

07212017, 05:28 PM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
Now for something different. Abstract algebra. In particular, group theory.
I start with a binary operation * over some set of entities S, an operation that is closed over it: S*S = S.
I add associativity and I get a semigroup: (a*b)*c = a*(b*c). Commutativity is optional.
I add an identity element and I get a monoid: element e such that a*e = e*a = a.
It is easy to prove that an identity element is unique  a monoid cannot contain more than one distinct identity.
I add inverses and I get a group: for all a in S, an inverse inv(a) satisfying a*inv(a) = inv(a)*a = e.
Commutative groups are usually called abelian groups.
One can easily make new groups from products of existing ones:
(a1,a2) * (b1,b2) = (a1*b1, a2*b2)

The integers under addition form a group, as do the rational numbers under addition, the real numbers under addition, the nonzero rational numbers under multiplication, the nonzero real numbers under multiplication, etc.
What are the identities and the inverses in those groups?
The group of integers under addition is called Z, and that of integers modulo n Z[sub]n[/sup] or Z/nZ or Z(n). One can picture the integers with a numberline, and for the modulo case, one can wrap part of that line around so that its ends join, making a circle.
The groups Z(n) are called cyclic groups, for obvious reasons. An element a generates it if every element in it is some power of a. That is, every element can be found from some a*a*...*a.
Every finite abelian (commutative) group is a product of cyclic groups, and cyclic groups themselves may be decomposed further with the help of the Chinese Remainder Theorem:
For n1 and n2 relatively prime (coprime): Z(n1*n2) = Z(n1)*Z(n2).
That means that groups Z(pn) for prime p cannot be reduced further in this fashion, and that every finite abelian group is thus the product of such groups.

Let us now consider subgroups. A group G has a group H as a subgroup if H has the same operation and if H's elements are all in G's elements. Now consider the order or number of elements of a group. Lagrange's theorem states that (order of H) evenly divides (order of G).
This constrains what subgroups a group can have. Every group has two trivial subgroups, itself and the identity group {e} or Z(1). The group Z(n) thus only has subgroups Z(m) where m evenly divides n. For n a prime, Z(n) only has the two trivial subgroups.
The order of an element a in a group is the number of a's n in an = a*a*...*a that give e such that no smaller number does so. It's easy to see that a generates cyclic group Z(n), and by Lagrange's theorem, n must divide the order of the group.
If a generates Z(n), then some power of a must also generate a cyclic group, either Z(n) itself or some Z(m) where m evenly divides n.

One can prove Lagrange's theorem by constructing cosets of a subgroup H of G:
Left cosets: a1*H, a2*H, a3*H, ...
Right cosets: H*a1, H*a2, H*a3, ...
Each side's cosets are disjoint and they all cover G.
If (left cosets) = (right cosets), then H is called a normal subgroup of G. In this case, one can define a coset multiplication law: (a1*H) * (a2*H) = (a3*H), and with the cosets, it forms a group. This is the quotient group or factor group of G and H: G/H.
I'll find these groups in a simple case: Z(4). Its elements are {e, a, a2, a3}. It has subgroup Z(2): {e, a2}. Its cosets are (e) = {e, a2} and (a) = {a, a3}, with multiplication law
(e) * (e) = (e)
(e) * (a) = (a)
(a) * (e) = (a)
(a) * (a) = (e)
Thus being Z(2). So, Z(4) / Z(2) = Z(2).
Likewise, the integer group Z has a normal subgroup, the even numbers: 2*Z. It has cosets 2*Z and the odd numbers 2*Z+1. From
even + even = even
even + odd = odd
odd + even = odd
odd + odd = even,
it is evident that Z / (2*Z) = Z(2).
In fact, every subgroup with only two cosets is a normal subgroup.
I think I'll stop here.

