20260818-851 The Source Forms Fix the Marked Localization Boundary Matrices
Question
Before choosing any primitive exact lift for the rank-twelve marked system, which maps in
0⟶Mq(9)j∗Mmark(12)ResWH1(W)(−1)⟶0
are already fixed by the source forms and Poincaré-residue orientation?
Frozen bases
Use the ordered marked basis
(Ω111,Ω101,Ω110,e1,…,e9)
and the source basis (e1,…,e9). The open restriction is therefore
j∗=(03×9I9).
No exact-lift convention enters this matrix.
Oriented wall residues
With fiber orientation da∧db, order each wall basis as its top-intersection class followed by its primitive wall class. Direct residue gives
ResW1(Ω111)=−t1,ResW1(Ω101)=−g1,
ResW2(Ω111)=+t2,ResW2(Ω110)=+g2,
and all other basis images vanish. In the stacked wall basis
(t1,g1,t2,g2)
the same-sheet top differential is
Restop=(1010).
Hence the two iterated residues of Ω111 cancel:
Restop(ResW1ResW2)=0.
Exact boundary complex
The stacked residue matrix has rank three, the top map has rank one, and
im(ResW1ResW2)=ker(Restop).
Choosing the three quotient basis vectors induced in order by
(Ω111,Ω101,Ω110)
makes the combined source-normalized residue matrix
ResW=(I303×9).
Consequently
rankj∗=9,rankResW=3,ResWj∗=0,
so the vector-space localization sequence is exact in these source-normalized bases.
Narrow result
The source forms uniquely fix j∗, both oriented wall-residue matrices, and the same-sheet top differential before any choice among the large exact-reduction nullspaces. These matrices are therefore mandatory boundary conditions on the rank-twelve Gauss--Manin connection.
This does not yet determine Ax(12) or Ay(12), prove horizontality, or establish compatibility with R∞. The next finite calculation is a wall-Laurent exact reduction with one common primitive normalization, constrained by the matrices above. Any candidate connection that violates their horizontality is rejected; no nullspace may be selected by sparsity or by searching for Q.