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Tag: Galois

Arnold’s trinities

Referring to the triple of exceptional Galois groups $L_2(5),L_2(7),L_2(11) $ and its connection to the Platonic solids I wrote : “It sure seems that surprises often come in triples…”. Briefly I considered replacing triples by trinities, but then, I didnt want to sound too mystic…

David Corfield of the n-category cafe and a dialogue on infinity (and perhaps other blogs I’m unaware of) pointed me to the paper Symplectization, complexification and mathematical trinities by Vladimir I. Arnold. (Update : here is a PDF-conversion of the paper)

The paper is a write-up of the second in a series of three lectures Arnold gave in june 1997 at the meeting in the Fields Institute dedicated to his 60th birthday. The goal of that lecture was to explain some mathematical dreams he had.

The next dream I want to present is an even more fantastic set of theorems and conjectures. Here I also have no theory and actually the ideas form a kind of religion rather than mathematics.
The key observation is that in mathematics one encounters many trinities. I shall present a list of examples. The main dream (or conjecture) is that all these trinities are united by some rectangular “commutative diagrams”.
I mean the existence of some “functorial” constructions connecting different trinities. The knowledge of the existence of these diagrams provides some new conjectures which might turn to be true theorems.

Follows a list of 12 trinities, many taken from Arnold’s field of expertise being differential geometry. I’ll restrict to the more algebraically inclined ones.

1 : “The first trinity everyone knows is”

where $\mathbb{H} $ are the Hamiltonian quaternions. The trinity on the left may be natural to differential geometers who see real and complex and hyper-Kaehler manifolds as distinct but related beasts, but I’m willing to bet that most algebraists would settle for the trinity on the right where $\mathbb{O} $ are the octonions.

2 : The next trinity is that of the exceptional Lie algebras E6, E7 and E8.

with corresponding Dynkin-Coxeter diagrams

Arnold has this to say about the apparent ubiquity of Dynkin diagrams in mathematics.

Manin told me once that the reason why we always encounter this list in many different mathematical classifications is its presence in the hardware of our brain (which is thus unable to discover a more complicated scheme).
I still hope there exists a better reason that once should be discovered.

Amen to that. I’m quite hopeful human evolution will overcome the limitations of Manin’s brain…

3 : Next comes the Platonic trinity of the tetrahedron, cube and dodecahedron



Clearly one can argue against this trinity as follows : a tetrahedron is a bunch of triangles such that there are exactly 3 of them meeting in each vertex, a cube is a bunch of squares, again 3 meeting in every vertex, a dodecahedron is a bunch of pentagons 3 meeting in every vertex… and we can continue the pattern. What should be a bunch a hexagons such that in each vertex exactly 3 of them meet? Well, only one possibility : it must be the hexagonal tiling (on the left below). And in normal Euclidian space we cannot have a bunch of septagons such that three of them meet in every vertex, but in hyperbolic geometry this is still possible and leads to the Klein quartic (on the right). Check out this wonderful post by John Baez for more on this.



4 : The trinity of the rotation symmetry groups of the three Platonics

where $A_n $ is the alternating group on n letters and $S_n $ is the symmetric group.

Clearly, any rotation of a Platonic solid takes vertices to vertices, edges to edges and faces to faces. For the tetrahedron we can easily see the 4 of the group $A_4 $, say the 4 vertices. But what is the 4 of $S_4 $ in the case of a cube? Well, a cube has 4 body-diagonals and they are permuted under the rotational symmetries. The most difficult case is to see the $5 $ of $A_5 $ in the dodecahedron. Well, here’s the solution to this riddle



there are exactly 5 inscribed cubes in a dodecahedron and they are permuted by the rotations in the same way as $A_5 $.

7 : The seventh trinity involves complex polynomials in one variable

the Laurant polynomials and the modular polynomials (that is, rational functions with three poles at 0,1 and $\infty $.

8 : The eight one is another beauty

Here ‘numbers’ are the ordinary complex numbers $\mathbb{C} $, the ‘trigonometric numbers’ are the quantum version of those (aka q-numbers) which is a one-parameter deformation and finally, the ‘elliptic numbers’ are a two-dimensional deformation. If you ever encountered a Sklyanin algebra this will sound familiar.

