RefacteKnowing what two operations do on a subspace and on the quotient does not determine what they do together. The missing information is how each lifted quotient vector spills into the subspace. Keeping track of that spill gives a small linear problem, including an explicit test for whether two answers describe the same operations in different coordinates.
Specify what is fixed
Let N = km and L = kn over a field k. Two fixed matrices A1, A2 commute on N, and two fixed matrices B1, B2 commute on L. We seek commuting operations on N ⊕ L whose restrictions to N and induced operations on L are these matrices.
The coordinates on N and L stay fixed. Only the chosen lifts of vectors of L into the larger space may change. This marking is part of the question; allowing arbitrary basis changes would ask for a different classification.
The unknown belongs in one block
Choose lifts of a basis of L. For a vector l, applying the larger operation to its lift must project to Bil. The difference between that result and the chosen lift of Bil therefore lies in N. Call it Cil. This is exactly the information left unspecified by the input.
In these coordinates, a vector is a pair (n,l), and its image is (Ain + Cil, Bil). Thus the matrix shape is forced:
Tᵢ = [ Aᵢ Cᵢ ], Cᵢ : L → N.
[ 0 Bᵢ ]
Rather than guess two whole matrices, keep only the two unknown rectangular blocks. There can be no product C1C2: each Ci goes from L to N, where the next operation is already fixed by Aj. This predicts that the commutativity test will be linear in the missing information.
Multiply before naming an obstruction
The diagonal blocks of T1T2 and T2T1 already agree. Their upper-right blocks agree exactly when
d₁(C₁,C₂) = A₁C₂ + C₁B₂ − A₂C₁ − C₂B₁ = 0.
The formula also has a direct reading. Acting first by operation 2 spills C2l into N, where operation 1 applies A1; the quotient part B2l spills a further C1B2l. Reversing the order gives the other two terms. Their difference is the defect.
Derive the allowed coordinate move
A different choice of lifts adds some Sl in N to each lift. The same S must be used for both operations: we are changing the coordinates of their shared space, not choosing a separate space for each matrix. This gives the matrix PS; its inverse subtracts the same vector:
Pₛ = [ I S ], Pₛ⁻¹ = [ I −S ].
[ 0 I ] [ 0 I ]
Directly multiplying PS−1TiPS gives
Cᵢ′ = Cᵢ + AᵢS − SBᵢ, d₀(S) = (A₁S−SB₁, A₂S−SB₂).
These moves add: PSPR = PS+R. A second multiplication verifies that they preserve compatibility:
d₁d₀(S) = (A₁A₂−A₂A₁)S − S(B₁B₂−B₂B₁) = 0.
Thus changing the lifts moves a compatible pair in a direction that keeps its defect zero. In particular, trying to remove the spill means solving both equations AiS − SBi = −Ci for one common S. Solving the two equations separately would not answer that question.
Turn the calculation into an algorithm
Use rectangular matrix units as coordinates. Apply d0 to each unit and record the results as columns. Do the same for d1. Row reduction now performs three tasks:
- Solve d1(C) = 0 to obtain every compatible coupling.
- To compare two couplings, solve d0(S) = C′−C. A solution gives the actual coordinate-change matrix. An inconsistent system certifies that the markings cannot be matched.
- Extend a basis of the image of d0 to a basis of the kernel of d1. The added vectors supply independent parameters for the classes.
The last step is legitimate because the identity just checked puts the image inside the kernel. The kernel of d0 records the coordinate moves fixing a given coupling. Thus we obtain parameters, equivalence witnesses and automorphisms from the same finite calculation.
A complete scalar example
Take one-dimensional N and L, and write Δi = Ai−Bi. The formulas become
Δ₁C₂ − Δ₂C₁ = 0, (C₁,C₂) ↦ (C₁,C₂) + S(Δ₁,Δ₂).
If (Δ1,Δ2) ≠ (0,0), every compatible pair is a scalar multiple of this same vector. A coordinate move removes it. For example, over the rationals, take Δ = (1,2). The pair (3,6) is compatible and S = −3 removes it. The pair (3,7) has defect 1 and does not define commuting operations.
If both differences vanish, every pair is compatible, while every allowed coordinate move acts trivially on the pair. Two parameters remain. Even in dimension one, the answer depends on both tests: does the pair commute, and can one shared change of lifts remove it?
What the procedure proves, and what must be added
The quotient ker(d1)/im(d0) classifies these couplings with the endpoints fixed. With more commuting operations, introduce one unknown block per operation and one compatibility equation per pair. If the input also requires a relation such as T1T2 = 0, expand and impose that relation too: commutativity does not imply it.
The finite procedure uses arithmetic and row reduction over the given field. A complexity bound would also have to specify how that field and its elements are represented.
Further context: the archived Tensorgami study relates this calculation to extension theories.