Lecture 2: Addition (and free abelian groups) of a series of preparatory lectures for the Fall 2013 online course MATH:7450 (22M:305) Topics in Topology:

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Lecture 2: Addition (and free abelian groups) of a series of preparatory lectures for the Fall 2013 online course MATH:7450 (22M:305) Topics in Topology: Scientific and Engineering Applications of Algebraic Topology Target Audience: Anyone interested in topological data analysis including graduate students, faculty, industrial researchers in bioinformatics, biology, computer science, cosmology, engineering, imaging, mathematics, neurology, physics, statistics, etc. Isabel K. Darcy Mathematics Department/Applied Mathematical & Computational Sciences University of Iowa

A free abelian group generated by the elements x 1, x 2, …, x k consists of all elements of the form n 1 x 1 + n 2 x 2 + … + n k x k where n i are integers. Z = The set of integers = { …, -2, -1, 0, 1, 2, …} = the set of all whole numbers (positive, negative, 0) Addition: (n 1 x 1 + n 2 x 2 + … + n k x k ) + (m 1 x 1 + m 2 x 2 + … + m k x k ) = (n 1 + m 1 ) x 1 + (n 2 + m 2 )x 2 + … + (n k + m k )x k

Formal sum: 4 cone flower + 2 rose + 3 cone flower + 1 rose = 7 cone flower + 3 rose Will add video clips when video becomes available.

A free abelian group generated by the elements x 1, x 2, …, x k consists of all elements of the form n 1 x 1 + n 2 x 2 + … + n k x k where n i are integers. Z = The set of integers = { …, -2, -1, 0, 1, 2, …} = the set of all whole numbers (positive, negative, 0) Addition: (n 1 x 1 + n 2 x 2 + … + n k x k ) + (m 1 x 1 + m 2 x 2 + … + m k x k ) = (n 1 + m 1 ) x 1 + (n 2 + m 2 )x 2 + … + (n k + m k )x k

A free abelian group generated by the elements x 1, x 2, …, x k consists of all elements of the form n 1 x 1 + n 2 x 2 + … + n k x k where n i are integers. Example: Z[x 1, x 2 ] 4x 1 + 2x 2 x 1 - 2x 2 -3x 1 kx 1 + nx 2 Z = The set of integers = { …, -2, -1, 0, 1, 2, …}

A free abelian group generated by the elements x 1, x 2, …, x k consists of all elements of the form n 1 x 1 + n 2 x 2 + … + n k x k where n i are integers. Example: Z[x, x ] 4 ix + 2 I x – 2 i -3 x k + n iii Z = The set of integers = { …, -2, -1, 0, 1, 2, …}

A free abelian group generated by the elements x 1, x 2, …, x k consists of all elements of the form n 1 x 1 + n 2 x 2 + … + n k x k where n i are integers. Example: Z[x 1, x 2 ] 4x 1 + 2x 2 x 1 - 2x 2 -3x 1 kx 1 + nx 2 Z = The set of integers = { …, -2, -1, 0, 1, 2, …}

Addition: (n 1 x 1 + n 2 x 2 + … + n k x k ) + (m 1 x 1 + m 2 x 2 + … + m k x k ) = (n 1 + m 1 ) x 1 + (n 2 + m 2 )x 2 + … + (n k + m k )x k Example: Z[x 1, x 2 ] (4x 1 + 2x 2 ) + (3x 1 + x 2 ) = 7x 1 + 3x 2 (4x 1 + 2x 2 ) + (x 1 - 2x 2 ) = 5x 1

Addition: (n 1 x 1 + n 2 x 2 + … + n k x k ) + (m 1 x 1 + m 2 x 2 + … + m k x k ) = (n 1 + m 1 ) x 1 + (n 2 + m 2 )x 2 + … + (n k + m k )x k Example: Z[x, x 2 ] (4x + 2x 2 ) + (3 x + x 2 ) = 7 x + 3x 2 (4x 1 + 2x 2 ) + (x 1 - 2x 2 ) = 5x 1

