v 2 The dimension of f The dimension

v 2 The dimension of f = The dimension of ei = The dimension of vi = The boundary of f = The boundary of e 1 = v 2 – v 1 The boundary of e 2 = v 3 – v 2 The boundary of e 4 = v 1 – v 4 The boundary of e 5 = e 1 v 1 e 4 f e 3 e 2 v 3 e 5 v 4 The boundary of e 1 + e 2 = The boundary of e 1 + e 4 = The boundary of e 1 + e 5 = The boundary of e 1 + e 2 + e 3 = The boundary of e 3 + e 4 + e 5 = The boundary of -e 3 + e 4 + e 5 = The boundary of e 1 + e 2 + e 4 + e 5 = Let X = e 1 + e 2 + e 3, let Y = -e 3 + e 4 + e 5, and let Z = e 1 + e 2 + e 4 + e 5. Show that Z = X + Y Answer: X + Y = (e 1 + e 2 + e 3) + (-e 3 + e 4 + e 5) = e 1 + e 2 + e 3 – e 3 + e 4 + e 5 = e 1 + e 2 + e 4 + e 5 = Z

v 2 Note z is a cycle if the boundary of z = 0. List three 1 -dimensional cycles: e 1 v 1 List four 0 -dimensional cycles: e 4 f e 3 e 2 v 3 e 5 v 4 Are there any 2 -dimensional cycles Consider the differences between Z and Z 2: When working over Z = the set of integers, one does not have multiplicative inverses. When working over Z 2 = {0, 1}, one has multiplicative inverses. Also, computationally, Z 2 is much easier to work with. v 2 For the following, we will work over Z 2 = {0, 1} e 1 The boundary of e 1 + e 2 + e 3 = The boundary of e 3 + e 4 + e 5 = -e 3 + e 4 + e 5 = |C 0| = ____, |C 1| = ____, |C 2| = ____, |C 3| = ____ 1. ) Find Ci, Bi, Zi 2. ) Find the matrix corresponding to each boundary map (from Ci to Ci-1) 3. ) Find Hi v 1 e 4 e 2 e 5 v 4 v 3 e 3