Persistent spatial shape barriers

Summary: For orthogonal higher circulation, three spatial restriction maps classify permanent, linear and quadratic frequency barriers, with a Dirac example using one interaction coefficient.

Ask whether a shape stays inaccessible throughout the response

Dynamical shape opening shows that a higher spatial pattern can have no static restriction into the lower sector and still drive it after the higher field responds. Across the particle hierarchy, the next question is which shapes remain inaccessible, and how strongly the others open at low frequency.

Consider a higher even-dimensional sector with a specified orthogonal circulation tensor $J$, so $J^T=-J$ and $J^2=-I$. The lower sector is a fixed spatial submanifold with local tangent inclusion $I$. Both belong to the one nested manifold of ordinary macroscopic spatial dimensions. Changing $J$ relative to this inclusion changes their physical relation; rotating both together only changes its coordinate description.

Use the normalized symmetric traceless spatial shape tensors of the full Dirac response. Write

$$\mathcal R(C)=I^TCI,\qquad \mathcal K(C)=\frac12[J,C],\qquad T=i\mathcal K. \tag{1}$$

The lower output contains all symmetric shapes, including its breathing component. A higher traceless tensor need not remain traceless below. The tensor components label physical field patterns; they are not further physical dimensions or additional interaction particles.

The complete response uses only three spatial maps

Split a symmetric tensor into parts commuting and anticommuting with $J$:

$$C_{\rm c}=\frac12(C-JCJ),\qquad C_{\rm a}=\frac12(C+JCJ). \tag{2}$$

The commuting part is annihilated by $\mathcal K$. On the anticommuting part, $\mathcal K C_{\rm a}=JC_{\rm a}$ and $\mathcal K^2C_{\rm a}=-C_{\rm a}$. Thus

$$\mathcal K^3=-\mathcal K,\qquad T^3=T. \tag{3}$$

The three eigenvalues of $T$ are $0,+1,-1$. For the radial Dirac background, let $A(z),B(z)$ be the complete bare response functions derived in the shape-opening note. They are respectively even and odd in frequency, and the bare response is $A I+B T$. Source feedback with the same scalar coefficient $g_0$ gives

$$\boxed{\chi(z)=a_0(z)I+a_1(z)T+a_2(z)T^2,} \tag{4}$$

where, writing $D(z)=1-g_0A(z)$,

$$\boxed{ \begin{aligned} a_0&=\frac{A}{D},\\ a_1&=\frac{B}{D^2-g_0^2B^2},\\ a_2&=\frac{g_0B^2}{D(D^2-g_0^2B^2)}. \end{aligned}} \tag{5}$$

These coefficients follow by evaluating the feedback response $(A+B\lambda)/[1-g_0(A+B\lambda)]$ at the three eigenvalues $\lambda=0,\pm1$. Equation (4) is exact away from its poles, not a low-frequency truncation.

Consequently the permanent geometric kernel is

$$\boxed{\mathcal N =\ker\mathcal R\cap\ker(\mathcal R\mathcal K) \cap\ker(\mathcal R\mathcal K^2).} \tag{6}$$

Every pattern in $\mathcal N$ has zero lower source throughout this response. It also stays invisible under every finite rotation generated by $J$, because $e^{t\mathcal K}$ is a polynomial in $I,\mathcal K,\mathcal K^2$.

Suppose $A(0)\ne0$, $B'(0)\ne0$, $g_0\ne0$ and $D(0)\ne0$. These conditions hold in the example below. Then (6) is also necessary for a fixed pattern to be invisible at every nearby frequency: successive zeroth, first and second derivatives of (4) recover its three restriction maps. Zeros at a single frequency can have a larger kernel.

The first nonzero map fixes the opening order

Put $b_1=B'(0)$ and $D_0=1-g_0A(0)$. Equations (4)–(5) give the following classification for a fixed source $c$:

The quadratic opening requires the field's own source feedback. The bare operator $A I+B T$ remains unable to see that third class at every frequency. Its feedback generates the $T^2$ term in (4).

The amplitudes are therefore constant, linear or quadratic at low frequency under the stated nondegeneracy conditions. Their squared amplitudes scale as $1,z^2,z^4$. This does not identify a physical rate: an actual incoming mode, occupied target state and flux are still needed. The lower object's response can introduce further zeros or resonances.

This classification follows a specified spatial source as its frequency changes. A freely oscillating mode must also satisfy its own spatial eigenproblem. In particular, for a pure shape helicity $Tc=\lambda c$,

$$\mathcal R\chi(z)c =\frac{A(z)+\lambda B(z)} {1-g_0[A(z)+\lambda B(z)]}\,\mathcal Rc. \tag{6a}$$

Such a pattern with $\mathcal Rc=0$ remains inaccessible throughout this response. The linear and quadratic openings above concern fixed sources which combine helicities: their relative responses change with frequency. Applying the result to an actual incoming particle therefore requires its mode map and spectrum, including how that source excites the higher field.

Incoming spatial shape modes derives the corresponding reciprocal equation for an actual incident moment and checks the full free shape eigenproblem. Every pure six-dimensional helicity retains at least half its moment norm on a five-dimensional hyperplane; the opening orders concern cancellation between several such components. Source frequencies in spatial coupling identifies the actual source clocks and includes the changing normalization when the bound frequency itself is varied.

