Separating the transverse modes of a mass core

The complete finite current hierarchy has a useful decomposition around one coherent particle core. Its many coefficient components do not each require a separate stability calculation. Every component transverse to the occupied coefficient ray has a complex-linear, self-adjoint spatial equation. The same is true of every initially empty lower grade. Only the occupied coefficient block retains the real/conjugate partner response studied in the spatial shape calculation.

This result assumes a fixed component frame and a spatially constant coherent coefficient ray. It concerns the formal linearized equation on the complete ordinary spatial domain. It does not prove nonlinear stability, settle the responding frame, or derive a physical spatial carrier for the prescribed simplex weights. Mass is the particle's selected oscillation frequency in equivalent units throughout.

Complete coefficient components

Use the density-normalized current action, with partner rank $s$, physical depth $d$, and $q=1/(s+1)>0$. At occupied grade $N$, write

$$\Psi_N(x,t)=e^{-i\omega_N t}\Phi(x)\otimes z^{\otimes N}, \qquad \Phi=(\phi,0)^T,\quad \phi=fI_s+i h\alpha_{\hat x},\quad \omega_N=\omega_0 D_{N,d},\quad D_{N,d}=\binom{N+d}{d}. \tag{1}$$

The core has $\phi\phi^\dagger=\phi^\dagger\phi=cI_s$, where $c=f^2+h^2>0$. All coefficient indices refer to the retained tensor description; they are not additional physical dimensions.

Choose a constant orthonormal coefficient basis with $z=e_0$. A normalized occupation vector contains $j$ indices outside that ray, with multiplicity

$$h_{j,d}=\binom{j+d-1}{d-1},\qquad 1+\sum_{j=1}^N h_{j,d}=D_{N,d}. \tag{2}$$

Each such coefficient retains an arbitrary complex $2s$-by-$s$ spinor/partner matrix disturbance. There is no restriction to the core's scalar/vector matrix form. A constant coefficient basis change preserves the fixed-frame spatial Dirac operator and all traced binding currents.

Two exact sums determine the whole transverse mass response

The normalized symmetric splitting of the graded current construction gives

$$\begin{aligned} \sum_{k=0}^N h_{k,d} \frac{\binom{N-j}{k}}{\binom Nk} &=\frac{D_{N,d}}{D_{j,d}},\\ \sum_{k=0}^N h_{k,d} \frac{\binom kj}{\binom Nj} &=D_{N,d}\frac d{d+j}. \end{aligned} \tag{3}$$

Unavailable binomials are zero. For the first identity, replace the binomial ratio by $\binom{N-k}{j}/\binom Nj$. The convolution of $\binom{k+d-1}{d-1}$ and $\binom{N-k}{j}$ is $\binom{N+d}{d+j}$. Dividing by $\binom Nj$ gives the stated answer. For the second, use

$$\binom{k+d-1}{d-1}\binom kj =\binom{d+j-1}{j}\binom{k+d-1}{d+j-1}$$

and apply the Hockey-Stick sum.

For $j\ge1$, varying the complete mass source now gives

$$\boxed{D S_{{\rm mass},j}[\eta] =\omega_N(a_j B_0\eta+b_jP_+\eta),\qquad a_j=1-q+\frac q{D_{j,d}},\quad b_j=\frac{qd}{d+j}.} \tag{4}$$

Here $B_0=\operatorname{diag}(I_s,-I_s)$ and $P_+=\operatorname{diag}(I_s,0)$. The term with $a_j$ comes from lifting the unvaried current gradient to the changed field. Its projector contraction is the first sum in (3). The varied current gradient supplies the second sum. The conformal-unitary relation in (1) reduces its full spinor/partner factor to $P_+$. Both contributions are necessary.

There is no conjugate-field contribution to (4): for a transverse coefficient the first current variation joins the occupied retained ray to an orthogonal retained component. Its return to the perturbation coefficient is complex-linear. Different occupations do not mix. The equality of complete core Hessians proved in the density note also transfers (4) to the previous regular root-current action.

