The Quaternion group

 The concept of a square root of a negative number dates back to ancient Greece, and is credited to Heron, whose name is immortalized in the method for finding the area of a triangle given the lengths of the three sides. The name imaginary number was coined by Rene Descartes in the 1600s, who did not like the concept and hoped he could kill it by mocking it. The concept did not die, and about a century later, Leonhard Euler became the first to use the lowercase letter i as the square root of -1.

 

Numbers of the form a + bi, where a and b are real numbers constitute the complex numbers C. The common way to represent the complex numbers as a picture is the complex plane, where the real numbers are represented by the x-axis and the imaginary numbers are lined up on the y-axis. The axes intersect at zero.

 

Just like the real numbers R, the complex numbers C are an abelian group under addition and C - {0} is an abelian group under multiplication. We have discussed that ({1, -1}, ✕) is a finite abelian group and a subgroup of the real numbers. Similarly, ({1, i, -1, -i}, ✕) is a finite abelian group and a subgroup of the complex numbers. Note that -i is also a square root of -1, so it has two square roots in much the same way any positive number has two square roots. For example, 3²=9 and (-3)²=9.

 

Enter William Rowan Hamilton (1805-1865), widely regarded as the greatest Irish mathematician of all time. He created a mathematical system he called the quaternions, which has not two square roots of -1, but six. The general form of a number in the quaternion system is a + bi + cj + dk, where a, b, c and d  are arbitrary real numbers and i, j and k are square roots of -1, as are -i, -j and -k.

 

The quaternion system under multiplication is non-abelian. Here are the rules for multiplying i, j and k.

 

ij = k      jk = i     ki = j

ji = -k     kj = -i    ik = -j

 

Here is the full Cayley table representation, where we can find ij in the square corresponding to row i, column j.

There are three subgroups of order 4, and since 8/2 = 4, these are all normal subgroups.

Here are the subgroups.

{1, i, -1, -i}

{1, j, -1, -j}

{1, k, -1, -k}

 

Since 8 is divisible by 8, 4, 2 and 1, we might expect there is a subgroup of order 2, and {1, -1} fills the bill.

The quaterions are a non-abelian group where all subgroups are normal. This cannot be said for any other non-abelian group we have studied.

I present group theory in terms of pure mathematics, but groups play a vital role in the study of modern physics, including the quaternions. This is a link to the Wikipedia page that tells more.

Commentary

I mentioned four mathematicians today, so let me present links to their biographies. Far be it for me to talk smack about any of these guys, but the general public is probably better acquainted with Rene Descartes than nay of the other names, because Descartes was also a philosopher. Very few mathematicians after the Greeks are famous for just their mathematical work. If we judged this quartet solely on their mathematics, Leonhard Euler (1707-1783) is the true giant. For the quantity and quality of his work, he could be compared to the composer Johann Sebastian Bach (1685-1750).

Link to the biography of William Rowan Hamilton.

 

Link to the biography of Leonhard Euler.

 

Link to the biography of Heron of Alexandria.

Link to the biography of Rene Descartes.

The Alternating Groups A_4 and A_5

 The symmetric group Sn is the group of all permutations of n elements and the alternating group An is the group of all even permutations of n elements. The order of Sn is n! and for n > 1, the order of An is n!/2. If we look at An as a subgroup of Sn, it is normal, as are all subgroups whose order is exactly half the order of the group.

Recall that a normal subgroup N of a group G has several properties, but here are the two we will look at in this post.

1. For finite groups, the order of N must divide the order of G. (This is true of non-normal subgroups as well.)

2. A normal subgroup N must be the union of conjugacy classes, and one of those classes must be the identity element, which is a singleton class because the identity commutes with all elements.

Let us start with A4.

 

The identity: 

(1)

 

The permutations of the form (ab)(cd): 

(12)(34), (13)(24), (14)(23)

 

The 3-cycles:

(123), (132), (124), (142), (134), (143), (234), (243)

 

The order of A4 is 12, and the cardinalities of the conjugacy classes are 1, 3 and 8 respectively. To add these numbers up when we must include the 1, only 1+3 = 4 will divide 12. As it happens, the identity and the double transpositions do form a subgroup isomorphic to the Klein four-group, and it is a normal subgroup of A4.

