Tag: representations

daddy wasn’t impressed

Published December 19, 2007 by lievenlb

A first year-first semester course on group theory has its hilarious moments. Whereas they can relate the two other pure math courses (linear algebra and analysis) _somewhat_ to what they’ve learned before, with group theory they appear to enter an entirely new and strange world. So, it is best to give them concrete examples : symmetry groups of regular polygons and Platonic solids, the symmetric group etc. One of the lesser traditional examples I like to give is Nim addition and its relation to combinatorial games.

For their first test they had (among other things) to find a winning move for the position below in the Lenstra’s turtle turning game. At each move a player must put one turtle on its back and may also turn over any single turtle to the left of it. This second turtle, unlike the first, may be turned either onto its feet or onto its back. The player wins who turns the last turtle upside-down.

So, all they needed to see was that one turtle on its feet at place n is equivalent to a Nim-heap of height n and use the fact that all elements have order two to show that any zero-move in the sum game can indeed be played by using the second-turtle alternative. (( for the curious : the answer is turning both 9 and 4 on their back ))

A week later, one of the girls asked at the start of the lecture :

Are there real-life applications of group-theory? I mean, my father asked me what I was learning at school and I told him we were playing games turning turtles. I have to say that he was not impressed at all!.

She may have had an hidden agenda to slow me down because I spend an hour talking about a lot of things ranging from codes to cryptography and from representations to elementary particles…

For test three (on group-actions) I asked them to prove (among other things) Wilson’s theorem that is

$~(p-1)! \equiv -1~\text{mod}~p $

for any prime number $p $. The hint being : take the trivial action of $S_p $ on a one-element set and use the orbit theorem. (they know the number of elements in an $S_n $-conjugacy class)

Fingers crossed, hopefully daddy approved…

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profinite groups survival guide

Published December 14, 2007 by lievenlb

Even if you don’t know the formal definition of a profinte group, you know at least one example which explains the concept : the Galois group of the algebraic numbers $Gal = Gal(\overline{\mathbb{Q}}/\mathbb{Q}) $ aka the absolute Galois group. By definition it is the group of all $\mathbb{Q} $-isomorphisms of the algebraic closure $\overline{\mathbb{Q}} $. Clearly, it is an object of fundamental importance for mathematics but in spite of this very little is known about it. For example, it obviously is an infinite group but, apart from the complex conjugation, try to give one (1!) other nontrivial element… On the other hand we know lots of finite quotients of $Gal $. For, take any finite Galois extension $\mathbb{Q} \subset K $, then its Galois group $G_K = Gal(K/\mathbb{Q}) $ is a finite group and there is a natural onto morphism $\pi_K~:~Gal \rightarrow G_K $ obtained by dividing out all $K $-automorphisms of $\overline{\mathbb{Q}} $. Moreover, all these projections fit together nicely. If we take a larger Galois extension $K \subset L $ then classical Galois theory tells us that there is a projection $\pi_{LK}~:~G_L \rightarrow G_K $ by dividing out the normal subgroup of all $K $-automorphisms of $L $ and these finite maps are compatible with those from the absolute Galois group, that is, for all such finite Galois extensions, the diagram below is commutative

[tex]\xymatrix{Gal \ar[rr]^{\pi_L} \ar[rd]_{\pi_K} & & G_L \ar[ld]^{\pi_{LK}} \\
& G_K &}[/tex]

By going to larger and larger finite Galois extensions $L $ we get closer and closer to the algebraic closure $\overline{Q} $ and hence a better and better finite approximation $G_L $ of the absolute Galois group $Gal $. Still with me? Congratulations, you just rediscovered the notion of a profinite group! Indeed, the Galois group is the projective limit

$Gal = \underset{\leftarrow}{lim}~G_L $

over all finite Galois extensions $L/\mathbb{Q} $. If the term ‘projective limit’ scares you off, it just means that all the projections $\pi_{KL} $ coming from finite Galois theory are compatible with those coming from the big Galois group as before. That’s it : profinite groups are just projective limits of finite groups.

