Past exam of the groups and representations course of the University of Oxford

A group representation on a vector space

V

is a group homomorphism

R : G \to GL (V)

, so

R (g h) = R (g) R (h)

and

R (e) = I

. A nonzero group representation is an irreducible representation if its only invariant subspaces are

0

and

V

. Here the representations and characters are finite-dimensional over the complex numbers.

The character of a representation is

χ_{R} (g) = tr R (g)

. Three useful properties are:

It is a class function: $χ_{R} (h g h^{- 1}) = χ_{R} (g)$ , by the cyclic property of the trace.
It is additive on a direct sum and multiplicative on a tensor product of group representations: $χ_{R \oplus S} = χ_{R} + χ_{S}$ and $χ_{R \otimes S} = χ_{R} χ_{S}$ .
The irreducible characters obey character orthogonality:
$⟨ χ_{r}, χ_{s} ⟩ = \frac{1}{∣ G ∣} \sum_{g \in G} χ_{r} (g)^{*} χ_{s} (g) = δ_{rs} .$
(1)

In particular,

χ_{R} (e) = dim V

and

χ_{R} (g^{- 1}) = χ_{R} (g)^{*}

, the latter following from unitarization of a finite-group representation. The character inner product

⟨ χ_{r}, χ_{R} ⟩

counts the multiplicity of an irreducible representation

r

R

Solution

For any

h \in G

, left multiplication permutes the elements of the finite group. Consequently averaging over a finite group gives

R (h) P v = \frac{1}{∣ G ∣} \sum_{g \in G} R (h g) v = \frac{1}{∣ G ∣} \sum_{k \in G} R (k) v = P v .

Thus

P v

belongs to the fixed-point subspace of a group action

V^{G}

; any nonzero such vector spans a trivial representation, or singlet. The zero vector is included in the invariance statement.

Conversely

Pw = w

for every

w \in V^{G}

, so

im P = V^{G}, P^{2} = P .

This is a projection operator onto all singlets, rather than necessarily onto a one-dimensional vector subspace. Its trace is the singlet multiplicity

∣ G ∣^{- 1} \sum_{g} χ_{R} (g)

Solution

Write

1

for the trivial representation,

ε

for the sign representation, and

2

for the given two-dimensional group representation. Their matrix representations on the generators of a group are

1 ε 2 r 11 diag (α, α^{- 1}) s 1 - 1 (0110)

with

α = e^{2 πi /3}

. The two eigenvalues of

r

2

are distinct, and

s

exchanges its two eigenlines; no line is invariant under both. Thus

2

is an irreducible representation. The dimensions already exhaust the sum of squares of irreducible representation dimensions, so this is the complete list.

Since

α + α^{- 1} = - 1

, the character table is

class class size χ_{1} χ_{ε} χ_{2} I 1112 {r, r^{2}} 211 - 1 {s, rs, r^{2} s} 31 - 1 0 .

Multiplying characters and using character orthogonality gives every tensor product of group representations:

\otimes 1 ε 2 11 ε 2 ε ε 12 222 1 \oplus ε \oplus 2 .

For example

χ_{2}^{2} = (4, 1, 0) = χ_{1} + χ_{ε} + χ_{2}

. Explicit Clebsch-Gordan coefficients can be read from two doublets

x = (x_{1}, x_{2})^{T}

y = (y_{1}, y_{2})^{T}

1 : x_{1} y_{2} + x_{2} y_{1}, ε : x_{1} y_{2} - x_{2} y_{1}, 2 : (x_{2} y_{2} x_{1} y_{1}) .

The first two combinations are invariant under

r

and respectively even and odd under

s

; the last transforms by the displayed doublet matrices.

Solution

Choose the doublet bases so that the column of barred fields

B = (\overset{ˉ}{ψ}_{L}^{1}, \overset{ˉ}{ψ}_{L}^{2})^{T}

and

R = (ψ_{R}^{1}, ψ_{R}^{2})^{T}

both transform by the doublet matrices in d. This choice follows the source's assignment of the barred doublet itself. Since

H

transforms in

ε

, its Yukawa coupling must multiply a fermion bilinear in

ε

There is exactly one such bilinear in

2 \otimes 2

: the Clebsch-Gordan coefficients in d give

B_{1} R_{2} - B_{2} R_{1}

. Equivalently, writing the upper block of the Yukawa matrix as

A

, invariance requires

D (r)^{T} A D (r) = A, D (s)^{T} A D (s) = - A .

The first equation sets

A_{11} = A_{22} = 0

, because

α^{2} \neq = 1

; the second sets

A_{21} = - A_{12}

A mixed doublet-singlet bilinear remains a doublet after multiplication by

H

, so all four mixed entries vanish. Finally

\overset{ˉ}{ψ}_{L}^{3} ψ_{R}^{3}

transforms in

ε \otimes 1 = ε

, and multiplication by

H

makes it invariant. Therefore

λ = 0 - a 0 a 00 00 b, a, b \in C .

The allowed Yukawa interaction is

H {a (\overset{ˉ}{ψ}_{L}^{1} ψ_{R}^{2} - \overset{ˉ}{ψ}_{L}^{2} ψ_{R}^{1}) + b \overset{ˉ}{ψ}_{L}^{3} ψ_{R}^{3}}

, together with its Hermitian conjugate when forming a real Lagrangian density.

The displayed texture depends on basis. If instead the unbarred left doublet is assigned

D

in the same ordering as the right doublet, its barred doublet has the dual representation

D^{*}

and the equivalent upper block is

a diag (1, - 1)

. Exchanging the two barred components converts between these descriptions. In either convention there are exactly two independent complex Yukawa couplings.

Solution

For a curve

O (t) = I + tX + O (t^{2})

in the special orthogonal group, differentiate

O (t)^{T} O (t) = I

at the identity. This gives

X^{T} + X = 0

. Conversely, the matrix exponential of any real skew-symmetric matrix is orthogonal and has determinant

det e^{tX} = e^{t tr X} = 1

. Hence

so (N) = {X \in M_{N} (R) : X^{T} = - X}, [X, Y] = X Y - Y X .

A basis is

L_{ij} = E_{ij} - E_{ji}

for

1 \leq i < j \leq N

, where

E_{ij}

is a matrix unit. If

L_{ji} = - L_{ij}

and

L_{ii} = 0

, their Lie brackets are

[L_{ij}, L_{k l}] = δ_{jk} L_{i l} - δ_{j l} L_{ik} - δ_{ik} L_{j l} + δ_{i l} L_{jk} .

There are

(2 N)

independent off-diagonal entries. A maximal torus rotates independently in mutually orthogonal two-planes, generated by

L_{12}, L_{34}, \dots

. Thus

dim so (N) = \frac{N ( N - 1 )}{2}, rank SO (N) = ⌊ \frac{N}{2} ⌋ .

The maximality follows by simultaneously decomposing commuting real skew-symmetric matrices into rotation two-planes, with at most one remaining fixed axis.

Solution

The unbroken group is the stabilizer subgroup of the vacuum expectation value. For a nonzero vector vacuum expectation value, an orthogonal transformation puts

⟨ ϕ ⟩ = v e_{N}

. A special orthogonal transformation fixing it has the block form

diag (A, 1)

with

A \in SO (N - 1)

. Consequently

SO (N) ⟶ SO (N - 1) (v \neq = 0) .

The zero vacuum expectation value leaves the entire special orthogonal group unbroken.

