CCFU Proof 20 — Parent Decomposition: P = S₁ ⊕_W S₂

2605245765172

CCFU Proof 20 — Parent Decomposition: P = S₁ ⊕_W S₂

05/24/2026

Given.
P = R⁵, sig(P) = (3,2) [Proof 5].

S₁ ⊂ P, dim S₁ = 4, sig(S₁) = (3,1) [Proof 4].

S₂ ⊂ P, dim S₂ = 4, sig(S₂) = (2,2) [Proofs 3, 16].

Step 1 — dim(S₁ ∩ S₂) = 3.

dim(S₁ ∩ S₂) ≥ dim S₁ + dim S₂ - dim P = 4+4-5 = 3.

If dim(S₁ ∩ S₂) = 4 then S₁ = S₂, but sig(S₁) = (3,1) ≠ (2,2) = sig(S₂).
Therefore dim(S₁ ∩ S₂) = 3.

It follows that dim(S₁ + S₂) = 4 + 4 - 3 = 5 = dim P, so S₁ + S₂ = P.
∎

Step 2 — W is nondegenerate.

S₁ is nondegenerate in P, so S₁⊥ is 1-dimensional.
sig(P) = (3,2) and sig(S₁) = (3,1), so S₁⊥ has sig(0,1): it is a negative line.

S₂ is nondegenerate in P, so S₂⊥ is 1-dimensional.
sig(P) = (3,2) and sig(S₂) = (2,2), so S₂⊥ has sig(1,0): it is a positive line.

Since a nonzero line cannot be both positive and negative, S₁⊥ and S₂⊥ are distinct.
Let u span S₁⊥ and v span S₂⊥.
Then u, v are linearly independent,
⟨u,u⟩ < 0 and ⟨v,v⟩ > 0, so the Gram determinant is

⟨u,u⟩ ⟨v,v⟩ - ⟨u,v⟩^2 < 0.

Therefore S₁⊥ + S₂⊥ is nondegenerate with sig(1,1).

W = S₁ ∩ S₂ = (S₁⊥ + S₂⊥)⊥.
The orthogonal complement of a nondegenerate subspace is nondegenerate.
Therefore W is nondegenerate.
∎

Step 3 — sig(W) = (2,1).
W is 3-dimensional and nondegenerate.

In S₁ with sig(3,1): a 3d nondegenerate subspace has sig ∈ (3,0), (2,1).

In S₂ with sig(2,2): a 3d nondegenerate subspace has sig ∈ (2,1), (1,2).

Common: sig(W) = (2,1).
∎

Step 4 — Orthogonal complements.
Since W is nondegenerate, define V₁ := S₁ ∩ W⊥ and V₂ := S₂ ∩ W⊥.
Then S₁ = W ⊕⊥ V₁ and S₂ = W ⊕⊥ V₂, with dim V₁ = dim V₂ = 1.

sig(W) + sig(V₁) = sig(S₁): (2,1) + sig(V₁) = (3,1) → sig(V₁) = (1,0).

sig(W) + sig(V₂) = sig(S₂): (2,1) + sig(V₂) = (2,2) → sig(V₂) = (0,1).

V₁ is positive.
V₂ is negative.
∎

Step 5 — Basis.

e₀, e₁: positive, in W = S₁ ∩ S₂ sig(2,0)

e₂: positive, in V₁ (S₁-only) sig(1,0)

e₃: negative, in W = S₁ ∩ S₂ sig(0,1)

e₄: negative, in V₂ (S₂-only) sig(0,1)

Choose unit vectors e₂ ∈ V₁ and e₄ ∈ V₂.
Since e₀,e₁,e₃ are orthonormal in W and V₁, V₂ ⊂ W⊥, the basis (e₀,e₁,e₂,e₃,e₄) satisfies

⟨eᵢ,eⱼ⟩ = 0 whenever one of i,j {0,1,3} and the other ∈ {2,4}.

The only potentially nonzero off-diagonal entry is α := ⟨e₂,e₄⟩, which is determined by the fixed subspaces S₁, S₂.

Verification: S₁ = span(e₀,e₁,e₂,e₃), sig = (3,1) ✓.
S₂ = span(e₀,e₁,e₃,e₄), sig = (2,2) ✓.

In an orthonormal basis of P (diagonalizing Q₅), the metric is diag(+1,+1,+1,-1,-1), but S₂ may not align with a coordinate subspace.
The adapted basis above keeps S₁ and S₂ as coordinate subspaces at the cost of one off-diagonal entry α.
∎

Conclusion.
S₁ and S₂ are not chosen—they are the linear closure
(Proof 4) and nonlinear closure (Proofs 3, 16) of C₂.
Hence

W=S₁∩ S₂

is fixed.
The proof shows that W is 3-dimensional,
nondegenerate, and has signature (2,1).
The orthogonal complement lines

V₁=S₁∩ W⊥, V₂=S₂∩ W⊥

are also fixed by S₁,S₂, with sig(V₁)=(1,0) and
sig(V₂)=(0,1).
In an adapted basis their mutual inner product

α=⟨e₂,e₄⟩

may appear as one off-diagonal metric entry; it is determined by
the fixed pair (S₁,S₂) and is not an extra choice.
For the downstream chain, however, no data from α is used:
Proof 21 depends only on W and constructs

V_W=Rτ ⊕ W ⊕ W*.

Thus no additional parameter enters the higher alternating closure.

[Dependencies: Proofs 3, 4, 5, 16.]