Examples of Manifolds

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1 Eamples of Manifolds Eample 1 Open Subset of IR n Anyopensubset, O, ofir n isamanifoldofdimension n. OnepossibleatlasisA = { O,ϕ id, whereϕ id istheidentitymap. Thatis, ϕ id =. Of course one possible choice of O is IR n itself. Eample 2 The Circle The circle S 1 = {,y IR 2 2 +y 2 = 1 is a manifold of dimension one. One possible atlas is A = {U 1,ϕ 1, U 1,ϕ 2 where U 1 = S 1 \{ 1,0 ϕ 1,y = arctan y with π < ϕ 1,y < π U 2 = S 1 \{1,0 ϕ 2,y = arctan y with 0 < ϕ 2,y < 2π ϕ 1 Eample 3 S n Then sphere S n = { = 1,, IR = 1 isamanifoldofdimensionn. One possibleatlasisa 1 = { U i,ϕ i, V i,ψ i 1 i where, for each 1 i, U i = { 1,, S n i > 0 ϕ i 1,, = 1,, 1,+1,, V i = { 1,, S n i < 0 ψ i 1,, = 1,, 1,+1,, So both ϕ i and ψ i project onto IR n, viewed as the hyperplane = 0. Another possible atlas is A 2 = { S n \{0,,0,1,ϕ, S n \{0,,0, 1,ψ where ϕ 1,, = 2 1 1,, ψ 1,, = ,, 2 n 1 2 n 1+ are the stereographic projections from the north and south poles, respectively. 0,,0,1 U 1 0,,0 ϕ Both ϕ and ψ have range IR n. So we can think of S n as IR n plus an additional single point at infinity. c Joel Feldman All rights reserved. March 9, 2008 Eamples of Manifolds 1

2 Eample 4 Surfaces Any smoothn dimensional surface inir n+m is ann dimensional manifold. When we say that M is an n dimensional surface in IR n+m, we mean that M is a subset of IR n+m with the property that for each z M, there are a neighbourhood U z of z in IR n+m n integers 1 j 1 < j 2 < < j n n+m and m C functions f k j1,, jn, k {1,,n+m\{j 1,,j n such that the point = 1,, n+m U z is in M if and onlf k = f k j1,, jn for all k {1,,n+m \ {j 1,,j n. That is, we may epress the part of M that is near z as 1 = f 1 j1, j2,, jn 2 = f 2 j1, j2,, jn. m = f m j1, j2,, jn where { i 1,,i m = {1,,n+m\{j 1,,j n for some C functions f 1,, f m. We may use j1, j2,, jn as coordinates for M in M U z. Of course, an atlas is A = { U z,ϕ z z M, with each ϕz = j1,, jn. M U z z j1,, jn Equivalently, M is an n dimensional surface in IR n+m, if, for each z M, there are a neighbourhood U z of z in IR n+m and m C functions g k : U z IR, such that the vectors { g k z 1 k m are linearlndependent such that the point U z is in M if and onlf g k = 0 for all 1 k m. To get from the implicit equations for M given by the g k s to the eplicit equations for M given by the f k s one need onlnvoke possible after renumbering the components of the Implicit Function Theorem Let m,n INandletU IR n+m beanopenset. Let g : U IR m bec withgz = 0 for some z U. Assume that det [ g i n+j z ] 1 i,j m 0. Write a = z 1,,z n and b = z,,z n+m. Then there eist open sets V IR n+m and W IR n with a W and z = a,b V such that for each W, there is a unique,y V such that g,y = 0. c Joel Feldman All rights reserved. March 9, 2008 Eamples of Manifolds 2

