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Tag: geometry

Anabelian vs. Noncommutative Geometry

Published December 12, 2007 by lieven

This is how my attention was drawn to what I have since termed
anabelian algebraic geometry, whose starting point was exactly a study
(limited for the moment to characteristic zero) of the action of absolute
Galois groups (particularly the groups $Gal(\overline{K}/K) $, where K is an extension of finite type of the prime field) on (profinite) geometric fundamental
groups of algebraic varieties (defined over K), and more particularly (breaking with a well-established tradition) fundamental groups which are very far
from abelian groups (and which for this reason I call anabelian). Among
these groups, and very close to the group $\hat{\pi}_{0,3} $ , there is the profinite compactification of the modular group $SL_2(\mathbb{Z}) $, whose quotient by its centre
$\{ \pm 1 \} $ contains the former as congruence subgroup mod 2, and can also be
interpreted as an oriented cartographic group, namely the one classifying triangulated oriented maps (i.e. those whose faces are all triangles or
monogons).

The above text is taken from Alexander Grothendieck‘s visionary text Sketch of a Programme. He was interested in the permutation representations of the modular group $\Gamma = PSL_2(\mathbb{Z}) $ as they correspond via Belyi-maps and his own notion of dessins d’enfants to smooth projective curves defined over $\overline{\mathbb{Q}} $. One can now study the action of the absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q}) $ on these curves and their associated dessins. Because every permutation representation of $\Gamma $ factors over a finite quotient this gives an action of the absolute Galois group as automorphisms on the profinite compactification

$\hat{\Gamma} = \underset{\leftarrow}{lim}~\Gamma/N $

where the limit is taken over all finite index normal subgroups $N \triangleleft PSL_2(\mathbb{Z}) $. In this way one realizes the absolute Galois group as a subgroup of the outer automorphism group of the profinite group $\hat{\Gamma} $. As a profinite group is a compact topological group one should study its continuous finite dimensional representations which are precisely those factoring through a finite quotient. In the case of $\hat{\Gamma} $ the simple continuous representations $\mathbf{simp}_c~\hat{\Gamma} $ are precisely the components of the permutation representations of the modular group. So in a sense, anabelian geometry is the study of these continuous simples together wirth the action of the absolute Galois group on it.

In noncommutative geometry we are interested in a related representation theoretic problem. We would love to know the simple finite dimensional representations $\mathbf{simp}~\Gamma $ of the modular group as this would give us all simples of the three string braid group $B_3 $. So a natural question presents itself : how are these two ‘geometrical’ objects $\mathbf{simp}_c~\hat{\Gamma} $ (anabelian) and $\mathbf{simp}~\Gamma $ (noncommutative) related and can we use one to get information about the other?

This is all rather vague so far, so let us work out a trivial case to get some intuition. Consider the profinite completion of the infinite Abelian group

$\hat{\mathbb{Z}} = \underset{\leftarrow}{lim}~\mathbb{Z}/n\mathbb{Z} = \prod_p \hat{\mathbb{Z}}_p $

As all simple representations of an Abelian group are one-dimensional and because all continuous ones factor through a finite quotient $\mathbb{Z}/n\mathbb{Z} $ we see that in this case

$\mathbf{simp}_c~\hat{\mathbb{Z}} = \mu_{\infty} $

is the set of all roots of unity. On the other hand, the simple representations of $\mathbb{Z} $ are also one-dimensional and are determined by the image of the generator so

$\mathbf{simp}~\mathbb{Z} = \mathbb{C} – { 0 } = \mathbb{C}^* $

Clearly we have an embedding $\mu_{\infty} \subset \mathbb{C}^* $ and the roots of unity are even dense in the Zariski topology. This might look a bit strange at first because clearly all roots of unity lie on the unit circle which ‘should be’ their closure in the complex plane, but that’s because we have a real-analytic intuition. Remember that the Zariski topology of $\mathbb{C}^*$ is just the cofinite topology, so any closed set containing the infinitely many roots of unity should be the whole space!

Let me give a pedantic alternative proof of this (but one which makes it almost trivial that a similar result should be true for most profinite completions…). If $c $ is the generator of $\mathbb{Z} $ then the different conjugacy classes are precisely the singletons $c^n $. Now suppose that there is a polynomial $a_0+a_1x+\ldots+a_mx^m $ vanishing on all the continuous simples of $\hat{\mathbb{Z}} $ then this means that the dimensions of the character-spaces of all finite quotients $\mathbb{Z}/n\mathbb{Z} $ should be bounded by $m $ (for consider $x $ as the character of $c $), which is clearly absurd.

Hence, whenever we have a finitely generated group $G $ for which there is no bound on the number of irreducibles for finite quotients, then morally the continuous simple space for the profinite completion

$\mathbf{simp}_c~\hat{G} \subset \mathbf{simp}~G $

should be dense in the Zariski topology on the noncommutative space of simple finite dimensional representations of $G $. In particular, this should be the case for the modular group $PSL_2(\mathbb{Z}) $.

There is just one tiny problem : unlike the case of $\mathbb{Z} $ for which this space is an ordinary (ie. commutative) affine variety $\mathbb{C}^* $, what do we mean by the “Zariski topology” on the noncommutative space $\mathbf{simp}~PSL_2(\mathbb{Z}) $ ? Next time we will clarify what this might be and show that indeed in this case the subset

$\mathbf{simp}_c~\hat{\Gamma} \subset \mathbf{simp}~\Gamma $

will be a Zariski closed subset!

