Exact local scalar gap on a round spatial sphere

Summary: The written local scalar Dirac interaction has an exact nonzero constant-current branch on a conditional round six-dimensional sphere, with complete spatial spectral stability but no derived simplex enhancement or particle mass identification.

A self-consistent local coefficient with a complete spatial solution

The local scalar Dirac interaction admits an exact smooth field with a nonzero constant scalar current on a round six-dimensional sphere. It uses the ordinary spinor bundle and includes both opposite spatial Dirac modes. There is no additional contact kernel in this construction. A harmonic decomposition of its full linearization proves the absence of exponentially growing spatial eigenmodes.

This is a conditional field-theory result. The round sphere is an assumed actual physical spatial geometry with undetermined radius $R$; its six directions are ordinary macroscopic spatial directions, without compactification. The calculation does not select that geometry, derive its projection from the master manifold, or identify the solution with an observed particle.

The scalar coefficient depends on the canonical norm, physical volume and radius. No simplex multiplier or occupied mass-band index follows. The field extends over the whole assumed sphere; localization of a particle and nonlinear orbital stability remain separate questions.

The local action and its two spatial Killing modes

Use the canonical conventions

$$\mathscr D^\dagger=\mathscr D,\qquad \beta^\dagger=\beta,\qquad \beta^2=I,\qquad \{\mathscr D,\beta\}=0. \tag{1}$$

Initially use unit radius, unit spatial measure and unit canonical norm. Write $h=3$ and take a positive dimensionless coefficient $g$:

$$E(\psi)=\langle\psi,\mathscr D\psi\rangle +\frac g2\int(\psi^\dagger\beta\psi)^2d\nu,\qquad i\partial_t\psi=\mathscr D\psi+gJ\beta\psi. \tag{2}$$

The real Killing-spinor construction supplies two pointwise orthonormal fields $\varepsilon$ and $\xi=\beta\varepsilon$, with

$$\begin{gathered} \mathscr D\varepsilon=h\varepsilon,\qquad \mathscr D\xi=-h\xi,\\ \varepsilon^\dagger\varepsilon=\xi^\dagger\xi=1,\qquad \varepsilon^\dagger\xi=0,\\ \varepsilon^\dagger c(X)\varepsilon=0,\qquad \varepsilon^\dagger c(X)\xi=0. \end{gathered} \tag{3}$$

They are actual spatial spinor fields with their full connection. The Killing-frame transfer gives the differential identities used below. The explicit real Clifford matrices in the companion current-response note prove all bilinear cancellations in (3).

Derive the nonzero scalar branch

Let $\chi=a\varepsilon+b\xi$, with $a,b$ real. Then $\|\chi(q)\|^2=a^2+b^2$ and $J=2ab$ at every point. The complete stationary equation becomes

$$\begin{pmatrix}h&m\\m&-h\end{pmatrix} \begin{pmatrix}a\\b\end{pmatrix} =\omega\begin{pmatrix}a\\b\end{pmatrix}, \qquad m=2gab,\qquad a^2+b^2=1. \tag{4}$$

For $m\ne0$, the eigenvector identity gives $2ab=m/\omega$. Combining it with self-consistency fixes $\omega=g$. Therefore a nonzero branch exists when $g>h$, with

$$\boxed{ m=\pm\sqrt{g^2-h^2},\qquad a=\sqrt{\frac{1+h/g}{2}},\qquad b=\operatorname{sgn}(m)\sqrt{\frac{1-h/g}{2}}. } \tag{5}$$

Substitution proves the full spatial solution

$$\boxed{ \psi(t,q)=e^{-igt}(a\varepsilon(q)+b\xi(q)),\qquad J=\frac mg,\qquad E=\frac g2+\frac{h^2}{2g}. } \tag{6}$$

The scalar nonlinearity is a self-consistent constant $m\beta$ on this background. Its two possible signs are not an antiparticle or electric-charge identification. The threshold $g=h$ joins the branch to the positive Killing mode with zero scalar current.

