= Solution
<Statistical homogeneity> forces a scalar two-point function to have momenta $(\mathbf k,-\mathbf k)$. <Statistical isotropy> then makes it a function only of $|\mathbf k|$, so it is parity even without using perturbation theory.
For a scalar three-point function, momentum conservation gives $\mathbf k_1+\mathbf k_2+\mathbf k_3=0$, so all three vectors lie in one plane. A rotation by $\pi$ around the normal to that plane sends every $\mathbf k_a$ to $-\mathbf k_a$. Rotational invariance therefore identifies a triangle with its parity reverse, proving nonperturbatively that the scalar <primordial bispectrum> is parity even.
A rotationally invariant local parity-odd three-scalar vertex must contain a <Levi-Civita symbol> contracted with three spatial momenta. Its momentum-space factor is proportional to
$$
\epsilon_{ijk}k_1^ik_2^jk_3^k
=\mathbf k_1\mathbin\cdot(\mathbf k_2\mathbin\times\mathbf k_3)=0,
$$
because momentum conservation makes the momenta linearly dependent. The equivalent position-space expression is a <total divergence>; its spatial integral vanishes under the stated <vanishing boundary condition>. Thus a parity-odd interaction of three scalar fields contributes nothing to the action.
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