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One-variable twisted bimodule Hochschild calculation

Example

Assume AC. Let R=Q[x] and let M=R as a left R-module with right action m⋅x=−xm, extended to all polynomials by m⋅f(x):=mf(−x). Then

HH0(R,M)≅Q,HH1(R,M)=0,HHj(R,M)=0(j≥2).

Facts & Assumptions

Given: AC, R=Q[x], the usual left R-module M=R, and the right action m⋅f(x)=mf(−x).

[F1]

Under AC, for a k-central R-bimodule M, Hochschild homology is isomorphic to the homology of the coefficient Koszul complex, whose one-variable differential is mθ1↦xm−mx (Polynomial Hochschild homology from the diagonal Koszul complex).

[F2]

A k-central bimodule has commuting left and right actions and equal induced scalar actions (Enveloping algebra and the bimodule–module dictionary).

[F3]

The assumed Axiom of Choice is the choice-function principle (The Axiom of Choice); it licenses the AC-qualified polynomial Hochschild theorem used at step 2.1.

Verification

technique · direct
1.1F2givenalgebra

Verify the twisted right action and the bimodule hypotheses. [F2, given] Define σ(f(x))=f(−x). Since σ(fg)=σ(f)σ(g), σ(1)=1, and σ2=id⁡, it is a unital ring automorphism of R. Put m⋅f=mσ(f). Then m⋅1=m and (m⋅f)⋅g=mσ(f)σ(g)=mσ(fg)=m⋅(fg), so this is a unital right action. In particular m⋅x=−mx=−xm. For r,f∈R, r(m⋅f)=rmσ(f)=(rm)⋅f, using commutativity of R; hence the left and right actions commute. Since σ(q)=q for q∈Q, the two scalar actions agree. Thus M is a Q-central R-bimodule, as required by [F1].

2.1F1F3step 1.1algebra

Write the one-variable coefficient complex and compute its map. [F1, F2, step 1.1, algebra] Under the AC premise [F3], [F1] computes HH∙(R,M) by the two-term complex Mθ1→dM. Its differential is d(mθ1)=xm−m⋅x=xm−(−xm)=2xm. There are no terms above degree one.

3.1F1step 2.1algebra

Compute the kernel and cokernel of multiplication by 2x. [F1, step 2.1, algebra] If 0≠f(x)=adxd+⋯ with ad≠0, then 2xf(x) has leading coefficient 2ad≠0 in Q, so d is injective. Because 2 is a unit, its image is 2xQ[x]=xQ[x]. Evaluation at zero ε:Q[x]→Q is surjective and has kernel xQ[x]: a polynomial with zero constant coefficient is divisible by x. Thus coker⁡d≅Q and ker⁡d=0, giving HH0(R,M)≅Q and HH1(R,M)=0. The terms above degree one vanish, so HHj(R,M)=0 for j≥2.

4.1

Check grading, endpoints, and the exact AC use. [F1, F2, step 2.1, step 3.1, given] If deg⁡intx=2, then f(x)↦f(−x) preserves degree, and the Koszul generator θ1 has internal degree 2; hence the degree-one term is the corresponding shift of M and d has internal degree zero. Degree zero is the cokernel computed in step 3.1; degree one is the kernel, with no incoming term from degree two; all higher terms are zero. AC is used only to apply the polynomial Hochschild theorem in step 2.1. The action, leading-term injectivity, and evaluation quotient are explicit and use no choice. The example is a computation, not an equivalence. [F1, F2, step 1.1, step 2.1, step 3.1, algebra] □

Source comparison

Weibel, An Introduction to Homological Algebra, §9.1.3 and Exercise 9.1.3, printed pp.302–304/PDF pp.2–4, gives the enveloping/bar framework and poses the polynomial Koszul computation as an exercise, but does not treat this twist. Khovanov, “Hochschild homology,” PDF p.1, lines 41–60, describes the polynomial Koszul complex and the coefficient maps xim−mxi. The automorphism twist and its kernel/cokernel calculation are proved explicitly above.

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