This trinity is based on a paper of Turaev and Frenkel and I must come back to it some time…

The paper has some other nice trinities (such as those among Whitney, Chern and Pontryagin classes) but as I cannot add anything sensible to it, let us include a few more algebraic trinities. The first one attributed by Arnold to John McKay

13 : A trinity parallel to the exceptional Lie algebra one is

between the 27 straight lines on a cubic surface, the 28 bitangents on a quartic plane curve and the 120 tritangent planes of a canonic sextic curve of genus 4.

14 : The exceptional Galois groups

explained last time.

15 : The associated curves with these groups as symmetry groups (as in the previous post)

where the ? refers to the mysterious genus 70 curve. I’ll check with one of the authors whether there is still an embargo on the content of this paper and if not come back to it in full detail.

16 : The three generations of sporadic groups

Do you have other trinities you’d like to worship?

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Galois’ last letter

“Ne pleure pas, Alfred ! J’ai besoin de tout mon courage pour mourir à vingt ans!”

We all remember the last words of Evariste Galois to his brother Alfred. Lesser known are the mathematical results contained in his last letter, written to his friend Auguste Chevalier, on the eve of his fatal duel. Here the final sentences :



Tu prieras publiquement Jacobi ou Gauss de donner leur avis non sur la verite, mais sur l’importance des theoremes.
Apres cela il se trouvera, j’espere, des gens qui trouvent leur profis a dechiffrer tout ce gachis.
Je t’embrasse avec effusion.
E. Galois, le 29 Mai 1832

A major result contained in this letter concerns the groups $L_2(p)=PSL_2(\mathbb{F}_p) $, that is the group of $2 \times 2 $ matrices with determinant equal to one over the finite field $\mathbb{F}_p $ modulo its center. $L_2(p) $ is known to be simple whenever $p \geq 5 $. Galois writes that $L_2(p) $ cannot have a non-trivial permutation representation on fewer than $p+1 $ symbols whenever $p > 11 $ and indicates the transitive permutation representation on exactly $p $ symbols in the three ‘exceptional’ cases $p=5,7,11 $.

Let $\alpha = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} $ and consider for $p=5,7,11 $ the involutions on $\mathbb{P}^1_{\mathbb{F}_p} = \mathbb{F}_p \cup { \infty } $ (on which $L_2(p) $ acts via Moebius transformations)

$\pi_5 = (0,\infty)(1,4)(2,3) \quad \pi_7=(0,\infty)(1,3)(2,6)(4,5) \quad \pi_{11}=(0,\infty)(1,6)(3,7)(9,10)(5,8)(4,2) $

(in fact, Galois uses the involution $~(0,\infty)(1,2)(3,6)(4,8)(5,10)(9,7) $ for $p=11 $), then $L_2(p) $ leaves invariant the set consisting of the $p $ involutions $\Pi = { \alpha^{-i} \pi_p \alpha^i~:~1 \leq i \leq p } $. After mentioning these involutions Galois merely writes :

Ainsi pour le cas de $p=5,7,11 $, l’equation modulaire s’abaisse au degre p.
En toute rigueur, cette reduction n’est pas possible dans les cas plus eleves.

Alternatively, one can deduce these permutation representation representations from group isomorphisms. As $L_2(5) \simeq A_5 $, the alternating group on 5 symbols, $L_2(5) $ clearly acts transitively on 5 symbols.

Similarly, for $p=7 $ we have $L_2(7) \simeq L_3(2) $ and so the group acts as automorphisms on the projective plane over the field on two elements $\mathbb{P}^2_{\mathbb{F}_2} $ aka the Fano plane, as depicted on the left.

This finite projective plane has 7 points and 7 lines and $L_3(2) $ acts transitively on them.