07222017, 11:48 PM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
Consider groups of matrices under matrix multiplication. Associativity: automatic. Identity: the identity matrix. Inverses: matrix inverses  the matrices must be nonsingular.
A group homomorphism is some function f of a group's elements that acts as a group and that satisfies f(a)*f(b) = f(a*b).
Its kernel is all group elements that map onto the identity of the new group.
A function that takes determinants of matrices is a homomorphism of a matrix group, with its kernel being all matrices with determinant 1.
det(A)*det(B) = det(A.B)
For a group of some subset of complex numbers under multiplication, like nth roots of unity, f(a) = an for integer n is a homomorphism.
Groups may be realized as sets of matrices, D(a) for element a. A matrix group can be realized with a different set of matrices, for instance. Such matrices are called representations, and those that cannot be reduced to combinations of representations irreducible ones or irreps.
More formally, consider some X that satisfies X.D(a) = D(a).X for all a in the group. If every such X is a multiple of the identity matrix, then the D's are an irrep.
There is a trivial rep for every group, the identity rep: D(a) = 1.
For abelian groups, all their irreps have dimension 1.
For nonabelian groups, at least one irrep has a dimension more than 1.
A rep of a group can be a rep of any of its quotient groups.
For group G,
a*G = (permutation of G)
So every group can be realized as a permutation group. The permutations form a representation:
D(a)cb = 1 if c = a*b, 0 otherwise
This is the "regular representation", and it is irreducible only for the identity group. It contains n copies of every irrep, where n is the irrep's dimension.
I'll now make a table of the irreps of a few groups.
Z(2):
e, a
1, 1
1, 1
Z(4):
e, a, a2, a3
1, 1, 1, 1
1, i, 1, i
1, 1, 1,1
1,i, 1, i
Z(2)*Z(2)  the 4group or Viergruppe
e, a, b, a*b
1, 1, 1, 1
1, 1, 1, 1
1, 1, 1, 1
1, 1, 1, 1

08152017, 08:29 PM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
Now for groups of rotations and reflections. For a ndimensional space, the group is called the orthogonal group O(n), and for reflections only, the special orthogonal group SO(n).
Members of O(n) have determinant values +1 or 1, while SO(n) has +1 only. The +1 means pure rotations, while 1 means rotations and reflections, sometimes called improper rotations. One might ask why not a count of reflections? Two reflections are equivalent to one rotation, so the 1 case means an odd number of reflections and +1 an even number.
So O(n) divides up into {pure rotations: O(n), rotations + reflections}
It is easy to show that the quotient group of this decomposition is Z(2):
rot * rot = rot
rot * rfl = rfl
rfl * rot = rfl
rfl * rfl = rot
This group's elements M satisfy M.MT = MT.M = I where the T means transpose.
For 1 dimension, it's very easy: SO(1) = {1} and O(1) = {1,1}
For 2 dimensions, the elements of SO(2) have the form {{cos(a), sin(a)}, {sin(a), cos(a)}} for all a between 0 and 2pi. The extra elements of O(2) have the form {{cos(a), sin(a)}, {sin(a), cos(a)}}.
For 3 dimensions, the elements of SO(3) are composed from a unit quaternion vector {q0,q1,q2,q3}, "unit" meaning
Σi=03 (qi)2 = 1
Those elements are Mij = ((q0)2  Σi=13 (qi)2) δij + 2*qi*qj + 2*q0*εijkqk
For O(3), the additional ones are  (SO(3) elements).
Note that M is a sort of square of q, that both q and q map onto the same M.
For SO(4), one must use a pair of quaternions, q1 and q2, with M being (linear in q1).(linear in q2). To get the O(4) extras, one has to multiply the SO(4) elements by something like diag({1,1,1,1}), since I is in SO(4).

