Author: lievenlb

F_un and braid groups

Published June 15, 2008 by lievenlb

Recall that an n-braid consists of n strictly descending elastic strings connecting n inputs at the top (named 1,2,…,n) to n outputs at the bottom (labeled 1,2,…,n) upto isotopy (meaning that we may pull and rearrange the strings in any way possible within 3-dimensional space). We can always change the braid slightly such that we can divide the interval between in- and output in a number of subintervals such that in each of those there is at most one crossing.

n-braids can be multiplied by putting them on top of each other and connecting the outputs of the first braid trivially to the inputs of the second. For example the 5-braid on the left can be written as $B=B_1.B_2 $ with $B_1 $ the braid on the top 3 subintervals and $B_2 $ the braid on the lower 5 subintervals.

In this way (and using our claim that there can be at most 1 crossing in each subinterval) we can write any n-braid as a word in the generators $\sigma_i $ (with $1 \leq i < n $) being the overcrossing between inputs i and i+1. Observe that the undercrossing is then the inverse $\sigma_i^{-1} $. For example, the braid on the left corresponds to the word

$\sigma_1^{-1}.\sigma_2^{-1}.\sigma_1^{-1}.\sigma_2.\sigma_3^{-1}.\sigma_4^{-1}.\sigma_3^{-1}.\sigma_4 $

Clearly there are relations among words in the generators. The easiest one we have already used implicitly namely that $\sigma_i.\sigma_i^{-1} $ is the trivial braid. Emil Artin proved in the 1930-ies that all such relations are consequences of two sets of ‘obvious’ relations. The first being commutation relations between crossings when the strings are far enough from each other. That is we have

$\sigma_i . \sigma_j = \sigma_j . \sigma_i $ whenever $|i-j| \geq 2 $

The second basic set of relations involves crossings using a common string

$\sigma_i.\sigma_{i+1}.\sigma_i = \sigma_{i+1}.\sigma_i.\sigma_{i+1} $

Starting with the 5-braid at the top, we can use these relations to reduce it to a simpler form. At each step we have outlined to region where the relations are applied

These beautiful braid-pictures were produced using the braid-metapost program written by Stijn Symens.

Tracing a string from an input to an output assigns to an n-braid a permutation on n letters. In the above example, the permutation is $~(1,2,4,5,3) $. As this permutation doesn’t change under applying basic reduction, this gives a group-morphism

$\mathbb{B}_n \rightarrow S_n $

from the braid group on n strings $\mathbb{B}_n $ to the symmetric group. We have seen before that the symmetric group $S_n $ has a F-un interpretation as the linear group $GL_n(\mathbb{F}_1) $ over the field with one element. Hence, we can ask whether there is also a F-un interpretation of the n-string braid group and of the above group-morphism.

Kapranov and Smirnov suggest in their paper that the n-string braid group $\mathbb{B}_n \simeq GL_n(\mathbb{F}_1[t]) $ is the general linear group over the polynomial ring $\mathbb{F}_1[t] $ over the field with one element and that the evaluation morphism (setting t=0)

$GL_n(\mathbb{F}_1[t]) \rightarrow GL_n(\mathbb{F}1) $ gives the groupmorphism $\mathbb{B}_n \rightarrow S_n $

The rationale behind this analogy is a theorem of Drinfeld‘s saying that over a finite field $\mathbb{F}_q $, the profinite completion of $GL_n(\mathbb{F}_q[t]) $ is embedded in the fundamental group of the space of q-polynomials of degree n in much the same way as the n-string braid group $\mathbb{B}_n $ is the fundamental group of the space of complex polynomials of degree n without multiple roots.