For a real symmetric tensor write its vacuum expectation value as a real symmetric matrix

Q

. The transformation is

Q \mapsto OQ O^{T}

. The spectral theorem for real symmetric matrices gives an orthonormal basis in which

Q = diag (q_{1} I_{m_{1}}, \dots, q_{p} I_{m_{p}}), q_{i} \neq = q_{j} (i \neq = j), \sum_{i} m_{i} = N .

Because

OQ O^{T} = Q

is equivalent to

OQ = QO

, an unbroken transformation preserves each eigenspace. The complete unbroken group is therefore

H = S (O (m_{1}) \times \dots \times O (m_{p})) = {(A_{1}, \dots, A_{p}) : i \prod det A_{i} = 1} .

Its identity component is

H^{0} = \prod_{i} SO (m_{i})

. This symmetric tensor vacuum stabilizer shows why the representation alone does not determine a unique breaking pattern:

If $Q = q I$ , all of $SO (N)$ survives, even when $q \neq = 0$ .
If all eigenvalues are distinct, $H ≅ (Z /2 Z)^{N - 1}$ , consisting of diagonal sign changes with total sign $+ 1$ ; no continuous symmetry survives.
If the two eigenvalue multiplicities are $m$ and $N - m$ , then $H = S (O (m) \times O (N - m))$ , with identity component $SO (m) \times SO (N - m)$ .

For example a rank-one tensor

Q = v e_{N} e_{N}^{T}

preserves

O (N - 1)

embedded by

A \mapsto diag (A, det A)

. Its continuous unbroken group is

SO (N - 1)

, but it also allows reversal of the distinguished axis. Restricting to a traceless symmetric tensor imposes

\sum_{i} m_{i} q_{i} = 0

and otherwise leaves the same classification.

Solution

The complexification of a Lie algebra gives

so (2 n, C)

of type Dn root system and

so (2 n + 1, C)

of type Bn root system. For

D_{n}

, the simple roots

α_{i} = e_{i} - e_{i + 1}

for

1 \leq i < n

and

α_{n} = e_{n - 1} + e_{n}

give a chain

1, \dots, n - 2

with two terminal nodes

n - 1, n

joined to

n - 2

. All edges are single. For

B_{n}

, use

α_{i} = e_{i} - e_{i + 1}

for

i < n

and

α_{n} = e_{n}

. The Bn Dynkin diagram is a chain with a double final edge and an arrow pointing to the short node

n

Figure 1.
Dynkin diagrams of the odd and even orthogonal Lie algebras, with the final arrow pointing toward the short root
. **Details**.

The usual family diagrams apply to

D_{n}

for

n \geq 4

and

B_{n}

for

n \geq 2

. The small cases are

D_{3} ≅ A_{3}

D_{2} ≅ A_{1} \oplus A_{1}

B_{1} ≅ A_{1}

;

so (2)

is one-dimensional and Abelian.

To fix Casimir normalization for orthogonal vector representations, use Hermitian Lie algebra generators

t_{ij} = i (E_{ij} - E_{ji}) / 2

, so the defining representation has Dynkin index

c (N) = 1

. With long roots of squared length two, the quadratic Casimir operator has Casimir eigenvalue

C_{2} (λ) = \frac{1}{2} (λ, λ + 2 ρ) = \frac{1}{2} λ^{T} G (λ + 21),

where

ρ

is the Weyl vector and

λ

in the final expression is the column of Dynkin labels. The defining fundamental representation has highest weight

(1, 0, \dots, 0)

in the stated family conventions.

For

D_{n}

, the row supplied in the source has sum

n - 1

and first entry one, so

C_{2} (2n) = \frac{1}{2} {1 + 2 (n - 1)} = \frac{2 n - 1}{2} .

For

B_{n}

, the supplied row has sum

n - \frac{1}{2}

, giving

C_{2} (2n + 1) = \frac{1}{2} {1 + 2 (n - \frac{1}{2})} = n .

Taking the trace of

\sum_{a} t_{a} t_{a} = C_{2} (R) I_{R}

gives the relation between the Dynkin index and Casimir eigenvalue,

c (R) dim g = C_{2} (R) dim R

. Therefore

C_{2} (N) = \frac{N - 1}{2}, c (N) = 1.

There is also a direct check: each coordinate appears in

N - 1

planes and

t_{ij}^{2} = (E_{ii} + E_{jj}) /2

, so

\sum_{i < j} t_{ij}^{2} = (N - 1) I /2

If the Casimir operator is instead defined by

C (λ) = (λ, λ + 2 ρ)

, every Lie algebra generator is larger by

2

and the corresponding answers are

C (N) = N - 1

and

c (N) = 2

. These are the same Lie algebra in a different normalization. At the exceptional small ranks the family Dynkin labels of the vector change, but the direct matrix computation remains valid.

Solution

Adopt the Minkowski metric

η = diag (- 1, + 1)

and the Clifford algebra convention

{γ_{μ}, γ_{ν}} = 2 η_{μν} I

. The source's Pauli matrices give

γ_{0} = (01 - 1 0), γ_{1} = (0110) .

They satisfy

γ_{0}^{2} = - I

γ_{1}^{2} = I

and

γ_{0} γ_{1} = - γ_{1} γ_{0} = - σ_{3}

, which verifies the required Clifford algebra. Raising the indices gives

γ^{0} = - γ_{0}

and

γ^{1} = γ_{1}

Fix the sign of the rapidity

θ

by taking the Lorentz boost

(t^{'} x^{'}) = Λ (θ) (t x), Λ (θ) = (cosh θ - sinh θ - sinh θ cosh θ) .

The corresponding Spinor representation of the Lorentz group is

S (θ) = exp (\frac{θ}{2} γ^{0} γ^{1}) = exp (\frac{θ}{2} σ_{3}) = diag (e^{θ /2}, e^{- θ /2}) .

For example

S^{- 1} γ^{0} S = γ^{0} cosh θ - γ^{1} sinh θ

and

S^{- 1} γ^{1} S = γ^{1} cosh θ - γ^{0} sinh θ

, establishing Lorentz covariance by

S^{- 1} γ^{μ} S = Λ^{μ}_{ν} γ^{ν}

Thus the Dirac spinor transforms as

ψ^{'} (x^{'}) = S (θ) ψ (x)

, with infinitesimal law

δ ψ = \frac{θ}{2} σ_{3} ψ, δ ψ_{+} = \frac{θ}{2} ψ_{+}, δ ψ_{-} = - \frac{θ}{2} ψ_{-} .

These variations compare the fields at corresponding spacetime points. At a fixed coordinate one adds the usual orbital term

- δ x^{μ} \partial_{μ} ψ

. Reversing the rapidity convention reverses both boost signs consistently. Infinitesimal transformations describe the proper orthochronous Lorentz group.

Solution

Choose the chirality operator

Γ = γ^{0} γ^{1} = σ_{3}, Γ^{2} = I, {Γ, γ^{μ}} = 0.

A Weyl spinor is a Dirac spinor restricted to one eigenspace of

Γ

. The chiral projectors are

P_{R} = (I + Γ) /2

and

P_{L} = (I - Γ) /2

. Consequently

ψ_{R} = P_{R} ψ = (ψ_{+} 0), ψ_{L} = P_{L} ψ = (0 ψ_{-}) .

The two independent Weyl spinors each contain one complex component. Since

[S (θ), Γ] = 0

, each chirality is preserved by every connected Lorentz transformation, with

ψ_{R} \mapsto e^{θ /2} ψ_{R}

and

ψ_{L} \mapsto e^{- θ /2} ψ_{L}

. A spatial reflection exchanges the two chiralities.