For $p=11 $ the geometrical object is a bit more involved. The set of non-squares in $\mathbb{F}_{11} $ is

${ 1,3,4,5,9 } $

and if we translate this set using the additive structure in $\mathbb{F}_{11} $ one obtains the following 11 five-element sets

${ 1,3,4,5,9 }, { 2,4,5,6,10 }, { 3,5,6,7,11 }, { 1,4,6,7,8 }, { 2,5,7,8,9 }, { 3,6,8,9,10 }, $

$ { 4,7,9,10,11 }, { 1,5,8,10,11 }, { 1,2,6,9,11 }, { 1,2,3,7,10 }, { 2,3,4,8,11 } $

and if we regard these sets as ‘lines’ we see that two distinct lines intersect in exactly 2 points and that any two distinct points lie on exactly two ‘lines’. That is, intersection sets up a bijection between the 55-element set of all pairs of distinct points and the 55-element set of all pairs of distinct ‘lines’. This is called the biplane geometry.

The subgroup of $S_{11} $ (acting on the eleven elements of $\mathbb{F}_{11} $) stabilizing this set of 11 5-element sets is precisely the group $L_2(11) $ giving the permutation representation on 11 objects.

An alternative statement of Galois’ result is that for $p > 11 $ there is no subgroup of $L_2(p) $ complementary to the cyclic subgroup

$C_p = { \begin{bmatrix} 1 & x \\ 0 & 1 \end{bmatrix}~:~x \in \mathbb{F}_p } $

That is, there is no subgroup such that set-theoretically $L_2(p) = F \times C_p $ (note this is of courese not a group-product, all it says is that any element can be written as $g=f.c $ with $f \in F, c \in C_p $.

However, in the three exceptional cases we do have complementary subgroups. In fact, set-theoretically we have

$L_2(5) = A_4 \times C_5 \qquad L_2(7) = S_4 \times C_7 \qquad L_2(11) = A_5 \times C_{11} $

and it is a truly amazing fact that the three groups appearing are precisely the three Platonic groups!

Recall that here are 5 Platonic (or Scottish) solids coming in three sorts when it comes to rotation-automorphism groups : the tetrahedron (group $A_4 $), the cube and octahedron (group $S_4 $) and the dodecahedron and icosahedron (group $A_5 $). The “4” in the cube are the four body diagonals and the “5” in the dodecahedron are the five inscribed cubes.

That is, our three ‘exceptional’ Galois-groups correspond to the three Platonic groups, which in turn correspond to the three exceptional Lie algebras $E_6,E_7,E_8 $ via McKay correspondence (wrt. their 2-fold covers). Maybe I’ll detail this latter connection another time. It sure seems that surprises often come in triples…

Finally, it is well known that $L_2(5) \simeq A_5 $ is the automorphism group of the icosahedron (or dodecahedron) and that $L_2(7) $ is the automorphism group of the Klein quartic.

So, one might ask : is there also a nice curve connected with the third group $L_2(11) $? Rumour has it that this is indeed the case and that the curve in question has genus 70… (to be continued).

Reference

Bertram Kostant, “The graph of the truncated icosahedron and the last letter of Galois”

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adeles and ideles

Before we can even attempt to describe the adelic description of the Bost-Connes Hecke algebra and its symmetries, we’d probably better recall the construction and properties of adeles and ideles. Let’s start with the p-adic numbers $\hat{\mathbb{Z}}_p $ and its field of fractions $\hat{\mathbb{Q}}_p $. For p a prime number we can look at the finite rings $\mathbb{Z}/p^n \mathbb{Z} $ of all integer classes modulo $p^n $. If two numbers define the same element in $\mathbb{Z}/p^n\mathbb{Z} $ (meaning that their difference is a multiple of $p^n $), then they certainly define the same class in any $\mathbb{Z}/p^k \mathbb{Z} $ when $k \leq n $, so we have a sequence of ringmorphisms between finite rings

$ \ldots \rightarrow^{\phi_{n+1}} \mathbb{Z}/p^n \mathbb{Z} \rightarrow^{\phi_n} \mathbb{Z}/p^{n-1}\mathbb{Z} \rightarrow^{\phi_{n-1}} \ldots \rightarrow^{\phi_3} \mathbb{Z}/p^2\mathbb{Z} \rightarrow^{\phi_2} \mathbb{Z}/p\mathbb{Z} $

The ring of p-adic integers $\hat{\mathbb{Z}}_p $ can now be defined as the collection of all (infinite) sequences of elements $~(\ldots,x_n,x_{n-1},\ldots,x_2,x_1) $ with $x_i \in \mathbb{Z}/p^i\mathbb{Z} $ such that
$\phi_i(x_i) = x_{i-1} $ for all natural numbers $i $. Addition and multiplication are defined componentswise and as all the maps $\phi_i $ are ringmorphisms, this produces no compatibility problems.

