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A parity symmetry in quantum field theory reverses spatial orientation, so its spatial matrix must lie in the determinant-$-1$ component of the orthogonal group. In two spatial dimensions, reversing both coordinates gives $-I_2$, which has determinant $+1$ and is a rotation through $\pi$ in the special orthogonal group in two dimensions. It cannot test a discrete orientation-reversing symmetry independent of ordinary rotations. Reflecting one coordinate, for example $\mathrm{diag}(-1,1)$, has determinant $-1$ and supplies the required parity symmetry in quantum field theory.

In three spatial dimensions, $-I_3$ has determinant $-1$, so simultaneous inversion of all three coordinates is the standard choice. A single-coordinate reflection also reverses orientation there, and differs from $-I_3$ by a proper rotation through $\pi$. Thus, in a rotationally invariant theory, it tests the same discrete symmetry. The word is in the source specifies conventional representatives: the essential distinction is orientation reversal, rather than a unique choice of axis. Both transformations preserve the ordinary spatial integration measure through the absolute value of their Jacobian determinants.

Use Haar sampling by Gaussian QR decomposition:

The matrix $G$ is invertible almost surely. Indeed, successively sampled Gaussian columns have probability zero of lying in the proper linear subspace spanned by the previous columns. Thus the diagonal signs used above exist with probability one. Equivalently, apply the Gram-Schmidt process to the columns, always dividing each residual by its positive Euclidean norm; this produces $M$ directly.

To prove uniformity, fix $H\in O(n)$. Each column of $G$ has density proportional to $e^{-\Vert x{\Vert }^2/2}$, so orthogonal transformations preserve its probability distribution. Column independence therefore gives $HG\stackrel{\mathrm{d}}{=}G$.

The QR decomposition with positive diagonal is unique: if $G=Q_1{R}_1=Q_2{R}_2$, then $Q_2^T{Q}_1=R_2{R}_1^{-1}$ is both an orthogonal matrix and an upper triangular matrix, hence diagonal with entries $\pm 1$. Its diagonal entries are positive, so it is the identity. It follows that the orthogonal factor of $HG=HMR$ is $HM$. ThereforeBy uniqueness of normalized Haar measure on the compact group $O(n)$, the returned matrix has exactly the required probability distribution. Reflection invariance gives each determinant sign probability $1/2$, so the algorithm samples both components of the orthogonal group. The positive-diagonal correction is essential when a numerical QR decomposition chooses its own signs.

Both the canonical kinetic term and $V({\varphi }^{\mathsf{T}}\varphi )$ are invariant under the orthogonal group $O(n)$. The stated nonzero vacuum expectation value has length $|v|$. Its stabilizer is the orthogonal group $O(n-1)$ acting on the perpendicular subspace. HenceThe two nonzero coordinates of this particular vacuum do not break more generators than an axis-aligned vacuum: the two choices are related by a common orthogonal transformation.

Explicitly, with $s={\varphi }^{\mathsf{T}}\varphi$,stationary point at nonzero $v$ requires $V^{\prime }(v^2)=0$. The $n-1$ transverse eigenvalues vanish, while the radial squared mass is $4v^2{V}^{\prime \prime }(v^2)$. If this curvature also vanishes, there is an additional accidental massless radial mode; the number of symmetry-protected Goldstone bosons remains $n-1$. If $v=0$, the symmetry is unbroken and none is required.

For a vector bundle carrying a smooth nondegenerate symmetric bilinear form of constant metric signature, a pseudo-orthonormal frame is a smooth local frame of a vector bundle whose Gram matrix is the constant diagonal signature matrix $\eta$. It identifies the metric bundle locally with a fixed model ${\mathbb{R}}^{p,q}$. Two such frames differ by a smooth map into the corresponding orthogonal group. Positive-definite signature is included as the ordinary orthonormal case. A global pseudo-orthonormal frame is a section of a fiber bundle of the orthonormal frame bundle and trivializes that principal bundle.

An orthogonally invariant random vector has a probability measure determined by its radial pushforward measure. Here $\sigma$ is normalized surface measure on a sphere, and the formula holds for bounded measurable $f$ and for nonnegative measurable $f$. It includes an atom of a measure at zero: the inner integral there is simply $f(0)$.

To prove the formula first for continuous bounded $f$, average $f(Qx)$ over normalized Haar measure of the compact orthogonal group. Right invariance makes this average depend only on $|x|$; averaging also over surface measure on a sphere identifies it with the spherical average. Invariance of $\mu$ and Fubini's theorem then give the identity. Uniqueness of finite Borel measures and measurable approximation extend it to the stated classes of $f$. No extra sphere-area factor belongs in an average against normalized surface measure on a sphere.