These groups come equipped with a natural topology : the Krull topology. Again, this notion is best clarified by considering the absolute Galois group. Now that we have $Gal $ we would like to extend the classical Galois correspondence between subgroups and subfields $\mathbb{Q} \subset K \subset \overline{\mathbb{Q}} $ and between normal subgroups and Galois subfields. For each finite Galois extension $K/\mathbb{Q} $ we have a normal subgroup of finite index, the kernel $U_K=Ker(\pi_K) $ of the projection map above. Let us take the set of all $U_K $ as a fundamental system of neighborhoods of the identity element in $Gal $. This defines a topology on $Gal $ and this is the Krull topology. As every open subgroup has finite index it is clear that this turns $Gal $ into a compact topological group. Its purpose is that we can now extend the finite Galois correspondence to Krull’s Galois theorem :

There is a bijective lattice inverting Galois correspondence between the set of all closed subgroups of $Gal $ and the set of all subfields $\mathbb{Q} \subset F \subset \overline{\mathbb{Q}} $. Finite field extensions correspond in this bijection to open subgroups and the usual normal subgroup and factor group correspondences hold!

So far we had a mysterious group such as $Gal $ and reconstructed it from all its finite quotients as a projective limit. Now we can reverse the situation : suppose we have a wellknown group such as the modular group $\Gamma = PSL_2(\mathbb{Z}) $, then we can look at the set of all its normal subgroups $U $ of finite index. For each of those we have a quotient map to a finite group $\pi_U~:~\Gamma \rightarrow G_U $ and clearly if $U \subset V $ we have a quotient map of finite groups $\pi_{UV}~:~G_U \rightarrow G_V $ compatible with the quotient maps from $\Gamma $

[tex]\xymatrix{\Gamma \ar[rr]^{\pi_U} \ar[rd]_{\pi_V} & & G_U \ar[ld]^{\pi_{UV}} \\
& G_V &}[/tex]

For the family of finite groups $G_U $ and groupmorphisms $\pi_{UV} $ we can ask for the ‘best’ group mapping to each of the $G_U $ compatible with the groupmaps $G_{UV} $. By ‘best’ we mean that any other group with this property will have a morphism to the best-one such that all quotient maps are compatible. This ‘best-one’ is called the projective limit

$\hat{\Gamma} = \underset{\leftarrow}{lim}~G_U $

and as a profinite group it has again a Krull topology making it into a compact group. Because the modular group $\Gamma $ had quotient maps to all the $G_U $ we know that there must be a groupmorphism to the best-one
$\phi~:~\Gamma \rightarrow \hat{\Gamma} $ and therefore we call $\hat{\Gamma} $ the profinite compactification (or profinite completion) of the modular group.

A final remark about finite dimensional representations. Every continuous complex representation of a profinite group like the absolute Galois group $Gal \rightarrow GL_n(\mathbb{C}) $ has finite image and this is why they are of little use for people studying the Galois group as it conjecturally reduces the study of these representations to ‘just’ all representations of all finite groups. Instead they consider representations to other topological fields such as p-adic numbers $Gal \rightarrow GL_n(\mathbb{Q}_p) $ and call these Galois representations.

For people interested in Grothendieck’s dessins d’enfants, however, continuous complex representations of the profinite compactification $\hat{\Gamma} $ is exactly their object of study and via the universal map $\phi~:~\Gamma \rightarrow \hat{\Gamma} $ above we have an embedding

$\mathbf{rep}_c~\hat{\Gamma} \rightarrow \mathbf{rep}~\Gamma $

of them in all finite dimensional representations of the modular group (
and we have a similar map restricted to simple representations). I hope this clarifies a bit obscure terms in the previous post. If not, drop a comment.

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Anabelian vs. Noncommutative Geometry

Published December 12, 2007 by lievenlb

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 finite type of the prime field) on (profinite) 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 profinite compactification 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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Iguanodon series of simple groups

Published November 7, 2007 by lievenlb

Bruce Westbury has a page on recent work on series of Lie groups including exceptional groups. Moreover, he did put his slides of a recent talk (probably at MPI) online.

Probably, someone considered a similar problem for simple groups. Are there natural constructions leading to a series of finite simple groups including some sporadic groups as special members ? In particular, does the following sequence appear somewhere ?

$L_2(7), M_{12}, A_{16}, M_{24}, A_{28}, A_{40}, A_{48}, A_{60}, \ldots $

Here, $L_2(7) $ is the simple group of order 168 (the automorphism group of the Klein quartic), $M_{12} $ and $M_{24} $ are the sporadic Mathieu groups and the $A_n $ are the alternating simple groups.

