Determine singular Kronecker block determinant images (no new κ)

research/kronecker-determinant-image/PROOF.md

@@ -0,0 +1,335 @@
+# Determinant image of an isotypic singular Kronecker pencil
+
+This note gives a self-contained proof for one positive right minimal index
+with arbitrary positive multiplicity, and obtains the left-block case by
+transposition. It addresses the corresponding open question in Nick Petalas's
+[pencil-determinant-image package at its pinned head](https://github.com/CrocSwap/integer-mult-bounds/blob/43142bedc7cb5c0f1f73ff3fea6d9bfd237949ee/research/pencil-determinant-image/README.md).
+The block and symmetric-power constructions are classical; no literature
+novelty is claimed. The addition here is the explicit determinant-image formula,
+its action conventions, and its bounded verification package. This is a scoped
+proof, not a new multiplication bound or a classification of arbitrary pencils.
+
+## 1. Pencil and action conventions
+
+Let K be any field, and let e,r be positive integers. Define
+
+```
+A_0 = [I_e | 0],       A_1 = [0 | I_e],
+L_e(s,t) = s A_0 + t A_1.
+```
+
+Thus L_e is e by e+1, with s on its main diagonal and t immediately above
+that diagonal. Set
+
+```
+T_0 = A_0 tensor I_r,   T_1 = A_1 tensor I_r,
+T(s,t) = L_e(s,t) tensor I_r.
+```
+
+The row dimension is er and the column dimension is (e+1)r. The tensor-product
+ordering groups each monomial index into an r-dimensional block. A simultaneous
+permutation of rows and columns identifies this pencil with the direct sum of
+r identical L_e blocks.
+
+For P in GL_(er)(K), Q in GL_((e+1)r)(K), and
+
+```
+R = [[a,b], [c,d]] in GL_2(K),
+```
+
+use exactly the action
+
+```
+(T_0,T_1) -> (P(a T_0 + b T_1)Q^T, P(c T_0 + d T_1)Q^T).
+```
+
+In particular the right factor in the action is Q^T, not Q^(-1). If
+z=(s,t)^T, the mixed pencil is T(R^T z), because
+
+```
+s(a T_0+b T_1) + t(c T_0+d T_1)
+    = (as+ct)T_0 + (bs+dt)T_1.
+```
+
+Let Stab(T) be the stabilizer under this action, and let
+
+```
+Delta(T) = { (det P, det Q, det R) : (P,Q,R) in Stab(T) }.
+```
+
+All equalities of pencils below are polynomial equalities in independent
+indeterminates s,t. No interpolation over the points of K is used.
+
+## 2. The exact image
+
+**Theorem.** For every field K and all e,r >= 1,
+
+```
+Delta(L_e tensor I_r)
+ = { (u^e z^[r e(e-1)/2],
+      u^[-(e+1)] z^[-r e(e+1)/2],
+      z) : u,z in K^* }
+ = { (x,y,z) in (K^*)^3 :
+       x^(e+1) y^e z^[r e(e+1)/2] = 1 }.
+```
+
+The half-products in these exponents are integers. They do not denote division
+inside K, so the statement includes characteristic two.
+
+For the transposed block the result is
+
+```
+Delta(L_e^T tensor I_r)
+ = { (x,y,z) in (K^*)^3 :
+       x^e y^(e+1) z^[r e(e+1)/2] = 1 }.
+```
+
+The proof also constructs a stabilizer element for every triple in the stated
+image, with any prescribed R having determinant z.
+
+## 3. A characteristic-safe lift of every slice transformation
+
+For k >= 0 let V_k be the homogeneous degree-k polynomials in X,Y, with ordered
+basis
+
+```
+X^k, X^(k-1)Y, ..., Y^k.
+```
+
+Let S_k(R) be the matrix of polynomial substitution
+
+```
+f(X,Y) -> f(aX+bY,cX+dY)
+```
+
+on coefficient columns in this basis. Equivalently, column i, with
+0 <= i <= k, consists of the coefficients of
+
+```
+(aX+bY)^(k-i) (cX+dY)^i.
+```
+
+In particular S_0(R)=[1] and S_1(R)=R^T. This convention matters for the
+transposes in the lift. Substitution by R^(-1) is inverse to substitution by R,
+so S_k(R) is invertible in every characteristic.
+
+Use ordinary monomials and binomial coefficients reduced in K. There are no
+normalized monomials, divided factorials, or assumptions that k! is invertible.
+The substitution identities remain valid when some binomial coefficients
+vanish in K.