Nested geometry changes the permanent kernel

First choose $J$ with unit rotations in coordinate planes $12,34,\ldots$ and take the lower $p$-plane to be the first $p$ axes. Its smallest $J$-invariant containing plane is

$$W=V+JV,\qquad s=\dim W=p+(p\bmod2). \tag{7}$$

For even $p$, $W=V$ and rotating the source cannot create another lower restriction. For odd $p$, the first rotation recovers the mixed components with the one additional axis of $W$; the second recovers that axis's diagonal component. Thus (6) is precisely the condition $C|_W=0$.

If $s<d$, the visible rank is $s(s+1)/2$. If $s=d$, the higher traceless condition reduces it by one and no nonzero shape remains hidden. The successive ranks of the stacked maps $(\mathcal R)$, $(\mathcal R,\mathcal R\mathcal K)$ and $(\mathcal R,\mathcal R\mathcal K,\mathcal R\mathcal K^2)$ are:

These are ranks among spatial quadrupole patterns, not counts of particle species or declarations that one particle fails to interact.

The tilted example uses

$$B=\operatorname{diag}\left(\frac35J_2,\frac5{13}J_2,0\right), \quad C=\sqrt{I+B^2}, \quad J_{10}=\begin{pmatrix}B&C\\-C&-B\end{pmatrix}, \quad J_2=\begin{pmatrix}0&1\\-1&0\end{pmatrix}. \tag{8}$$

Here $C=\operatorname{diag}(4/5,4/5,12/13,12/13,1)$, so the circulation is exactly rational and $J_{10}^2=-I$. The lower planes are again the first five or six axes. Independent exact rational matrix ranks verify the last two rows without a floating-point rank threshold.

For comparison, $B=0$ makes the lower five-plane totally real relative to $J_{10}$: $JV$ is perpendicular to $V$. Its successive ranks are $15,30,44$, leaving a ten-dimensional permanent kernel. Thus depth and shared support alone do not fix the surviving shapes. The circulation relative to the actual inclusion matters.

A quadratic barrier using one common scalar coefficient

Take the coordinate-plane geometry for $5\subset10$, and the unit higher shape

$$C_* =\frac1{\sqrt2}(e_6e_6^T-e_7e_7^T). \tag{9}$$

Its full lower restriction and first rotational restriction vanish. The second does not: $\|\mathcal R\mathcal K^2C_*\|=1/(2\sqrt2)$. The axis labels describe the source relative to this particular lower submanifold; they do not divide the ten spatial directions into different kinds.

Solve the isolated five- and ten-dimensional scalar Dirac profiles with one common $g_0=6.5$ and bare scalar coefficient $m=0.5$. The source strengths are determined by self-consistency, giving approximately

$$a_5=1.031769686,\qquad a_{10}=0.909053520. \tag{10}$$

Both profiles' widths and frequencies therefore come from the same interaction. They are the appropriate backgrounds for the leading weak-contact response. A finite pair interaction would require the readjusted pair background in Nested Dirac pair.

Retain the lower breathing source

The second rotational restriction of (9) has a lower trace. In addition to the lower traceless response, use its full radial breathing response. Let $Q$ be an orthonormal basis perpendicular to the normalized stationary radial profile $\phi$. It imposes the fixed norm and removes its phase zero mode. Define

$$L=Q^T(H-E)Q,\qquad u=Q^T\frac{\beta r^2}{\sqrt p}\phi, \qquad \chi_{b,0}(z)=u^T(z-L)^{-1}u+u^T(-z-L)^{-1}u. \tag{11}$$

The breathing susceptibility is $\chi_b=\chi_{b,0}/(1-g_0\chi_{b,0})$. The source $\beta r^2/\sqrt p$ corresponds to the normalized lower tensor $I_p/\sqrt p$. Both signs of the radial Dirac spectrum remain in $L$; the calculation does not replace this source by a positive oscillator.

For the radial background, breathing and traceless quadrupoles are angularly orthogonal. The complete lower symmetric response is therefore

$$\chi_a^{\rm sym}(z,J_a)= \operatorname{diag}\big(\chi_b(z),\chi_a(z,J_a)\big). \tag{12}$$

The leading weak-contact transfer from (9) is $\epsilon\chi_a^{\rm sym}\mathcal R\chi_{10}C_*$, including both objects' responses. For a lower circulation with unit planes $12,34$:

The coefficient of frequency squared tends to $1.67090763\times10^{-6}$ in these model units. Its breathing component is $8.23623413\times10^{-7}$ and cannot be discarded. Changing the lower circulation while holding the higher shape fixed changes the full complex response, including its phase and direction; its norm alone can conceal much of that difference.

This constructs a stronger spatial matching barrier within the same interaction model. It does not assign (9) to a photon or derive a neutrino rate. Its use for physical particles requires their actual occupied shapes and incident mode maps.

Verification

nested_shape_visibility_checks.py compares coefficient evolution with rotations of actual spatial tensors, checks the complete feedback polynomial against the full matrix response, and verifies permanent kernels under finite rotations. Exact rational calculations independently certify the tilted geometry's ranks.

Both stationary profiles use the same scalar coefficient. Full constrained radial source response and independent profile changes verify the breathing channel. The measured quadratic transfer agrees with its independently calculated inverse-radial-operator coefficient. Common rotations transform the source, polarization and inclusion together and preserve every result. All checks remain active under Python optimization.

Related

• Dynamical opening of spatial shape channels — supplies the complete polarized response functions and the linear opening example.
• Polarized Dirac shape response — derives the spatial angular source and its ordinary/conjugate feedback.
• Self-consistent scalar Dirac moment modes — supplies the bound radial modes, source variation and stability limits.
• Self-consistent nested Dirac pair — supplies the reciprocal background when the pair interaction is finite.