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 $\mathbb{F}_2$.