+
+The determinant is
+
+```
+det S_k(R) = (det R)^[k(k+1)/2].                    (1)
+```
+
+For completeness, this follows on generators of GL_2(K). For a diagonal
+R=diag(alpha,beta), S_k is diagonal with entries alpha^(k-i) beta^i,
+whose product is (alpha beta)^[k(k+1)/2]. Upper and lower unitriangular
+matrices act triangularly with all diagonal entries one. For the matrix
+interchanging X,Y, S_k reverses k+1 basis vectors, with determinant
+(-1)^[k(k+1)/2]. These matrices generate GL_2(K) by row elimination.
+With the stated coefficient-column convention, composition satisfies
+S_k(R)S_k(U)=S_k(UR). Determinants are multiplicative in either order,
+proving (1) for all R. This proof also works in characteristic two.
+
+Let B_e(s,t)=L_e(s,t)^T. It is the coefficient matrix of multiplication by
+sX+tY from V_(e-1) to V_e. Substituting the two linear forms into this product
+gives
+
+```
+S_e(R) B_e(s,t) = B_e(as+ct,bs+dt) S_(e-1)(R).
+```
+
+Transposing and rearranging proves
+
+```
+L_e(as+ct,bs+dt)
+    = S_(e-1)(R)^(-T) L_e(s,t) S_e(R)^T.           (2)
+```
+
+Consequently a stabilizer lift of every R is
+
+```
+P_R = S_(e-1)(R)^T tensor I_r,
+Q_R = S_e(R)^(-1) tensor I_r.                      (3)
+```
+
+Indeed (2) says P_R T(R^T z) Q_R^T = T(z). There is no residual sign and
+no restriction on R beyond invertibility.
+
+## 4. The complete fixed-slice stabilizer
+
+Suppose R=I_2 and (P,Q,I_2) fixes T. Put C=Q^(-T). The two equations are
+
+```
+P T_0 = T_0 C,       P T_1 = T_1 C.               (4)
+```
+
+Write P as an e by e block matrix and C as an e+1 by e+1 block matrix,
+with every block of size r by r. Use indices starting at zero. The first
+equation of (4) says that the first e block rows of C equal [P | 0]. The
+second says that the last e block rows of C equal [0 | P]. In their overlap,
+for 1 <= i <= e-1, one obtains
+
+```
+P_(i,0) = 0,
+P_(i,j) = P_(i-1,j-1)       for 1 <= j <= e-1,
+P_(i-1,e-1) = 0.                                  (5)
+```
+
+The recurrence makes each block diagonal constant. The first boundary
+condition kills every diagonal below the main diagonal; the last kills every
+diagonal above it. Thus
+
+```
+P = I_e tensor D,       C = I_(e+1) tensor D
+```
+
+for one r by r matrix D. When e=1 there is no overlapping block row,
+but the two equations directly give C=diag(P,P), which is the same conclusion.
+Invertibility of P forces D to be invertible. Conversely every invertible D
+satisfies (4). Therefore the fixed-slice stabilizer is exactly
+
+```
+(P,Q) = (I_e tensor D, I_(e+1) tensor D^(-T)),
+       D in GL_r(K).                              (6)
+```
+
+In particular arbitrary mixing between the r equal blocks is already included;
+it contributes one full GL_r(K), not just r independent scalar factors.
+
+## 5. All stabilizers and their determinants
+
+Fix R. Both a given stabilizer (P,Q,R) and the lift (P_R,Q_R,R) send
+T(R^T z) to T(z). Their quotient on the row and column factors therefore
+fixes the unmixed pencil T(z). Equation (6) applies. Since the factors acting
+on the monomial index commute with those acting on the multiplicity index,
+every stabilizer with slice factor R is uniquely of the form
+
+```
+P = S_(e-1)(R)^T tensor D,
+Q = S_e(R)^(-1) tensor D^(-T),       D in GL_r(K).  (7)
+```
+
+More explicitly, putting H=P P_R^(-1) and J=Q Q_R^(-1) gives
+H T(z) J^T=T(z), so (6) gives H=I_e tensor D and
+J=I_(e+1) tensor D^(-T). This proves necessity as well as sufficiency of (7).
+
+Let u=det D and z=det R. Applying (1) and
+det(A tensor B)=(det A)^(dim B)(det B)^(dim A) to (7) gives
+
+```
+det P = u^e z^[r e(e-1)/2],
+det Q = u^[-(e+1)] z^[-r e(e+1)/2].               (8)
+```
+
+The determinant maps GL_r(K)->K^* and GL_2(K)->K^* are surjective,
+as witnessed by diagonal matrices. Hence every u,z in K^* occurs, proving
+the parametrized form of the theorem. Eliminating u from (8) gives
+
+```
+(det P)^(e+1) (det Q)^e (det R)^[r e(e+1)/2] = 1.
