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test

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test

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Units, Constants and Useful Formulas

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Last modified: December 6, 2019

Basic Representation Theory --------------------------- In essence, a representation makes an abstract object like a group more concrete. It is the algebraic action of the group elements on a set (basis) where the 'things' that perform the action are invertible n x n matrices. These matrices 'act' on the set by matrix multiplication. A representation gives a matrix for each element, and so another representation with a an element acting on a different basis is possible. We will use the dihedral Group, D4 to illustrate the concepts involved. The Dihedral Group ------------------ The dihedral group is the group of symmetries of a regular polygon, which includes rotations and reflections. Dihedral groups play an important role in group theory, geometry, and chemistry. Consider an arbitrary basis set {1,2,3,4} (i.e., elements on which symmetry operations are performed). 2 1 ------- | | | O | | | ------- 3 4 There are 8 symmetry operations that we can perform. Do nothing. Rotations by 90°, 180° and 270° Relections about the horizontal axix through O. Relections about the vertical axix through O. Relections about the right diagonal through O. Relections about the left diagonal through O. These operations can be represented by the following matrices: Identity: - - - - - - | 1 0 0 0 || 1 | | 1 | I = | 0 1 0 0 || 2 | = | 2 | | 0 0 1 0 || 3 | | 3 | | 0 0 0 1 || 4 | | 4 | - - - - - - 1 2 3 4 Cycle notation: 1234 => (1)(2)(3)(4) 1234 Rotation by 90°: - - - - - - | 0 0 0 1 || 1 | | 4 | r = | 1 0 0 0 || 2 | = | 1 | | 0 1 0 0 || 3 | | 2 | | 0 0 1 0 || 4 | | 3 | - - - - - - 1 2 3 4 Cycle notation: 1234 => (1432) 4123 Rotation by 180°: - - - - - -   | 0 0 1 0 || 1 | | 3 | r2 = | 0 0 0 1 || 2 | = | 4 |   | 1 0 0 0 || 3 | | 1 |   | 0 1 0 0 || 4 | | 2 | - - - - - -   1 2 3 4 Cycle notation: 1234 => (13)(24) 3412 Rotation by 270°: - - - - - -   | 0 1 0 0 || 1 | | 2 | r3 = | 0 0 1 0 || 2 | = | 3 |   | 0 0 0 1 || 3 | | 4 |   | 1 0 0 0 || 4 | | 1 | - - - - - -   1 2 3 4 Cycle notation: 1234 => (1234) 2341 Reflection about horizontal axis: - - - - - -   | 0 0 0 1 || 1 | | 4 | dr2 = | 0 0 1 0 || 2 | = | 3 |   | 0 1 0 0 || 3 | | 2 |   | 1 0 0 0 || 4 | | 1 | - - - - - -   1 2 3 4 Cycle notation: 1234 => (14)(23) 4321 Reflection about vertical axis: - - - - - - | 0 1 0 0 || 1 | | 2 | d = | 1 0 0 0 || 2 | = | 1 | | 0 0 0 1 || 3 | | 4 | | 0 0 1 0 || 4 | | 3 | - - - - - - 1 2 3 4 Cycle notation: 1234 => (12)(34) 2143 Reflection about left diagonal: - - - - - -   | 0 0 1 0 || 1 | | 3 | dr3 = | 0 1 0 0 || 2 | = | 2 |   | 1 0 0 0 || 3 | | 1 |   | 0 0 0 1 || 4 | | 4 | - - - - - -   1 2 3 4 Cycle notation: 1234 => (13)(2)(4) ≡ (13) 3214 Reflection about right diagonal: - - - - - - | 1 0 0 0 || 1 | | 1 | dr = | 0 0 0 1 || 2 | = | 4 | | 0 0 1 0 || 3 | | 3 | | 0 1 0 0 || 4 | | 2 | - - - - - - 1 2 3 4 Cycle notation: 1234 => (1)(24)(3) ≡ (24) 1432 The Cayley Table ---------------- The Cayley table for D4 is: Conjugacy classes defines a partition of the group. To find the conjugacy classes we compute π-1σπ = ρ π-1 is found by looking along the row π and reading off the column corresponding to the element I (i.e. ππ-1 = I). Therefore: D4 has a total of 5 conjugacy classes: {I} = (1234) {r2} = (13)(24) {r,r3} = (1234),(1234) {d,dr2} = (12)(34),(14)(23) {dr,dr3} = (24),(13) If 2 matrices are in the same conjugacy class they are similar, have the same trace, eigenvalues, period (order) and determinant. Therefore; {I} {r2} {r,r3} {d,dr2} {dr,dr3} ------------------------------------- Evalues | 1 1,-1 1,-1,i,-i 1,-1 1,-1 } In Determ. | 1 1 -1 1 -1 } 4D Trace | 4 0 0 0 2 } rep. Period* | 1 2 4 2 2 } * often called the order of an element = m = gm = e Reducibility of a Representation -------------------------------- One of the biggest topics in representation theory is the reducibility of a representation. In other words, given a representation, could we encode the same information about the group with less dimensions (smaller basis). We will present the critical formulas and derive some. Let us label the different representations by a superscript (r), (s), .... Therefore D(r)(g) represents the element g in the representation r. The Character of a Representation --------------------------------- A given group can have many different representations. The matrices of a representation are basis dependent. We can eliminate this basis dependence by defining the CHARACTER Χ(r) of D(r)(g) as: χ(r) = Tr(D(r)(g)) The independency can be shown by using the cyclicity of the trace: Tr(D'(g)) = Tr(SD(g)S-1) = Tr(D(g)SS-1) = Tr(D(g) Therefore, all elements in the same conjugacy class have the same character that is independent of the basis chosen. For D4 we had, from above: {I} {r2} {r,r3} {d,dr2} {dr,dr3} ------------------------------------- χ(g) | 4 0 0 0 2 Block Diagonal Form ------------------- A representation is COMPLETELY REDUCIBLE if it can be reduced to a representation whose matrix elements have the form: - - | D(r)(g) | D(g) ≡ | D(s)(g) | | ...     | | D(z)(g) | - - Where D(r)(g), D(s)(g) etc. are n x n matrices that are irreducible representations that cannot be broken up into smaller pieces. This is called the BLOCK DIAGONAL form. The square matrices representing the diagonal elements can be of any size including 1 × 1. A representation in block diagonal form can be written as the DIRECT SUM: D(r)(g) ⊕ D(s)(g) ⊕ ... ⊕ D(z)(g) Block Diagonalization --------------------- Consider the matrices: - - | 1 1 1 1 | S = (1/2)| 1 -1 i -i | | 1 1 -1 -1 | | 1 -1 -i i | - - and, - -    | 1 1 1 1 | S-1 = (1/2)| 1 -1 1 -1 |    | 1 -i -1 i |    | 1 i -1 -i | - - We won't discuss how S is determined but instead focus on the end results. Block diagonalization is accomplished by applying the transformation S-1AS = AB We get: r' = S-1rS: - - - -    | 0 0 0 1 | | 1 0 0 0 | S-1| 1 0 0 0 |S = | 0 -1 0 0 |    | 0 1 0 0 | | 0 0 -i 0 |    | 0 0 1 0 | | 0 0 0 i | - - - - r2' = S-1r2S: - - | 1 0 0 0 | | 0 1 0 0 | | 0 0 -1 0 | | 0 0 0 -1 | - - r3' = S-1r3S: - - | 1 0 0 0 | | 0 -1 0 0 | | 0 0 i 0 | | 0 0 0 -i | - - d' = S-1dS: - - - -    | 0 1 0 0 | | 1 0 0 0 | S-1| 1 0 0 0 |S = | 0 -1 0 0 |    | 0 0 0 1 | | 0 0 0 -i |    | 0 0 1 0 | | 0 0 i 0 | - - - - dr' = S-1(dr)S: - - | 1 0 0 0 | | 0 1 0 0 | | 0 0 0 1 | | 0 0 1 0 | - - dr2' = S-1(dr2)S: - - | 1 0 0 0 | | 0 -1 0 0 | | 0 0 0 i | | 0 0 -i 0 | - - dr3' = S-1(dr3)S: - - | 1 0 0 0 | | 0 1 0 0 | | 0 0 0 -1 | | 0 0 -1 0 | - - It is easy to verify that these matrices follows the D4 rules of the Cayley table. For example, dr'.dr' = I r'.r' = r2' and so on. Therefore, the matrices after the transformation still form a representation of the same group. The 1's on the diagonal are just the trivial representation. It is the one dimensional representation where all elements of G act as the identity. Great Orthogonality Theorem --------------------------- The Great Orthogonality Theorem states: (1/|G|)Σχ(r)†(g)χ(s)(g) = δrs ≡ <χ(r)(g)|χ(s)(g)> g or, (1/|G|)Σncχ(r)†(c)χ(s)(c) = δrs ≡ <χ(r)(c)|χ(s)(c)> c Where nc is the number of elements in the conjugacy class, c. Consider the block diagonal matrices for D4 from above. D(s)(g)     | D(r)(g) | D(t)(g) |    | | - v    v v - | 1 | | ±1 | - - | | | a b | | | | c d | | | - - | - - Orthogonality of χ(r)(g) and χ(s)(g): I r r2 r3 d dr dr2 dr3 (1/8){1.1 + 1.(-1) + 1.1 + 1.(-1) + 1.(-1) + 1.1 + 1.(-1) + 1.1} = 0 Orthogonality of χ(r)(g) and χ(t)(g): I r r2 r3 d dr dr2 dr3 (1/8){1.2 + 1.0 + 1.(-2) + 1.0 + 1.0 + 1.0 + 1.0 + 1.0} = 0 Orthogonality of χ(t)(g) and χ(t)(g): I r r2 r3 d dr dr2 dr3 (1/8){2.2 + 0.0 + (-2).(-2) + 0.0 + 0.0 + 0.0 + 0.0 + 0.0} = 1 Test of Reducibility -------------------- If nr = the number of times the irreducible representation Dr(c) appears in the block diagonal matrix representation we can write: χ(g) = Σnrχr(g) r For example, χ(r') = 1.(1).+ 1.(-1) + 1.(0) = 0 χ(dr') = 1.(1).+ 1.(1) + 1.(0) = 2 etc. Therefore, we can write: <χ(g)|χ(g)> = (1/|G|)ΣΣnrχ(r)†(g)nsχ(s)(g) rs Now, ΣΣnrχ(r)†(g)nsχ(s)(g) = |G|ΣΣnrnsδrs rs rs Therefore, <χ(g)|χ(g)> = |G|ΣΣnrnsδrs rs For r = s this becomes: <χ(g)|χ(g)> = |G|Σ(nr)2 r If Σ(nr)2 = 1 then we know that only one of the r nr's is equal to 1 with all others equal to 0, the given representation D(g) contains the irreducible representation r once and only once. In other words, D(g) is irreducible; in fact, it is just D(r)(g). In contrast, if Σ(nr)2 > 1 then we know that D(g) is r reducible. For example, if Σ(nr)2 = 3 then the only r possibility is for 3 different irreducible representations to each occur once. We can also write the above equation as: Σncχ(c)χ(c) = |G|Σ(nr)2 c r Where nc is the number of elements in each conjugacy class. For D4 we get: (1/|G|)Σχ(g)χ(g) = Σ(nr)2 g r LHS = (1/8)[{(Tr(I))2} + {Tr(r2))2} + {(Tr(r))2 + (Tr(r3))2| + {(Tr(d))2| + (Tr(dr2))2} + {(Tr(dr))2 + (Tr(dr3))2}] = (1/8)[{42} + {02} + {02 + 02} + {02 + 22} + {02 + 22}] = 24/8 = 3 = 12 + 12 + 12 = 1 dimensional + 1 dimensional + 2 dimensional Since the dimension = 4 the original representation must break apart into 2 one dimensional and 1 two dimensional irreducible representations, i.e., = 1 dimensional + 1 