Construct every tensor chain

On 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 xⁱy^j, 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]/(x^r,y^s).

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 E^av to a known multiple of E^a−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]/(x^r), 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 x^r to the nonzero vector rx^r−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 c_r = 0 makes the same rule respect the top truncation. On a surviving basis vector, the difference FE − EF is now multiplication by c_i+1 − c_i, including at the endpoints.

A nonzero constant discrepancy is impossible: summing it across the row would contradict c_r − 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 E^av 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 x^jy^k−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 x^jy^k−j−1 in Fv_k 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 m_k = r + s − 2 − 2k. Then Hv_k = −m_kv_k. Inductively commuting F past each forward step gives

FEᵃvₖ = a(mₖ−a+1)Eᵃ⁻¹vₖ.

For the induction, the next exchange adds the weight m_k − 2a; adding this to a(m_k−a+1) gives (a+1)(m_k−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 E^qv_k ≠ 0. Apply the formula to the next, zero vector. It gives 0 = (q+1)(m_k−q)E^qv_k. Thus q = m_k. 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 E^d−kv_k. Choose the least k occurring with nonzero coefficient and apply F^d−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 v_k. This contradicts the relation. Different total degrees separate, so all the constructed vectors are independent.

Writing h = min(r,s), their number is Σ_k=0^h−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.

This is the classical tensor-chain rule. Further geometric interpretations appear in the archived Tensorgami study.