Constant linear coordinates and coupled dilation
Exact meaning of decoupling
Let \(q,r,a,b>0\), with contraction \(<1\) imposed when discussing the normalized entire solution. For an arbitrary analytic vector input \(X=(u,v)\) with \(u(0)=v(0)=0\), define the functional vector field
A nonzero constant row \(c=(c_1,c_2)\in\mathbb C^2\) is an autonomous scalar projection if some scalar functional \(\mathcal H\) satisfies
The functional \(\mathcal H\) may use the whole scalar input. Thus this definition allows any scalar dilation equation, not only a prescribed quadratic ansatz. Equality (S2) means that two vector inputs with the same scalar projection have the same projected vector field. It is an identity of functional differential operators. It does not assert an identity only along the distinguished normalized solution.
A constant invertible change \(W=PX\) completely decouples the operator if both rows of \(P\) are autonomous scalar projections. No assumption about the form of their scalar functionals is needed.
Classification of autonomous linear projections
Theorem H.1.
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For positive nonzero scales, a nonzero row \(c\) satisfies (S2) if and only if
In that exceptional case the projected vector field is identically zero. Consequently, no invertible constant linear change completely decouples (S1), including when some or all scales coincide. If \((q,r)\ne(a,b)\), there is not even one nonzero autonomous scalar projection. Four pairwise distinct contraction scales are a special case of this stronger result.
Proof.
Subtract the constant forcing. Define
If (S2) holds, \(Q_c\) is constant on every fiber of \((u,v)\mapsto c_1u+c_2v\). Fix \(z_0\ne0\). Polynomial interpolation gives a scalar polynomial \(f\) with \(f(0)=0\) and \(f(sz_0)=1\) at every distinct scale \(s\in\{q,r,a,b\}\). For inputs \(u=xf,v=yf\), where \(x,y\in\mathbb C\),
It must depend only on \(c_1x+c_2y\). If \(c_1+c_2\ne0\), differentiating the polynomial \(xy\) along the fiber direction \((c_2,-c_1)\) would give
which forces \(c=0\), a contradiction. Therefore \(c_2=-c_1\), and scaling the row reduces the question to \(c=(1,-1)\).
For this row, put \(u=w+g,v=g\). This input and \((w,0)\) have the same projection \(w\). The second input has \(Q_c[w,0]=0\), so autonomy forces
for all zero-constant analytic \(w,g\). Setting \(w=0\) and subtracting that instance of (S5) leaves
Taking successively \((w,g)=(z,z),(z^2,z),(z,z^2)\) yields \(qr=ab\), \(q^2r=a^2b\), and \(qr^2=ab^2\). Nonzero scales then give \(q=a\) and \(r=b\). Conversely these equalities make the two components of (S1) identical on every input, so \((1,-1)\mathcal F[X]=0\). The only possible autonomous rows therefore span a one-dimensional space, which cannot be the two rows of an invertible matrix. \(\square\)
The same proof works on odd analytic inputs for positive scales: choose \(f(z)=z p(z^2)\) by interpolation at the distinct positive scale squares, and use \((z,z),(z^3,z),(z,z^3)\) in (S6). Positivity then resolves the equalities of squares. The theorem does not depend on allowing even inputs.
Tensor explanation for four distinct scales
The preceding proof also makes a coordinate invariant obstruction explicit. Under an invertible state change, write the original coordinates as \(u=\ell_1(W),v=\ell_2(W)\), where \(\ell_1,\ell_2\) are independent linear forms. At the pair of distinct nodes \((qz,rz)\), the first quadratic term has bilinear coefficient
It has rank one, but is not symmetric: \(B-B^{\mathsf T}=\ell_1\wedge\ell_2\ne0\). At four distinct nonzero nodes, polynomial interpolation lets their vector values vary independently, so terms supported at \((az,bz)\) cannot cancel this coefficient. A scalar autonomous channel at the same node pair would require \(B=\kappa\ell\otimes\ell\), a symmetric rank-one tensor, after projecting the output onto that channel. For any output row with a nonzero coefficient on this term, the alternating part remains nonzero. The other node pair gives the same obstruction if the first output coefficient vanishes.
This is not a rank-only argument: the tensor \(\ell_1\otimes\ell_2\) already has rank one. Its nonzero alternating part is the decisive invariant at distinct nodes. When the nodes of a product coincide, the symmetric quadratic form \(uv\) has matrix
of rank two, whereas a square of one linear coordinate has rank one. Rank is then the appropriate obstruction. The fiber proof in Theorem H.1 handles node coincidences and cancellations without mixing these two cases.
Particular solutions and degenerate parameters
Theorem H.1 is an operator statement. It does not exclude coincident components for particular initial data, algebraic relations along a solution, elimination into a scalar equation of higher order, nonlinear coordinate changes, or identities with other named special functions.
For contraction scales, the normalized analytic germ is unique. If \((a,b)=(q,r)\), its components coincide because their derivatives are identical and their initial values agree. The common component is the scalar family \(T'=1+T(qz)T(rz),T(0)=0\). The same coincidence holds if \((a,b)=(r,q)\): \((T,T)\) solves both equations because multiplication is commutative, and uniqueness identifies the normalized solution. For \(q\ne r\), this reversed-pair case has no autonomous scalar operator projection by Theorem H.1, despite having \(U=V\) along its normalized solution. This is a concrete reason the two notions must remain distinct.
In fact, equality of normalized components occurs exactly when the unordered scale pairs agree:
To prove necessity, write \(U=V=T=z+c_1z^3+\cdots\). Then \(3c_1=qr=ab>0\). Comparing the \(z^4\) coefficients of \(T(qz)T(rz)=T(az)T(bz)\) gives \(c_1qr(q^2+r^2)=c_1ab(a^2+b^2)\), so both products and sums of squares agree. Positive scales then have equal sums and are the roots of the same quadratic. Sufficiency is the uniqueness argument above. For example, \((q,r,a,b)=(1/2,1/3,1/3,1/2)\) has coincident normalized components and no nonzero autonomous operator projection.
At \(q=r=a=b=s>0\), the variables \(w=u+v,d=u-v\) give
This triangular system is not a completely decoupled operator. The normalized solution lies in the invariant subspace \(d=0\).
Zero scales lie outside Theorem H.1. On zero-constant inputs, a product vanishes identically if either of its scales is zero. If both products vanish, both equations are already \(u'=v'=1\). If only the first product vanishes, the projection \(u\) is autonomous, regardless of whether the second pair agrees with the first. Thus extending Theorem H.1 to zero scales would be false. Scale one and positive scales greater than one do not affect the algebraic operator theorem, although the contraction-domain entire growth theorems do not automatically extend to them. Negative or complex scales and odd-input coincidences caused by signs are not classified here.
Claim status
Theorem H.1 and (S7) are proved above. The proof is elementary and independent of the coupled growth theorem. Historical priority for these precise operator statements is not asserted. The theorem provides a specified nondecoupling result, not a universal assertion that the normalized functions cannot reduce to any previously studied function class.