This is a local spatial realization of a scalar self-consistency mechanism. Unlike a separately assumed global scalar gap, (6) satisfies the pointwise local equation on the specified ordinary spinor bundle.

Retain the complete spatial variation

Put $\psi=e^{-igt}(\chi+\eta)$. The full real-linear perturbation equation is

$$i\partial_t\eta= (\mathscr D+m\beta-g)\eta +2g\,\beta\chi\,\operatorname{Re}(\chi^\dagger\beta\eta). \tag{7}$$

The last term varies the scalar current; it cannot be omitted by treating $m$ as a fixed external mass.

Take any real normalized scalar harmonic $u$ of degree $\ell\ge1$, with $\Delta u=\Lambda u$, $\Lambda=\ell(\ell+5)$. The relevant complex spatial subspace is

$$\mathcal V_u=\operatorname{span}_{\mathbb C} \{u\varepsilon,\ u\xi,\ c(du)\varepsilon,\ c(du)\xi\}. \tag{8}$$

Both spinor modes, their derivative partners and arbitrary complex relative phases are included. The exact Dirac identities are

$$\begin{aligned} \mathscr D(u\varepsilon)&=h\,u\varepsilon+c(du)\varepsilon,\\ \mathscr D(u\xi)&=-h\,u\xi+c(du)\xi,\\ \mathscr D(c(du)\varepsilon)&=\Lambda u\varepsilon+(1-h)c(du)\varepsilon,\\ \mathscr D(c(du)\xi)&=\Lambda u\xi+(h-1)c(du)\xi. \end{aligned} \tag{9}$$

Its physical Gram matrix is $\operatorname{diag}(1,1,\Lambda,\Lambda)$. The derivative columns are normalized by dividing by $\sqrt{\Lambda}$. In this orthonormal basis, let

$$L_\ell= \begin{pmatrix} h-g&m&\sqrt{\Lambda}&0\\ m&-h-g&0&\sqrt{\Lambda}\\ \sqrt{\Lambda}&0&1-h-g&-m\\ 0&\sqrt{\Lambda}&-m&h-1-g \end{pmatrix}, \tag{10}$$

and

$$B= \begin{pmatrix} g-h&m&0&0\\ m&g+h&0&0\\ 0&0&0&0\\ 0&0&0&0 \end{pmatrix} =2g\,vv^\dagger,\qquad v=(b,a,0,0)^T. \tag{11}$$

The pointwise bilinears in (3) make the current variation $2u\operatorname{Re}(bz_0+az_1)$ for coefficients $z$ in (8). Thus (7) preserves this subspace and, writing $z=x+iy$, becomes

$$\dot x=L_\ell y,\qquad \dot y=-(L_\ell+B)x. \tag{12}$$

The quadratic energy is $z^\dagger L_\ell z+x^TBx$. The four squared frequencies are eigenvalues of $L_\ell(L_\ell+B)$. Both matrices in (12) retain the complete canonical normalization.

Prove that every harmonic block has nonnegative squared frequencies

For $\ell\ge2$, define

$$\begin{aligned} r_-&=\sqrt{(\ell+h-1)^2+m^2},& r_+&=\sqrt{(\ell+h)^2+m^2},\\ p_\pm&=r_\pm-g,& n_\pm&=r_\pm+g. \end{aligned} \tag{13}$$

The four eigenvalues of $L_\ell$ are $p_-,p_+,-n_-,-n_+$. The first two are positive. They come from the two neighboring complete Dirac levels in (9), with the constant background scalar term retained.

A useful number between their squared eigenvalues is the scalar Laplacian value itself:

$$\boxed{ p_-^2<p_+^2<\Lambda<n_-^2<n_+^2. } \tag{14}$$

To check the two middle inequalities, use $m^2=g^2-h^2$. The upper positive root obeys $p_+<\sqrt{\Lambda}$ because $\ell<2g\sqrt{\Lambda}$. The lower negative magnitude obeys $n_->\sqrt{\Lambda}$ because $\ell+2h-1<2g\sqrt{\Lambda}$. Both inequalities hold for $h=3$, $g>3$, $\ell\ge1$: the relevant ratios are less than one and at most $\sqrt6$, respectively, whereas $2g>6$.