I’ve stumbled upon this series playing around with Farey sequences and their associated ‘dessins d’enfants’ (I’ll come back to the details of the construction another time) and have dubbed this sequence the Iguanodon series because the shape of the doodle leading to its first few terms

reminded me of the Iguanodons of Bernissart (btw. this sketch outlines the construction to the experts). Conjecturally, all groups appearing in this sequence are simple and probably all of them (except for the first few) will be alternating.

I did verify that none of the known low-dimensional permutation representations of other sporadic groups appear in the series. However, there are plenty of similar sequences one can construct from the Farey sequences, and it would be nice if one of them would contain the Conway group $Co_1 $. (to be continued)

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M-geometry (3)

Published September 18, 2007 by lievenlb

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 lievenlb

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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M-geometry (1)

Published September 15, 2007 by lievenlb

Take an affine $\mathbb{C} $-algebra A (not necessarily commutative). We will assign to it a strange object called the tangent-quiver $\vec{t}~A $, compute it in a few examples and later show how it connects with existing theory and how it can be used. This series of posts can be seen as the promised notes of my talks at the GAMAP-workshop but in reverse order… If some of the LaTeX-pictures are not in the desired spots, please size and resize your browser-window and they will find their intended positions.

A vertex $v $ of $\vec{t}~A $ corresponds to the isomorphism class of a finite dimensional simple A-representations $S_v $ and between any two such vertices, say $v $ and $w $, the number of directed arrows from $v $ to $w $ is given by the dimension of the Ext-space

$dim_{\mathbb{C}}~Ext^1_A(S_v,S_w) $

Recall that this Ext-space counts the equivalence classes of short exact sequences of A-representations

[tex]\xymatrix{0 \ar[r] & S_w \ar[r] & V \ar[r] & S_v \ar[r] & 0}[/tex]

where two such sequences (say with middle terms V resp. W) are equivalent if there is an A-isomorphism $V \rightarrow^{\phi} W $ making the diagram below commutative

[tex]\xymatrix{0 \ar[r] & S_w \ar[r] \ar[d]^{id_{S_w}} & V \ar[r] \ar[d]^{\phi} & S_v \ar[r] \ar[d]^{id_{S_v}} & 0 \\\
0 \ar[r] & S_w \ar[r] & W \ar[r] & S_v \ar[r] & 0}[/tex]

The Ext-space measures how many non-split extensions there are between the two simples and is always a finite dimensional vectorspace. So the tangent quiver $\vec{t}~A $ has the property that in all vertices there are at most finitely many loops and between any two vertices there are a finite number of directed arrows, but in principle a vertex may be the origin of arrows connecting it to infinitely many other vertices.

Right, now let us at least motivate the terminology. Let $X $ be a (commutative) affine variety with coordinate ring $A = \mathbb{C}[X] $ then what is $\vec{t}~A $ in this case? To begin, as $\mathbb{C}[X] $ is commutative, all its finite dimensional simple representations are one-dimensional and there is one such for every point $x \in X $. Therefore, the vertices of $\vec{t}~A $ correspond to the points of the affine variety $X $. The simple A-representation $S_x $ corresponding to a point $x $ is just evaluating polynomials in $x $. Moreover, if $x \not= y $ then there are no non-split extensions between $S_x $ and $S_y $ (a commutative semi-local algebra splits as a direct sum of locals), therefore in $\vec{t}~A $ there can only be loops and no genuine arrows between different vertices. Finally, the number of loops in the vertex corresponding to the point $x $ can be computed using the fact that the self-extensions can be identified with the tangent space at $x $, that is

$dim_{\mathbb{C}}~Ext^1_{\mathbb{C}[X]}(S_x,S_x) = dim_{\mathbb{C}}~T_x~X $

That is, if $A=\mathbb{C}[X] $ is the coordinate ring of an affine variety $X $, then the quiver $\vec{t}~A $ is the set of points of $X $ having in each point $x $ as many loops as the dimension of the tangent space $T_x~X $. So, in this case, the quiver $\vec{t}~A $ contains all information about tangent spaces to the variety and that’s why we call it the tangent quiver.

Let’s go into the noncommutative wilderness. A first, quite trivial, example is the group algebra $A = \mathbb{C} G $ of a finite group $G $, then the simple A-representations are just the irreducible G-representations and as the group algebra is semi-simple every short exact sequence splits so all Ext-spaces are zero. That is, in this case the tangent quiver $\vec{t}~A $ in just a finite set of vertices (as many as there are irreducible G-representations) and no arrows nor loops.

