Question

Suppose all even-indexed linear transformations in the chain complexC•are 0 transformations and all odd-indexed transformations are bijections.Compute the homology of this complex

Answer 1

The definition of a C-linear function   implies that l(z) = az + bz¯ , so l(iz) = i(az − bz¯). Therefore, l(iz) = il(z) if and only if  iaz − ibz¯ = iaz + ibz

Therefore if l(iz) = il(z) for all z ∈ C then b = 0 and hence l is C-linear. We set a = a1 + ia2, b = b1 + ib2, and also z = x + iy, w = u + iv. We may represent an R-linear function w = az + bz¯ as two real equations  

  u = (a1 + b1)x − (a2 − b2)y, v = (a2 + b2)x + (a1 − b1)y.

Therefore geometrically an R-linear function is an affine transform of a plane y = Ax with the matrix

the composition of dilation by |a| and rotation by the angle α. Such transformations preserve angles and map squares onto squares. We note that preservation of angles characterizes C-linear transformations

The homology of this chain complex is called the singular homology of X, and is a commonly used invariant of a topological space. Chain complexes are studied in homological algebra, but are used in several areas of mathematics, including abstract algebra, Galois theory, differential geometry and algebraic geometry

A chain homotopy offers a way to relate two chain maps that induce the same map on homology groups, even though the maps may be different. Given two chain complexes A and B, and two chain maps f,g : A → B, a chain homotopy is a sequence of homomorphisms hn : An → Bn+1 such that hdA + dBh = f − g. The maps may be written out in a diagram as follows, but this diagram is not commutative.