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If a Lagrangian density depending on first and second derivatives has no undifferentiated field, the scalar-field shift symmetry $\delta \varphi =ϵ$ gives the displayed Noether current. Its divergence is minus the Euler-Lagrange field equation with second derivatives, so it is conserved on shell. The spatial integral of $j^0$ is constant only if the boundary current vanishes. For $\mathcal{L}=\rho {\varphi }_t^2/2+\alpha {\varphi }_x^2/2-\beta {\varphi }_{xx}^2/2$, it gives $j^0=\rho {\varphi }_t$, $j^x=\alpha {\varphi }_x+\beta {\varphi }_{xxx}$. Fixed transverse endpoint positions break the global shift symmetry of the boundary problem; support forces can then change the total transverse momentum even though the bulk continuity equation still holds.

The initial field values are ${\varphi }_i/M_{\mathrm{Pl}}=\sqrt{122n}$, approximately $15.6$, $19.1$ and $22.1$ for the three powers. A simple monomial inflation potential over this super-Planckian field range needs theoretical protection. Generic Planck-suppressed corrections to inflaton potentials, such as $c_p{\varphi }^p/M_{\mathrm{Pl}}^{p-4}$, need not be small there; they can change the slope and curvature and spoil slow-roll inflation. Radiative corrections and possible couplings to other particles likewise require control. A symmetry, such as an approximate scalar-field shift symmetry, or a specified ultraviolet completion could supply that protection, but it is not part of the bare monomial model.

A large field value is not by itself a proof that the energy density is Planckian: a sufficiently small $\lambda$ can keep $V\ll M_{\mathrm{Pl}}^4$. The issue is control of the effective field theory and stability of the flat potential over its field range, rather than simply comparing the field value with a mass scale.

There is also a global potential issue for $n=3$: $V=\lambda {\varphi }^3$ continued over all real $\varphi$ is unbounded below and has no stable minimum at zero. Restricting to $\varphi >0$ does not specify what happens when the field reaches that boundary, so a completion is needed for post-inflationary evolution and reheating. The even powers have a stable minimum but still need interactions that transfer the inflaton energy to a hot bath. These interactions and the resulting reheating history also affect the mapping between a pivot scale and the assumed 60 number of e-folds. Therefore the concise theoretical concerns are control of large-field corrections, a consistent stable completion, and a specified reheating mechanism.

At the Gaussian fixed point, $[{\varphi }_i]=(d-2)/2$, so all three quartic couplings have engineering dimension $d-4[{\varphi }_i]=4-d$. ThusTake $A>0$ and $ϵ>0$, as in the perturbative epsilon expansion. Define $x=A{g}_1/ϵ$, $y=A{g}_2/ϵ$, $z=A\lambda /ϵ$. The renormalization-group fixed point equations becomeIf $z=0$, each of $x,y$ is independently $0$ or $1/36$, giving four renormalization-group fixed points. If $z\ne 0$, subtracting the first two equations givesThe branch $x+y=1/36$ forces $z=1/12$; inserting this in the first equation yields a repeated root $x=y=1/72$. Hence every nonzero-$z$ solution has $x=y$. Then $z=1/8-3x$, andhas roots $x=1/40$ and $x=1/72$. The complete list of six coupled Ising fixed points near four dimensions, in coordinates $(g_1,g_2,\lambda )$, isThe internal symmetries of these renormalization-group fixed points are:

The last point displays the field-rotation equivalence of decoupled Ising theories. With the orthogonal transformation ${\psi }_{\pm }=({\varphi }_1\pm {\varphi }_2)/\sqrt{2}$,It is the same pair of decoupled Ising models written in fields rotated by $45^{\circ }$, but remains a distinct coordinate solution of the stated renormalization-group beta functions. Here $D_4$ denotes the dihedral group of order eight; an alternative convention calls it $D_8$. We have described linear internal transformations preserving the gradient energy; free massless sectors also have constant scalar-field shift symmetries. At $ϵ=0$, the six coordinates coalesce at the Gaussian fixed point.

The operator ${\dot{\varphi }}^2\varphi$ is not invariant under a scalar-field shift symmetry because one field is undifferentiated. An arbitrary large coefficient would therefore radiatively generate equally unsuppressed nonderivative operators, including a mass and a steep scalar potential, and would spoil the approximate shift symmetry and flat potential needed for single-field slow-roll inflation. In a technically natural slow-roll model its coefficient must consequently be small, so it does not produce parametrically large primordial non-Gaussianity. Treated merely as an effective spectator-field interaction, it does generate the tree-level primordial bispectrum computed below, whose size is controlled by the dimensionless interaction strength at the Hubble scale; taking that strength large abandons the controlled slow-roll premise.

The constant scalar-field shift symmetry has momentum-space actionFor the canonical commutation relation $[\varphi (\mathbf{x}),\Pi (\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$, its Noether charge can be writtenIndeed, $i[Q,\varphi (\mathbf{x})]=c$.

With constant $\nu$, the bulk action is invariant under time translations, translations in the two spatial coordinates, and rotations or reflections within the substrate plane. The Laplacian has no preferred direction, giving the planar $O(2)$ symmetry. It also has scalar-field shift symmetry $h\mapsto h+a$, expressing the arbitrary origin of height, and the discrete symmetry $h\mapsto -h$ because its residual is squared.

There is an additional exact tilt symmetry $h\mapsto h+\mathbf{b}\cdot \mathbf{x}$: the added slope is time independent and has zero Laplacian. More generally, adding any function $f$ satisfying $f_t=\nu {\nabla }^{2}f$ leaves the residual unchanged. Such bulk transformations need not preserve prescribed boundary conditions or a restricted physical interpretation of the height.

In two spatial dimensions the action also has the anisotropic scale invarianceThe spacetime measure scales as $b^4$ and the squared residual as $b^{-4}$. The time and length scales are therefore related diffusively, rather than by a Lorentz transformation.

Ordinary time reversal needs a boundary qualification. The identityshows that the density is time-reversal invariant up to a total divergence. Thus the bulk stationary equation $h_{tt}-{\nu }^2{\nabla }^{4}h=0$ is invariant under $t\mapsto -t$, although the printed density is not pointwise unchanged. The time reversal of the squared diffusion-residual action exchanges the growing and decaying branches; the selected first-order diffusion equation alone is not time-reversal invariant at fixed positive $\nu$.

A constant shift $\varphi \mapsto \varphi +ϵ$ leaves every derivative unchanged, and the density has no undifferentiated $\varphi$. Thus it is exactly invariant under this scalar-field shift symmetry. Localize the shift parameter temporarily to $ϵ(x,t)$:Integrating the last term by parts once identifies the Noether current for a constant shift with second derivatives:The Euler-Lagrange field equation is $\rho {\varphi }_{tt}+\alpha {\varphi }_{xx}+\beta {\varphi }_{xxxx}=0$, precisely the continuity equation ${\partial }_t{j}^0+{\partial }_x{j}^x=0$. Therefore the conserved quantity, when the endpoint current vanishes or the field has suitable periodic/decaying conditions, isthe total transverse momentum. More explicitly $dQ/dt=-[\alpha {\varphi }_x+\beta {\varphi }_{xxx}{]}_{\mathrm{ends}}$. The fixed endpoints introduced later can exert transverse forces and break the global shift symmetry of the boundary problem, so vanishing endpoint current must not be inferred merely from fixed displacement.

If a scalar vacuum is invariant under an exact scalar-field shift symmetry, the Ward identity makes the leading soft bispectrum vanish: