F1.4ZA1 TOPOLOGY EXAMPLES 8

1. The barycentric subdivision of a square S is the triangulation T :

with a vertex for each vertex and edge of S and one for the square itself. There is an ede for each inclusion vertex edge, vertex square or edge square; and a triangle for each chain of inclusions vertex⊂ edge ⊂square. ⊂ ⊂ ⊂ Count the number of simplices of T in each dimension, and hence verify that χ(T ) = 1.

2. 1 Let T be the barycentric subdivision of a 3-dimensional cube, in other words the 3- dimensional version of the square in Question 1. Thus a k-simplex of T corresponds to chain of proper inclusions of k + 1 of the objects: vertices, edges, faces, cube. [For example, since the cube has 8 vertices, 12 edges and 6 faces, T has 8 + 12 + 6 + 1 = 27 vertices (or 0-simplices) in T .] Calculate the number of simplices of T of dimensions 1, 2 and 3, and hence calculate χ(T ).

3. Consider the torus as being formed from a large square by identification of opposite sides. Subdivide the torus into triangles by first subdividing the large square into four smaller squares:

and then replacing each smaller square by its barycentric subdivision (see Question 1). Verify that this gives a triangulation of the torus, and that this triangulation has euler characteristic 0.

4. (A 3-dimensional version of Question 3.) The 3-dimensional torus S1 S1 S1 is a 3-dimensional manifold that can be formed from a large cube by identifying× × each pair of opposite faces. Following Question 3, we can form a triangulation T of this manifold by first subdividing the large cube into 8 smaller cubes, which we then replace by their barycentric subdivisions (as described in Question 2). Calculate the number of simplices of T in each dimension, and hence calculate χ(T ). 1Homework question 5. Prove that (the geometric realisation K of) a simplicial complex K is compact if and only if K has a finite number of simplices.| | [Hint: show that the set of vertices of K is closed in K and has the discrete topology.] | |

6. If we define real vector spaces Cn and linear transformations dn : Cn+1 Cn as follows, → show in each case that (Cn, dn) is a chain complex:

2 2 4 (a) Cn = R for all n 0, and dn =   for all n 0. ≥ 1 2 ≥ − − 1 1 0 1 1 1 3  −    (b) Cn = R for all n 0, dn = 0 1 1 for n odd, and dn = 1 1 1 ≥ −  1 0 1   1 1 1  for n even. −

7. Calculate the homology groups of the chain-complexes in Question 6.

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