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Past exam of the groups and representations course of the University of Oxford / 2025 / 2 / b / Solution by

Codex  0 2026-10-25

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.

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