08162017, 01:59 AM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
There don't seem to be any simple expressions for SO(5) and higher.
There's a complexnumber version of orthogonal matrices called unitary ones. For these, we need the "Hermitian conjugate" of a matrix:
M+ = MT*
Complex conjugate of transpose (or transpose of complex conjugate). A matrix equal to its Hermitian conjugate is called "Hermitian" (M = M+), and if equal to minus its H.C. "antiHermitian" (M =  M+). If a matrix satisfies
M.M+ = M+.M = I
then its called "unitarity". Its determinant has absolute value 1.
For orthogonal matrices, O(n) = {1,1} * SO(n) only for odd n, but for unitary matrices, U(n) = U(1) * SU(n) for all n.
So SU(1) is {1} and U(1) is {ei*a for all a in 0 to 2pi}. In fact, U(1) ~ SO(2).
There is a simple expression for SU(2) matrices in terms of quaternions: {{q0 + i*q3, i*q1 + q2}, {i*q1  q2, q0  i*q3}}
This is equal to q0*(identity matrix) + i*q1,2,3.(Pauli matrices of quantum mechanics)
Thus, quaternions can be expressed as complex 2*2 matrices of their coefficients.
So 3D and 4D rotations are related to these matrices: SO(3) ~ SU(2) / Z(2). Likewise, SO(4) ~ SU(2) * SU(2) / Z(2).

There is an additional sort, "symplectic", defined with the help of antisymmetric matrix J, a matrix that can be conveniently expressed as {{0, I}, {I, 0}} for dimension 2n in terms of dimensionn parts. Elements:
M.J.M+ = J
The group is called Sp(2n).
There is a nice series of smalldimension matrixgroup interrelationships:
SO(2), U(1)
SO(3), SU(2), Sp(2)
SO(4), SU(2)*SU(2)
SO(5), Sp(4)
SO(6), SU(4)

08162017, 04:33 AM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
There don't seem to be any simple ways of expressing SU(3) values and higher, just as for SO(5) and higher.
I'll now take on the finite subgroups of O(n) and SO(n) for small n. For n = 1, rotations and reflections of a line (1space), it is very easy.
For n = 2 (2space or plane), the symmetry groups for a psided regular polygon are C(p) (cyclic) for pure rotations and D(p) ("dihedral") for rotations and reflections. I have some demos of these groups in these pages:  C(p) contains all rotation angles 2pi*{0, 1/p, 2/p, ..., (p1)/p}
 D(p) contains C(p) and reflections with reflection lines separated by angle pi/p.
As an example, let's consider a square. Its symmetry group is D(4), with rotation angles 0d, 90d, 180d, and 270d, and reflection lines parallel to pairs of sides and along the two diagonals.
Turn it into a rectangle. Its symmetry goes down to D(2), with rotation angles 0d and 180d, and reflection lines parallel to pairs of sides.
Then a parallelogram. Its symmetry goes down to C(2) with rotation angles 0d and 180d.
Alternately, then a trapezoid with the inclined sides inclined by the same amount. Its symmetry goes down to D(1) with rotation angle 0d and a reflection line between the two tilted sides.
No symmetry is C(1), rotation angle 0d  the identity group.

08162017, 05:39 AM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
So in summary, there are two infinite families of finite 2D rotationreflection groups: C(p) and D(p).
Not surprisingly, the 3D case is much more complicated than the 2D case.
The first type of group is the axial or prismatic groups. They are extensions of the 2D case, with inplane rotations and reflections, and also acrossplane reflections or flips. Here they are, in Schoenflies notation:
C(p): 2D C(p).
C(p,h): C(p) and it with flipping
S(2p): C(2p) with elements alternately flipped and nonflipped
C(p,v): 2D D(p)
D(p): C(p,v) with the inplane reflected ones also flipped
D(p,h): C(p,v) and it with flipping
D(p,d): C(2p,v) with elements alternately flipped and nonflipped
There are also some quasispherical or polyhedral symmetry groups.
T: rotations of a regular tetrahedron
Th: T and it with inversion (multiply by 1)
Td: rotations and reflections of a regular tetrahedron
O: rotations of a regular octahedron
Oh: rotations and reflections of a regular octahedron: O and it with inversion
I: rotations of a regular icosahedron
Ih: rotations and reflections of a regular icosahedron: I and it with inversion
There is a nice symmetry here: 7 infinite families of axial groups and 7 quasispherical groups.
But the groups C(1), S(2), D(2), and D(2,h), though axial, also have quasispherical properties.
C(1): identity group: {I}
S(2): identity and inversion: {I, I}
D(2): diagonal matrices for rotation: {{1,1,1}, {1,1,1}, {1,1,1}, {1,1,1}}
D(2,h): diagonal matrices for rotation and reflection: {{1,1,1}, {1,1,1}, {1,1,1}, {1,1,1}, {1,1,1}, {1,1,1}, {1,1,1}, {1,1,1}}