Now you may ask whether there are examples of tangent quivers having arrows apart from loops. So, take another easy finite dimensional example : the path algebra $A = \mathbb{C} Q $ of a finite quiver $Q $ without oriented cycles. Recall that the path algebra is the vectorspace having as basis all vertices and all oriented paths in the quiver Q (and as there are no cycles, this basis is finite) and multiplication is induced by concatenation of paths. Here an easy example. Suppose the quiver Q looks like

[tex]\xymatrix{\vtx{} \ar[r] & \vtx{} \ar[r] & \vtx{}}[/tex]

then the path algebra is 6 dimensional as there are 3 vertices, 2 paths of length one (the arrows) and one path of length two (going from the leftmost to the rightmost vertex). The concatenation rule shows that the three vertices will give three idempotents in A and one easily verifies that the path algebra can be identified with upper-triangular $3 \times 3 $ matrices

$\mathbb{C} Q \simeq \begin{bmatrix} \mathbb{C} & \mathbb{C} & \mathbb{C} \\\ 0 & \mathbb{C} & \mathbb{C} \\\ 0 & 0 & \mathbb{C} \end{bmatrix} $

where the diagonal components correspond to the vertices, the first offdiagonal components to the two arrows and the corner component corresponds to the unique path of length two. Right, for a general finite quiver without oriented cycles is the quite easy to see that all finite dimensional simples are one-dimensional and correspond to the vertex-idempotents, that is every simple is of the form $S_v = e_v \mathbb{C} Q e_v $ where $e_v $ is the vertex idempotent. No doubt, you can guess what the tangent quiver $\vec{t}~A = \vec{t}~\mathbb{C} Q $ will be, can’t you?

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Vacation reading

Published July 26, 2007 by lievenlb

Im in the process of writing/revising/extending the course notes for next year and will therefore pack more math-books than normal.

These are for a 3rd year Bachelor course on Algebraic Geometry and a 1st year Master course on Algebraic and Differential Geometry. The bachelor course was based this year partly on Miles Reid’s Undergraduate Algebraic Geometry and partly on David Mumford’s Red Book, but this turned out to be too heavy going. Next year I’ll be happy if they know enough on algebraic curves. The backbone of these two courses will be Fulton’s old but excellent Algebraic curves. It’s self contained (unlike Hartshorne’s book that assumes a prior course on commutative algebra), contains a lot of exercises and goes on to the Brill-Noether proof of Riemann-Roch. Still, Id like to extend it with the introductory chapter and the chapters on Riemann surfaces from Complex Algebraic Curves by Frances Kirwan, a bit on elliptic and modular functions from Elliptic curves by Henry McKean and Victor Moll and the adelic proof of Riemann-Roch and applications of it to the construction of algebraic codes from Algebraic curves over finite fields by Carlos Moreno. If time allows Id love to include also the chapter on zeta functions but I fear this will be difficult.

These are to spice up a 2nd year Bachelor course on Representations of Finite Groups with a tiny bit of Galois representations, both as motivation and to wet their appetite for elliptic curves and algebraic geometry. Ive received Fearless Symmetry by Avner Ash and Robert Gross only yesterday and find it hard to stop reading. It attempts to explain Galois representations and generalized reciprocity laws to a general audience and from what I read so far, they really do a terrific job. Another excellent elementary introduction to elliptic curves and Galois representations is in Invitation to the Mathematics of Fermat-Wiles by Yves Hellegouarch. On a gossipy note, the appendix “The origin of the elliptic approach to Fermat’s last theorem” is fun reading. Finally, Ill also take Introduction to Fermat’s Last Theorem by Alf van der Poorten along simply because I love his writing style.

These are included just for fun. The Poincare Conjecture by Donal O’Shea because I know far too little about it, Letters to a Young Mathematician by Ian Stewart because I like the concept of the book and finally The sensual (quadratic) form by John Conway because I need to have at all times at least one Conway-book nearby.

