Evariste Galois (1811-1832) must rank pretty high on the all-time

list of moving last words. Galois was mortally wounded in a duel he

fought with Perscheux d\’Herbinville on May 30th 1832, the reason for

the duel not being clear but certainly linked to a girl called

Stephanie, whose name appears several times as a marginal note in

Galois\’ manuscripts (see illustration). When he died in the arms of his

younger brother Alfred he reportedly said “Ne pleure pas, j\’ai besoin

de tout mon courage pour mourir ‚àö‚Ä† 20 ans”. In this series I\’ll

start with a pretty concrete problem in Galois theory and explain its

elegant solution by Aidan Schofield and Michel Van den Bergh.

Next, I\’ll rephrase the problem in non-commutative geometry lingo,

generalise it to absurd levels and finally I\’ll introduce a coalgebra

(yes, a co-algebra…) that explains it all. But, it will take some time

to get there. Start with your favourite basefield $k$ of

characteristic zero (take $k = \mathbb{Q}$ if you have no strong

preference of your own). Take three elements $a,b,c$ none of which

squares, then what conditions (if any) must be imposed on $a,b,c$ and $n

\in \mathbb{N}$ to construct a central simple algebra $\Sigma$ of

dimension $n^2$ over the function field of an algebraic $k$-variety such

that the three quadratic fieldextensions $k\sqrt{a}, k\sqrt{b}$ and

$k\sqrt{c}$ embed into $\Sigma$? Aidan and Michel show in \’Division

algebra coproducts of index $n$\’ (Trans. Amer. Math. Soc. 341 (1994),

505-517) that the only condition needed is that $n$ is an even number.

In fact, they work a lot harder to prove that one can even take $\Sigma$

to be a division algebra. They start with the *algebra free
product* $A = k\sqrt{a} \ast k\sqrt{b} \ast k\sqrt{c}$ which is a pretty

monstrous algebra. Take three letters $x,y,z$ and consider all

non-commutative words in $x,y$ and $z$

*without repetition*(that is, no

two consecutive $x,y$ or $z$\’s). These words form a $k$-basis for $A$

and the multiplication is induced by concatenation of words subject to

the simplifying relations $x.x=a,y.y=b$ and $z.z=c$.

Next, they look

at the affine $k$-varieties $\mathbf{rep}(n) A$ of $n$-dimensional

$k$-representations of $A$ and their irreducible components. In the

parlance of $\mathbf{geometry@n}$, these irreducible components correspond

to the minimal primes of the level $n$-approximation algebra $\int(n) A$.

Aidan and Michel worry a bit about reducedness of these components but

nowadays we know that $A$ is an example of a non-commutative manifold (a

la Cuntz-Quillen or Kontsevich-Rosenberg) and hence all representation

varieties $\mathbf{rep}*n A$ are smooth varieties (whence reduced) though
they may have several connected components. To determine the number of
irreducible (which in this case, is the same as connected) components
they use _Galois descent*, that is, they consider the algebra $A

\otimes_k \overline{k}$ where $\overline{k}$ is the algebraic closure of

$k$. The algebra $A \otimes_k \overline{k}$ is the group-algebra of the

group free product $\mathbb{Z}/2\mathbb{Z} \ast \mathbb{Z}/2\mathbb{Z}

\ast \mathbb{Z}/2\mathbb{Z}$. (to be continued…) A digression : I

cannot resist the temptation to mention the

*tetrahedral snake problem*

in relation to such groups. If one would have started with $4$ quadratic

fieldextensions one would get the free product $G =

\mathbb{Z}/2\mathbb{Z} \ast \mathbb{Z}/2\mathbb{Z} \ast

\mathbb{Z}/2\mathbb{Z} \ast \mathbb{Z}/2\mathbb{Z}$. Take a supply of

tetrahedra and glue them together along common faces so that any

tertrahedron is glued to maximum two others. In this way one forms a

tetrahedral-snake and the problem asks whether it is possible to make

such a snake having the property that the orientation of the

\’tail-tetrahedron\’ in $\mathbb{R}^3$ is exactly the same as the

orientation of the \’head-tetrahedron\’. This is

*not*possible and the

proof of it uses the fact that there are no non-trivial relations

between the four generators $x,y,z,u$ of $\mathbb{Z}/2\mathbb{Z} \ast

\mathbb{Z}/2\mathbb{Z} \ast \mathbb{Z}/2\mathbb{Z} \ast

\mathbb{Z}/2\mathbb{Z}$ which correspond to reflections wrt. a face of

the tetrahedron (in fact, there are no relations between these

reflections other than each has order two, so the subgroup generated by

these four reflections is the group $G$). More details can be found in

Stan Wagon\’s excellent book The Banach-tarski paradox, p.68-71.

## Comments