RefacteOn a rectangular grid of monomials, one operation adds the rightward and upward moves. Finding its chains by repeated expansion soon becomes unwieldy. We will design a backward operation that measures how many forward steps have been taken, then use it to construct the chains and separate them.
A grid, a move, and the answer to construct
Work over a field of characteristic zero. Fix positive integers r and s. The vector space has basis xiyj, with 0 ≤ i < r and 0 ≤ j < s. Set a monomial equal to zero whenever either exponent reaches its bound. Thus we can write the space as M = k[x,y]/(xr,ys).
The forward move is E f = (x + y)f. A chain starts at a vector v and follows v, Ev, E2v, … until the next vector is zero. We seek starting vectors whose chains, taken together, form a basis.
E(xⁱyʲ) = xⁱ⁺¹yʲ + xⁱyʲ⁺¹.
Invent a backward move with a simple discrepancy
Suppose a backward move F could send Eav to a known multiple of Ea−1v. It would tell us where the chain stops and let us distinguish its vectors from other chains. We do not yet know v or F, but this gives the auxiliary operation a concrete job.
Begin with one row, k[x]/(xr), where E raises the exponent. The ordinary derivative looks like a candidate for lowering it. At the top edge, however, it would send the zero relation xr to the nonzero vector rxr−1. Equivalently, defining the derivative only on the surviving basis gives a constant discrepancy in the interior and an exceptional one at the top.
Allow the backward coefficients to adjust to both ends:
F(xⁱ) = cᵢxⁱ⁻¹, c₀ = cᵣ = 0.
Here c0 = 0 stops the move below the bottom. The formal endpoint value cr = 0 makes the same rule respect the top truncation. On a surviving basis vector, the difference FE − EF is now multiplication by ci+1 − ci, including at the endpoints.
A nonzero constant discrepancy is impossible: summing it across the row would contradict cr − c0 = 0. The next simplest choice is an affine function of i. Its change under a forward step is constant, so commuting F past repeated forward steps will add an arithmetic progression. Choose its drop to be two; another nonzero drop merely rescales F:
cᵢ₊₁ − cᵢ = A − 2i, 0 = cᵣ − c₀ = rA − r(r−1), so A = r−1 and cᵢ = i(r−i).
The affine-discrepancy requirement is a design choice, not a consequence of the original problem. The endpoint equations then determine its intercept and every coefficient. For the grid, add the two backward moves; this keeps the work in each coordinate independent:
F(xⁱyʲ) = i(r−i)xⁱ⁻¹yʲ + j(s−j)xⁱyʲ⁻¹. H(xⁱyʲ) = (2(i+j) − (r+s−2))xⁱyʲ. FE − EF = −H, HE − EH = 2E.
Terms with zero coefficients are omitted. The two identities follow by expanding on a monomial; moves in different coordinates commute. They are the only operator identities the argument needs.
Turn the search for starting vectors into a recurrence
To make the backward move count forward steps from a chosen start, impose Fv = 0. Then commuting F through Eav will leave no unknown backward term at the end. Also keep v in a single total degree, where the diagonal operation H acts by one scalar.
In degree k below both truncation bounds, every monomial xjyk−j is available. There are k+1 unknown coefficients and k equations in degree k−1. This suggests looking for one new start in each degree 0 ≤ k < min(r,s). Write the general vector in that degree and fix its scale:
vₖ = Σⱼ₌₀ᵏ aⱼ xʲyᵏ⁻ʲ, a₀ = 1.
The coefficient of xjyk−j−1 in Fvk receives exactly two contributions. Make them cancel:
aⱼ₊₁(j+1)(r−j−1) + aⱼ(k−j)(s−k+j) = 0, aⱼ₊₁ = −aⱼ (k−j)(s−k+j) / ((j+1)(r−j−1)).
Every denominator is nonzero because j < k < r, and the field has characteristic zero. The recurrence determines all coefficients from a0; taking a0 = 1 gives a nonzero start. This constructs the proposed starts. The length and basis arguments below will decide whether they suffice.
The backward move determines the length
Put mk = r + s − 2 − 2k. Then Hvk = −mkvk. Inductively commuting F past each forward step gives
FEᵃvₖ = a(mₖ−a+1)Eᵃ⁻¹vₖ.
For the induction, the next exchange adds the weight mk − 2a; adding this to a(mk−a+1) gives (a+1)(mk−a). This is why the arithmetic progression was useful.
Since E raises degree in a bounded grid, it is nilpotent. Let q be the last exponent with Eqvk ≠ 0. Apply the formula to the next, zero vector. It gives 0 = (q+1)(mk−q)Eqvk. Thus q = mk. This chain has exactly r+s−1−2k vectors.
Prove that nothing overlaps and nothing is missing
In a relation of total degree d, the possible vectors are Ed−kvk. Choose the least k occurring with nonzero coefficient and apply Fd−k. Every larger-index term dies: it has fewer forward steps to undo before reaching a start killed by F. The chosen term becomes a nonzero scalar times vk. This contradicts the relation. Different total degrees separate, so all the constructed vectors are independent.
Writing h = min(r,s), their number is Σk=0h−1(r+s−1−2k) = h(r+s−h) = rs. That is the dimension of the grid. The chains therefore form a basis.
Run it on a two-by-three grid
For (r,s) = (2,3), the starts are 1 and y−2x. The complete basis is displayed by two chains:
1 → x+y → 2xy+y² → 3xy² → 0, y−2x → y²−xy → 0.
The construction uses characteristic zero in an essential way. In characteristic two, on a two-by-two grid, (x+y)² = 0. The characteristic-zero answer of lengths three and one cannot survive there.
The resulting change of basis
The original problem asked for a change of basis for the combined move. The backward-move ansatz replaced that search by endpoint equations, a coefficient recurrence, and a short cancellation argument. We obtain the vectors as well as the lengths. In this basis, the combined move has nilpotent blocks of lengths r+s−1, r+s−3, …, |r−s|+1.
In matrix language, let Jℓ denote a nilpotent Jordan block of size ℓ. The separate moves on the two factors are single Jordan blocks, so the combined move is their Kronecker sum. The constructed basis proves the similarity
Jᵣ ⊗ Iₛ + Iᵣ ⊗ Jₛ ∼ ⊕ₖ₌₀ʰ⁻¹ Jᵣ₊ₛ₋₁₋₂ₖ, h = min(r,s).
The same basis also handles blocks with scalar eigenvalues: adding αI to the first factor and βI to the second adds α+β to every resulting block. This is the tensor-product rule for the action of an operator that acts by the sum of its two factor actions.
The lengths answer further questions without expanding another power. A chain of length L contributes max(L−a, 0) to the rank of Ea, and min(a, L) to its nullity. Hence, for every integer a ≥ 0,
rank(Eᵃ) = Σₖ₌₀ʰ⁻¹ max(r+s−1−2k−a, 0).
For the two-by-three grid the ranks are 6, 4, 2, 1, 0 for a = 0, 1, 2, 3, 4. The nilpotency index is r+s−1, and dim ker E = min(r,s). The backward move has therefore made both the basis and the behavior of all iterates explicit.
Sources and the classical rule
The operators are the standard finite-dimensional sl2 construction, and the block decomposition is its Clebsch–Gordan rule. See Pavel Etingof and collaborators, Introduction to Representation Theory (American Mathematical Society, 2011), §§2.14–2.15, especially Problem 2.15.1(m). The construction and basis proof here are explicit over any field of characteristic zero; no complete-reducibility theorem is assumed. Further geometric interpretations appear in the archived Tensorgami study.