IDOS Energy, Coherence, and Supporting Dynamics — Part 76

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$$z_{tt}+G_{\rm phys}z_t+\mathscr A z=0,\qquad \mathscr A=\mathcal T^*\mathcal T\ge0,\qquad G_{\rm phys}^*=-G_{\rm phys},\qquad \mathcal D(\mathscr A)=H^2(\mathbb R^{10}), \tag{N2}$$

with physical Hilbert space $L^2(dx)$ and form domain $H^1$. The density reconstruction is bounded in this phase space. Finite residual energy alone is a weaker condition than the inhomogeneous physical norm used here.

Exhaustive Static Kernel And The Common-Phase Obstruction

The positive occupied radial and empty forms set $u=h_l=0$ in a static zero mode. The complete flow square has remainder at least $(223/424)w\Omega|v|^2$, so $v=0$ and $\nabla\chi=0$. Frame-gradient and phase squares then give constant $F$ and

$$\theta_k(x)=\alpha_k+(F\xi_k)\cdot x. \tag{N3}$$

For $\eta=0$ all $D$ constants $\alpha_k$ are allowed. For $\eta>0$ the phase edges force $\alpha_k=\alpha_0+k\cdot\varpi$, a space of dimension $d+1$. To check the support kernel, put $S=W^{-1/4}$. Its first residual gives $Z=(L/2)S^{-5}\nabla S$; the other two become

$$B_{\rm sup}/W=(L/2)S^{-5}(\nabla^2S-2I/L^2),\qquad G_{\rm sup}=S^{-10}(L^2|\nabla S|^2/4-S+1).$$

Their linearized zero conditions are $\nabla^2\delta S=0$ and $x\cdot\nabla\delta S-\delta S=0$. Hence $\delta S$ is linear with no constant term, exactly $Y=-D_sa$ for a translation $a\in\mathbb R^{10}$. There is no extra support size or amplitude zero mode. These modes are tangents to the exact translated, rotated and rephased branch, and their physical $H^2$ norms are finite. All 45 frame rotations are retained, including $r_d=(10-d)(9-d)/2$ rotations in unused directions. Consequently

$$\dim\ker\mathscr A=55+D\quad(\eta=0),\qquad \dim\ker\mathscr A=56+d\quad(\eta>0). \tag{N4}$$

The integrated gyro pairing of any two static modes vanishes, but the gyro itself annihilates only the unused-frame rotations. In particular, writing $\bar\alpha=\mathbb E[\alpha]$ and $\gamma=\alpha-\bar\alpha+(F\xi)\cdot x$, its covector rows are

$$(Gk)_\chi=-a\cdot\nabla w,\quad (Gk)_u=-2w\gamma,\quad (Gk)_Y=-w'(W_0)\bar\alpha e_W,\quad (Gk)_v=\mu_sD_s^TD_sa-\varepsilon wB\gamma. \tag{N5}$$

The frame and horizontal-phase rows vanish. Integration of the possible support/common-phase pairing gives $\int\partial_iw\,dx=0$. This necessary orthogonality does not imply that $-G_{\rm phys}k$ lies in the range of $\mathscr A$ at its essential zero threshold.

A generalized partner means a finite physical solution $z(t)=tk+z_1$ with $\mathscr A k=0$ and $\mathscr A z_1=-G_{\rm phys}k$. A nonzero $\bar\alpha$ would require the independent support equation $K_sY=\bar\alpha w'(W_0)e_W$. In its exterior,

$$w'(W_0)=2\sigma\epsilon_p^8L^8|x|^{-8}[1+O(|x|^{-2})],\qquad \mu_s=C_s\sigma\epsilon_p^8[1+O(|x|^{-2})].$$

Take nonnegative nonzero $\varphi\in C_c^\infty(1<|x|<2)$, $f_R=R^{-5}\varphi(x/R)$ and

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