dimensional + 2 dimensional Schur's Lemma ------------- Schur's Lemma dictates how the blocks should be arranged for the representation to be irreducible. The lemma states that for D'(g) to be an irreducible representation of a finite group G there needs to be some matrix A such that: AD(g) = D(g)A for all g Where A = λI for some constant λ. Consider a group of matrices. All of these matrices will naturally commute with the identity matrix. However, it may also be possible to find another matrix, A, that commutes with all of the matrices in the group. If this is the case then the representation is reducible. If the only matrix that works is the identity then the representation is irreducible. Now we can find a unitary matrix, U, which transforms A into a diagonal matrix, AD i.e. AD = U-1AU and DD(g) = U-1D(g)U If we apply this to both sides of ADD(g) = DD(g)A we get: U-1ADUU-1DD(g)U = U-1DD(g)UU-1ADU U-1ADDD(g)U = U-1DD(g)ADU Multiplying by U from the left and U-1 from the right gives: ADDD(g) = DD(g)AD Now if DD(g) is irreducible it is of block diagonal form. So we have a diagonal matrix AD multiplying a block diagonal matrix. The only way the equality holds is if AD is some multiple of the identity matrix. Finding Other Irreducible Representations ---------------------------------------- Postulate: # of inequivalent irreps = # of conjugacy classes. Proof: Any finite group G has an h dimensional representation, known as the REGULAR representation. To obtain the regular representation we start with the Cayley table and rearrange it so that the group elements are listed in the first column and the corresponding inverses are listed in the first row. The matrix of the regular representation of the element is then obtained from this table by replacing the corresponding every appearance of g with 1, and filling up the rest of the corresponding matrix with zeros. That is, for example: What does this mean? By left multiplication, each element of G acts as a permutation of the elements of G. We can write this as: φ(g): x -> gx for all x ∈ g For r we have: r.{I,r3,r2,r,d,dr,dr2,dr3} which is: - - - - - - | 0 0 0 1 0 0 0 0 || I   | | r   | | 1 0 0 0 0 0 0 0 || r3 | = | I   | | 0 1 0 0 0 0 0 0 || r2 | | r3 | | 0 0 1 0 0 0 0 0 || r   | | r2 | | 0 0 0 0 0 0 0 1 || d   | | dr3 | | 0 0 0 0 1 0 0 0 || dr  | | d   | | 0 0 0 0 0 1 0 0 || dr2 | | dr  | | 0 0 0 0 0 0 1 0 || dr3 | | dr2 | - - - - - - Therefore, φ(g) is a homomorphism (a structure preserving map between two algebraic structures of the same type). The characters for the regular representation are easy to compute since none of the matrices except I have any diagonal entries. The character χ(I) is equal to the dimension of the representation, which in our case is 8 for the regular representation. Therefore, χ(reg)(g) = { 0 g ≠ e       { |G| g = e We can now apply the reducibility test from before: Σχ(g)χ(g) = |G|Σ(nr)2 g r or, Σncχ(c)χ(c) = |G|Σ(nr)2 c r Thus, Σncχ(c)χ(c) = (χ(I))2 = |G|Σ(nr)2 c r From which we get: Σ(nr)2 = 8 r Therefore, the regular representation is reducible. 