, \beta] = \beta^2 + \beta$. This algebra is clearly a lot more complicated than the simple field extension. One should note that there are no other field extensions of degree 2 of $\mathbb{F}_2$.