Because A4 has a normal subgroup that is not either the whole group or the identity, A4 is not simple.

 

A5 is much larger, with 60 elements. Again we will list the conjugacy classes.

The identity: 

(1)

(1 total)

 

The permutations of the form (ab)(cd): 

(12)(34), (13)(24), (14)(23), 

(12)(35), (13)(25), (15)(23),

(12)(45), (14)(25), (15)(24),

(13)(45), (14)(35), (15)(34),

(23)(45), (24)(35), (25)(34)

(15 total)

 

The 3-cycles:

(123), (132), (124), (142), (125), (152),

(134), (143), (135), (153), (145), (154),

(234), (243), (235), (253), (245), (254),

(345), (354)

(20 total)

The 5-cycles:

(12345), (12354), (12435), (12453), (12534), (12543), 

(13245), (13254), (13425), (13452), (13524), (13542), 

(14235), (14253), (14325), (14352), (14523), (14532), 

(15234), (15243), (15324), (15342), (15423), (15432)

(24 total)

Now the numbers we must use to sum to a divisor of 60 are 1, which is mandatory, and 15, 20 and 24. Clearly, the sum of any three will be greater than 30 and less than 60, and there are no proper divisors of 60 greater than 30. Our two number sums are

1+15 = 16

1+20 = 21

1+24 = 25

None of these numbers divide 60 evenly, so the union of these classes cannot be a subgroup.

A5 is our first example of a simple non-abelian group. This is an important fact in Abel's proof of the insolubility of the quintic.

The invariants of conjugacy classes

 There is a direct link between groups represented by permutations and groups represented by permutation matrices. For example, if we have the permutation (12)(45) and we assume it belongs to S5, it can also be represented by the 5x5 matrix

0 1 0 0 0

1 0 0 0 0

0 0 1 0 0

0 0 0 0 1

0 0 0 1 0

In Sn, the conjugacy classes are all permutations with the same cycle structure, so the definition is necessary and sufficient. If we deal with subgroups of Sn, two conjugates must have the same cycle structure, but having the same cycle structure is not sufficient to state that the elements are conjugate. For example, the permutation (1234) generates a four element abelian group.

 

(1234)    

(1234)² = (13)(24)   

(1234)³ = (1432)

(1234)⁴ = (1)   

 

(1234) and (1432) have the same cycle structure, but this group is abelian and every element is in a conjugacy class by itself. In S4, (24)(1234)(24) = (1432), but (24) isn't available here.

 

A group of matrices also has invariants under conjugacy, the determinant, which was discussed in the last post and the trace, which is the sum of the elements along the main diagonal. Let's look at a four element group of 2x2 matrices that represent rotations of 0°, 90°, 180° and 270° in the xy-plane. This is an abelian group isomorphic to the earlier permutation group, so every conjugacy class is a singleton.

0° matrix

1  0 

0  1

determinant = 1, trace = 2

90° matrix

0 -1

1  0

determinant = 1, trace = 0

180° matrix

-1  0 

 0 -1

determinant = 1, trace = -2

270° matrix

0   1

-1  0

determinant = 1, trace = 0

Note that the 90° matrix and the 270° matrix have the same determinant and trace, which means there is a 2x2 non-singular matrix M such that M(90° matrix)M⁻¹ = 270° matrix. One such matrix is

 

-1  0

 0  1

 

which is not one of the elements of our defined group.

 

Commentary

 

As I have stated earlier, group theory is a generalization of the concept of symmetry. It is taught as an abstract field and the applications seem remote, with the possible exception of Rubik's Cube solutions. In fact, having symmetry in a physical problem makes it easier to solve. According to my professor Stu Smith, all solved differential equations rely on symmetry except for one. The differential equation that solves the solitary wave, also known as a soliton, does not have symmetry.

Most waves have peaks and valleys, and when the hit other waves, they can add to each other or cancel each other out, but a soliton only is almost all peak with a tiny valley. The bigger a soliton is, the faster it moves. In the ocean, we call the biggest solitons tsunamis.

 

A basic tenet of physics is big + fast = fuck you up. This is why a tsunami can wreak havoc when it hits land thousands of miles away from the source, because it will nearly the same speed and size it had when it was formed.