8 can be factores as follows: 8 = 12 + 12 + 12 + 12 + 22 + 22 Multiplicity ------------ The number of times an irreducible representation D(s)(g) appears in the decomposition of D(reg) is called the multiplicity. It is given by <χ(reg)(g)|χ(s)(g)>. <χ(reg)(g)|χ(s)(g)> = (1/|G|)Σχ(reg)†(g)χ(s)(g)   g = (1/|G|)χ(reg)†(I)χ(s)(I) = (|G|/|G|)(I)χ(s)(I) = χ(s)(I) = Tr(I(s)) = dim(D(s)(g)) Therefore, the number of times the same irreducible representation can appear in the block diagonal form of the regular representationis is equal to its dimension. We know that the regular representation can be decomposed into a direct sum of all the distinct irreducible representations of the group, D(si)(g), and their multiplicity is equal to the number of times they appear in the direct sum. Therefore, we can write the character of the regular representation as a sum of the irreducible characters, weighted by their multiplicity: χ(reg)(g) = Σdim(D(si)(g))χ(si)(g)      i We have found from before that the number of irreducible representations in the regular representation can be factored as: 8 = 12 + 12 + 12 + 12 + 22 + 22 Using the above equation we can now write the the above in terms of their multiplicity as: 8 = 1(12) + 1(12) + 1(12) + 1(12) + 2(22) Therefore, there are 5 inequivalent irreducible representations of the regular representation. By induction, we see that these 5 inequivalent irreducible representations correspond to the 5 conjugacy classes of D4. This enables us to deduce: The number of irreducible representations of a group is equal to the number of conjugacy classes of the group. The sum of the squares of the dimensions of all distinct irreducible representations is equal to the order of the group: Σ(dr)2 = |G| r Proof: <χ(reg)(I)|χ(reg)(I)> = (χ(reg)(I))2 = |G|2 But, <χ(reg)(I)|χ(reg)(I)> also = |G|Σ(nr)2 Therefore, by comparison, Σ(nr)2 = |G| r Now, neglecting the multiplicity: n1 n2 n3 n4 n5 n6 Σ(nr)2 = 12 + 12 + 12 + 12 + 22 + 22 = 8 r Now write the dimensions of each matrix in the same way: d1 d2 d3 d4 d5 d6 Σ(dr)2 = 12 + 12 + 12 + 12 + 22 + 22 = 8 r We conclude that dr = nr Character Tables ---------------- If we consider the irreducible representations as vectors whose components are characters of the representation then 2 different IRREDUCIBLE representations are orthogonal. This dictates the construction of character tables. We know that the number of conjugacy classes is equal to the number of irreducible representations. Therefore, character tables are always square. The Character table of D4 turns out to be: {I} {r2} {r,r3} {d,dr2} {dr,dr3} -------------------------------- D1(g) | 1 1 1 1 1 } D2(g) | 1 1 1 -1 -1 } In arbitrary D3(g) | 1 1 -1 1 -1 } basis (A,B,C,D) D4(g) | 1 1 -1 -1 1 } D5(g) | 2 -2 0 0 0 } ^ ^ ^ ^ ^ / | \ | | / | \ |____ |______ / | \ | | - - - - - - - - - - | 1 0 | | -1 0 | | -i 0 | | 0 -i | | 0 1 | | 0 1 | | 0 -1 | | 0 i | | i 0 | | 1 0 | - - - - - - - - - - - - - - - - | i 0 | | 0 i | | 0 -1 | | 0 -i | | -i 0 | | -1 0 | - - - - - - To see the row and column orthogonalities we need to expand the table as follows: I r2 r r3 d dr2 dr dr3 ------------------------------- D1(g) | 1 1 1 1 1 1 1 1 D2(g) | 1 1 1 1 -1 -1 -1 -1 D3(g) | 1 1 -1 -1 1 1 -1 -1 D4(g) | 1 1 -1 -1 -1 -1 1 1 D5(g) | 2 -2 0 0 0 0 0 0 Using the Great Orthogonality Theorem, <χ(r)(g)|χ(s)(g)> = (1/|G|)Σχ(r)†(g)χ(s)(g) = δrs      g we get: 1st and 2nd rows: (1/8)(1.1 + 1.1 + 1.1 + 1.1 + 1.