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Hochschild chains and Hochschild homology with coefficients

Definition

Let k be a field, A a unital associative k-algebra, and M a k-central A-bimodule (Enveloping algebra and the bimodule–module dictionary). For n≥1 put

Cn(A,M):=M⊗kA⊗kn,

and put C0(A,M):=M, using the canonical tensor-unit identification M⊗kk≅M. For n≥1 and 0≤i≤n, define the faces on elementary tensors by

δi(n)(m⊗a1⊗⋯⊗an)={(ma1)⊗a2⊗⋯⊗an,i=0,m⊗a1⊗⋯⊗(aiai+1)⊗⋯⊗an,0<i<n,(anm)⊗a1⊗⋯⊗an−1,i=n.

These are well-defined k-linear maps by the multilinear universal property of the finite tensor product. Set C−1(A,M)=0 and b0=0. The Hochschild boundary is

bn:=∑i=0n(−1)iδi(n):Cn(A,M)⟶Cn−1(A,M)(n≥1).

In particular, b1(m⊗a)=ma−am. For n≥2 the faces satisfy

δi(n−1)δj(n)=δj−1(n−1)δi(n)(0≤i<j≤n),

so the alternating-sum boundaries satisfy bn−1bn=0 for n≥2; b0b1=0 because b0=0. Thus C∙(A,M) is a chain complex of k-modules. Its nth homology object is the Hochschild homology with coefficients in M,

HHn(A,M):=Hn(C∙(A,M)).

Facts & Assumptions

Given: A field k, a unital associative k-algebra A, and a k-central A-bimodule M.

[F1]

The left and right A-actions on M commute, and their scalar actions agree because M is k-central (Enveloping algebra and the bimodule–module dictionary).

[F2]

A finite tensor product over a commutative ring represents multilinear maps into a module, so a multilinear face formula induces a unique linear map on the tensor product (Finite iterated tensor products represent multilinear maps independently of parenthesization).

[F3]

The canonical map M⊗kk→M, m⊗λ↦mλ, is an isomorphism with inverse m↦m⊗1 (The regular module is a tensor unit: R⊗RN≅N and M⊗RR≅M).

[F4]

Disjoint adjacent multiplication faces satisfy the reindexed face identity (The bar boundary squares to zero and is augmented).

[F5]

Overlapping adjacent multiplication faces satisfy the face identity by associativity (The bar boundary squares to zero and is augmented).

[F6]

The homology object Hn(C) is defined when C∙ is a chain complex (Homology object of a chain complex).

Proof

technique · direct
1.1F2F3givenalgebra

Each endpoint formula is multilinear because the bimodule actions are k-bilinear, and each interior formula is multilinear because multiplication in A is k-bilinear; hence [F2] induces the displayed k-linear face maps and every face sends a zero input to zero. The tensor-unit isomorphism [F3] identifies the degree-zero term with M.

1.2givenalgebra

The degree-zero boundary is zero by definition, while in degree one the two faces are δ0(1)(m⊗a)=ma and δ1(1)(m⊗a)=am, so b1(m⊗a)=ma−am and b0b1=0.

1.3F4F5givenalgebra

If 1≤i<j≤n−1, both faces are internal adjacent multiplications among the A-slots. For disjoint slots their operations commute and reindex as in [F4]; for overlapping slots the two composites multiply the same triple, and associativity gives [F5]. Thus δi(n−1)δj(n)=δj−1(n−1)δi(n) for all such internal pairs.

1.4F1givenalgebra

For the adjacent first pair i=0,j=1, the two composites have first coefficients (ma1)a2 and m(a1a2), which agree by the right module law. For i=n−1,j=n, they have first coefficients an−1(anm) and (an−1an)m, which agree by the left module law. For the pair i=0,j=n, they have first coefficients (anm)a1 and an(ma1), which agree because the left and right actions commute by [F1]. The untouched slots agree in their original order in all three cases.

1.5givenalgebra

The remaining pairs with one endpoint face and one internal face act on disjoint data: for i=0 and 2≤j<n, the first face acts on m,a1 while the other multiplies aj,aj+1; for 1≤i<n−1 and j=n, the internal face multiplies ai,ai+1 while the last face acts by an on m. In either order the same multiplication/action is applied to each of these disjoint slots, with the same remaining tensor factors and order, proving the face identity. Together with 1.3 and 1.4 this covers every 0≤i<j≤n.

1.6F2F3givenalgebra

In the degenerate case A=k, the canonical tensor-unit identifications turn each face into the identity on M; hence bn=∑i=0n(−1)iid⁡M, which is the identity for even n and zero for odd n. Consecutive composites therefore vanish in this case too.

2.1step 1.3step 1.4step 1.5algebra

Expand bn−1bn as the sum of terms (−1)i+jδi(n−1)δj(n). For every i<j, the face identity from 1.3--1.5 pairs this term with the term indexed by (j−1,i); their composites agree and their signs are opposite because (−1)i+j=−(−1)(j−1)+i. Every term with first index r≥s is uniquely such a partner, obtained from i=s,j=r+1; thus every term occurs in exactly one pair. Hence bn−1bn=0 for n≥2; the case n=1 was checked in 1.2.

3.1step 1.1step 1.2step 2.1F6∎

The maps are k-linear by 1.1 and their consecutive composites vanish by 1.2 and 2.1, so they form a chain complex. The definition of homology applies by [F6], giving HHn(A,M)=Hn(C∙(A,M)).

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