The left/right names depend on the sign convention for

Γ

. Choosing

γ_{0} γ_{1} = - σ_{3}

as the chirality operator exchanges the labels

L

and

R

; the two invariant one-component representations are the same.

These are component counts before imposing an equation of motion. In particular the simultaneous reality and chirality constraints are compatible in this spacetime dimension, unlike the familiar four-dimensional case.

d

Solution

Because the present

γ_{0}

is anti-Hermitian, choose the Dirac adjoint with Hermitian intertwining matrix

A = i γ_{0}

\overset{ˉ}{ψ} = ψ^{†} A, A = i γ_{0} = (0 i - i 0) .

This convention makes the Dirac scalar bilinear

\overset{ˉ}{ψ} ψ

Hermitian. For real

m

, a Lorentz invariant mass term is

L_{m} = - m \overset{ˉ}{ψ} ψ = im (ψ_{+}^{*} ψ_{-} - ψ_{-}^{*} ψ_{+}) .

To check infinitesimal Lorentz invariance, put

K = σ_{3} /2

. Then

K^{†} = K

and

K^{†} A + A K = {K, A} = 0

. Using a,

δ L_{m} = - m θ ψ^{†} (K^{†} A + A K) ψ = 0.

Equivalently,

δ (ψ_{+}^{*} ψ_{-}) = θ ψ_{+}^{*} ψ_{-} /2 - θ ψ_{+}^{*} ψ_{-} /2 = 0

, and similarly for its complex conjugate. Exponentiation gives

S^{†} A S = A

and therefore

\overset{ˉ}{ψ}^{'} (x^{'}) = \overset{ˉ}{ψ} (x) S^{- 1}

At fixed coordinates

L_{m}

has only the orbital variation of a Lorentz scalar; the spacetime measure is invariant under the connected Lorentz group, so the integrated action is invariant. The mass term pairs the two opposite chiralities; retaining only one Weyl spinor removes this Dirac mass. An equivalent spinor phase convention gives a real multiple of

ψ_{+}^{*} ψ_{-} + ψ_{-}^{*} ψ_{+}

instead.

4

a

Solution

For the A5 root system, take simple roots

α_{i} = e_{i} - e_{i + 1}

i = 1, \dots, 5

, in the hyperplane

\sum_{i = 1}^{6} x_{i} = 0

. Their squared lengths are two, adjacent simple roots have inner product

- 1

, and all other pairs are orthogonal. Hence the An Dynkin diagram is the five-node single-edge chain

1 \circ - 2 \circ - 3 \circ - 4 \circ - 5 \circ .

Its Cartan matrix is

A (A_{5}) = 2 - 1 000 - 1 2 - 1 00 0 - 1 2 - 1 0 00 - 1 2 - 1 000 - 1 2 .

The complex Lie algebra is

sl_{6} (C)

; the corresponding compact special unitary group is

S U (6)

, whose Lie algebra is its compact real form.

b

Solution

Let

V = C^{6}

be the defining fundamental representation of

S U (6)

. Its weights are

w_{i} = e_{i} - \frac{1}{6} \sum_{j = 1}^{6} e_{j}

. Pairing with the simple coroots gives Dynkin labels

(w_{i})_{a} = δ_{ia} - δ_{i, a + 1}

. Thus the complete list is

w_{1} w_{2} w_{3} w_{4} w_{5} w_{6} a_{1} 1 - 1 0000 a_{2} 01 - 1 000 a_{3} 001 - 1 00 a_{4} 0001 - 1 0 a_{5} 00001 - 1 .

Successive lowering operators subtract the corresponding rows of the Cartan matrix in a. Each weight multiplicity is one.

The complex-conjugate representation

\overline{6} ≅ V^{*}

is the dual representation and has the negatives of these weights. Its highest weight is

- w_{6} = (0, 0, 0, 0, 1)

, and its entire weight decomposition is

Wt (\overline{6}) = ⎩ ⎨ ⎧ (- 1, 0, 0, 0, 0), (1, - 1, 0, 0, 0), (0, 1, - 1, 0, 0), (0, 0, 1, - 1, 0), (0, 0, 0, 1, - 1), (0, 0, 0, 0, 1) ⎭ ⎬ ⎫ .

Every entry has weight multiplicity one.

The two-index antisymmetric representation is

Λ^{2} V = 15

, of dimension

(2 6) = 15

and highest weight

w_{1} + w_{2} = (0, 1, 0, 0, 0)

. A weight basis is

e_{i} \land e_{j}

for

i < j

, carrying weight

w_{i} + w_{j}

. The fifteen sums, all distinct and all of weight multiplicity one, are

ij 121314151623242526343536454656 a_{1} 01111 - 1 - 1 - 1 - 1 000000 a_{2} 1 - 1 0000111 - 1 - 1 - 1 000 a_{3} 01 - 1 001 - 1 00011 - 1 - 1 0 a_{4} 001 - 1 001 - 1 01 - 1 001 - 1 a_{5} 0001 - 1 001 - 1 01 - 1 1 - 1 0 .

The exclusion of

i = j

comes from the exterior square, where

e_{i} \land e_{i} = 0

c

Solution

fundamental representation:

For the standard A4 root system inside the A5 root system, let

p

delete the fifth Dynkin label. Put

v_{i} = p (w_{i})

for

i \leq 5

. These are precisely the five weights of the defining

S U (5)

(v_{1}, v_{2}, v_{3}, v_{4}, v_{5}) = ((1, 0, 0, 0), (- 1, 1, 0, 0), (0, - 1, 1, 0), (0, 0, - 1, 1), (0, 0, 0, - 1)) .

Since

p (w_{6}) = 0

, the six projected weights of

\overline{6}

are

- v_{1}, \dots, - v_{5}, 0

. The first five form the dual representation

\overline{5}

, with highest weight

(0, 0, 0, 1)

; the last is a singlet. Thus

\overline{6} ↓_{S U (5)} = \overline{5} \oplus 1 .

For

15

, split the fifteen weights in b into the ten pairs with

1 \leq i < j \leq 5

and the five pairs

i 6

. The former project to

v_{i} + v_{j}

, the complete weight decomposition of

Λ^{2} 5 = 10

, with highest weight

(0, 1, 0, 0)

. The latter project to

v_{i}

, the complete weight decomposition of

5

. Therefore

15 ↓_{S U (5)} = 10 \oplus 5 .

This restriction of a representation can also be checked without a table:

V ↓_{S U (5)} = 5 \oplus 1

, so its dual representation is

\overline{5} \oplus 1

, while its exterior square is

Λ^{2} 5 \oplus (5 \otimes 1)

. The dimensions match

6 = 5 + 1

and

15 = 10 + 5

d

Solution

One generation of left-handed Standard Model fermions fits into

\overline{5} \oplus 10

of an SU(5) grand unified theory. With hypercharge defined by

Q = T_{3} + Y

, their restriction of a representation to

S U (3)_{c} \times S U (2)_{L} \times U (1)_{Y}

\overline{5} 10 ⟶ (\overline{3}, 1)_{1/3} \oplus (1, 2)_{- 1/2} ⟶ (3, 2)_{1/6} \oplus (\overline{3}, 1)_{- 2/3} \oplus (1, 1)_{1} carrying d^{c}, L, carrying Q, u^{c}, e^{c} .

Here

Q

in the multiplet list denotes the left-handed quark doublet, and the superscript

c

labels left-handed charge-conjugate fields of the corresponding right-handed fermions.

