Homological Algebra – Does A∞ Algebra Compensate for any Loss of Information in the Study of Chain Complexes? 1.0


In an abelian category, homological algebra is the homotopy theory of chain complexes up to quasi-isomorphism of chain complexes.  When considering nonnegatively graded chain complexes, homological algebra may be viewed as a linearized version of the homotopy theory of homotopy types or infinite groupoids. When considering unbounded chain complexes, it may be viewed as a linearized and stabilized version. Conversely, we may view homotopical algebra as a nonabelian generalization of homological algebra.

Suppose we have a topological space X and a “multiplication map” m2 : X × X → X. This map may or may not be associative; imposing associativity is an extra condition. An A space imposes a weaker structure, which requires m2 to be associative up to homotopy, along with “higher order” versions of this. Indeed, there are very standard situations where one has natural multiplication maps which are not associative, but obey certain weaker conditions.

The standard example is when X is the loop space of another space M, i.e., if m0 ∈ M is a chosen base point,

X = {x : [0,1] → M |x continuous, x(0) = x(1) = m0}.

Composition of loops is then defined, with

x2x1(t) = x2(2t), when 0 ≤ t ≤ 1/2

= x1(2t−1), when  1/2 ≤ t ≤ 1

However, this composition is not associative, but x3(x2x1) and (x1x2)x3 are homotopic loops.

Screen Shot 2019-06-06 at 5.45.25 AM

On the left, we first traverse x3 from time 0 to time 1/2, then traverse x2 from time 1/2 to time 3/4, and then x1 from time 3/4 to time 1. On the right, we first traverse x3 from time 0 to time 1/4, x2 from time 1/4 to time 1/2, and then x1 from time 1/2 to time 1. By continuously deforming these times, we can homotop one of the loops to the other. This homotopy can be represented by a map

m3 : [0, 1] × X × X × X → X such that

{0} × X × X × X → X is given by (x3, x2, x1) 􏰀→ m2(x3, m2(x2, x1)) and

{1} × X × X × X → X is given by (x3, x2, x1) 􏰀→ m2(m2(x3, x2), x1)

What, if we have four elements x1, . . . , x4 of X? Then there are a number of different ways of putting brackets in their product, and these are related by the homotopies defined by m3. Indeed, we can relate

((x4x3)x2)x1 and x4(x3(x2x1))

in two different ways:

((x4x3)x2)x1 ∼ (x4x3)(x2x1) ∼ x4(x3(x2x1))


((x4x3)x2)x1 ∼ (x4(x3x2))x1 ∼ x4((x3x2)x1) ∼ x4(x3(x2x1)).

Here each ∼ represents a homotopy given by m3.

Schematically, this is represented by a polygon, S4, with each vertex labelled by one of the ways of associating x4x3x2x1, and the edges represent homotopies between them

Screen Shot 2019-06-06 at 6.02.33 AM

The homotopies myield a map ∂S4 × X4 → X which is defined using appropriate combinations of m2 and m3 on each edge of the boundary of S4. For example, restricting to the edge with vertices ((x4x3)x2)x1 and (x4(x3x2))x1, this map is given by (s, x4, . . . , x1) 􏰀→ m2(m3(s, x4, x3, x2), x1).

Thus the conditionality on the structure becomes: this map extend across S4, giving a map

m4 : S4 × X4 → X.

As homological algebra seeks to study complexes by taking quotient modules to obtain the homology, the question arises as to whether any information is lost in this process. This is equivalent to asking whether it is possible to reconstruct the original complex (up to quasi-isomorphism) given its homology or whether some additional structure is needed in order to be able to do this. The additional structure that is needed is an A-structure constructed on the homology of the complex…