+```
+
+## 6. Constructive converse without root extraction
+
+Suppose x,y,z are nonzero and satisfy
+
+```
+x^(e+1) y^e z^[r e(e+1)/2] = 1.                  (9)
+```
+
+Define
+
+```
+u = (xy)^(-1) z^(-re).                            (10)
+```
+
+There is no need to take an e-th root or an (e+1)-st root. To check (10),
+put A=r e(e-1)/2 and B=r e(e+1)/2. Then A+B=re^2 and
+2B=re(e+1). Equation (9) gives
+
+```
+u^e z^A
+  = x^(-e) y^(-e) z^(-re^2+A)
+  = x z^(B-re^2+A) = x,
+
+u^[-(e+1)] z^(-B)
+  = x^(e+1) y^(e+1) z^[re(e+1)-B]
+  = y z^[re(e+1)-2B] = y.
+```
+
+Take D=diag(u,1,...,1), and take any R with determinant z, for example
+diag(z,1). Equations (7) then give an explicit stabilizer with determinant
+triple (x,y,z). This proves that (9) is the only constraint, over any field.
+Conversely (8) implies xy=u^(-1)z^(-re), so (10) also proves uniqueness
+of u for a given image triple.
+
+## 7. Transposition and the finite-field orbit count
+
+Transposing the two slice equations exchanges P and Q and leaves R unchanged.
+Thus (P,Q,R) stabilizes (T_0,T_1) if and only if (Q,P,R) stabilizes
+(T_0^T,T_1^T). The determinant image for the left block is exactly the right
+block's image with its first two coordinates exchanged. This proves the
+transposed formula in Section 2.
+
+For K=F_q, the parametrization (8) is injective in (u,z), by (10), so
+
+```
+|Delta(T)| = (q-1)^2.
+```
+
+Let G=GL_(er)(F_q) x GL_((e+1)r)(F_q) x GL_2(F_q) and let N be the
+product of its three special linear subgroups. The determinant map identifies
+G/N with (F_q^*)^3. Because N is normal, the N-orbits within the single
+G-orbit of T are indexed by
+
+```
+(F_q^*)^3 / Delta(T).
+```
+
+The character
+
+```
+chi(x,y,z) = x^(e+1) y^e z^[r e(e+1)/2]
+```
+
+has kernel Delta(T) and is surjective: chi(w,w^(-1),1)=w.
+Consequently there are exactly q-1 such special-linear orbits. For the left
+block exchange x and y in this character. This is an orbit statement about
+the displayed isotypic pencil only.
+
+## 8. Boundaries of the result and checks that can falsify mistakes
+
+The theorem covers e>=1 and r>=1 only. The degree-zero right and left blocks
+have shapes 0 by 1 and 1 by 0, respectively. They represent unused columns
+or rows and require separate treatment of zero-dimensional factors; the lift
+above would also call for the undefined V_(-1). Adjoining such blocks to a
+positive-index pencil introduces extra fixed-slice maps that are outside the
+recurrence proved here. This exclusion is a scope boundary, not a claim that
+the displayed equation has been disproved under some extension of conventions.
+
+Different positive minimal indices, mixtures of right and left blocks, and
+regular/singular coupling are also outside this theorem. Their off-diagonal
+homomorphisms have not been classified here. No result is claimed for the
+separate regular-pencil question with no invertible rational slice.
+
+Useful discriminating finite checks include:
+
+- For e=1,r=1, the lift reads P=[u] and Q=u^(-1)R^(-T). The equation is
+  x^2 y z=1, which checks the sign and the inverse transpose on the right.
+- For e=2,r=1 over F_3, the image equation is x=z with y unrestricted.
+  Dropping the slice determinant exponent would instead require x=1.
+- For e=1,r=2 over F_3, the equation is y=1 with x,z unrestricted,
+  whereas e=1,r=1 gives y=z. This detects an omitted multiplicity factor.
+- In characteristic two, e at least two exercises vanishing binomial
+  coefficients in S_e. F_4 provides nontrivial determinant characters;
+  testing only F_2 cannot distinguish determinant exponents.
+- Directly solve the stabilizer slice equations in small dimensions instead
+  of checking only matrices generated by (7). Such a check can falsify the
+  completeness assertion in Sections 4 and 5.
+
+Finite checks support the implementation and expose convention errors. They
+do not replace the arbitrary-field proof above or settle any excluded case.