The results of c therefore give

\overline{6} \oplus 15 ⟶ one ordinary generation \overline{5} \oplus 10 \oplus additional multiplets 1 \oplus 5 .

Thus the required matter representations are present. The extra singlet has Standard Model representation

(1, 1)_{0}

and can describe the left-handed charge-conjugate field of a right-handed neutrino. Its gauge singlet character permits a Majorana mass term; a Yukawa interaction with a lepton doublet and a suitable Higgs field can then supply neutrino masses. The branching alone does not fix those interactions or their scales.

The extra

5

has

5 ⟶ (3, 1)_{- 1/3} \oplus (1, 2)_{1/2} .

If it describes Weyl fermions, these are additional coloured and electroweak fermions in the conjugate Standard Model representations to the ordinary

\overline{5}

. With an additional

\overline{5}

, they can form a vectorlike gauge spectrum and acquire invariant masses. Pairing them with the only ordinary

\overline{5}

would also remove that generation's

d^{c}, L

fields, so a realistic heavy pairing needs extra matter. If the

5

instead describes a scalar field, it has the quantum numbers of the usual

S U (5)

Higgs field: a weak doublet and a coloured triplet. A single fermionic

15

does not turn its extra fermions into a scalar Higgs field merely by branching.

There is a further consistency condition if these are the entire chiral matter content of an

S U (6)

gauge theory. Normalize the cubic anomaly coefficient by

A (6) = 1

. The dual representation has

A (\overline{6}) = - 1

, and the two-index antisymmetric representation has

A (15) = 6 - 4 = 2

. Thus one copy of each has nonzero gauge anomaly,

A_{total} = 1

. Their inclusion of the desired

S U (5)

multiplets is therefore not, by itself, a complete consistent chiral unified theory.

For example, anomaly cancellation holds for

15 \oplus 2 \overline{6}

, whose restriction of a representation is

15 \oplus 2 \overline{6} ⟶ (10 \oplus \overline{5}) \oplus (5 \oplus \overline{5}) \oplus 21 .

This leaves one ordinary generation, a potentially massive vectorlike gauge spectrum, and two possible singlet neutrinos. This anomaly-free matter choice is also used in explicit

S U (6)

model constructions.^[ref] These interpretations require additional choices of matter statistics, spontaneous symmetry breaking, and interactions; the weight systems determine the multiplet content.

2024 (A15090W1)

Source PDF: mmathphys.physics.ox.ac.uk/sites/g/files/vzsiac2991/files/mmathphys/documents/media/a15090w1.pdf

1

a

Solution

A group representation on a vector space

V

is a group homomorphism

R : G \to GL (V)

R (g h) = R (g) R (h)

and

R (e) = I

. A reducible representation has a nonzero proper invariant subspace

W \subset V

, meaning

R (g) W \subseteq W

for every

g

. Over the complex numbers, a finite-dimensional representation of a finite group is a direct sum of irreducible representations, by Maschke's theorem.

The character of a representation is

χ_{R} (g) = tr R (g)

. The cyclic property of the trace makes it a class function:

χ_{R} (h g h^{- 1}) = tr (R (h) R (g) R (h)^{- 1}) = χ_{R} (g) .

Thus a character is specified by one number per conjugacy class, much less information than all the representation matrices. Moreover,

χ_{R \oplus S} = χ_{R} + χ_{S}, χ_{R \otimes S} = χ_{R} χ_{S}, m_{r} = \frac{1}{∣ G ∣} g \in G \sum \overline{χ_{r} (g)} χ_{R} (g) .

Here character orthogonality says that

m_{r}

is the multiplicity of the irreducible representation

r

. Consequently characters determine finite-dimensional complex representations up to equivalence of group representations and compute tensor product of group representations decompositions without constructing a new basis explicitly.

b

Solution

Compose permutations from right to left. The three transpositions are

τ_{1} = (23)

τ_{2} = (13)

and

τ_{3} = (12)

. In cycle notation, direct composition gives

τ_{1} τ_{2} = (123), τ_{2} τ_{1} = (132), τ_{i}^{2} = e .

The complete symmetric group is therefore

S_{3} = {e, τ_{1}, τ_{2}, τ_{3}, τ_{1} τ_{2}, τ_{2} τ_{1}} .

Conjugation relabels the entries of a cycle, so the conjugacy classes are precisely

{e}, {τ_{1}, τ_{2}, τ_{3}}, {τ_{1} τ_{2}, τ_{2} τ_{1}} .

Their sizes are

1, 3, 2

, respectively. The differing orders of group elements also show that these three sets cannot merge under conjugation.

c

Solution

There is one complex irreducible representation for each conjugacy class, hence three. The trivial representation

1

and sign representation

ε

are distinct and one-dimensional. The sum of squares of irreducible representation dimensions leaves

1^{2} + 1^{2} + d^{2} = 6, d = 2.

Call the remaining irreducible representation

2

. The permutation representation on

C^{3}

has character equal to the number of fixed objects, namely

(3, 1, 0)

on the ordered conjugacy classes

(e, transpositions, three-cycles)

. Removing its constant-vector trivial representation gives

(2, 0, - 1)

, whose character inner product with itself is one. This constructs

2

and gives the character table

class class size χ_{1} χ_{ε} χ_{2} e 1112 (12) 31 - 1 0 (123) 211 - 1 .

To obtain every Clebsch-Gordan decomposition of group representations, multiply the characters pointwise. In particular,

χ_{ε}^{2} = χ_{1}

χ_{ε} χ_{2} = χ_{2}

and

χ_{2}^{2} = (4, 0, 1) = χ_{1} + χ_{ε} + χ_{2} .

Thus all ordered pairs of irreducible representations are covered by

\otimes 1 ε 2 11 ε 2 ε ε 12 222 1 \oplus ε \oplus 2 .

d

Solution

All three displayed transposition matrices have trace zero. Multiplying two of them gives

R (τ_{1} τ_{2}) = diag (α, α^{- 1}), R (τ_{2} τ_{1}) = diag (α^{- 1}, α),

since

α^{3} = 1

. Their traces are

α + α^{- 1} = - 1

, whereas

tr R (e) = 2

. Therefore

χ_{R} = (2, 0, - 1), ⟨ χ_{R}, χ_{R} ⟩ = \frac{4 + 3 \cdot 0 + 2 \cdot 1}{6} = 1.

By character orthogonality,

R

is the irreducible representation

2

from c.

There is also a direct invariant subspace argument. The two distinct eigenvalues of

R (τ_{1} τ_{2})

force any invariant line to be one of the two coordinate lines. But

R (τ_{3})

exchanges those lines. Neither is invariant under the whole group, so this two-dimensional representation is irreducible.

e

Solution

The scalar fields commute, so a degree-

d

expression in one doublet belongs to a symmetric power, rather than an arbitrary tensor product. Put

x = ϕ_{1}

y = ϕ_{2}

. The generators of a group

r = τ_{1} τ_{2}

and

s = τ_{3}

act by

r : (x, y) \mapsto (αx, α^{- 1} y), s : (x, y) \mapsto (y, x) .

