Algebra Seminar - Shashank Singh; University of Iowa Department of Mathematics

Title: Reconstruction of a Presented Tensor Action from an Extension with a Crossing

Abstract: This talk shows how a presented tensor action on an abelian category 𝒜 is reconstructed from an extension with a crossing. The extension is a short exact sequence of endofunctors 0 → T → E → Id → 0 on 𝒜, with maps ι: T → E and π: E → Id and with T an autoequivalence. The crossing is a natural isomorphism κ: ET → TE satisfying (Tπ)κ = πT, which records how E commutes with T. On the copy of T² inside ET, κ acts through a central unit c of 𝒜, κ(ιT) = (Tι)(cT²), which we assume commutes with T and E.

For m ≥ 2 write [m]c = 1 + c + ⋯ + cm−1. The relation [n]c = 0 implies cn = 1 and is exactly the condition for the t-derivative ∂t(xk) = [k]t xk−1, with t acting as c, to descend to M[x]/xnM[x]; the induced operators D and X (from ∂t and multiplication by x) then satisfy DX − tXD = I. This relation does not make the lower [m]c invertible for 1 ≤ m < n. If d ≥ 2 is the least index with [d]c not invertible, then d ∣ n, and on the full subcategory of objects killed by [d]c the degree layers of M[x]/xnM[x] split under ∂t into n/d chains of length d, while endomorphisms U, V with VU = cUV satisfy (U + V)n = (Ud + Vd)n/d.

Invertibility of [1]c, …, [n−1]c is a separate requirement: it is what allows each chain to be normalized. Call the case d = n admissible. There Rn = ℤ[t, [2]t−1, …, [n−1]t−1]/([n]t) is the universal ring with an element t satisfying these conditions, and κ gives a ring map Rn → 𝒵(𝒜), t ↦ c, compatible with T and E. Let 𝒞 be the Rn-linear monoidal category generated by T, E, ι, π, κ subject to πι = 0, (Tπ)κ = πT, κ(ιT) = (Tι)(t·idT²) and further polynomial relations. Actions of 𝒞 on 𝒜 correspond, up to isomorphisms preserving ι, π and κ, to extensions of 𝒜 with a crossing satisfying those relations. Two algebraic consequences follow from admissibility. The Gaussian binomials C(n,k)c vanish for 0 < k < n, so (U + V)n = Un + Vn whenever VU = cUV. Over every Rn-algebra R, ∂t maps each degree layer of R[x]/(xn) isomorphically onto the previous one, and in the basis xk/[k]t! it is the standard shift ek ↦ ek−1.

Wednesday, October 14, 2026 3:30pm
MacLean Hall
214
2 West Washington Street, Iowa City, IA 52240
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