08162017, 06:21 AM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
The quaternion versions of the 3D rotation groups are very interestinglooking.
The cyclic one:
QC = all {cos(a), 0, 0, sin(a)}
where a finite one can have an odd number of a's as well as an even number.
QC(2m) > C(m)
QC(2m+1) > C(2m+1)
The dihedral one:
QD = QC + all {0, cos(a), sin(a), 0}
where a finite one must have an even number of a's.
QD(m) > D(m)
QD(2) is {+1, 0, 0, 0}, {0, + 1, 0, 0}, {0, 0, +1, 0}, {0, 0, 0, +1}
Tetrahedral: QT = QD(2) + all sign combinations of {+1, +1, +1, +1}/2
Octahedral: QO = QT + all permutations and sign combinations of {+1, +1, 0, 0}/sqrt(2)
Icosahedral: QI = QT + all *even* permutations and all sign combinations of {+ (sqrt(5)+1)/4, + 1/2, + (sqrt(5)1)/4, 0}

08172017, 12:57 AM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
The most general group for n dimensions is the "general linear" group, the group of every possible invertible matrix with the values in some suitable domain. GL(n,R) is for the real numbers, GL(n,C) is for the complex numbers, and GL(n,F) is for algebraic field F in general.
It has some subgroups.
Identifying multiples of some element M with each other, making sets {a*M}, one gets the "projective linear" group PGL(n,F). It is the quotient group of GL(n,F) with {a*I} where a is nonzero and in F.
The "special linear" group SL(n,F) has all elements of GL(n,F) with determinant 1, and the "projective special linear" group PSL(n,F) is like PGL(n,F), but based on SL(n,F), and with an = 1.

There is a relationship between GL(n,R) and U(n), and also between SL(n,R) and SU(n), but it's a rather long story.

08172017, 10:40 AM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
So far, I've been doing "point groups" with elements R turning x into x':
x' = R.x
Adding a translation / shift / displacement D gives
x' = R.x + D
This is the Euclidean group E(n) or Euc(n), named after the famous geometer. Its elements may also be expressed in matrix form: {{R, D}, {0, 1}}.
It looks related to O(n+1), and it is  O(n+1) is the symmetry group of a hypersphere with a nD surface. 1D: circle, 2D: sphere, ... A hyperbolic surface also has similar symmetries, though it's a bit of a long story.
Let's now consider different numbers of dimensions of variation of D. If zero dimensions, then we are back at point groups again. If less than n, then R splits into R1 * R2, where R1 is for D's dimensions and R2 is for the others.
It's easy to show that the Donly group (R = identity matrix) is a normal subgroup of the Euclidean group. Its quotient group is the Ronly group.

08172017, 02:16 PM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
An important class of subgroups of the Euclidean group is lattice groups, groups where D is restricted to points on a lattice. That is,
D = D0(R) + k1*v1 + k2*v2 + ...
where the lattice vectors v1, v2, ... are linearly independent.
For 1D, I'll distinguish the point groups as C: {1} and D: {1, 1}. The only nonvanishing lattice is v*k, and we can set v = 1 without loss of generality. Thus, the two lattice groups are:
C(lat): x' = x + k
D(lat): C(lat) with x' = x + k
For 2D space and a 1D lattice, one has the 7 "frieze groups".
Code:
b b b b
b b b b
p p p p
b b b b
p p p p
bdbdbdbd
b b b b
q q q q
bdbdbdbd
pqpqpqpq
bdbdbdbd
qpqpqpqp
The 7 infinite families of 3D point groups are all wraparound versions of these groups.
For a 2D lattice, one has the 17 "wallpaper groups".
Code:
b bq bd bp <bd>
bd bd bp /bd
pq qp dq pq/
There are 3 square ones and 5 triangular and hexagonal ones, and these ones are much more difficult to do with ASCII art.
The wallpaper groups' possible point groups are C1, C2, C3, C4, C6, D1, D2, D3, D4, D6, a result of the "crystallographic restriction theorem". Notice how it skips over the pentagonal ones, C5 and D5.
The frieze groups' possible point groups are even more restricted: C1, C2, D1, D2.
The five possible lattice types for wallpaper groups are parallelogram, rectangular, rhombic, square, and triangular/hexagonal. Of these, the rhombic one can be formed from the rectangular one by adding points in the rectangle centers to the rectangle vertices. That makes the rhombic lattice a facecentered variant of the rectangular lattice. So combining these two makes four.
It is easy to find examples of all of these groups online.

