By Michael Spivak

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Extra info for A Comprehensive Introduction to Differential Geometry, Vol. 4, 3rd Edition

Example text

Since eDi = cos θi − sin θi sin θi cos θi when Di is a 2 × 2 skew symmetric matrix and Wi = Wi = P −1 , where We can compute Dk (k + 1)! 1 cos(θi t)dt 0 1 sin(θi t)dt 0 that is, Wi = 1 0 − sin(θi t)dt 1 cos(θi t)dt 0 1 θi = 1 θi 1 Di t e dt, 0 we get sin(θi t) |10 cos(θi t) |10 , − cos(θi t) |10 sin(θi t) |10 −(1 − cos θi ) sin θi , sin θi 1 − cos θi and Wi = 1 when Di = 0. Now, in the ﬁrst case, the determinant is 2 1 (sin θi )2 + (1 − cos θi )2 = 2 (1 − cos θi ), 2 θi θi 38 CHAPTER 1. INTRODUCTION TO MANIFOLDS AND LIE GROUPS which is nonzero, since θi = k2π for all k ∈ Z.

To compute V , since Ω = P D P = P DP −1 , observe that V Ωk (k + 1)! = In + k≥1 P Dk P −1 (k + 1)! = In + k≥1 In + = P k≥1 −1 = PWP W = In + k≥1 W = In + k≥1 by computing Dk . (k + 1)! Dk = (k + 1)! 1 0 eDt dt,   ... W1   W2 . .   W =  .. . ..   . .  . . Wp by blocks. Since eDi = cos θi − sin θi sin θi cos θi when Di is a 2 × 2 skew symmetric matrix and Wi = Wi = P −1 , where We can compute Dk (k + 1)! 1 cos(θi t)dt 0 1 sin(θi t)dt 0 that is, Wi = 1 0 − sin(θi t)dt 1 cos(θi t)dt 0 1 θi = 1 θi 1 Di t e dt, 0 we get sin(θi t) |10 cos(θi t) |10 , − cos(θi t) |10 sin(θi t) |10 −(1 − cos θi ) sin θi , sin θi 1 − cos θi and Wi = 1 when Di = 0.

23 can be found in Gallot, Hulin and Lafontaine [60] or Berger and Gostiaux [17]. 11. This equivalence is also proved in Gallot, Hulin and Lafontaine [60] and Berger and Gostiaux [17]. 52 CHAPTER 1. INTRODUCTION TO MANIFOLDS AND LIE GROUPS For example, we can show again that the sphere S n = {x ∈ Rn+1 | x 2 2 − 1 = 0} is an n-dimensional manifold in Rn+1 . Indeed, the map f : Rn+1 → R given by f (x) = x 22 −1 is a submersion (for x = 0) since n+1 df (x)(y) = 2 xk yk . k=1 We can also show that the rotation group, SO(n), is an n2 R .