Applied Topology for Dimensionality Reduction in Data Analysis: Cohomology II

1. MATH:7450 (22M:305) Topics in Topology: Scientific and Engineering Applications of Algebraic Topology Oct 23, 2013: Cohomology II Fall 2013 course offered through the University of Iowa Division of Continuing Education Isabel K. Darcy, Department of Mathematics Applied Mathematical and Computational Sciences, University of Iowa http://www.math.uiowa.edu/~idarcy/AppliedTopology.html

2. Dimensionality Reduction: • Given dataset D RN • Want: embedding f: D  Rn where n << N • which “preserves” the structure of the data. • Many reduction methods: • f1: D  R,f2: D  R, …fn: D  R • (f1, f2, … fn): D  Rn U

3. Goal f1: D  S1,f2: D  S1, …fn: D  S1 Note S1 is a ring: q1, q2 in S1 implies q1+ q2 and q1q2 are in S1

5. circle courtesy of knotplot.com

6. circle courtesy of knotplot.com

7. Cohomology and circular coordinates: Let X = finite simplicial complex. X0 = set of 0-simplices = vertices. X1 = set of 1-simplices = edges. X2 = set of 2-simplices = faces. . . . v2 e1 e2 v1 v3 e3

8. Cohomology and circular coordinates: Let X = finite simplicial complex, R a ring. C0(X, R) = set of 0-cochains = { f : X0 R } C1(X, R) = set of 1-cochains = { f : X1 R } C2(X, R) = set of 2-cochains = { f : X2 R } . . .

9. coboundarymaps d0:C0 = { f : X0 R } → C1 = { f : X1 R } f : X0 R →d0f : X1 R (d0f )(ab) = f (b)− f (a) d1:C1 = {α : X1 R } → C2= {α : X2 R } α : X1 R →d1α: X2 R (d1α)(abc) = α(bc)− α(ac)+α(ab)

10. Let α ∈ Ci , di : Ci= { f : Xi R } → Ci+1 = { f : Xi+1 R } α is a cocycle if diα = 0. I.e., α is in Ker(d1) α is a coboundary if there exists f in Ci-1 s.t. di-1f = α I.e., α is in Im(di-1) Note di di-1f = 0 for any f ∈ Ci-1. di di-1f = 0 implies Im (di-1) ⊆ Ker (di). Hi(X;A ) = Ker(di)/ Im(di-1).

11. C0(X, R) = set of 0-cochains = { f : X0 R } = { f | f(v1) = r1, f(v2) = r2, f(v3) = r3; (r1, r2, r3) in R3 } C1(X, R) = set of 1-cochains = {α : X1 R } = {α| α(e1) = r1, α(e2) = r2, α(e3) = r3; (r1, r2, r3) in R3 } Hi(X;A ) = Ker (di)/ Im (di-1). 0  C0 C1 0 v2 e1 e2 v1 v3 e3

12. C0(X, R) = set of 0-cochains = { f : X0 R } = { f | f(v1) = r1, f(v2) = r2, f(v3) = r3; (r1, r2, r3) in R3 } 0  C0  C1 H0(X;A ) = Ker (d0 )/ Im (d-1) = v2 e1 e2 v1 v3 e3

13. C0(X, R) = set of 0-cochains = { f : X0 R } = { f | f(v1) = r1, f(v2) = r2, f(v1) = r3; (r1, r2, r3) in R3 } 0  C0  C1 H0(X;A ) = Ker (d0 )/ Im (d-1) = Ker (d0 ) v2 e1 e2 v1 v3 e3

14. 0  C0  C1 Ker (d0 ) = ? Suppose d0f = 0 d0f (v1v2) = f(v2) – f(v1) = 0 implies f(v1) = f(v2). d0f (v2v3) = f(v3) – f(v2) = 0 implies f(v2) = f(v3). d0f (v3v1) = f(v1) – f(v3) = 0 implies f(v3) = f(v1). v2 e1 e2 Thus f is the constant map v1 v3 e3

15. 0  C0  C1 Ker (d0 ) = { f | f(vi) = r for all i} Suppose d0f = 0 d0f (v1v2) = f(v2) – f(v1) = 0 implies f(v1) = f(v2). d0f (v2v3) = f(v3) – f(v2) = 0 implies f(v2) = f(v3). d0f (v3v1) = f(v1) – f(v3) = 0 implies f(v3) = f(v1). v2 e1 e2 Thus f is the constant map v1 v3 e3