 

Determinants and finite groups of real-valued nxn matrices

 Every square matrix has a determinant, a number that corresponds to the area, volume or hypervolume of the shape formed by the n-vectors defined by the rows of the matrix. The columns can also be used, and though the shape would be different, the determinant is unchanged.

 

Example: for any 2x2 matrix

 

a b

c d

 

the determinant equals ad - bc.

 

This is the area of a parallelogram defined by the points (0,0), (a,b), (c,d) and (a+c,b+d), or by the points (0,0), (a,c), (b,d) and (a+b,c+d).

 

We get a positive area if the vector (c d) is on the left hand side of the vector (a b) in the two-dimensional plane. Conversely, a negative area means the vector (c d) is on the right hand side of the vector (a b) in the two-dimensional place. A determinant of zero means there is a scalar k such that c = ak and d = bk.

 

There is a problem with determinants. The number of terms increases factorially, which is a growth rate even faster than exponential growth. 3x3 matrices have 6 terms in the determinant, 4x4 have 24, 5x5 have 120, etc. I will show the 3x3 formula, but I will avoid the general formula for larger cases.

Example: for any 3x3 matrix

a b c

d e f

g h i

 

the determinant equals aei + bfg + cdh - afh - bdi - ceg.

 

Why these terms and why add some while subtracting others?

Every term in an nxn determinant is the product of n entries, with exactly 1 term in every row and every column. The term aei corresponds to the main diagonal matrix

1 0 0

0 1 0

0 0 1

This is also the identity matrix. Every other term corresponds to the positions of the 1 entries in one of the permutation matrices. If the permutation consists of an even number of transpositions from the identity, the corresponding term is added to the total, and the terms that correspond to odd permutations are subtracted from the total.

Thinking in the language of group theory, the group even and odd under addition is mapped onto the group 1 and -1 under multiplication.

 

A matrix M is singular if det(M) = 0.

 

There is a standard method for finding the inverse of a matrix and one of the steps is to divide by the determinant. We can't divide by zero, which means singular matrices do not have inverses. 

Stated without proof: det(MN) = det(M)det(N).

If the singular nxn matrices are removed from the set of all nxn matrices, we have a group, and determinant is a homomorphism from the group of non-singular nxn matrices under multiplication to the group (ℝ-{0}, ✕).

If a matrix M has real entries, det(M) will also be real, since the operations used to compute the determinant are multiplication, addition and subtraction. This brings us to a simply proved statement.

Lemma: If G is a finite group of nxn matrices under multiplication, the determinants of the matrices must be either 1 or -1.

 

Proof by contradiction: Assume one of our matrices M has a determinant x, not equal to 1 or -1. This would mean det(M²)=x² and det(M³)=x³, and so on, ad infinitum. It could not be a finite group, so instead we must assume det(M) must be either 1 or -1 for the group to be finite.

 

Stated without proof: If B is a matrix created by switching two rows of the matrix A, det(B)=-det(A).

 

Since every nxn permutation matrix can be created by multiple transpositions of rows of the identity matrix In, and the determinant of In = 1, all permutation matrices have a determinant of either 1 or -1. Having a lot of zeros in a matrix can make the computation of the determinant much simpler, since any product that includes a zero must be zero. Permutation matrices have so many zeros that there is only term that is non-zero and that term is the product 1ⁿ=1, which will be multipled by 1 or -1, depending on the parity of the number of row switches away from the idenity.

Theorem: If two matrices M and N are conjugates, det(M)=det(N).

Proof: Conjugacy means N = XMX⁻¹. If det(X)=x, then
det(X⁻¹)=1/x.

 

det(N)= det(XMX⁻¹) = x*det(M)*1/x= det(M).

 

The determinant of a matrix is invariant under conjugacy. 

 

Stu Smith, who taught me the majority of my postgraduate math classes, said that one should always pay attention to invariance, because it usually implied something of a physical nature. 

 

Like Tracewell, Stu Smith is someone I tried to emulate when I taught, but I admit I fell short on both counts. I was never as enthusiastic as Ted Tracewell, and I was was never as erudite as Stu Smith.