One can put a topology on $\hat{\mathbb{Z}}_p $ making it into a compact ring. Here’s the trick : all components $\mathbb{Z}/p^n \mathbb{Z} $ are finite so they are compact if we equip these sets with the discrete topology (all subsets are opens). But then, Tychonov’s product theorem asserts that the product-space $\prod_n \mathbb{Z}/n \mathbb{Z} $ with the product topology is again a compact topological space. As $\hat{\mathbb{Z}}_p $ is a closed subset, it is compact too.

By construction, the ring $\hat{\mathbb{Z}}_p $ is a domain and hence has a field of fraction which we will denote by $\hat{\mathbb{Q}}_p $. These rings give the p-local information of the rational numbers $\mathbb{Q} $. We will now ‘glue together’ these local data over all possible prime numbers $p $ into adeles. So, forget the above infinite product used to define the p-adics, below we will work with another infinite product, one factor for each prime number.

The adeles $\mathcal{A} $ are the restricted product of the $\hat{\mathbb{Q}}_p $ over $\hat{\mathbb{Z}}_p $ for all prime numbers p. By ‘restricted’ we mean that elements of $\mathcal{A} $ are exactly those infinite vectors $a=(a_2,a_3,a_5,a_7,a_{11},\ldots ) = (a_p)_p \in \prod_p \hat{\mathbb{Q}}_p $ such that all but finitely of the components $a_p \in \hat{\mathbb{Z}}_p $. Addition and multiplication are defined component-wise and the restriction condition is compatible with both adition and multiplication. So, $\mathcal{A} $ is the adele ring. Note that most people call this $\mathcal{A} $ the finite Adeles as we didn’t consider infinite places, i will distinguish between the two notions by writing adeles resp. Adeles for the finite resp. the full blown version. The adele ring $\mathcal{A} $ has as a subring the infinite product $\mathcal{R} = \prod_p \hat{\mathbb{Z}}_p $. If you think of $\mathcal{A} $ as a version of $\mathbb{Q} $ then $\mathcal{R} $ corresponds to $\mathbb{Z} $ (and next time we will see that there is a lot more to this analogy).

The ideles are the group of invertible elements of the ring $\mathcal{A} $, that is, $\mathcal{I} = \mathcal{A}^{\ast} $. That s, an element is an infinite vector $i = (i_2,i_3,i_5,\ldots) = (i_p)_p $ with all $i_p \in \hat{\mathbb{Q}}_p^* $ and for all but finitely many primes we have that $i_p \in \hat{\mathbb{Z}}_p^* $.

As we will have to do explicit calculations with ideles and adeles we need to recall some facts about the structure of the unit groups $\hat{\mathbb{Z}}_p^* $ and $\hat{\mathbb{Q}}_p^* $. If we denote $U = \hat{\mathbb{Z}}_p^* $, then projecting it to the unit group of each of its components we get for each natural number n an exact sequence of groups

$1 \rightarrow U_n \rightarrow U \rightarrow (\mathbb{Z}/p^n \mathbb{Z})^* \rightarrow 1 $. In particular, we have that $U/U_1 \simeq (\mathbb{Z}/p\mathbb{Z})^* \simeq \mathbb{Z}/(p-1)\mathbb{Z} $ as the group of units of the finite field $\mathbb{F}_p $ is cyclic of order p-1. But then, the induced exact sequence of finite abalian groups below splits

$1 \rightarrow U_1/U_n \rightarrow U/U_n \rightarrow \mathbb{F}_p^* \rightarrow 1 $ and as the unit group $U = \underset{\leftarrow}{lim} U/U_n $ we deduce that $U = U_1 \times V $ where $\mathbb{F}_p^* \simeq V = { x \in U | x^{p-1}=1 } $ is the specified unique subgroup of $U $ of order p-1. All that remains is to determine the structure of $U_1 $. If $p \not= 2 $, take $\alpha = 1 + p \in U_1 – U_2 $ and let $\alpha_n \in U_1/U_n $ denote the image of $\alpha $, then one verifies that $\alpha_n $ is a cyclic generator of order $p^{n-1} $ of $U_1/U_n $.