Addition: (n 1 x 1 + n 2 x 2 + … + n k x k ) + (m 1 x 1 + m 2 x 2 + … + m k x k ) = (n 1 + m 1 ) x 1 + (n 2 + m 2 )x 2 + … + (n k + m k )x k Example: Z[x 1, x 2 ] (4x 1 + 2x 2 ) + (3x 1 + x 2 ) = 7x 1 + 3x 2 (4x 1 + 2x 2 ) + (x 1 - 2x 2 ) = 5x 1

v = vertexe = edgef = face Example: 4 vertices + 5 edges + 1 faces 4v + 5e + f.

v = vertexe = edge Example 2: 4 vertices + 5 edges 4v + 5e

v 1 + v 2 + v 3 + v 4 + e 1 + e 2 + e 3 + e 4 + e 5 e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4

v 1 + v 2 + v 3 + v 4 in Z[v 1, v 2,, v 3, v 4 ] e 1 + e 2 + e 3 + e 4 + e 5 in Z[e 1, e 2,, e 3, e 4, e 5 ] e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4

Technical note: In graph theory, the cycle also includes vertices. I.e, this cycle in graph theory is the path v 1, e 1, v 2, e 2, v 3, e 3, v 1,. Since we are interested in simplicial complexes (see later lecture), we only need the edges, so e 1 + e 2 + e 3 is a cycle. Note that e 1 + e 2 + e 3 is a cycle. e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4

Technical note: In graph theory, the cycle also includes vertices. I.e, the cycle in graph theory is the path v 1, e 1, v 2, e 2, v 3, e 3, v 1,. Since we are interested in simplicial complexes (see later lecture), we only need the edges, so e 1 + e 2 + e 3 is a cycle. Note that e 3 + e 4 + e 5 is a cycle. Note that e 1 + e 2 + e 3 is a cycle. e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4

Note that – e 3 + e 5 + e 4 is a cycle. e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4

Note that e 1 + e 2 + e 3 is a cycle. Note that – e 3 + e 4 + e 5 is a cycle. e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4 eiei Objects: oriented edges

Note that e 1 + e 2 + e 3 is a cycle. Note that – e 3 + e 5 + e 4 is a cycle. e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4 eiei Objects: oriented edges in Z[e 1, e 2, e 3, e 4, e 5 ]

Note that e 1 + e 2 + e 3 is a cycle. Note that – e 3 + e 5 + e 4 is a cycle. e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4 eiei Objects: oriented edges in Z[e 1, e 2, e 3, e 4, e 5 ] – e i

(e 1 + e 2 + e 3 ) + (–e 3 + e 5 + e 4 ) = e 1 + e 2 + e 5 + e 4 e2e2 e1e1 e5e5 e4e4 e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4

e 1 + e 2 + e 3 + e 1 + e 2 + e 5 + e 4 = 2e 1 + 2e 2 + e 3 + e 4 + e 5 e 1 + e 2 + e 5 + e 4 + e 1 + e 2 + e 3 = 2e 1 + 2e 2 + e 3 + e 4 + e 5 e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4 e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4

The boundary of e 1 = v 2 – v 1 e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4

The boundary of e 2 = v 3 – v 2 The boundary of e 3 = v 1 – v 3 e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4 The boundary of e 1 + e 2 + e 3 = v 2 – v 1 + v 3 – v 2 + v 1 – v 3 = 0

e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4 Add a face

e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4 Add an oriented face

e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4 e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4

e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4 Note that the boundary of this face is the cycle e 1 + e 2 + e 3

e2e2 e1e1 e5e5 e4e4 e3e3 v1v1 v2v2 v3v3 v4v4 Simplicial complex

v2v2 e2e2 e1e1 e3e3 v1v1 v3v3 2-simplex = oriented face = (v i, v j, v k ) Note that the boundary of this face is the cycle e 1 + e 2 + e 3 1-simplex = oriented edge = (v j, v k ) Note that the boundary of this edge is v k – v j e vjvj vkvk 0-simplex = vertex = v

3-simplex = (v 1, v 2, v 3, v 4 ) = tetrahedron 4-simplex = (v 1, v 2, v 3, v 4, v 5 ) v4v4 v3v3 v1v1 v2v2