Lemma 10 ([Odd-path support]). Let k=2t+1 with t≥0 . On positions 1,…,k , put alternating pairs in the first layer on (2,3),(4,5),…,(k-1,k) and the singleton letter 1 at position 1 . In the second layer put alternating pairs on …

Identify a pair-letter covector with a matrix Y by evaluation Y(e_a⊗e_c)=Y_ ac , and put J= 0, 1; -1, 0 . Expanding the pair alternations and successively contracting the indices along the path gives ⟨u,Y_1⊗⋯⊗Y_k⟩ , =[Y_1JY_2^ T JY_3⋯ JY_ k-1 ^ T JY_k]_ 11 ; , =(-1)^t [Y_1(Y_2)Y_3⋯(Y_ k-1 )Y_k]_ 11 , because JY^ T J=-(Y) for a two-by-two matrix. The formula also holds for k=1 , with no intervening factors. Use the ordered matrix basis M_1=I, M_2=Z= 1, 0; 0, -1 , M_3=T= 0, 1; 1, 0 , M_4=J=ZT. The last three matrices are traceless, invertible, and pairwise anticommuting. Adjugation fixes I and negates the other three. For a segment of length a≤4 beginning at global position p , write D_j(Y)=Y for odd j and D_j(Y)=(Y) for even j . Its alternated matrix is A_ a,p =∑_ π∈S_a (π) D_p(M_ π(1) )⋯D_ p+a-1 (M_ π(a) ). Let e be the number of even positions in this segment, and let ε=1 when p is even and ε=0 otherwise. Then A_ a,p =a!(-1)^ e-ε M_1⋯M_a. Indeed, if I occupies the local slot j , moving it there contributes (-1)^ j-1 to the permutation sign, while reordering all remaining matrices contributes precisely their permutation sign by anticommutation. The adjugations contribute (-1)^ e-ε_j , where ε_j records whether p+j-1 is even. Since ε_j≡ε+j-12 , the total sign is (-1)^ e-ε for every summand. This proves , whose right side is invertible. For example, at length three the formula gives -6J for both starting parities: the canonical product is IZT=J , and e-ε=1 in either case. Consequently, alternating independently within the prescribed consecutive segments in gives (-1)^t[B]_ 11 , where B is the ordered product of their invertible matrices A_ a,p . Choose an eigenvector of B with nonzero eigenvalue β , and a matrix Q∈_2() taking it to the first coordinate vector. Replace every M_i by QM_iQ^ -1 . Since adjugation commutes with conjugation, the new contraction is (-1)^t[QBQ^ -1 ]_ 11 =(-1)^tβ≠0 . The endpoint letters of u have remained fixed throughout. The dual tensor is a polytabloid for the partition with the specified column lengths, regardless of their order along the path. Lemma 2 ([Dual-polytabloid test]) proves the support assertion. For any one placement its contraction is a polynomial in the entries of the common basis and is not identically zero by what we just proved. The nonvanishing loci of finitely many such polynomials have nonempty intersection with the open set of ordered bases, since _4() is irreducible. This proves the last assertion.