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From Galois to NOG

Published December 22, 2004 by lievenlb

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.

curvatures

Published September 16, 2004 by lievenlb

[Last
time][1] we saw that the algebra $(\Omega_V~C Q,Circ)$ of relative
differential forms and equipped with the Fedosov product is again the
path algebra of a quiver $\tilde{Q}$ obtained by doubling up the arrows
of $Q$ . In our basic example the algebra map $C \tilde{Q} \rightarrow \Omega_V~C Q$ is clarified by the following picture of $\tilde{Q}$
$\xymatrix{\vtx{} \ar@/^/[rr]^{a=u+du} \ar@/_/[rr]_{b=u-du} & & \vtx{} \ar@(u,ur)^{x=v+dv} \ar@(d,dr)_{y=v-dv}}$ (which
generalizes in the obvious way to arbitrary quivers). But what about the
other direction $\Omega_V~C Q \rightarrow C \tilde{Q}$ ? There are two
embeddings $i,j : C Q \rightarrow C \tilde{Q}$ defined by $i : (u,v) \rightarrow (a,x)$ and $j : (u,v) \rightarrow (b,y)$ giving maps
$\forall a \in C Q~:~p(a) = \frac{1}{2}(i(a)+j(a))~\quad~q(a) = \frac{1}{2}(i(a)-j(a))$ Using these maps, the isomorphism $\Omega_V~C Q \rightarrow C \tilde{Q}$ is determined by $a_0 da_1 \ldots da_n \rightarrow p(a_0)q(a_1) \ldots q(a_n)$ In particular, $p$ gives the
natural embedding (with the ordinary multiplication on differential
forms) $C Q \rightarrow \Omega_V~C Q$ of functions as degree zero
differential forms. However, $p$ is no longer an algebra map for the
Fedosov product on $\Omega_V~C Q$ as $p(ab) = p(a)Circ p(b) + q(a) Circ q(b)$ . In Cuntz-Quillen terminology, $\omega(a,b) = q(a) Circ q(b)$ is
the _curvature_ of the based linear map $p$ . I\’d better define
this a bit more formal for any algebra $A$ and then say what is special
for formally smooth algebras (non-commutative manifolds). If $A,B$ are
$V = C \times \ldots \times C$ -algebras, then a $V$ -linear map $A \rightarrow^l B$ is said to be a _based linear map_ if $l | V = id_V$ .
The _curvature_ of $l$ measures the obstruction to $l$ being an algebra
map, that is $\forall a,b \in A~:~\omega(a,b) = l(ab)-l(a)l(b)$ and
the curvature is said to be _nilpotent_ if there is an integer $n$ such
that all possible products $\omega(a_1,b_1)\omega(a_2,b_2) \ldots \omega(a_n,b_n) = 0$ For any algebra $A$ there is a universal algebra
$R(A)$ turning based linear maps into algebra maps. That is, there is a
fixed based linear map $A \rightarrow^p R(A)$ such that for every based
linear map $A \rightarrow^l B$ there is an algebra map $R(A) \rightarrow B$ making the diagram commute $\xymatrix{A \ar[r]^l \ar[d]^p & B \\\ R(A) \ar[ru] &}$ In fact, Cuntz and Quillen show that $R(A) \simeq (\Omega_V^{ev}~A,Circ)$ the algebra of even differential forms
equipped with the Fedosov product and that $p$ is the natural inclusion
of $A$ as degree zero forms (as above). Recall that $A$ is said to be
_formally smooth_ if every $V$ -algebra map $A \rightarrow^f B/I$ where
$I$ is a nilpotent ideal, can be lifted to an algebra morphism $A \rightarrow B$ . We can always lift $f$ as a based linear map, say
$\tilde{f}$ and because $I$ is nilpotent, the curvature of $\tilde{f}$
is also nilpotent. To get a _uniform_ way to construct algebra lifts
modulo nilpotent ideals it would therefore suffice for a formally smooth
algebra to have an _algebra map_ $A \rightarrow \hat{R}(A)$ where
$\hat{R}(A)$ is the $\mathfrak{m}$ -adic completion of $R(A)$ for the
ideal $\mathfrak{m}$ which is the kernel of the algebra map $R(A) \rightarrow A$ corresponding to the based linear map $id_A : A \rightarrow A$ . Indeed, there is an algebra map $R(A) \rightarrow B$
determined by $\tilde{f}$ and hence also an algebra map $\hat{R}(A) \rightarrow B$ and composing this with the (yet to be constructed)
algebra map $A \rightarrow \hat{R}(A)$ this would give the required lift
$A \rightarrow B$ . In order to construct the algebra map $A \rightarrow \hat{R}(A)$ (say in the case of path algebras of quivers) we
will need the Yang-Mills derivation and its associated flow.