The rank-one determinant formula gives the secular function

$$F(z)= \frac{\det[L_\ell(L_\ell+B)-zI]} {\det[L_\ell^2-zI]} =1+2g\sum_j\frac{\lambda_j|v_j|^2}{\lambda_j^2-z}. \tag{15}$$

Here $\lambda_j$ are the four eigenvalues of $L_\ell$ and $v_j$ are the components of $v$ in its orthonormal eigenbasis. Direct substitution of (10)–(11), using $m^2=g^2-h^2$, yields

$$\boxed{ \det[L_\ell(L_\ell+B)-\Lambda I] =\Lambda(2h-1) \left[(4g^2-1)\Lambda+4(g^2-h^2)\right]>0. } \tag{16}$$

The denominator of (15) is also positive at $z=\Lambda$, by (14). Hence $F(\Lambda)>0$.

The residues at the two positive kinetic poles have positive $\lambda_j|v_j|^2$; those at the two negative kinetic poles have negative values. Their one-sided signs, together with $F(\Lambda)>0$, force a root in each interval

$$(p_-^2,p_+^2),\quad (p_+^2,\Lambda),\quad (\Lambda,n_-^2),\quad (n_-^2,n_+^2). \tag{17}$$

For example, $F$ rises from $-\infty$ to $+\infty$ between the first two poles, is negative just above the second pole, and is positive at $\Lambda$. The two negative-residue poles give the other two sign changes. The numerator has degree four, so these account for all four squared frequencies. If a residue vanishes, continuity from nearby nonzero residues gives the same nonnegative conclusion.

The $\ell=1$ block is the continuous limit of this argument as the lower positive kinetic eigenvalue reaches zero. Its squared frequencies remain nonnegative and include a zero. For $\ell=0$, the derivative columns in (8) vanish; the separate two-complex-component block has

$$\Omega^2=0,\qquad \Omega^2=4m^2. \tag{18}$$

Thus every coupled harmonic block has no exponential eigenmode. This conclusion follows from the analytic pole ordering and determinant sign, not an extrapolation of a finite harmonic scan.

Why the conclusion covers the full spatial field

Let $T_\pm f=c(df)\varepsilon_\pm$, with $\varepsilon_+=\varepsilon$ and $\varepsilon_-=\xi$. The bilinear identities (3), the Dirac identities (9) and self-adjointness imply

$$\begin{aligned} \langle f\varepsilon_s,g\varepsilon_t\rangle &=\delta_{st}\langle f,g\rangle,\\ \langle f\varepsilon_s,T_tg\rangle&=0,\\ \langle T_sf,T_tg\rangle &=\delta_{st}\langle f,\Delta g\rangle. \end{aligned} \tag{19}$$

For the last identity, expand $\langle\mathscr D(f\varepsilon_s),\mathscr D(g\varepsilon_t)\rangle$ in two ways, moving one Dirac operator by its adjoint. The mixed zeroth- and first-derivative inner products vanish by (3); the remaining term is exactly the last line of (19). Mixed derivative products for different harmonics need not vanish pointwise. Their spatial integrals vanish by this identity.

Consequently the spaces (8), over a real orthonormal basis of all scalar harmonics and with their complex coefficients retained, are mutually orthogonal. Their closed sum $\mathcal V$ is reducing for $\mathscr D$ and $\beta$, as (9) shows.

The complete scalar-source range in (7) lies in $\mathcal V$: it consists of $\beta\chi$ times a real spatial function. For a perturbation orthogonal to $\mathcal V$, orthogonality to every complex multiple of $f\beta\chi$ forces $(\beta\chi)^\dagger\eta=0$ almost everywhere. The source then vanishes. This complement evolves under the self-adjoint operator $\mathscr D+m\beta-g$ and cannot add exponential modes.

