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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 tobecause 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.

A vanishing boundary condition sets a specified boundary quantity to zero. The quantity must be named: zero boundary value is a homogeneous Dirichlet boundary condition, while zero outward normal derivative is a homogeneous Neumann boundary condition. These are different restrictions.

Which boundary terms disappear under integration by parts depends on the operator and the permitted test functions. For a second-order weak problem, zero test-function trace eliminates ${\int }_{\partial U}({\partial }_{n}u)v$. A fourth-order problem can also involve derivative traces, so a zero value alone does not remove every boundary term. On $[0,1]$, $u(x)=x(1-x)$ vanishes at both endpoints but has nonzero endpoint first derivatives; the clamped boundary condition would require those slopes to vanish as well.