A monomial

x^{a} y^{b}

is fixed by

r

precisely when

a - b

is divisible by three. Imposing

s

then pairs the exponents. Through degree four the resulting holomorphic polynomial invariants are

degree 01234 basis 1 none x y x^{3} + y^{3} x^{2} y^{2}

In particular, two explicit invariant terms are

ϕ_{1} ϕ_{2}

and

ϕ_{1}^{3} + ϕ_{2}^{3}

. Their invariance follows directly from

α^{3} = 1

and the interchange of the two components. A quartic example is

(ϕ_{1} ϕ_{2})^{2}

For a physical real scalar potential, complex conjugate fields are allowed as well. The Hermitian inner product

ϕ^{†} ϕ

and its square are also invariant, while a holomorphic term can enter with its complex conjugate, for example

b (ϕ_{1}^{3} + ϕ_{2}^{3}) + \overset{ˉ}{b} (\overset{ˉ}{ϕ}_{1}^{3} + \overset{ˉ}{ϕ}_{2}^{3})

. There is still no invariant linear term, because neither the doublet nor its conjugate representation contains a trivial representation. Thus the possible nonconstant total degrees are

S_{3} : 2, 3, 4.

The constant term is unrestricted.

For the fundamental representation of the special unitary group

S U (2)

, its central element

- I

sends

ϕ \mapsto - ϕ

and

\overset{ˉ}{ϕ} \mapsto - \overset{ˉ}{ϕ}

. Every odd-degree polynomial changes sign, excluding degrees one and three. Further,

S U (2)

acts transitively on the unit sphere in

C^{2}

: a unit vector

(u, v)^{T}

is the first column of

(u v - \overset{v}{ˉ} \overset{u}{ˉ}) \in S U (2) .

An invariant polynomial must therefore depend only on

ϕ^{†} ϕ

. Through quartic order the general real scalar potential is

V = V_{0} + m^{2} ϕ^{†} ϕ + λ (ϕ^{†} ϕ)^{2}, V_{0}, m^{2}, λ \in R .

(5)

Hence the possible nonconstant degrees are

2, 4

; the cubic term permitted by

S_{3}

is excluded by

S U (2)

If the notation

V (ϕ)

is instead read as restricting to holomorphic polynomials alone, the distinction matters: a single commuting

S U (2)

doublet has no nonconstant holomorphic polynomial invariant. The candidate contraction

ϵ_{ab} ϕ_{a} ϕ_{b} = ϕ_{1} ϕ_{2} - ϕ_{2} ϕ_{1}

vanishes identically. The quadratic and quartic

S U (2)

terms above require complex conjugate fields. This also explains why counting a trivial representation in

2 \otimes 2

alone would give an incorrect holomorphic answer.

2

a

Solution

Use the mathematical Lie algebra convention

U (t) = I + tX + O (t^{2})

. Differentiating

U^{†} U = I

and

det U = 1

t = 0

gives

X^{†} = - X

and

tr X = 0

. Conversely, for such an

X

, the matrix exponential is unitary and has determinant

det e^{tX} = e^{t tr X} = 1

. Thus

su (4) = {X \in M_{4} (C) : X^{†} = - X, tr X = 0}, [X, Y] = X Y - Y X .

The four diagonal entries are purely imaginary, and the six entries above the diagonal are arbitrary complex numbers determining the entries below it. This gives

4 + 2 \cdot 6 = 16

real parameters before imposing the one real trace constraint, so

dim_{R} su (4) = 15

A Cartan subalgebra of this compact Lie algebra is a maximal commuting Lie subalgebra; choose

h = {i diag (h_{1}, h_{2}, h_{3}, h_{4}) : h_{i} \in R, \sum_{i} h_{i} = 0} .

A simple basis is

i H_{1}, i H_{2}, i H_{3}

, where

H_{1} = diag (1, - 1, 0, 0), H_{2} = diag (0, 1, - 1, 0), H_{3} = diag (0, 0, 1, - 1) .

Any matrix commuting with every diagonal element is itself diagonal, which proves maximality. Commuting skew-Hermitian matrices can also be simultaneously diagonalized, so no larger commuting Lie subalgebra is possible. Consequently

dim su (4) = 15, rank su (4) = dim h = 3.

The Hermitian quantum generator convention instead uses the Hermitian

H_{a}

and

U = exp (i θ^{a} H_{a})

. The two conventions describe the same Lie algebra.

b

Solution

Use the Hermitian

H_{a}

from a to record real weight coordinates:

H_{a} e_{j} = (w_{j})_{a} e_{j}

. Reading off the diagonal entries gives

w_{1} = (1, 0, 0), w_{2} = (- 1, 1, 0), w_{3} = (0, - 1, 1), w_{4} = (0, 0, - 1) .

These are weights in the simple-coroot coordinate system, not coordinates in an orthonormal basis. In the skew-Hermitian convention they mean

(i H_{a}) e_{j} = i (w_{j})_{a} e_{j}

. Intrinsically, the weight on

i diag (h_{1}, h_{2}, h_{3}, h_{4})

is the functional

h \mapsto h_{j}

, with

\sum_{j} h_{j} = 0

The conjugate representation acts by

U^{*}

, so a diagonal transformation with entries

e^{i h_{j}}

becomes one with entries

e^{- i h_{j}}

. Thus its four weights are

\overset{w}{ˉ}_{1} = (- 1, 0, 0), \overset{w}{ˉ}_{2} = (1, - 1, 0), \overset{w}{ˉ}_{3} = (0, 1, - 1), \overset{w}{ˉ}_{4} = (0, 0, 1) .

All have multiplicity one. In both representations the sum of the weights is zero, consistent with the vanishing trace of every Lie algebra generator.

c

Solution

Write

[r_{1}, r_{2}, \dots]

for a Young diagram with row lengths

r_{i}

, and depict an empty box by

□

. Appending the box of the fundamental representation to another single box gives a row or a column:

□ \otimes □ = (□ □) \oplus □ □, [1] \otimes [1] = [2] \oplus [1, 1] .

The corresponding symmetric tensors and antisymmetric tensors have dimensions

4 \cdot 5/2 = 10

and

4 \cdot 3/2 = 6

, so

4 \otimes 4 = 10_{s} \oplus 6_{a} .

For the conjugate representation, the invariant volume tensor identifies

\overset{ˉ}{4} ≅ Λ^{3} 4

, represented by a column of three boxes. Adding one box gives

[1] \otimes [1, 1, 1] = [2, 1, 1] \oplus [1, 1, 1, 1], □ \otimes □ □ □ = □ □ □ □ \oplus □ □ □ □ .

A column of four boxes is the determinant representation, which is trivial on

S U (4)

. The remaining Young diagram has Dynkin labels

(1, 0, 1)

and dimension fifteen. For example the hook-content formula gives

dim [2, 1, 1] = \frac{4 \cdot 5 \cdot 3 \cdot 2}{4 \cdot 1 \cdot 2 \cdot 1} = 15.

Equivalently, a mixed tensor

M^{i}_{j}

transforms by

M \mapsto U M U^{†}

and splits into its trace and traceless matrix parts. Therefore

4 \otimes \overset{ˉ}{4} = 1 \oplus 15,

(5)

where

15

is the Adjoint representation.

d

Solution

The embedded

S U (3)

acts on the first three coordinates and fixes the fourth. This immediately gives the branching rules

4 ↓ S U (3) = 3 \oplus 1, \overset{ˉ}{4} ↓ S U (3) = \overset{ˉ}{3} \oplus 1 .

Distribute the tensor product of group representations over these direct sums and split

3 \otimes \overset{ˉ}{3} = 8 \oplus 1

by taking the trace:

(4 \otimes \overset{ˉ}{4}) ↓ S U (3) = 8 \oplus 3 \oplus \overset{ˉ}{3} \oplus 1 \oplus 1 .