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Hexagonal Moonshine (3)

Published July 13, 2007 by lievenlb

Hexagons keep on popping up in the representation theory of the modular group and its close associates. We have seen before that singularities in 2-dimensional representation varieties of the three string braid group $B_3 $ are ‘clanned together’ in hexagons and last time Ive mentioned (in passing) that the representation theory of the modular group is controlled by the double quiver of the extended Dynkin diagram $\tilde{A_5} $, which is an hexagon…

Today we’re off to find representations of the extended modular group $\tilde{\Gamma} = PGL_2(\mathbb{Z}) $, which is obtained by adding to the modular group (see this post for a proof of generation)

$\Gamma = \langle U=\begin{bmatrix} 0 & -1 \\\ 1 & 0 \end{bmatrix},V=\begin{bmatrix} 0 & 1 \\\ -1 & 1 \end{bmatrix} \rangle $ the matrix $R=\begin{bmatrix} 0 & 1 \\\ 1 & 0 \end{bmatrix} $

In terms of generators and relations, one easily verfifies that

$\tilde{\Gamma} = \langle~U,V,R~|~U^2=R^2=V^3=(RU)^2=(RV)^2=1~\rangle $

and therefore $\tilde{\Gamma} $ is the amalgamated free product of the dihedral groups $D_2 $ and $D_3 $ over their common subgroup $C_2 = \langle~R~\rangle $, that is

$\tilde{\Gamma} = \langle U,R | U^2=R^2=(RU)^2=1 \rangle \ast_{\langle R | R^2=1 \rangle} \langle V,R | V^3=R^2=(RV)^2=1 \rangle = D_2 \ast_{C_2} D_3 $

From this description it is easy to find all n-dimensional $\tilde{\Gamma} $-representations $V $ and relate them to quiver-representations. $D_2 = C_2 \times C_2 $ and hence has 4 1-dimensonal simples $S_1,S_2,S_3,S_4 $. Restricting $V\downarrow_{D_2} $ to the subgroup $D_2 $ it decomposes as

$V \downarrow_{D_2} \simeq S_1^{\oplus a_1} \oplus S_2^{\oplus a_2} \oplus S_3^{\oplus a_3} \oplus S_4^{\oplus a_4} $ with $a_1+a_2+a_3+a_4=n $

Similarly, because $D_3=S_3 $ has two one-dimensional representations $T,S $ (the trivial and the sign representation) and one simple 2-dimensional representation $W $, restricting $V $ to this subgroup gives a decomposition

$V \downarrow_{D_3} \simeq T^{b_1} \oplus S^{\oplus b_2} \oplus W^{\oplus b_3} $, this time with $b_1+b_2+2b_3=n $

Restricting both decompositions further down to the common subgroup $C_2 $ one obtains a $C_2 $-isomorphism $V \downarrow_{D_2} \rightarrow^{\phi} V \downarrow_{D_3} $ which implies also that the above numbers must be chosen such that $a_1+a_3=b_1+b_3 $ and $a_2+a_4=b_2+b_3 $. We can summarize all this info about $V $ in a representation of the quiver

Here, the vertex spaces on the left are the iso-typical factors of $V \downarrow_{D_2} $ and those on the right those of $V \downarrow_{D_3} $ and the arrows give the block-components of the $C_2 $-isomorphism $\phi $. The nice things is that one can also reverse this process to get all $\tilde{\Gamma} $-representations from $\theta $-semistable representations of this quiver (having the additional condition that the square matrix made of the arrows is invertible) and isomorphisms of group-representation correspond to those of quiver-representations!

This proves that for all n the varieties of n-dimensional representations $\mathbf{rep}_n~\tilde{\Gamma} $ are smooth (but have several components corresponding to the different dimension vectors $~(a_1,a_2,a_3,a_4;b_1,b_2,b_3) $ such that $\sum a_i = n = b_1+b_2+2b_3 $.

The basic principle of _M-geometry_ is that a lot of the representation theory follows from the ‘clan’ (see this post) determined by the simples of smallest dimensions. In the case of the extended modular group $\tilde{\Gamma} $ it follows that there are exactly 4 one-dimensional simples and exactly 4 2-dimensional simples, corresponding to the dimension vectors

$\begin{cases} a=(0,0,0,1;0,1,0) \\\ b=(0,1,0,0;0,1,0) \\\ c=(1,0,0,0;1,0,0) \\\ d=(0,0,1,0;1,0,0) \end{cases} $ resp. $\begin{cases} e=(0,1,1,0;0,0,1) \\\ f=(1,0,0,1;0,0,1) \\\ g=(0,0,1,1;0,0,1) \\\ h=(1,1,0,0;0,0,1) \end{cases} $

If one calculates the ‘clan’ of these 8 simples one obtains the double quiver of the graph on the left. Note that a and b appear twice, so one should glue the left and right hand sides together as a Moebius-strip. That is, the clan determining the representation theory of the extended modular group is a Moebius strip made of two hexagons!

