# Construct every tensor chain Henry Shin · Refacte · Mechanism calculus How can we construct every chain for the combined operator? Begin with: Finite-dimensional linear algebra and polynomial arithmetic; characteristic zero is explicit. 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 x^{i}y^{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, E^{2}v, … until the next vector is zero. We seek starting vectors whose chains, taken together, form a basis. ```text E(xⁱyʲ) = xⁱ⁺¹yʲ + xⁱyʲ⁺¹. ``` ## Invent a backward move with a simple discrepancy Suppose a backward move F could send E^{a}v to a known multiple of E^{a−1}v. 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: ```text F(xⁱ) = cᵢxⁱ⁻¹, c₀ = cᵣ = 0. ``` Here c_{0} = 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} − c_{0} = 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: ```text 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: ```text 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^{a}v 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^{j}y^{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: ```text vₖ = Σⱼ₌₀ᵏ aⱼ xʲyᵏ⁻ʲ, a₀ = 1. ``` The coefficient of x^{j}y^{k−j−1} in Fv_{k} receives exactly two contributions. Make them cancel: ```text 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 a_{0}; taking a_{0} = 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_{k}v_{k}. Inductively commuting F past each forward step gives ```text 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^{q}v_{k} ≠ 0. Apply the formula to the next, zero vector. It gives 0 = (q+1)(m_{k}−q)E^{q}v_{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−k}v_{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: ```text 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](https://refacte.com/notes/fourier-symmetry-and-reconstruction.html#jordan-tensor-calculus).