Commentary

In the middle of the 19th Century, a great number of mathematicians were working on ways to simplify the calculation of large matrix determinants, including Leopold Kronecker, who was introduced in the last post, and Charles Dodgson, better known to the general public as Lewis Carroll. 

 

I learned a story in a math class, I forget what professor, that Queen Victoria loved Alice in Wonderland, published in 1865, after the death of her husband Prince Albert in 1861, an event that left her nearly inconsolable. She wrote to Lewis Carroll and asked him to send her the next book he published. Following her wishes, she sent him a book on determinants.

 

The story is true until the punch line. Dodgson knew the Queen wanted his next Lewis Carroll book Alice Through the Looking Glass, and he sent it to her as soon as it was published in 1871. He did have a quirky sense of humor, but he would never disrespect his monarch in such a fashion.

Cayley tables in a special form and nxn permutation matrices representing a group with n elements

To begin, let me introduce the Kronecker delta function, 𝞭(x,y), where x and y are both elements of some set S.

If two things are the same, the delta function returns a 1. If not, the function returns a zero. 

 

Easy peasy, lemon squeezy.

 

Next, a Cayley table in special form. So far, the top row and left most column were transposes of each other, the same group element in position k of the top row as in position k of the leftmost column.

 

This time, we will line up the group elements with their inverses, so if x is in position k in the top row, position k in the leftmost column will be x⁻¹.

Now, we will do the Kronecker delta of the Cayley table with the identity element (1).

   

This is what the identity matrix should look like, ones along the mail diagonal, zeros everyplace else, which is to say we use the Kronecker delta on i and j for every entry mij in matrix M.

 

Remarkably enough, if we do the Kronecker delta on our Cayley table with any element in S3, we will get the 6x6 matrix that represents that element.

Matrix multiplication agrees with the combination operator of the permutations. More than that, if we want to turn the identity matrix into an even permutation mtarix such as (123) or (132), it will take an even number of row swaps or column swaps. To change the identity into (12), (13) or (23), we will do an odd number of column swaps. 

 

The biography of Leopold Kronecker is remarkable. An excellent student, he was also from a wealthy family and had little interest in teaching, only research at the top level. He was finally persuaded to teach at the age of 39, but students found his lectures hard to follow.

 

His most famous quote is "God made the integers, all the rest is the work of man." He had a visceral hatred of the new concepts of infinity presented by Georg Cantor, and did not like the definition of any irrational real number being defined as the limit of some infinite series of rational numbers. It should be noted that Gauss, born long before Kronecker and Cantor, did do research into infinity and decided not to publish, fearing it would just cause controvesry.

 

The next post will be about invariants in conjugacy classes of permutation matrices.

 

Permutation matrices

 The symmetric group Sn for n > 0 is the set of permutations of n distinct elements. We have seen the cycle notation representation, but we can also use permutation matrices, which are matrices whose entries are all ones and zeros, with the rule that every row and every column has a single one in it, all the other entries equal to zero.

Lemma: The identity matrix for In for n x n matrices is a permutation matrix.

Proof: By definition, the identity matrix has ones on the main diagonal and zeros in every other position.

Here is a table showing how S3 can be written as 3x3 permutation matrices. For example the permutation (123) is represented by the matrix which moves column 1 of the identity to column 2, column 2 of the identity to column 3, and column 3 of the identity to column 1.

In some books, rows are moved instead columns, but this way, multiplying the matrices left to right gives the same result as combining the permutations left to right.

 

Cycle notation is more compact than permutation matrix representation, but multiplying 0-1 matrices is a very easy operation, especially since we know if we get a one in a row or column, we can immediately fill in zeros in the rest of both the row and column.

Next time, we will use the Cayley table in a special form that will let us turn any group of order n into a permutation group representation of nxn matrices.

The Symmetric Group S_n

My blog only scratches the surface of the topic of group theory, as you can well imagine. For example, the topic of today's post, The Symmetric Group, can fill a graduate level text, like this one by Bruce Sagan. Sagan got his doctorate at MIT, but his undergraduate career was at Cal State Hayward, my alma mater. Go Pioneers!

 

The symmetric group Sn is the group of permutations on n distinct elements. If n > 2, Sn is non-abelian.