[1]: http://www.matrix.ua.ac.be/index.php?p=354

differential forms

Published September 14, 2004 by lievenlb

The
previous post in this sequence was [(co)tangent bundles][1]. Let $A$ be
a $V$ -algebra where $V = C \times \ldots \times C$ is the subalgebra
generated by a complete set of orthogonal idempotents in $A$ (in case $A = C Q$ is a path algebra, $V$ will be the subalgebra generated by the
vertex-idempotents, see the post on [path algebras][2] for more
details). With $\overline{A}$ we denote the bimodule quotient
$\overline{A} = A/V$ Then, we can define the _non-commutative
(relative) differential n-forms_ to be $\Omega^n_V~A = A \otimes_V \overline{A} \otimes_V \ldots \otimes_V \overline{A}$ with $n$ factors
$\overline{A}$ . To get the connection with usual differential forms let
us denote the tensor $a_0 \otimes a_1 \otimes \ldots \otimes a_n = (a_0,a_1,\ldots,a_n) = a_0 da_1 \ldots da_n$ On $\Omega_V~A = \oplus_n~\Omega^n_V~A$ one defines an algebra structure via the
multiplication $(a_0da_1 \ldots da_n)(a_{n+1}da_{n+2} \ldots da_k)$ $= \sum_{i=1}^n (-1)^{n-i} a_0da_1 \ldots d(a_ia_{i+1}) \ldots da_k$
$\Omega_V~A$ is a _differential graded algebra_ with differential $d : \Omega^n_V~A \rightarrow \Omega^{n+1}_V~A$ defined by $d(a_0 da_1 \ldots da_n) = da_0 da_1 \ldots da_n$ This may seem fairly abstract but in
case $A = C Q$ is a path algebra, then the bimodule $\Omega^n_V~A$ has a
$V$ -generating set consisting of precisely the elements $p_0 dp_1 \ldots dp_n$ with all $p_i$ non-zero paths in $A$ and such that
$p_0p_1 \ldots p_n$ is also a non-zero path. One can put another
algebra multiplication on $\Omega_V~A$ which Cuntz and Quillen call the
_Fedosov product_ defined for an $n$ -form $\omega$ and a form $\mu$ by
$\omega Circ \mu = \omega \mu -(-1)^n d\omega d\mu$ There is an
important relation between the two structures, the degree of a
differential form puts a filtration on $\Omega_V~A$ (with Fedosov
product) such that the _associated graded algebra_ is $\Omega_V~A$ with
the usual product. One can visualize the Fedosov product easily in the
case of path algebras because $\Omega_V~C Q$ with the Fedosov product is
again the path algebra of the quiver obtained by doubling up all the
arrows of $Q$ . In our basic example when $Q$ is the quiver
$\xymatrix{\vtx{} \ar[rr]^u & & \vtx{} \ar@(ur,dr)^v}$ the
algebra of non-commutative differential forms equipped with the Fedosov
product is isomorphic to the path algebra of the quiver
$\xymatrix{\vtx{} \ar@/^/[rr]^{a=u+du} \ar@/_/[rr]_{b=u-du} & & \vtx{} \ar@(u,ur)^{x=v+dv} \ar@(d,dr)_{y=v-dv}}$ with the
indicated identification of arrows with elements from $\Omega_V~C Q$ .
Note however that we usually embed the algebra $C Q$ as the degree zero
differential forms in $\Omega_V~C Q$ with the usual multiplication and
that this embedding is no longer an algebra map (but a based linear map)
for the Fedosov product. For this reason, Cuntz and Quillen invent a
Yang-Mills type argument to “flow” this linear map to an algebra
embedding, but to motivate this we will have to say some things about
[curvatures][3].

[1]: http://www.neverendingbooks.org/index.php?p=352
[2]: http://www.neverendingbooks.org/index.php?p=349
[3]: http://www.neverendingbooks.org/index.php?p=353

Tag: Cuntz

From Galois to NOG

curvatures

differential forms