08172017, 02:51 PM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
In three dimensions, there are many more solutions, and more complicated ones at that ( Space group  Wikipedia). There are 32 3D "crystallographic point groups":
C1, c2, C3, C4, C6
C1h, C2h, C3h, C4h, C6h
S2, S4, S6
C1v=C1h, C2v, C3v, C4v, C6v
D1=C2, C2, C3, C4, C6
D1h=C2v, D2h, D3h, D4h, D6h
D1d=C2h, D2d, D3d
T, Th, Td, O, Oh
The lack of icosahedral groups is related to the lack of pentagonal groups.
There are 14 kinds of crystal lattices, "Bravais lattices", as they are called. Grouping together plain, facecentered, and bodycentered variants gives 7 "crystal systems". Grouping together rhombohedral and hexagonal crystal lattices gives 6 "crystal families": triclinic, monoclinic, orthorhombic, tetragonal, hexagonal (with rhombohedral), and cubic.  Triclinic: the "unit cell" is a 3D parallelogram with sides different lengths.
 Monoclinic: like triclinic, with two pairs of faces rectangular.
 Orthorhombic: a 3D rectangle.
 Tetragonal: like orthorhombic, but one pair of faces square.
 Rhombohedral: all sides equally long, all angles at two vertices equal.
 Hexagonal: like monoclinic, but parallelogram has 60d and 120d angles.
 Cubic: as it says.
There is a total of 230 3D space groups, or 219 if one counts mirror images together.

Quasicrystal  Wikipedia
QuasiPeriodic Crystals—The Long Road from Discovery to Acceptance
Introduction to Quasicrystals
Such was the success of this spacegroup paradigm that when fivefoldsymmetric diffraction patterns were observed for some materials, they seemed impossible. But what was going on there was that these materials had pentagonal / icosahedral shortrange order, even if they had no longrange order and were thus not true crystals.
Some scientists refused to believe that there is a such a thing as quasicrystals, including a very eminent one: Linus Pauling. But their existence is now accepted, and their discoverers have received a Nobel prize for that.

08182017, 04:13 AM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
Let's look at more operations. Consider an analog of addition, +, and an analog of multiplication, *, acting on a set. This is a "ring" if it satisfies these conditions:  Addition forms an abelian group (identity 0, inverse of a: a)
 Multiplication is associative
 Multiplication has an identity (1)
 Multiplication is distributive over addition
Some mathematicians define a ring as not necessarily having a multiplicative identity. Relative to a ring with one, such a ring is sometimes called a rng ("rung"; no i) or a pseudoring. But I'll use the multiplicative identity here.
The additive identity, 0, is the multiplicative zero.
A ring with 0 = 1 is the "zero ring", with only one element.
Every ring contains either the integer ring, Z, or else the ring of integers modulo some number n, Z(n). This subring commutes with all the other elements in the ring.