One can locate every summand using the block matrix

M = (A c b d) ⟼ (U_{3} A U_{3}^{†} c U_{3}^{†} U_{3} b d) .

The traceless part of

A

8

b

3

c

\overset{ˉ}{3}

, and

tr A, d

are independent trivial representations. Removing the overall trace instead gives

15 ↓ S U (3) = 8 \oplus 3 \oplus \overset{ˉ}{3} \oplus 1

. The dimension check is

16 = 8 + 3 + 3 + 1 + 1

e

Solution

A meson has quark model flavour content

q_{i} \overset{q}{ˉ}^{j}

, so four quark flavours give sixteen flavour states in

4 \otimes \overset{ˉ}{4} = 15 \oplus 1 .

This counts a flavour set for a specified spin and orbital assignment; it does not count every excited meson. Three quark flavours instead give the familiar nine states

3 \otimes \overset{ˉ}{3} = 8 \oplus 1

Taking the fourth quark to be the charm quark, the first nine combinations

q_{α} \overset{q}{ˉ}^{β}

, with

α, β \in {u, d, s}

, are the light sector. The extra seven are

q_{α} \overset{c}{ˉ}

3

c \overset{q}{ˉ}^{α}

\overset{ˉ}{3}

, and

c \overset{c}{ˉ}

1

under light-flavour

S U (3)

. For the lowest pseudoscalar mesons, the open-charm states are the three D meson/Ds meson states and their antiparticles, and the hidden-charm state is the eta c meson. The two neutral trivial representation states can mix when the flavour symmetry is broken.

The intended limitation is the accuracy of flavour symmetry: the charm quark is much heavier than the three light quarks, so a quark mass term is far from

S U (4)

invariant. Exact flavour

S U (4)

would relate light and charmed states with the same spin, contrary to their large observed mass splittings. However, saying that a four-flavour quark model is never considered would be too strong: flavour

S U (4)

is used as classification bookkeeping, with much stronger symmetry breaking than light-flavour

S U (3)

.^[ref] This is a flavour special unitary group, distinct from the three-colour gauge group of Quantum chromodynamics.

3

a

Solution

For a real linear change of coordinates

x^{'} = A x

, one has

d x^{'} = A d x

. The Euclidean metric therefore transforms to

g^{'} = d x^{' T} d x^{'} = d x^{T} A^{T} A d x .

It agrees with

d x^{T} d x

for every displacement precisely when

A^{T} A = I

. For example, equality of the associated quadratic forms determines their symmetric coefficient matrices by polarization identity. The orthogonal group is defined by this condition, so

Stab_{GL (6, R)} (g) = O (6) = {A : A^{T} A = I} .

Taking determinants gives

(det A)^{2} = 1

. Thus these matrices are invertible automatically; both signs of the determinant are allowed, since the metric alone does not specify an orientation.

b

Solution

Write

x = (x_{1}, x_{2}, x_{3})^{T}

v = (x_{4}, x_{5}, x_{6})^{T}

and

z = x + i v

. Here products of differentials in a metric tensor are symmetric products, unlike the wedge products used later. Then

g = k = 1 \sum 3 d z_{k} d \overset{z}{ˉ}_{k} = \frac{1}{2} k = 1 \sum 3 (d z_{k} \otimes d \overset{z}{ˉ}_{k} + d \overset{z}{ˉ}_{k} \otimes d z_{k}) .

Indeed

d z_{k} d \overset{z}{ˉ}_{k} = d x_{k}^{2} + d x_{k + 3}^{2}

. A transformation

z^{'} = U z

preserves this metric whenever

U^{†} U = I

, because

\sum_{k} d \overset{z}{ˉ}_{k}^{'} d z_{k}^{'} = d \overset{z}{ˉ}^{T} U^{†} U d z = d \overset{z}{ˉ}^{T} d z .

Thus the entire unitary group

U (3)

preserves it, including its special unitary group

S U (3)

To obtain a real matrix embedding, write

U = A + i B

with

A, B

real. Then

ρ : S U (3) ↪ SO (6), ρ (U) = (A B - B A) .

The unitarity equations imply

A^{T} A + B^{T} B = I

and

A^{T} B = B^{T} A

, hence

ρ (U)^{T} ρ (U) = I

. Multiplication of the block matrices gives

ρ (U V) = ρ (U) ρ (V)

, and the blocks recover

U

, proving injectivity. Finally

det_{R} ρ (U) = det_{C} U det_{C} \overset{ˉ}{U} = ∣ det_{C} U ∣^{2} = 1.

For this last identity, over the complex numbers the change of basis from

(x, v)

(z, \overset{z}{ˉ})

conjugates

ρ (U)

diag (U, \overset{ˉ}{U})

. This proves that the image lies in the special orthogonal group.

c

Solution

The change of variables in b diagonalizes the complexified action into

z \mapsto U z

and

\overset{z}{ˉ} \mapsto \overset{ˉ}{U} \overset{z}{ˉ}

. Consequently the conventional complex branching rule is

6_{SO (6)} ↓ S U (3) = 3 \oplus \overset{ˉ}{3} .

Precisely, this is a statement about the complexification of a real vector space carrying the real defining representation. Over the real numbers, the six-dimensional space is the realification of a complex representation

3

; the conjugate summands above are not separate three-dimensional real invariant subspaces.

The block embedding

SO (6) ↪ SO (7)

fixes the seventh coordinate. It adds a trivial representation, giving

7_{SO (7)} ↓ S U (3) = 3 \oplus \overset{ˉ}{3} \oplus 1 .

The dimensions are complex dimensions after complexification of a real vector space, or equivalently

6 + 1

as real dimensions before it.

d

Solution

Let

ρ_{3} (P) = P \otimes P \otimes P

act on the alternating coefficient tensor

φ

, and set

H = {P \in SO (7) : ρ_{3} (P) φ = φ} .

The coefficient convention here is the one in the source's hint; using the usual covariant pullback gives the same stabilizer subgroup, since an invertible transformation preserves a differential form if and only if its inverse does. The identity matrix fixes

φ

. If

P, Q

fix it, then

ρ_{3} (PQ) φ = ρ_{3} (P) ρ_{3} (Q) φ = φ

. Multiplying

ρ_{3} (P) φ = φ

ρ_{3} (P^{- 1})

shows that

P^{- 1}

fixes it as well. Thus

H

is a subgroup of

SO (7)

. Its defining equations are polynomial, so it is closed and hence a Matrix Lie group.

For

P (t) = I + tT + O (t^{2}) \in H

, differentiating

P^{T} P = I

gives

T^{T} = - T

. Differentiating the three factors in

ρ_{3} (P (t)) φ = φ

gives

h = {T \in so (7) : n \sum (T_{kn} φ_{n l m} + T_{l n} φ_{knm} + T_{mn} φ_{k l n}) = 0 for all k, l, m} .

The skew-symmetric matrix condition records the ambient special orthogonal group; it must be included when differentiating that group's definition.

These linear conditions are sufficient as well as necessary. On the tensor product, the infinitesimal action is

D (T) = T \otimes I \otimes I + I \otimes T \otimes I + I \otimes I \otimes T .