However, one should not focuss too much on the hexagons (that is, the extended Dynkin diagram $\tilde{A_5} $) here. The two ‘backbones’ (e–f and g–h) have their vertices corresponding to 2-dimensional simples whereas the topand bottom vertices correspond to one-dimensional simples. Hence, the correct way to look at this clan is as two copies of the double quiver of the extended Dynkin diagram $\tilde{D_5} $ glued over their leaf vertices to form a Moebius strip. Remark that the components of the sotropic root of $\tilde{D_5} $ give the dimensions of the corresponding $\tilde{\Gamma} $ simples.

The remarkable ubiquity of (extended) Dynkins never ceases to amaze!

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Hexagonal Moonshine (2)

Published July 11, 2007 by lievenlb

Delving into finite dimensional representations of the modular group $\Gamma = PSL_2(\mathbb{Z}) $ it is perhaps not too surprising to discover numerical connections with modular functions. Here, one such strange observation and a possible explanation.

Using the _fact_ that the modular group $\Gamma = PSL_2(\mathbb{Z}) $ is the free group product $C_2 \ast C_3 $ it is fairly easy to see that the variety of all n-dimensional representations $\mathbf{rep}_n~\Gamma $ is smooth (though it contains several connected components). Some of these components will contain simple representations, some will not. Anyway, we are not interested in all n-dimensional representations but in the isomorphism classes of such representations. The best algebraic approximation to this problem is by studying the quotient varieties

$\mathbf{iss}_n~\Gamma = \mathbf{rep}_n~\Gamma // GL(n) $

under the action of $GL(n) $ by basechange. Geometric invariant theory tells us that the points of this quotient variety correspond to isoclasses of semi-simple n-dimensional representations (whence the notation $\mathbf{iss}_n $). Again, these quotient varieties split into several connected components, some of which will have an open subset of points corresponding to simple representations.

It is a natural idea to compute the codimension of subvariety of proper semi-simples in the component of maximal dimension containing simple representations. _M-geometry_ allows you to reduce such calculation to quiver-problems. Anyway, if one does this for small values of n one obtains the following sequence of codimension-numbers (starting with $n=1 $

0,1,1,1,1,3,1,3,3,3,3,5,3,5,5,5,5,7,5,7,…

which doesnt seem too exciting before you feed it to Sloan’s integer sequences encyclopedia when one discovers that it is precisely sequence A063195 which gives the dimensions of weight 2n _cuspidal newforms_
for $\Gamma_0(6) $…

The optimistic “moonshine”-interpretation of this might be that these newforms can be viewed somehow as functions on the varieties of finite dimensional $\Gamma $-representations having the property that they pick out generic simple representations as their non-zeroes.

Be that as it may (one never knows in these matters), here a more down-to-earth explanation. The sequence A063195 obviously has a 6-periodicity behaviour so it suffices to understand why the codimension-sequence should have a similar feature (modulo computing the first few terms of it and observing the coincidence with the first few terms of A063195).

The modular group has exactly 6 one-dimensional representations and if one calculates their clan as in hexagonal moonshine (1) one obtains the hexagonal quiver

[tex]\xymatrix{& \vtx{S_1} \ar@/^/[dl] \ar@/^/[dr] & \\ \vtx{S_6} \ar@/^/[ur] \ar@/^/[d] & & \vtx{S_2} \ar@/^/[ul] \ar@/^/[d] \\ \vtx{S_5} \ar@/^/[u] \ar@/^/[dr] & & \vtx{S_3} \ar@/^/[u] \ar@/^/[dl] \\ & \vtx{S_4} \ar@/^/[ur] \ar@/^/[ul] & }[/tex]

M-geometry tells us that this clan contains enough information to determine the components of $\mathbf{rep}_n~\Gamma $ that contain simple representations. They correspond to dimension-vectors of this hexagonal quiver, say

$~(a_1,a_2,a_3,a_4,a_5,a_6) $

such that $a_i \leq a_{i-1}+a_{i+1} $. Moreover, the component is of maximal dimension if the components $a_i $ are evenly spread over the six vertices.
This then explains that the codimension sequence we are interested in must satisfy 6-periodicity.

Reference

This post corrects the erroneous statement made in math.AG/0610587 that the codimension sequence are the dimensions of weight 2n modular forms. The day the paper hit the arXiv I informed the author of the mistakes he made and told him how they could be corrected. Having waited 9 months I’ve given up hope that a revision/correction is imminent.