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One-variable diagonal Hochschild calculation

Example

Assume AC. Let k be a field and R=k[x] with deg⁡intx=2, regarded as its regular bimodule. Then, as graded k-vector spaces,

HH0(R,R)≅R,HH1(R,R)≅R{2},HHj(R,R)=0(j≥2).

Facts & Assumptions

Given: AC, a field k, the one-variable polynomial algebra R=k[x], the regular bimodule, and deg⁡intx=2.

[F1]

The diagonal complex has degree-p terms free over Re on increasing wedge symbols, differential given by alternating deletion with coefficients ui=xiL−xiR, and no terms above degree n (The polynomial diagonal Koszul bimodule complex).

[F2]

Under AC, the polynomial Hochschild theorem identifies HHj(R,M) with the homology of the coefficient diagonal Koszul complex, naturally in M, and preserves the internal grading (Polynomial Hochschild homology from the diagonal Koszul complex).

[F3]

When deg⁡intxi=2, each θi has internal degree 2 and the degree-p diagonal term has shift {2p} (The polynomial diagonal Koszul bimodule complex).

[F4]

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 1.1.

Verification

technique · direct
1.1F1F2F4given

Specialize the diagonal complex to one variable. [F1, F3, given] It has the two terms Reθ1≅Re{2} in homological degree one and Re in degree zero, with differential d(θ1)=u1=x⊗1−1⊗x and augmentation μ:Re→R. Thus the displayed augmented complex is 0⟶Re{2}→ x⊗1−1⊗x Re→ μ R⟶0. The polynomial Hochschild theorem applies to this diagonal complex under AC [F4].

2.1F2F3step 1.1algebra

Tensor with the regular bimodule and compute the only differential. [F2, F3, step 1.1, algebra] The theorem gives the coefficient complex R{2}→dR. Writing the degree-one generator as θ1, for m∈R its differential is d(mθ1)=xm−mx=0, since the left and right actions on the regular bimodule agree. This is a degree-zero map because x and θ1 both have internal degree 2.

3.1F1F2step 2.1algebra

Read the homology of the two-term zero-differential complex. [F1, F2, step 2.1, algebra] There is no term above degree one. Since d=0, every degree-one element is a cycle and there are no degree-one boundaries, so H1=R{2}. In degree zero, there are no degree-zero boundaries and all of R is a cycle, so H0=R. For j≥2 the chain term is zero, giving HHj=0. Applying [F2] gives the displayed Hochschild groups.

4.1

Check the endpoints, shift, and exact use of AC. [F1, F2, F3, step 2.1, step 3.1, given] The empty wedge is the degree-zero basis and carries shift {0}; the top wedge θ1 has internal degree 2 and gives R{2}. The outgoing degree-zero differential is zero by convention, the incoming degree-one map is zero by the explicit calculation, and there are no terms in degrees j≥2. AC is used only to invoke the polynomial Hochschild theorem in step 1.1; the one-variable differential and homology calculation use no choice. The example is a computation, not an equivalence. [F1, F2, F3, step 1.1, step 2.1, step 3.1, algebra] □

Source comparison

Weibel, An Introduction to Homological Algebra, Exercise 9.1.3, printed p.304/PDF p.4, asks for the polynomial calculation with the Koszul resolution but does not supply its proof. Khovanov, “Hochschild homology,” PDF p.1, lines 33–60, states the polynomial diagonal Koszul complex and its coefficient differential. The displayed terms, zero map, and homology groups are calculated directly above.

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