 

An n-cycle is a permutation that rearranges n elements. (xyz), (123), (cat) are all 3-cycles on different sets of 3 symbols.

 

A 2-cycle is often called a transposition. 

 

Let me list the elements of S1, S2, S3, and S4 using cyclic notation The order of Sn is n!, which means S5 has 120 elements. I am not spending all morning listing those.

 

S1: (1) 

There is only one way to have a one element group, it's just the identity.

 

S2: (1), (12)

There is also a unique up to isomorphism two element group. It's the same structure as (Z2, +) and ({1, -1}, ✕). There's the identity element e and the non-identity element x where x*x=e.

 

S3: (1), (12), (13), (23), (123), (132)

S3 is the smallest non-abelian group, because a transposition doesn't commute with anything except the identity (1) and itself, because every transposition is its own inverse. Some examples.

(12)(123) = (13), while (123)(12) = (23)

(12)(13) = (123), while (13)(12) = (132)

 

Stated without proof: Every permutation can be represented as a combination of transpositions. 

 

Even permutations are the combination of an even number of transpositions, while odd permutations are the combination of an odd number of transpositions. The combinations that create a permutation are not unique, but these combinations must share the same parity, either even or odd.

 

A slightly annoying convention to get used to. If n is odd, the n-cycle is an even permutation, and if n is even, the n-cycle is odd.

 

Stated without proof: If n>1, Sn can be split into two equal sized partitions, the even permutations and the odd permutations.

 

Lemma: The set of all even permutations is a normal subgroup of Sn known as An, the alternating group on n elements. 

Proof: If we combine an even permutation with an even permutation, the result must also be an even permutation, so the subset is closed under the combination operator. The identity (1) is even, and every even permutation's inverse must also be even, since combining them makes the even permutation (1). This is the same as saying if even + x = even, x must itself be even.

And now S4, split into cycle structures. The even permutations will be written in red, the odd in black.

The identity: (1)

The 2-cycles: (12), (13), (14), (23), (24), (34)

The 3-cycles: (123), (124), (132), (134), (142), (143), (234), (243)

 

The 4-cycles: (1234), (1243), (1324), (1342), (1423), (1432)

 

The pairs of transpositions: (12)(34), (13)(24), (14)(23)

 

Stated without proof: In Sn, the conjugacy classes are defined as all permutations with the same cycle structure.

 

When we look at subgroups of Sn, just having the same cycle structure does not promise conjugacy in the subgroup. Remember that a conjugate of a must be of the form xax⁻¹, and x might not be in the subgroup. So in subgroups of Sn, it is still necessary that conjugates must be of the same cycle structure, but it is not sufficient. In the post introducing permutation groups, D4 was presented. The elements are

 

Rotations: (1), (1234), (13)(24), (1432)

Reflections: (12)(34), (14)(23), (13), (24)

 

Note that all the pairs of transpositions in S4 are also in
D4, but the 180° rotation commutes with everything in D4, so it stands alone in its conjugacy class, while (12)(34) and (14)(23) are still conjugates, because (13)(12)(34)(13)=(14)(23).

  

Let's consider A4. This is a non-abelian group of order 12. All the pairs of transpositions are still conjugate to one another, but the 3-cycles are split into two conjugacy classes. The inverse of a 3-cycle (abc) is (acb), and the elements needed to make such permutations conjugate to their inverses are transpositions such as (ab), (ac) and (bc), which aren't even permutations. So the conjugacy classes here are: 

The identity: (1)

3-cycle conjugacy class #1: (123), (124), (134), (234)

3-cycle conjugacy class #1: (132), (142), (143), (243)

The pairs of transpositions: (12)(34), (13)(24), (14)(23)

 

Remember that a normal subgroup must be a union of conjugacy classes and its order must divide the order of the group. In A4, the only way to meet both those criteria to get a proper normal subgroup is to add the pairs of transpositions to the the identity, {(1), (12)(34), (13)(24), (14)(23)}.

 

One last fun fact: A4 is isomorphic to the symmetries of a regular tetrahedron. The 3-cycles are the same as spinning one of the four faces, and the pairs of transpositions correspond to swapping one edge {a, b} with the edge connecting the other two vertices {c, d}.

The next post will be about permutation matrices.