But then, if we denote the isomorphism $\theta_n~:~\mathbb{Z}/p^{n-1} \mathbb{Z} \rightarrow U_1/U_n $ between the ADDITIVE group $\mathbb{Z}/p^{n-1} \mathbb{Z} $ and the MULTIPLICATIVE group $U_1/U_n $ by the map $z \mapsto \alpha_n^z $, then we have a compatible commutative diagram

[tex]\xymatrix{\mathbb{Z}/p^n \mathbb{Z} \ar[r]^{\theta_{n+1}} \ar[d] & U_1/U_{n+1} \ar[d] \\
\mathbb{Z}/p^{n-1} \mathbb{Z} \ar[r]^{\theta_n} & U_1/U_n}[/tex]

and as $U_1 = \underset{\leftarrow}{lim}~U_1/U_n $ this gives an isomorphism between the multiplicative group $U_1 $ and the additive group of $\hat{\mathbb{Z}}_p $. In case $p=2 $ we have to start with an element $\alpha \in U_2 – U_3 $ and repeat the above trick. Summarizing we have the following structural information about the unit group of p-adic integers

$\hat{\mathbb{Z}}_p^* \simeq \begin{cases} \hat{\mathbb{Z}}_{p,+} \times \mathbb{Z}/(p-1)\mathbb{Z}~(p \not= 2) \\ \hat{\mathbb{Z}}_{2,+} \times \mathbb{Z}/2 \mathbb{Z}~(p=2) \end{cases}$

Because every unit in $\hat{\mathbb{Q}}_p^* $ can be written as $p^n u $ with $u \in \hat{\mathbb{Z}}_p^* $ we deduce from this also the structure of the unit group of the p-adic field

$\hat{\mathbb{Q}}_p^* \simeq \begin{cases} \mathbb{Z} \times \hat{\mathbb{Z}}_{p,+} \times \mathbb{Z}/(p-1)\mathbb{Z}~(p \not= 2) \\ \mathbb{Z} \times \hat{\mathbb{Z}}_{2,+} \times \mathbb{Z}/2 \mathbb{Z}~(p=2) \end{cases} $

Right, now let us start to make the connection with the apparently abstract ringtheoretical post from last time where we introduced semigroup crystalline graded rings without explaining why we wanted that level of generality.

Consider the semigroup $\mathcal{I} \cap \mathcal{R} $, that is all ideles $i = (i_p)_p $ with all $i_p = p^{n_p} u_p $ with $u_p \in \hat{\mathbb{Z}}_p^* $ and $n_p \in \mathbb{N} $ with $n_p=0 $ for all but finitely many primes p. Then, we have an exact sequence of semigroups

$1 \rightarrow \mathcal{G} \rightarrow \mathcal{I} \cap \mathcal{R} \rightarrow^{\pi} \mathbb{N}^+_{\times} \rightarrow 1 $ where the map is defined (with above notation) $\pi(i) = \prod_p p^{n_p} $ and exactness follows from the above structural results when we take $\mathcal{G} = \prod_p \hat{\mathbb{Z}}_p^* $.

This gives a glimpse of where we are heading. Last time we identified the Bost-Connes Hecke algebra $\mathcal{H} $ as a bi-crystalline group graded algebra determined by a $\mathbb{N}^+_{\times} $-semigroup crystalline graded algebra over the group algebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $. Next, we will entend this construction starting from a $\mathcal{I} \cap \mathcal{R} $-semigroup crystalline graded algebra over the same group algebra. The upshot is that we will have a natural action by automorphisms of the group $\mathcal{G} $ on the Bost-Connes algebra. And… the group $\mathcal{G} = \prod_p \hat{\mathbb{Z}}_p^* $ is the Galois group of the cyclotomic field extension $\mathbb{Q}^{cyc} $!

But, in order to begin to understand this, we will need to brush up our rusty knowledge of algebraic number theory…

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