Wikipedia lists a chain of ringlike objects starting with commutative rings and ending with algebraic fields.  Commutative rings: multiplication is commutative.
 Integral domain: no zero divisors. a*b = 0 implies a = 0 or b = 0. An integral domain has the "cancellation property": a*b = a*c implies b = c for all a. It's easy to prove that this follows from the lack of zero divisors.
 Integrally closed domains  rather complicated.
 A GCD domain has the property that any two elements have a nonzero greatest common divisor: a and b have g such that a = g*a1 and b = g*b1.
 A unique factorization domain has the property that every element other than 0 can be written as a unique product of a unit element (1, 1, etc.) and some "prime elements". Here is a ring that is not a UFD: the ring of numbers with the form a + b*sqrt(5), where a and b are integers. 6 = 2*3 = (1 + sqrt(5))*(1  sqrt(5)).
 A principal ideal domain has the property that every ideal is principal, generated by only one element. A left ideal of a ring: (ideal)*(ring) = (ideal). A right ideal: (ring)*(ideal) = (ideal). If both sides, then a twosided ideal. An ideal of a commutative ring is always a twosided ideal. The integer ring Z has ideals n*Z, generated by n.
 A Euclidean domain has a generalization of Euclid's GCD algorithm in it.
 An algebraic field has the property that multiplication over all elements but 0 is an abelian group.
Though the integers do not form a field, the rational numbers, the algebraic numbers, the real numbers, and various other sorts of numbers are all fields.
A finite field is a field with a finite number of elements. Sometimes called Galois fields, after the famous mathematician Évariste Galois (no, he didn't write down all his mathematical discoveries the night before his fatal duel).

08182017, 08:02 AM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
I'll now discuss finite fields in more detail. The theory of them I find very elegant.
There is only one for each order or number of elements: GF(pn), where p is a prime number. The simplest one is GF(p), and that one is Z(p). For the rest, we introduce variable x. Their elements are polynomials in x with coefficients in Z(p) with degree from 0 to n1.
It may not seem possible to do multiplication, since that may give orders greater than n1. But there is a way out of that difficulty. Interpret x as the solution of equation P(x) = 0, where P is a degreen polynomial that cannot be factored in the field. It is thus an "irreducible" or "primitive" polynomial, and one can use any of the possible ones  their results are isomorphic.
So after doing multiplication, one must divide by P(x) and take the remainder.
The addition group of GF(pn) is (Z(p))n, while the multiplication group of its nonzero elements is Z(pn11).
A field GF(pn) has subfields GF(pm), where m evenly divides n. GF(p) is always present, of course.
Here are some examples:
GF(p)  irreducible polynomials x + k (0 <= k < p)
GF(4): 0, 1, x, x+1  irreducible polynomial x2 + x + 1
GF(8): 0, 1, x, x+1, x2, x2+1, x2+x, x2+x+1  irreducible polynomials x3+x+1, x3+x2+1
GF(9): 0, 1, 2, x, x+1, x+2, 2x, 2x+1, 2x+2  irreducible polynomials x2+1, x2+x+2, x2+2x+2

08182017, 10:16 AM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
I'll now turn to automorphisms, mapping of objects onto themselves. Automorphisms of algebraic fields turn out to have some very interesting applications.
For GF(p), the only automorphism is the trivial one, the identity one. For GF(4), there is also x > x+1. It interchanges x and x+1, and forms a version of Z(2). More generally, the "Frobenius automorphism" always exists. It is x > xp, and it generates a cyclic group of n automorphisms: Z(n). That GF(4) automorphism is a version of that field's Frobenius one.
Now for infinite fields. The simplest one is Q, the rational numbers. It only has the identity automorphism. That is also true of the real algebraic numbers, and of the real numbers. But complex numbers have a nontrivial automorphism: complex conjugation, i > i. Not surprisingly, the automorphism group here is Z(2).
Let's now extend fields with polynomial roots. First, Q with the roots of
x2  3x + 2 = 0
Those roots are 1 and 2, so we get Q again. Now try
x2  2 = 0
One gets + sqrt(2), and sqrt(2) is not a rational number. Thus, we get extension field Q(sqrt(2)) with elements a + b*sqrt(2).
It has the automorphism sqrt(2) > sqrt(2) and its automorphism group is Z(2).
Let's now take on cube roots. x3  2 = 0 has roots 21/3, w*21/3, and w2*21/3, where w = (1 + sqrt(3))/2. Its automorphism group is D(3), the trianglesymmetry group.
The general case is more complicated. In general, the roots of xn = a are wk*a1/n, where w is a "primitive" root of unity: wn = 1 but that is not true for any smaller positive power.
If a = 1, then it's easy. w has automorphisms w > wb, where b is relatively prime to n. Thus, the automorphism group of the nth roots of unity is Z*(n), the integermultiply group modulo n. Since it is a group, it does not contain 0 or any divisors of n. Since it is abelian, it is a product of cyclic groups.
If a1/n is not a rational number, then xk = x0*(x1/x0)k giving automorphisms xk = xa * (xa+b/xa)k for some a and b. This is equivalent to
(new k) = a + b*k mod n
Here, a is from 0 to n1, and b is relatively prime to n. This group has a normal subgroup (all a's, b=1: Z(n)) with quotient group (b's: Z*(n)).
In summary, for nth roots, the extension symmetry group can be broken down into combinations of cyclic groups, both as products and as quotient groups. This turns out to be important for determining what equations can be solved with what techniques.
For square roots, the group is Z(2), while for cube roots in general, it is D(3), with normal subgroup Z(3) and quotient group Z(2).