Its three terms commute, so

ρ_{3} (e^{tT}) = e^{t D (T)}

. If

D (T) φ = 0

, every positive power of

D (T)

annihilates

φ

, giving

e^{t D (T)} φ = φ

. Also

T \in so (7)

implies

e^{tT} \in SO (7)

. Therefore

e^{tT} \in H

, proving the asserted Lie algebra description. For the given positive three-form this is the compact G2 Lie algebra

g_{2}

e

Solution

For each complex coordinate, exchanging the factors of a wedge product of differential forms reverses its sign, giving

d z_{k} \land d \overset{z}{ˉ}_{k} = (d x_{k} + i d x_{k + 3}) \land (d x_{k} - i d x_{k + 3}) = - 2 i d x_{k} \land d x_{k + 3} .

Consequently the Kähler form is

ω = d x_{14} + d x_{25} + d x_{36}

, and

ω \land d x_{7} = d x_{147} + d x_{257} + d x_{367}

. Expanding the holomorphic volume form gives

Re ((d x_{1} + i d x_{4}) \land (d x_{2} + i d x_{5}) \land (d x_{3} + i d x_{6})) = d x_{123} - d x_{156} - d x_{426} - d x_{453} = d x_{123} - d x_{156} + d x_{246} - d x_{345} .

The sign of

d x_{426}

changes on exchanging

4, 2

, while reordering

4, 5, 3

3, 4, 5

requires two exchanges. This verifies every sign in

ϕ = ω \land d y + Re Ω, y = x_{7} .

Under

z \mapsto U z

, unitarity preserves

ω

ω^{'} = \frac{i}{2} \sum_{a, b} (\sum_{k} U_{ka} \overset{ˉ}{U}_{kb}) d z_{a} \land d \overset{z}{ˉ}_{b} = ω .

The determinant is the factor by which a transformation acts on a top exterior power, so

Ω^{'} = (det U) Ω = Ω

for

U \in S U (3)

. Extend the real embedding by fixing

y

. It preserves

ω

d y

and

Ω

, and therefore

ϕ

. Thus

U ⟼ diag (ρ (U), 1) embeds S U (3) into G_{2} .

(5)

The induced seven-dimensional representation is exactly the restriction already computed in c. In the conventional complex notation its branching rule is

7_{G_{2}} ↓ S U (3) = 3 \oplus \overset{ˉ}{3} \oplus 1 .

(6)

Over the real numbers, the six-dimensional summand is the underlying real space of the complex fundamental representation, and the last real line is the fixed

y

axis.

4

a

Solution

The image is in the unitary group, since

(z U)^{†} (z U) = \overset{z}{ˉ} z U^{†} U = I

for

z \in U (1)

and

U \in S U (n)

. The direct product of groups has multiplication

(z, U) (w, V) = (z w, U V)

. Scalar matrices commute with every matrix, so

f ((z, U) (w, V)) = z w U V = (z U) (w V) = f (z, U) f (w, V) .

Thus

f

is a group homomorphism; in particular

f (1, I) = I

b

Solution

First prove surjectivity. For any

A \in U (n)

, its determinant has modulus one. Choose

z \in U (1)

with

z^{n} = det A

and define

U = z^{- 1} A

. Then

U^{†} U = I

and

det U = z^{- n} det A = 1

, so

A = f (z, U)

Next,

f (z, U) = I

is equivalent to

U = z^{- 1} I

. This lies in

S U (n)

precisely when

z^{n} = 1

. Therefore the kernel of a group homomorphism is the central cyclic group

ker f = {(ζ^{k}, ζ^{- k} I_{n}) : k = 0, \dots, n - 1} ≅ Z_{n}, ζ = e^{2 πi / n} .

Its generator is

(ζ, ζ^{- 1} I_{n})

, which has order

n

. The inverse phases in the two factors are essential: this is an explicitly embedded subgroup, not a quotient by scalar matrices in just one factor. The first isomorphism theorem now gives

U (n) ≅ (U (1) \times S U (n)) / Z_{n} .

This also holds as a Lie group isomorphism: the induced map is smooth and a continuous bijection from a compact quotient space to the Hausdorff unitary group, and its differential is the Lie algebra isomorphism

(i t, X) \mapsto i t I + X

. For

n = 1

the kernel of a group homomorphism is trivial, consistently with

S U (1) = {1}

c

Solution

Use the standard chain ordering of the five simple roots of

S U (6)

. A Young diagram of row lengths

r_{1} \geq \dots \geq r_{6} = 0

has highest weight Dynkin labels

a_{i} = r_{i} - r_{i + 1}

. A full column of six boxes can be removed because it is the trivial determinant representation. The required data are

representation 6 \overset{ˉ}{6} 211535 (a_{1}, a_{2}, a_{3}, a_{4}, a_{5}) (1, 0, 0, 0, 0) (0, 0, 0, 0, 1) (2, 0, 0, 0, 0) (0, 1, 0, 0, 0) (1, 0, 0, 0, 1) row lengths [1] [1, 1, 1, 1, 1] [2] [1, 1] [2, 1, 1, 1, 1] tensor v^{i} w_{i} S^{ij} = S^{ji} A^{ij} = - A^{ji} M^{i}_{j}, M^{i}_{i} = 0 dim 66211535 .

In the same order, the Young tableaux are

□, □ □ □ □ □, □ □, □ □, □ □ □ □ □ □ .

Each upper index transforms by

U

, and each lower index by

U^{- 1}

. In particular

M \mapsto U M U^{- 1}

is the complex Adjoint representation. The column of five boxes can be represented either by

w_{i}

or by the totally antisymmetric tensor

B^{i_{1} \dots i_{5}}

, using the invariant six-index alternating tensor:

w_{j} = \frac{1}{5 !} ϵ_{j i_{1} \dots i_{5}} B^{i_{1} \dots i_{5}} .

Thus it really is the conjugate representation, rather than a second independent six-dimensional choice. Similarly converting the lower index of

M^{i}_{j}

into five alternating upper indices gives the displayed hook Young diagram; the removed trace would be its full six-box column component.

The dimension counts are

dim Sym^{2} C^{6} = \frac{6 \cdot 7}{2} = 21, dim Λ^{2} C^{6} = \frac{6 \cdot 5}{2} = 15, dim End_{0} (C^{6}) = 6^{2} - 1 = 35.

The first two count independent entries of symmetric tensors and antisymmetric tensors; the last removes one trace from a mixed tensor. The hook-content formula independently gives these dimensions from the Young diagrams.

d

Solution

Let index

0

denote the first coordinate, which the embedded

S U (5)

fixes, and let

a, b = 1, \dots, 5

denote the remaining coordinates. The fundamental representation and its conjugate representation immediately give

6 ↓ S U (5) = 1 \oplus 5, \overset{ˉ}{6} ↓ S U (5) = 1 \oplus \overset{ˉ}{5} .

For a symmetric tensor,

S^{00}

is a trivial representation,

S^{0 a} = S^{a 0}

is a

5

, and

S^{ab}

is the fifteen-dimensional symmetric square of a group representation of

5

. For an antisymmetric tensor,

A^{00} = 0

A^{0 a} = - A^{a 0}

is a

5

, and

A^{ab}

is the ten-dimensional exterior power

Λ^{2} 5

. Hence

21 ↓ S U (5) = 15 \oplus 5 \oplus 1, 15 ↓ S U (5) = 10 \oplus 5 .

Here the left-hand

15

is the antisymmetric tensor of

S U (6)

, while the right-hand

15

in the first equation is the symmetric tensor of

S U (5)

; equal dimensions alone do not identify representations of different groups.

Finally decompose the traceless block matrix

M = (a c b D), a + tr D = 0.