And, now that we know the basics of absolute linear algebra, we can give an absolute braid-group representation

$\mathbb{B}_n = GL_n(\mathbb{F}_1[t]) \rightarrow GL_n(\mathbb{F}_{1^n}) $

obtained by sending each generator $\sigma_i $ to the matrix over $\mathbb{F}_{1^n} $ (remember that $\mathbb{F}_{1^n} = (\mu_n)^{\bullet} $ where $\mu_n = \langle \epsilon_n \rangle $ are the n-th roots of unity)

$\sigma_i \mapsto \begin{bmatrix}
1_{i-1} & & & \\
& 0 & \epsilon_n & \\
& \epsilon_n^{-1} & 0 & \\
& & & 1_{n-1-i} \end{bmatrix} $

and it is easy to see that these matrices do indeed satisfy Artin’s defining relations for $\mathbb{B}_n $.

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Galois’ last letter

Published June 12, 2008 by lievenlb

“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”

Absolute linear algebra

Published June 10, 2008 by lievenlb

Today we will define some basic linear algebra over the absolute fields $\mathbb{F}_{1^n} $ following the Kapranov-Smirnov document. Recall from last time that $\mathbb{F}_{1^n} = \mu_n^{\bullet} $ and that a d-dimensional vectorspace over this field is a pointed set $V^{\bullet} $ where $V $ is a free $\mu_n $-set consisting of n.d elements. Note that in absolute linear algebra we are not allowed to have addition of vectors and have to define everything in terms of scalar multiplication (or if you want, the $\mu_n $-action). In the hope of keeping you awake, we will include an F-un interpretation of the power residue symbol.

Direct sums of vectorspaces are defined via $V^{\bullet} \oplus W^{\bullet} = (V \bigsqcup W)^{\bullet} $, that is, correspond to the disjoint union of free $\mu_n $-sets. Consequently we have that $dim(V^{\bullet} \oplus W^{\bullet}) = dim(V^{\bullet}) + dim(W^{\bullet}) $.

For tensor-product we start with $V^{\bullet} \times W^{\bullet} = (V \times W)^{\bullet} $ the vectorspace cooresponding to the Cartesian product of free $\mu_n $-sets. If the dimensions of $V^{\bullet} $ and $W^{\bullet} $ are respectively d and e, then $V \times W $ consists of n.d.n.e elements, so is of dimension n.d.e. In order to have a sensible notion of tensor-products we have to eliminate the n-factor. We do this by identifying $~(x,y) $ with $(\epsilon_n x, \epsilon^{-1} y) $ and call the corresponding vectorspace $V^{\bullet} \otimes W^{\bullet} $. If we denote the image of $~(x,y) $ by $x \otimes w $ then the identification merely says we can pull the $\mu_n $-action through the tensor-sign, as we’d like to do. With this definition we do indeed have that $dim(V^{\bullet} \otimes W^{\bullet}) = dim(V^{\bullet}) dim(W^{\bullet}) $.

Recall that any linear automorphism $A $ of an $\mathbb{F}_{1^n} $ vectorspace $V^{\bullet} $ with basis ${ b_1,\ldots,b_d } $ (representants of the different $\mu_n $-orbits) is of the form $A(b_i) = \epsilon_n^{k_i} b_{\sigma(i)} $ for some powers of the primitive n-th root of unity $\epsilon_n $ and some permutation $\sigma \in S_d $. We define the determinant $det(A) = \prod_{i=1}^d \epsilon_n^{k_i} $. One verifies that the determinant is multiplicative and independent of the choice of basis.

For example, scalar-multiplication by $\epsilon_n $ gives an automorphism on any $d $-dimensional $\mathbb{F}_{1^n} $-vectorspace $V^{\bullet} $ and the corresponding determinant clearly equals $det = \epsilon_n^d $. That is, the det-functor remembers the dimension modulo n. These mod-n features are a recurrent theme in absolute linear algebra. Another example, which will become relevant when we come to reciprocity laws :

Take $n=2 $. Then, a $\mathbb{F}_{1^2} $ vectorspace $V^{\bullet} $ of dimension d is a set consisting of 2d elements $V $ equipped with a free involution. Any linear automorphism $A~:~V^{\bullet} \rightarrow V^{\bullet} $ is represented by a $d \times d $ matrix having one nonzero entry in every row and column being equal to +1 or -1. Hence, the determinant $det(A) \in \{ +1,-1 \} $.