08182017, 10:57 AM


[thanks] whisperer


Join Date: Jul 2004
Location: England/Miisaland
Gender: Male


Re: Math trivia
I think we're beyond "trivia".
I learned all this stuff and have forgotten it.

08182017, 01:52 PM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
What I described for extending the rational numbers also works for any superset of those numbers, and the extensionautomorphism groups are called "Galois groups".
Évariste Galois also proved an important theorem. Consider a field F with extension E. Its Galois group is the automorphism group of E while keeping F fixed, and is denoted Gal(E/F). Consider an extension of E called D. Then the Galois groups are related:
Gal(D/F) has normal subgroup Gal(D/E) with quotient group Gal(E/F)

A polynomial equation has maximum Galois group Sym(n), the symmetric group, the group of all permutations of n symbols. In this case, the n roots.
So if it is to be solved with radicals (nth roots), then its Galois group must be decomposable into cyclic groups. That is, "solvable". This is true in general for n = 2, 3, and 4, but not for n = 5 and higher.
So one can't solve a general quintic or higher using nth roots.

In the tradition of Euclidean geometry, only ruler and compass are allowed, and they are equivalent to doing arithmetic and square roots. This means that the Galois group of a polynomial equation for a constructible object ought to be decomposable to Z(2)'s only among Z(prime)'s.
Two traditional problems, duplication of the cube and trisection of the angle, both involve solving cubic equations with fullscale Galois groups: D(3). It decomposes into Z(2) and Z(3), and the latter makes rulerandcompass constructions impossible.
For constructing an ngon, we must find primitive nth roots of unity, and for that, we must consider how Z*(n) decomposes. Its order is (Euler phi function)(n), the product of pm1*(p1) for primepower factor pm of n.
So for ruler and compass, the only constructible ngons are those with n having form 2m * (product of distinct primes with form 2k+1)
For 2k+1 to be prime, k must have no odd factors, making the primes 22^k+1, the Fermat primes.
Only five of them are known: 3, 5, 17, 257, 65537, and there is a conjecture that states that these are the only "Fermat numbers" that are prime. Every other one that has been tested has turned out to be composite.

Archimedes introduced a neusis, or marked ruler, and it is possible to solve cubic and quartic equations with it  anything whose Galoisgroup decompositions include both Z(2) and Z(3).
This makes possible construction of duplication of the cube and trisection of the angle, and also ngons with n having form 2m1 * 3[/sup]m2[/sup] * (product of distinct primes with form 2k1*3k2+1)
Those primes are called "Pierpoint primes", and here are the first few of them:
2, 3, 5, 7, 13, 17, 19, 37, 73, 97, 109, 163, 193, 257, 433, 487, 577, 769, 1153, 1297, 1459, 2593, 2917, 3457, 3889, 10369, 12289, 17497, 18433, 39367, 52489, 65537, 139969, 147457, 209953, 331777, 472393, 629857, 746497, 786433, 839809, 995329, ...
Though some very large ones have been found, it is not known whether or not there is an infinite number of them.