16. C0(X, R) = set of 0-cochains = { f : X0 R } = { f | f(v1) = r1, f(v2) = r2, f(v1) = r3; (r1, r2, r3) in R3 } 0  C0  C1 H0(X;A ) = Ker (d0 )/ Im (d-1) = Ker (d0) = { f | f(vi) = r for all i } v2 e1 e2 v1 v3 e3

17. C1(X, R) = set of 1-cochains = {α : X1 R } = {α | α (e1) = r1, α(e2) = r2, α(e3) = r3; (r1, r2, r3) in R3 } C0  C1  0 H1(X;R ) = Ker (d1)/ Im (d0 ) = v2 e1 e2 v1 v3 e3

18. C1(X, R) = set of 1-cochains = {α : X1 R } = {α | α (e1) = r1, α(e2) = r2, α(e1) = r3; (r1, r2, r3) in R3 } C0  C1  0 H1(X;R ) = Ker (d1)/ Im (d0 ) = C1/ Im (d0 ) v2 e1 e2 v1 v3 e3

19. C1(X, R) = set of 1-cochains = {α : X1 R } = {α | α (e1) = r1, α(e2) = r2, α(e1) = r3; (r1, r2, r3) in R3 } C0  C1  0 H1(X;R ) = Ker (d1)/ Im (d0 ) = C1/ Im (d0 ) Suppose f is in C0, then d0f in C1 is defined by: d0f (v1v2) = f(v2) – f(v1) d0f (v2v3) = f(v3) – f(v2) d0f (v3v1) = f(v1) – f(v3) v2 e1 e2 v1 v3 e3

20. C1(X, R) = set of 1-cochains = {α : X1 R } = {α | α (e1) = r1, α(e2) = r2, α(e1) = r3; (r1, r2, r3) in R3 } C0  C1  0 H1(X;R ) = Ker (d1)/ Im (d0 ) = C1/ Im (d0 ) Suppose f is in C0, then d0f in C1 is defined by: d0f (v1v2) = f(v2) – f(v1) d0f (v2v3) = f(v3) – f(v2) d0f (v3v1) = f(v1) – f(v3) Im(d0 ) = { f : X1R | f(e1) = r1, f(e2) = r2, f(e3) = -r1 - r2} v2 e1 e2 v1 v3 e3

21. C1(X, R) = set of 1-cochains = {α : X1 R } = {α | α (e1) = r1, α(e2) = r2, α(e1) = r3; (r1, r2, r3) in R3 } Im(d0 ) = { f : X1R | f(e1) = r1, f(e2) = r2, f(e3) = -r1 - r2} C0 C1  0 H1(X;R ) = Ker (d1)/ Im (d0 ) = C1/ Im (d0) = { f | f(e1) = r1, f(e2) = r2, f(e3) = r3}/ Im (d0 ) = { f | f(e1) = r1, f(e2) = r2, f(e3) = r3} { f : X1R | f(e1) = r1, f(e2) = r2, f(e3) = -r1 - r2}

22. Persistent Cohomology e1 < e2 < … < em X(e1) < X(e2) < … < X(em) Ci(X(ej), R)= { f : Xi(ej)  R }  …  Ci(X(e1), R) Ci(X(e2, R))  … Ci(X(em, R)) Hi(X(e1), R) Hi(X(e2, R))  … Hi(X(em, R)) R R R

23. Proposition 1: Let α ∈ C1(X;Z) be a cocycle. Then there exists a continuous function θ : X → S1 which maps each vertex to 0, and each edge ab around the entire circle with winding number α(ab). v2 → e1 e2 v1 v3 e3

24. Proposition 1: Let α ∈ C1(X;Z) be a cocycle. Then there exists a continuous function θ : X → S1 which maps each vertex to 0, and each edge ab around the entire circle with winding number α(ab). →

25. Proposition 1: Let α ∈ C1(X;Z) be a cocycle. Then there exists a continuous function θ : X → S1 which maps each vertex to 0, and each edge ab around the entire circle with winding number α(ab). v2 → e1 e2 v1 v3 e3

26. Proposition 2: 2 Let α ∈ C1(X;R) be a cocycle. Suppose there existsα ∈ C1(X;Z) and f ∈ C0(X;R) such that α = α + d0f . Then there exists a continuous function θ : X → S1 which maps each edge ablinearly to an interval of length α(ab), measured with sign. v2 → e1 e2 v1 v3 e3

27. θ: X → R/Z = S1