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M-geometry (3)

Published September 18, 2007 by lieven

For any finite dimensional A-representation S we defined before a character $\chi(S) $ which is an linear functional on the noncommutative functions $\mathfrak{g}_A = A/[A,A]_{vect} $ and defined via

$\chi_a(S) = Tr(a | S) $ for all $a \in A $

We would like to have enough such characters to separate simples, that is we would like to have an embedding

$\mathbf{simp}~A \hookrightarrow \mathfrak{g}_A^* $

from the set of all finite dimensional simple A-representations $\mathbf{simp}~A $ into the linear dual of $\mathfrak{g}_A^* $. This is a consequence of the celebrated Artin-Procesi theorem.

Michael Artin was the first person to approach representation theory via algebraic geometry and geometric invariant theory. In his 1969 classical paper “On Azumaya algebras and finite dimensional representations of rings” he introduced the affine scheme $\mathbf{rep}_n~A $ of all n-dimensional representations of A on which the group $GL_n $ acts via basechange, the orbits of which are exactly the isomorphism classes of representations. He went on to use the Hilbert criterium in invariant theory to prove that the closed orbits for this action are exactly the isomorphism classes of semi-simple -dimensional representations. Invariant theory tells us that there are enough invariant polynomials to separate closed orbits, so we would be done if the caracters would generate the ring of invariant polynmials, a statement first conjectured in this paper.

Claudio Procesi was able to prove this conjecture in his 1976 paper “The invariant theory of $n \times n $ matrices” in which he reformulated the fundamental theorems on $GL_n $-invariants to show that the ring of invariant polynomials of m $n \times n $ matrices under simultaneous conjugation is generated by traces of words in the matrices (and even managed to limit the number of letters in the words required to $n^2+1 $). Using the properties of the Reynolds operator in invariant theory it then follows that the same applies to the $GL_n $-action on the representation schemes $\mathbf{rep}_n~A $.

So, let us reformulate their result a bit. Assume the affine $\mathbb{C} $-algebra A is generated by the elements $a_1,\ldots,a_m $ then we define a necklace to be an equivalence class of words in the $a_i $, where two words are equivalent iff they are the same upto cyclic permutation of letters. For example $a_1a_2^2a_1a_3 $ and $a_2a_1a_3a_1a_2 $ determine the same necklace. Remark that traces of different words corresponding to the same necklace have the same value and that the noncommutative functions $\mathfrak{g}_A $ are spanned by necklaces.

The Artin-Procesi theorem then asserts that if S and T are non-isomorphic simple A-representations, then $\chi(S) \not= \chi(T) $ as elements of $\mathfrak{g}_A^* $ and even that they differ on a necklace in the generators of A of length at most $n^2+1 $. Phrased differently, the array of characters of simples evaluated at necklaces is a substitute for the clasical character-table in finite group theory.

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M-geometry (2)

Published September 17, 2007 by lieven

Last time we introduced the tangent quiver $\vec{t}~A $ of an affine algebra A to be a quiver on the isoclasses of simple finite dimensional representations. When $A=\mathbb{C}[X] $ is the coordinate ring of an affine variety, these vertices are just the points of the variety $X $ and this set has the extra structure of being endowed with the Zariski topology. For a general, possibly noncommutative algebra, we would like to equip the vertices of $\vec{t}~A $ also with a topology.

In the commutative case, the Zariski topology has as its closed sets the common zeroes of a set of polynomials on $X $, so we need to generalize the notion of ‘functions’ the the noncommutative world. The NC-mantra states that we should view the algebra A as the ring of functions on a (usually virtual) noncommutative space. And, face it, for a commutative variety $X $ the algebra $A=\mathbb{C}[X] $ does indeed do the job. Still, this is a red herring.

Let’s consider the easiest noncommutative case, that of the group algebra $\mathbb{C} G $ of a finite group $G $. In this case, the vertices of the tangent quiver $\vec{t}~A $ are the irreducible representations of $G $ and no sane person would consider the full group algebra to be the algebra of functions on this set. However, we do have a good alternative in this case : characters which allow us to separate the irreducibles and are a lot more manageable than the full group algebra. For example, if $G $ is the monster group then the group algebra has dimension approx $8 \times 10^{53} $ whereas there are just 194 characters to consider…

But, can we extend characters to arbitrary noncommutative algebras? and, more important, are there enough of these to separate the simple representations? The first question is easy enough to answer, after all characters are just traces so we can define for every element $a \in A $ and any finite dimensional simple A-representation $S $ the character

$\chi_a(S) = Tr(a | S) $

where $a | S $ is the matrix describing the action of a on S. But, you might say, characters are then just linear functionals on the algebra A so it is natural to view A as the function algebra, right? Wrong! Traces have the nice property that $Tr(ab)=Tr(ba) $ and so they vanish on all commutators $[a,b]=ab-ba $ of A, so characters only carry information of the quotient space

$\mathfrak{g}_A = \frac{A}{[A,A]_{vect}} $

where $[A,A]_{vect} $ is the vectorspace spanned by all commutators (and not the ideal…). If one is too focussed on commutative geometry one misses this essential simplification as clearly for $A=\mathbb{C}[X] $ being a commutative algebra,

$[\mathbb{C}[X],\mathbb{C}[X]]_{vect}=0 $ and therefore in this case $\mathfrak{g}_{\mathbb{C}[X]} = \mathbb{C}[X] $

Ok, but are there enough characters (that is, linear functionals on $\mathfrak{g}_A $, that is elements of the dual space $\mathfrak{g}_A^* $) to separate the simple representations? And, why do I (ab)use Lie-algebra notation $\mathfrak{g}_A $ to denote the vectorspace $A/[A,A]_{vect} $???

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