This proves complete spatial spectral stability of the specified branch. Zero-frequency and possible Jordan directions are retained. It is not a theorem of nonlinear orbital stability or uniformly bounded motion under arbitrary finite perturbations.

Test the conformal current term on this stable local branch

The derivative-current candidate can also be tested on the nonzero branch (6). Add $\eta\langle J,\widehat P_6J\rangle/2$ to (2), with dimensionless $\eta$. Since its background current is constant and $\widehat P_6 1=0$, the exact solution and its scalar coefficient remain unchanged. The full response does change.

On degree $\ell$, replace $B$ in (12) by $c_\ell B$, where

$$c_\ell=1+\frac{\eta}{g}S(\ell,6). \tag{20}$$

No kinetic term is dropped. Exact determinants give, writing $m^2=g^2-h^2$,

$$\begin{aligned} \operatorname{tr}[L_\ell(L_\ell+c_\ell B)] &=8g^2-4h+2+4\Lambda-4h^2c_\ell,\\ \det L_\ell&=\Lambda(\Lambda-2h),\\ \det(L_\ell+c_\ell B) &=\Lambda(\Lambda-2h) +c_\ell\left[4m^2(\Lambda-2h+1)+2h\Lambda\right]. \end{aligned} \tag{21}$$

For $\ell\ge2$ the kinetic determinant and the square bracket are positive. If $\eta>0$, the sixth-degree growth of $c_\ell$ eventually makes the squared-frequency trace negative. If $\eta<0$, it eventually makes $\det(L_\ell+c_\ell B)$ negative; the product determinant is then negative too. Each case gives an exponential mode.

Thus every nonzero coefficient of this added conformal current term destabilizes the exact branch at sufficiently high degree in the unrestricted continuum model. This conclusion retains a nonzero self-consistent local scalar coefficient and the full spatial response. It concerns this specified extension; it does not exclude a different master-derived kinetic or nonlocal interaction.

Physical normalization and what remains unselected

Restore radius $R$, canonical norm $Q$ and physical spatial volume $\mathcal V=16\pi^3R^6/15$. In units $\hbar=c=1$, put

$$g=\frac{R\kappa Q}{\mathcal V},\qquad g>3. \tag{22}$$

The physical field is $\sqrt{Q/\mathcal V}\,\chi$ and the exact frequency and scalar coefficient are

$$\boxed{ \omega=\frac{\kappa Q}{\mathcal V},\qquad M=\pm\sqrt{\left(\frac{\kappa Q}{\mathcal V}\right)^2-\frac9{R^2}}. } \tag{23}$$

The total energy is

$$\boxed{ E=\frac{\kappa Q^2}{2\mathcal V} +\frac{9\mathcal V}{2\kappa R^2},\qquad \frac{\partial E}{\partial Q}=\omega. } \tag{24}$$

The homogeneous collective frequency is $2|M|$; the other physical response frequencies are those above divided by $R$. The distinction between frequency, total energy and scalar coefficient agrees with the canonical mass-reduction analysis.

For illustration at unit radius and norm:

The action has supplied a self-consistent scalar coefficient and a spectrally stable spatial branch under specified geometry. It has not supplied the simplex count, the charged-lepton mode selections, a common six- and ten-dimensional scale, or an identification with measured rest masses. No alternative particle dimensionalities or mode assignments have been introduced.

Verification

round_local_scalar_gap_checks.py checks the full spatial stationary equation and every complex component of its nonlinear variation. Zonal and angular harmonics test all four columns in (8). Independent physical quadrature checks scalar and gradient Gram integrals, while the general orthogonality statement follows from the adjoint proof (19).

Exact symbolic determinants verify (16), both neighboring kinetic levels, the homogeneous response (18), and the derivative-current extension (21). Independent complex action variations and eight-real-component generators verify the canonical blocks and the four positive root intervals. Physical normalization checks retain the sphere's actual volume and the distinction in (23)–(24).