(-1)) + 1.(-1) + 1.(-1) + 1.(-1)) = 0 2nd and 3rd rows: (1/8)(1.1 + 1.1 + 1.(-1) + 1.(-1) + (-1).1 + (-1).1 + (-1)(-1) + (-1).(-1)) = 0 1st and 3rd rows: (1/8)(1.1 + 1.1 + 1.(-1) + 1.(-1) + 1.1 + 1.1 + 1.(-1) + 1.(-1)) = 0 and so on. 1st and 2nd cols: (1/8)(1.1 + 1.1 + 1.1 + 1.1 + 2.(-2)) = 0 2nd and 3rd cols: (1/8)(1.1 + 1.1 + 1.(-1) + 1.(-1) + (-2).0) = 0 1st and 3rd cols: (1/8)(1.1 + 1.1 + 1.(-1) + 1.(-1) + 2.0) = 0 and so on. Note if Dr(g) = Ds(g) we get: Σncχ(r)†χ(s) = (1/8)(1.(2)2 + 1.(-2)2 + 2.(0)2 + 2.(0)2 c + 2.(0)2) = 1 We can test the reducibility of this representation as follows: (1/|G|)Σncχ(c)χ(c) = Σ(nr)2 c r LHS = (1/8)[(2)2 + (-2)2] = 1 RHS = 12 Since the dimension = 2 the 2 x 2 representation cannot be broken down further and is therefore irreducible. As an alternative proof of this we can use Schur's Lemma to show that AD(g) = D(g)A if and only if A = λI. For example consider: - - - - - - - - | a b || 0 -i | - | 0 -i || a b | | c d || i 0 | | i 0 || c d | - - - - - - - - - - = | (bi + ci) (-ai + di) | | (di - ai) (-ci - bi) | - - For the RHS to equal 0 we must have: a = d and b = -c. Therefore, we can write: I: - - - - - - - - | a b || 1 0 | - | 1 0 || a b | = 0 for | -b a || 0 1 | | 0 1 || -b a | any b - - - - - - - - r: - - - - - - - - | a b || -i 0 | - | -i 0 || a b | = 0 for | -b a || 0 i | | 0 i || -b a | b = 0 only - - - - - - - - r2: - - - - - - - - | a b || -1 0 | - | -1 0 || a b | = 0 for | -b a || 0 -1 | | 0 -1 || -b a | any b - - - - - - - - r3: - - - - - - - - | a b || i 0 | - | i 0 || a b | = 0 for | -b a || 0 -i | | 0 -i || -b a | b = 0 only - - - - - - - - d: - - - - - - - - | a b || 0 -i | - | 0 -i || a b | = 0 for | -b a || i 0 | | i 0 || -b a | any b - - - - - - - - dr: - - - - - - - - | a b || 0 1 | - | 0 1 || a b | = 0 for | -b a || 1 0 | | 1 0 || -b a | b = 0 only - - - - - - - - dr2: - - - - - - - - | a b || 0 i | - | 0 i || a b | = 0 for | -b a || -i 0 | | -i 0 || -b a | any b - - - - - - - - dr3: - - - - - - - - | a b || 0 -1 | - | 0 -1 || a b | = 0 for | -b a || -1 0 | | -1 0 || -b a | b = 0 only - - - - - - - - Since this has to work for ALL of the matrices we conclude that A is λI and Schur's Lemma holds. Ultimately, the elements in the 2 x 2 matrices of the irreducible representation will depend on the basis chosen. There are many possible bases that we could use. However, if we choose the normal Cartesian basis and pick a (-1,1), the matrices take the form: {I} {R2} {R,R3} {dL,dR} {rx,ry} ------------------------------- D5(g) | 2 -2 0 0 0 ^ ^ ^ ^ ^ / | \ | | / | \ |____ |______ / | \ | | - - - - - - - - - - | 1 0 | | -1 0 | | 0 1 | | 0 -1 | | 1 0 | | 0 1 | | 0 -1 | | -1 0 | | -1 0 | | 0 -1 | - - - - - - - - - - - - - - - - | 0 -1 | | 0 1 | | -1 0 | | 1 0 | | 1 0 | | 0 1 | - - - - - - As in the arbitrary basis case, these elements can be written as products of the following matrices. - - R = | 0 1 | ≡ r | -1 0 | - - and, - - D = | 0 -1 | ≡ dL | -1 0 | - - For example, - - - - - - DR = | 0 -1 || 0 1 | = | 1 0 | ≡ rx | -1 0 || -1 0 | | 0 -1 | - - - - - - DR2 = dR DR3 = ry We can now think of the transformation matrices as acting on the vector, (-1,1), in the Cartesian basis (x,y) as follows: (-1,1) (1,1) 2 1 ------- | | I = | o | | (0,0) | ------- 3 4 (-1,-1) (1,-1) Rotation by 90°: 1 4 ------- | | R = | o | | | ------- 2 3 - - - - - - | 0 1 || -1 | = | 1 | | -1 0 || 1 | | 1 | - - - - - - Cycle notation: 1234 => (1234) 2341 Rotation by 180°: 4 3 -------   | | R2 = | o |   | | ------- 1 2 - - - - - - | -1 0 || -1 | = | 1 | | 0 -1 || 1 | | -1 | - - - - - - Cycle notation: 1234 => (13)(24) 3412 Rotation by 270°: 3 2 -------   | | R3 = | o |   | | ------- 4 1 - - - - - - | 0 -1 || -1 | = | -1 | | 1 0 || 1 | | -1 | - - - - - - Cycle notation: 1234 => (1234) 2341 Reflection about x axis: 3 4 -------   | | rx = | o |   | | ------- 2 1 - - - - - - | 1 0 || -1 | = | -1 | | 0 -1 || 1 | | -1 | - - - - - - Cycle notation: 1234 => (14)(23) 4321 Reflection about y axis: 2 3 -------   | | ry = | o |   | | ------- 1 4 - - - - - - | -1 0 || -1 | = | 1 | | 0 1 || 1 | | 1 | - - - - - - Cycle notation: 1234 => (13)(2)(4) 3214 Reflection about right diagonal: 4 1 -------   | | dR = | o |   | | ------- 3 2 - - - - - - | 0 1 || -1 | = | 1 | | 1 0 || 1 | | -1 | - - - - - - Cycle notation: 1234 => (1)(24)(3) 1432 Reflection about left diagonal: 3 2 -------   | | dL = | o |   | | ------- 4 1 - - - - - - | 0 -1 || 1 | = | -1 | | -1 0 || 1 | | -1 | - - - - - - Cycle notation: 1234 = (13)(2)(4) 3214 The matrices in each class are similar. They have the same eigenvalues, determinant, trace and period. {I} {R2} {R,R3} {dL,dR} {rx,ry} ------------------------------------- Evalues | 1 -1 i,-i -1,1 -1,1 Determ. | 1 -1 1 -1 1 Trace | 2 -2 0 0 0 Period* | 1 2 4 2 2 * often called the order of an element = m = gm = e As before, by applying Schur's Lemma it is easy to show that this representation is irreducible. Change of Basis --------------- Consider from before the reflection about right diagonal: 4 1 -------   | | dR = | o |   | | ------- 3 2 - - - - - - | 0 1 || -1 | = | 1 | | 1 0 || 1 | | -1 | - - - - - - Consider now if we rotate the coordinate system clockwise by 45° so that x -> x' and y -> y'. In the x', y' basis the point (-1,1) becomes (-1,0) and its reflection across the right diagonal becomes (1,0). We now look for a 2 x 2 matrix that accomplishes this. We get: - - - - - - | -1 0 || -1 | = | 1 | | 0 1 || 0 | | 0 | - - - - - - Therefore, a reflection across the right diagonal in the x,y basis becomes a reflection across y' in the x',y' basis. - - - - dR 1 : | 1 | -1 : | -1 |   | 1 | | 1 | - - - - P(λ) = det(dR - λI) = | 0 1 - λ | = -λ2 + 2λ - 1 | 1 - λ 0 | - - - - rY' 1 : | 0 | -1 : | 1 |    | 1 | | 0 | - - - - P(λ) = det(dR - λI) = | -(1 - λ) 0 | = -λ2 + 2λ - 1 | 0 1 - λ | The matrices have the same rank, the same trace, the same determinant, the same eigenvalues, and the same characteristic polynomial. This means that the matrices are similar and, therefore, the representations are indeed equivalent.