At $j=1$, $a_1+b_1=1$. The upper rotating-frame mass shift therefore vanishes, as required for an infinitesimal change of the coherent ray. For $j\ge2$,

$$1-a_j-b_j=q\left(1-\frac1{D_{j,d}}-\frac d{d+j}\right)>0. \tag{5}$$

This is a signed disturbance-frequency shift. It is not a negative inertia of the complete particle.

Binding and spatial derivatives preserve this reduction

The binding currents trace the coefficient indices. Orthogonality to $z^{\otimes N}$ consequently gives $\delta J=\delta K=\delta\rho=0$ for every transverse disturbance. The complete binding derivative is simply

$$D S_{{\rm bind},j}[\eta] =m_cB_1\eta\beta-\kappa B_1\eta F(x),\qquad F(x)=\frac{\operatorname{Tr}(\beta K_0)K_0} {\operatorname{Tr}K_0^2},\quad K_0=\phi^\dagger\beta\phi. \tag{6}$$

This includes the density-denominator derivatives: their first variations vanish in precisely these directions. It is not a rule to omit them in other variations. Since $K_0^2=c^2I_s$, $F=tK_0/c$, with $t=(f^2-h^2)/c$. Thus $F$ is Hermitian and $\|F\|\le1$ everywhere, including the decaying spatial tail.

The shape source adds the complex-linear Hermitian projector $\mu P_{\perp,+}\eta=(\mu P_\perp\eta_+,0)^T$. The coefficient basis is spatially constant, so the actual Dirac derivative also preserves each block. In the rotating frame the whole operator is

$$\boxed{ H_j\eta=-iA_a\partial_a\eta +\omega_N(a_jB_0+b_jP_+-I)\eta +m_cB_1\eta\beta-\kappa B_1\eta F(x) +\mu P_{\perp,+}\eta.} \tag{7}$$

Left and right Hermitian matrix multiplication commute as operators on the matrix field. All nondifferential terms in (7) are therefore bounded and Hermitian on the canonical matrix $L^2$ space. The free spatial Dirac operator is self-adjoint on $H^1(\mathbb R^d)$, and adding this bounded Hermitian multiplication preserves self-adjointness on the same domain. The evolution $i\partial_t\eta=H_j\eta$ is unitary:

$$\|\eta(t)\|_{L^2}=\|\eta(0)\|_{L^2}. \tag{8}$$

This covers all ordinary spatial angles and wavelengths for these coefficient blocks. It is stronger than a finite angular or frozen-point scan. It does not apply to the remaining occupied coefficient block.

Initially empty lower grades

Suppose $M<N$ is initially empty. Its first current variation is zero because the current is quadratic in that grade's field. Its complete mass Hessian is consequently

$$D S_{{\rm mass},M,j}[\eta] =\omega_0D_{M,d}\left(1-q+\frac q{D_{j,d}}\right)B_0\eta, \qquad 0\le j\le M. \tag{9}$$

There is no $b_j$ term. Equations (6) and the shape projector still apply. These blocks are also complex-linear and self-adjoint, and they do not linearly force the occupied grade. In particular their coefficients are independent of the larger target grade. This is a spatial response version of the common-target identity.

The formal finite-tower linearization through the highest occupied grade therefore consists of the one occupied coefficient block, the transverse blocks (7), and the lower-grade blocks (9). Empty channels above the highest occupied grade can have a Lipschitz source without a derivative at zero; this decomposition does not assert otherwise. A spatially varying coefficient ray or a responding frame requires its connection terms to be examined separately.

Verification

transverse_core_checks.py checks (3) with exact rational arithmetic at every maintained particle grade. It independently evaluates the complete current and binding Hessians on all coefficient, spinor and partner components at small grades in depths 3, 4, 5, 6 and 10. It retains arbitrary complex matrix variations, different physical radial directions and all initially empty lower grades. The 657-check report records its complete scope and source hashes. The spatial self-adjointness argument is analytic; the executable report is not a nonlinear evolution or a test of a derived physical carrier.