3 If the y above is denoted f, then f : W IR m is C, fa = b and g,f = 0 for all W. The n sphere S n is the n dimensional surface in IR given implicitly by the equation g 1,, = = 0. In a neighbourhood of the north pole for eample, the northern hemisphere, S n is given eplicitly by the equation = n. If you think of the set, M 3, of all 3 3 real matrices as IR 9 because a 3 3 matri has 9 matri elements then SO3 = { A M 3 A t A = 1l, deta = 1 is a 3 dimensional surface 1 in IR 9. SO3 is the group of all rotations about the origin in IR 3 and is also the set of all orientations of a rigid body, Eample 5 The Torus The torus T 2 is the two dimensional surface T 2 = {,y,z IR 3 2 +y z 2 = 1 4 in IR 3. In cylindrical coordinates = rcosθ, y = rsinθ, z = z, the equation of the torus is r z 2 = 1 4. Fi any θ, say θ 0. Recall that the set of all points in IR 3 that have θ = θ 0 is like one page in an open book. It is a half plane that starts at the z ais. The intersection of the torus with that half plane is a circle of radius 1 2 centred on r = 1, z = 0. As ϕ runs from 0 to 2π, the point r = cosϕ, z = 1 2 sinϕ, z θ 0 y θ = θ 0 runs over that circle. If we now run θ from 0 to 2π, the circle on the page sweeps out the whole torus. So, as ϕ runs from 0 to 2π and θ runs from 0 to 2π, the point,y,z = cosϕcosθ, cosϕsinθ, 1 2 sinϕ runs over the whole torus. So we may build coordinate patches for T 2 using θ and ϕ with ranges 0,2π or π,π as coordinates. 1 Note that A t A = 1l forces deta { 1,1. If you fi any B SO3, then, just by continuity, all matrices A that obey A t A = 1l and are sufficiently close to B automatically obey deta = 1. So the equation deta = 1 is redundant. Since A t A is automatically symmetric, the requirement A t A = 1l gives at most 6 independent equations. In fact they are independent. c Joel Feldman All rights reserved. March 9, 2008 Eamples of Manifolds 3

4 Eample 6 The Cartesian Product If M is a manifold of dimension m with atlas A and N is a manifold of dimension n with atlas B then M N = {,y M, y N is an m + n dimensional manifold with atlas { U V,ϕ ψ U,ϕ A, V,ψ B where ϕ ψ,y = ϕ,ψy For eample, IR m IR n = IR m+n, S 1 IR is a cylinder, S 1 S 1 is a torus and the configuration space of a rigid bods IR 3 SO3 with the IR 3 components giving the location of the centre of mass of the body and the SO3 components giving the orientation. Eample 7 The Möbius Strip Take a length of ribbon. Put a half twist in it and glue the ends together. The result is a Möbius Strip. Mathematically, you can think of it as the set [0,1] 1,1 but with the points 0,t and 1, t identified i.e. pretend that they are the same point for all 1 < t < 1. We can view the Möbius Strip as a manifold with the set points M = [0,1 1, 1 and the two patch atlas A = {U 1,ϕ 1, U 1,ϕ 2 where U 1 = 1 8, 7 ϕ 8 1,y =,y U 2 = [ { 0, 4 1 3,f 0 < 1 4,1 4 ϕ 2,y = 1, f 3 4 < < 1 The range of ϕ 2 is 1 4, 1 4 1,1. Eample 8 Projective n space, IP n The projective n space, IP n, is the set of all lines through the origin in IR. If IR is nonzero, then there is a unique line L through the origin in IR that contains. Namely L = { λ λ IR. If, y IR are both nonzero, then L = L f and onlf there is a λ IR \ {0 such that y = λ. One choice of atlas for IP n is A = { U i,ϕ i 1 i with U i = { L IR, 0 ϕl = 1,, 1, +1,, IR n Observe that if ϕ i is well defined, because if, y IR are both nonzero and L = L y, then, for each 1 i, either both and are zero or both and are nonzero and in the latter case 1,, 1, +1,, = y1,, 1, +1,, y c Joel Feldman All rights reserved. March 9, 2008 Eamples of Manifolds 4

5 EachlinethroughtheorigininIR intersectstheunitspheres n = { IR = 1 in eactly two points and the two points are antipodal i.e. and. So you can think of IP n as S n but with antipodal points identified: IP = { {, S n Each line L IP n that is not horizontal i.e. with 0 intersects the northern hemisphere { IR = 1, 0 in eactly one point. Each line L IP n that is horizontal i.e. with = 0 intersects the northern hemisphere in eactly two points and the two points are antipodal. Bgnoring, you can think of the northern hemisphere as the closed unit disk { IR n 1 in IR n. So you can think of IP n as the closed unit ball in IR n but with antipodal points on the boundary = 1 identified. In the case of three dimensions, you can also think of SO3 as being the closed unit disk { IR 3 1 IR 3 but with antipodal points on the boundary = 1 identified. This is because, geometrically, each element of SO3 is a matri which implements a rotation by some angle about some ais through the origin in IR 3. We can associate each ωˆω IR 3, where ˆΩ is a unit vector and ω IR, with the rotation by an angle πω about the ais ˆΩ. But then any two ω s that differ by an even integer give the same rotation. So the set of all rotations is associated with { ωˆω ω 1, ˆΩ IR 3, ˆΩ = 1 but with 1ˆΩ and 1ˆΩ identified. Thus SO3 and IP 3 are diffeomorphic. c Joel Feldman All rights reserved. March 9, 2008 Eamples of Manifolds 5