The traceless part of

D

is the Adjoint representation

24

, the column

c

5

, the row

b

\overset{ˉ}{5}

, and the one remaining diagonal trivial representation is generated by

diag (- 5, 1, 1, 1, 1, 1)

. Thus

35 ↓ S U (5) = 24 \oplus 5 \oplus \overset{ˉ}{5} \oplus 1 .

All dimensions match:

6 = 1 + 5

21 = 15 + 5 + 1

15 = 10 + 5

and

35 = 24 + 5 + 5 + 1

e

Solution

Among the representations in c, the

10

S U (5)

occurs in the

15

S U (6)

, while the simplest source of

\overset{ˉ}{5}

\overset{ˉ}{6}

. Thus the economical direct sum containing the prescribed family is

R = \overset{ˉ}{6} \oplus 15 ↓ S U (5) (\overset{ˉ}{5} \oplus 10) \oplus 5 \oplus 1 .

The extra

S U (5)

multiplets are one

5

and one trivial representation. Calling

R

a representation does not require it to be an irreducible representation; none of the five individual choices in c contains both required

S U (5)

summands.

There is a further physical qualification if this is to be the entire chiral matter content of a consistent grand unified theory. Merely containing one Standard Model family does not guarantee cancellation of the cubic

S U (6)

gauge anomaly. In the convention

A (6) = 1

, one has

A (\overset{ˉ}{6}) = - 1

and

A (15) = 2

, so the displayed sum has

A (R) = 1

. With these types of representations, the minimal anomaly-free completion is instead

R_{af} = 2 \overset{ˉ}{6} \oplus 15 ↓ S U (5) (\overset{ˉ}{5} \oplus 10) \oplus (5 \oplus \overset{ˉ}{5}) \oplus 21 .

Beyond one family this adds a vectorlike gauge spectrum

5 \oplus \overset{ˉ}{5}

and two trivial representations. The conjugate pair can receive a mass after the appropriate symmetry breaking, without removing the remaining family.^[ref] The first boxed answer addresses representation containment; the second includes the extra consistency condition for a chiral gauge theory.

H

of a group

G

is a nonempty subset closed under the inherited multiplication and inverses; equivalently,

x y^{- 1} \in H

whenever

x, y \in H

. A normal subgroup satisfies

g H g^{- 1} = H

for every

g \in G

A group representation on a vector space

V

is a group homomorphism

R : G \to GL (V)

, so

R (g h) = R (g) R (h)

and

R (e) = I_{V}

. An irreducible representation has

V \neq = 0

and no invariant subspaces other than

0

and

V

: there is no proper nonzero

W

with

R (g) W \subseteq W

for every

g \in G

. We use finite-dimensional complex representations, with continuity for Lie groups, throughout.

b

Solution

Branching of a representation describes the irreducible representation content of a restriction of a representation:

R_{H} ≅ ⨁_{λ} m_{λ} S_{λ} .

This direct decomposition applies when the restriction of a representation is a semisimple representation, as for finite groups or continuous finite-dimensional representations of compact groups. For an arbitrary group and subgroup, complete reducibility need not hold. A vector subspace preserved by every

R (h)

for

h \in H

need not be preserved by the additional

R (g)

for

g \in / H

, so an irreducible representation can become reducible.

For a concrete example, take the defining representation of

G = S U (2)

C^{2}

and

H = {diag (z, z^{- 1}) : ∣ z ∣ = 1}

, a subgroup isomorphic to the circle group. The defining representation is irreducible:

S U (2)

acts transitively on unit vectors, so no complex line is preserved by all its elements. Under

H

, the two coordinate lines are invariant subspaces, and

2_{U (1)} = 1_{+ 1} \oplus 1_{- 1} .

Here the two one-dimensional representations act by

z

and

z^{- 1}

c

Solution

Let

ζ = e^{2 πi / n}

. The group homomorphism

Z_{n} \to U (1)

given by

[k] \mapsto ζ^{k}

is injective, identifying the finite cyclic group with the subgroup of

n

th roots of unity in the circle group.

The continuous irreducible representations of the circle group are the one-dimensional representations

R_{Q} (e^{i θ}) = e^{i Qθ}

with

Q \in Z

. One way to see this is to make the representation unitary by averaging, simultaneously diagonalize its commuting unitary matrices, and impose

R_{Q} (e^{i (θ + 2 π)}) = R_{Q} (e^{i θ})

. Its restriction of a representation is

R_{Q} (ζ^{k}) = e^{2 πi Q k / n} = e^{2 πi q k / n}, q = Q (mod n) .

Thus each irreducible representation remains one-dimensional, and representations whose integer labels differ by a multiple of

n

become equivalent. Every irreducible representation of the finite cyclic group occurs this way.

d

Solution

Choose a normalization in which the circle group has period

2 π

and the U(1) charges

Q_{i}

of the complex scalar fields are integers. By restriction of a representation,

q_{i} = Q_{i} (mod n) .

Write

v_{i} = ⟨ ϕ_{i} ⟩

. A vacuum expectation value

v_{i} \neq = 0

is preserved by

e^{i θ}

precisely when

e^{i Q_{i} θ} = 1

. Therefore the specified finite cyclic group survives spontaneous symmetry breaking exactly when every condensing complex scalar field obeys

Q_{i} \equiv 0 (mod n) .

Only U(1) charges of fields with nonzero vacuum expectation values enter this test; in the paper all

N

fields condense.

Suppose at least one condensing U(1) charge is nonzero and put

d = g cd (∣ Q_{i} ∣ : v_{i} \neq = 0)

. Since each such U(1) charge is a multiple of

d

, every

d

th root of unity preserves the vacuum expectation values. Conversely, the Bezout identity expresses

d

as an integer combination of these U(1) charges, so a common stabilizer obeys

e^{i d θ} = 1

. The discrete remnant of a circle symmetry is consequently

H_{unbroken} = Z_{d}, n_{m a x} = d .

A nontrivial retained finite cyclic group exists exactly when

d > 1

, and its possible orders are the divisors

n > 1

d

. For example, condensing U(1) charges

6

and

10

leave

Z_{2}

. If

d = 1

, the stabilizer is trivial. If all condensing U(1) charges vanish, the entire circle group is unbroken and there is no largest finite

n

; this case does not break the symmetry assumed in the question.

e

Solution

Normalize the circle group as in solution and count all Weyl fermions as left-handed; a right-handed Weyl fermion is replaced by its left-handed charge conjugate. The restriction of a representation gives

q_{a} = Q_{a} (mod n) .

The triangle coefficient of the cubic U(1) gauge anomaly is proportional to

\sum_{a} Q_{a}^{3}

. In a theory with this single gauge group, cancellation of its local pure gauge anomaly requires

a = 1 \sum M Q_{a}^{3} = 0.

If absence of the mixed gauge-gravitational anomaly is also required, as when coupling to a general gravitational background, one additionally needs

\sum_{a} Q_{a} = 0

. Additional gauge group factors would require their mixed gauge anomalies to cancel as well.

Choose integer representatives and write

Q_{a} = q_{a} + n k_{a}

. Expanding the cubic gives

0 = \sum_{a} Q_{a}^{3} = \sum_{a} q_{a}^{3} + n \sum_{a} (3 q_{a}^{2} k_{a} + 3 n q_{a} k_{a}^{2} + n^{2} k_{a}^{3}) .

Hence the inherited necessary congruence, and its gravitational counterpart when imposed, are

a \sum q_{a}^{3} \equiv 0 (mod n), a \sum q_{a} \equiv 0 (mod n) (if mixed gravitational cancellation is imposed) .