On the other hand, by definition, the linear automorphism $A $ determines a permutation $\sigma_A \in S_{2d} $ on the 2d non-zero elements of $V^{\bullet} $. The connection between these two interpretations is that $det(A) = sgn(\sigma_A) $ the determinant gives the sign of the permutation!

For a prime power $q=p^k $ with $q \equiv 1~mod(n) $, we have seen that the roots of unity $\mu_n \subset \mathbb{F}_q^* $ and hence that $\mathbb{F}_q $ is a vectorspace over $\mathbb{F}_{1^n} $. For any field-unit $a \in \mathbb{F}_q^* $ we have the power residue symbol

$\begin{pmatrix} a \\ \mathbb{F}_q \end{pmatrix}_n = a^{\frac{q-1}{n}} \in \mu_n $

On the other hand, multiplication by $a $ is a linear automorphism on the $\mathbb{F}_{1^n} $-vectorspace $\mathbb{F}_q $ and hence we can look at its F-un determinant $det(a \times) $. The F-un interpretation of a classical lemma by Gauss asserts that the power residue symbol equals $det(a \times) $.

An $\mathbb{F}_{1^n} $-subspace $W^{\bullet} $ of a vectorspace $V^{\bullet} $ is a subset $W \subset V $ consisting of full $\mu_n $-orbits. Normally, in defining a quotient space we would say that two V-vectors are equivalent when their difference belongs to W and take equivalence classes. However, in absolute linear algebra we are not allowed to take linear combinations of vectors…

The only way out is to define $~(V/W)^{\bullet} $ to correspond to the free $\mu_n $-set $~(V/W) $ obtained by identifying all elements of W with the zero-element in $V^{\bullet} $. But… this will screw-up things if we want to interpret $\mathbb{F}_q $-vectorspaces as $\mathbb{F}_{1^n} $-spaces whenever $q \equiv 1~mod(n) $.

For this reason, Kapranov and Smirnov invent the notion of an equivalence $f~:~X^{\bullet} \rightarrow Y^{\bullet} $ between $\mathbb{F}_{1^n} $-spaces to be a linear map (note that this means a set-theoretic map $X \rightarrow Y^{\bullet} $ such that the invers image of 0 consists of full $\mu_n $-orbits and is a $\mu_n $-map elsewhere) satisfying the properties that $f^{-1}(0) = 0 $ and for every element $y \in Y $ we have that the number of pre-images $f^{-1}(y) $ is congruent to 1 modulo n. Observe that under an equivalence $f~:~X^{\bullet} \rightarrow Y^{\bullet} $ we have that $dim(X^{\bullet}) \equiv dim(Y^{\bullet})~mod(n) $.

This then allows us to define an exact sequence of $\mathbb{F}_{1^n} $-vectorspaces to be

[tex]\xymatrix{0 \ar[r] & V_1^{\bullet} \ar[r]^{\alpha} & V^{\bullet} \ar[r]^{\beta} & V_2^{\bullet} \ar[r] & 0}[/tex]

with $\alpha $ a set-theoretic inclusion, the composition $\beta \circ \alpha $ to be the zero-map and with the additional assumption that the map induced by $\beta $

$~(V/V_1)^{\bullet} \rightarrow V_2^{\bullet} $

is an equivalence. For an exact sequence of spaces as above we have the congruence relation on their dimensions $dim(V_1)+dim(V_2) \equiv dim(V)~mod(n) $.

More importantly, if as before $q \equiv 1~mod(n) $ and we use the embedding $\mu_n \subset \mathbb{F}_q^* $ to turn usual $\mathbb{F}_q $-vectorspaces into absolute $\mathbb{F}_{1^n} $-spaces, then an ordinary exact sequence of $\mathbb{F}_q $-vectorspaces remains exact in the above definition.

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