Interactions source an omitted Kaluza–Klein harmonic

The circle integrals are ${\int }_0^{2\pi R}{\cos }^2(y\mathrm{/}R)dy=\pi R$, ${\int }_0^{2\pi R}{\sin }^2(y\mathrm{/}R)dy=\pi R$, and ${\int }_0^{2\pi R}{\cos }^4(y\mathrm{/}R)dy=3\pi R\mathrm{/}4$. Substituting the ansatz into [A1] therefore gives $S_r=-\pi R{\int }_{\mathrm{\Omega }}{d}^{4}x[\frac{1}{2}(\mathrm{\partial }u{)}^2-\frac{1}{2}{\mu }^2{u}^2+\frac{3\lambda }{16}{u}^4]\mathrm{.}$ The quartic coefficient follows from $(\lambda \mathrm{/}4)(3\pi R\mathrm{/}4)=3\lambda \pi R\mathrm{/}16$.

Varying $S_r$ yields ${\mathrm{\square }}_{4}u+{\mu }^{2}u-(3\lambda \mathrm{/}4)u^3=0$. Its constant solutions include $u=0$ and $u=u_0$ with $u_0^2=4{\mu }^2\mathrm{/}(3\lambda )$, which is real and nonzero since $\mu ,\lambda >0$. The reduced potential grows as $+3\lambda u^4\mathrm{/}16$ at large $\mathrm{\mid }u\mathrm{\mid }$, so this stationary point is not an artifact of an unbounded quartic action.

Varying the full action gives ${\mathrm{\square }}_4\mathrm{\Phi }+{\mathrm{\partial }}_y^2\mathrm{\Phi }-m_0^2\mathrm{\Phi }-\lambda {\mathrm{\Phi }}^3=0$. At $u=u_0$, ${\mathrm{\square }}_4\mathrm{\Phi }+{\mathrm{\partial }}_y^2\mathrm{\Phi }-m_0^2\mathrm{\Phi }-\lambda {\mathrm{\Phi }}^3={\mu }^2{u}_0\cos (y\mathrm{/}R)-\lambda u_0^3{\cos }^3(y\mathrm{/}R)=-\frac{\lambda u_0^3}{4}\cos (3y\mathrm{/}R),$ using ${\cos }^{3}z=(3\cos z+\cos 3z)\mathrm{/}4$ and $\lambda u_0^2=4{\mu }^2\mathrm{/}3$. This residual is not identically zero, so the reduced solution fails the full equation in the omitted third harmonic.

By [F2], one reduced solution that does not uplift refutes the asserted consistent-truncation claim. The interaction creates the omitted harmonic even though the quadratic circle modes are orthogonal; this example establishes no general impossibility theorem for nonlinear or symmetry-protected truncations.