08182017, 11:27 PM


puzzler


Join Date: Aug 2004
Location: UK


Re: Math trivia
In 1856, Norman Robert Pogson defined the system we use to measure the apparent brightness of stars, as seen from here on Earth.
He came up with the wonderfully arcane system where a higher number means a dimmer star, and each whole number of the scale indicates a star that is 2.512 times dimmer than the number before. The number 2.512 is now known as Pogson's ratio in his honor.
So a magnitude 4 star is 2.512 times dimmer than a magnitude 3 star, and a magnitude 5 star is another 2.512 times dimmer  so that's a total of 2.512 squared or 6.31 times dimmer. On this scale, the brightest object in the sky, the Sun, has an apparent magnitude of 27.
In fairness to Pogson, he didn't invent this system just to be awkward  he was just trying to rationalize a system devised by the Ancient Greeks who had a scale from 1 to 6 for stars in the night sky where magnitude 1 stars were the brightest and magnitude 6 stars were the faintest that could be seen with the naked eye  which, of course, was the only observing instrument available to the Greeks at that time.
The Greeks reckoned that each magnitude represented a halving in the brightness of the stars, which would make a magnitude 6 star 1/32 as bright as a magnitude 1 star, but by Pogson's time there were instruments that could measure the actual amount of light received from stars and he found that the stars called magnitude 6 by the Greeks were nearer a hundred times dimmer than the brightest magnitude 1 stars.
So Pogson decreed that a magnitude 1 star was exactly 100 times brighter than a magnitude 6 one, and that's where his ratio of 2.512 comes from  to make the 6 different magnitudes cover a range of one to one hundredth, the ratio between each magnitude number has to be the fifth root of a hundred.
Last edited by ceptimus; 08182017 at 11:39 PM.

08192017, 06:29 AM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
I'd mentioned two of three problems that mathematicians had been unable to solve by rulerandcompass methods. Here they are, along with the third one.  Duplication of the cube: find a cube size that gives twice the volume of some cube. Equivalent to finding 21/3.
 Trisecting the angle: find an angle that is 1/3 of some angle. Equivalent to solving 4*x3  3*x = cos(a) for x given a, where x = cos(a/3).
 Squaring the circle: find a square whose area is equal to the area of some circle. Equivalent to finding sqrt(pi).
The first two were shown to be insoluble by that complicated Galoisgroup method that I had posted on just before this. It's moreorless that cube roots can't be turned into square roots.
The third one is a result of pi being transcendental, and the proof of that is rather complicated: Lindemann–Weierstrass theorem  Wikipedia
Also from antiquity is what seems to me like a conjecture of unsolvability. Euclid's fifth postulate or parallel postulate:
Quote:
If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.

For over 2000 years, that cumbersomelooking axiom stood out like a sore thumb in Euclid's Elements, with numerous mathematicians trying to prove it from Euclid's other axioms, and failing. A common feature of such attempted proofs was including something equivalent to Euclid's fifth.
But in the mid 19th cy., some mathematicians proved that it was independent of Euclid's other axioms, and that one could substitute some contrary ones without causing inconsistency. That was the beginning of nonEuclidean geometry.

08192017, 06:46 AM


Member


Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male


Re: Math trivia
The LindemannWeierstrass theorem states that for a set of distinct (complex) algebraic numbers {ai}, there is no set of (complex) algebraic numbers {bi} with at least one nonzero such that
b1*ea1 + b2*ea2 + ... + bn*ean = 0
These powers of e are thus linearly independent over the algebraic numbers. e is the base of the natural logarithms, 2.718...
Let's first see about e. Is it algebraic? Consider ai values 0 and 1: b0 + b1*e. Since this is always nonzero, that means that e is not algebraic, that is, transcendental.
Now let's see about pi. We use DeMoivre's theorem: 1 + epi*i = 0. Since 1 is obviously algebraic, this means that pi*i cannot be algebraic. Since i is also algebraic, that means that pi is transcendental.

Currently Active Users Viewing This Thread: 1 (0 members and 1 guests)


Thread Tools 

Display Modes 
Linear Mode

Posting Rules

You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts
HTML code is Off



All times are GMT +1. The time now is 08:25 AM.



