Riemannian Geometry – Lecture 16 Computing Sectional Curvatures

Dr. Emma Carberry

September 14, 2015

Stereographic projection of the sphere

Example 16.1. We write Sn or Sn(1) for the unit sphere in Rn+1, and Sn(r) for the sphere of radius r > 0. Stereographic projection φ from the north pole N = (0,..., 0, r) gives local coordinates on Sn(r) \{N}

Example 16.1 (Continued).

Example 16.1 (Continued). Example 16.1 (Continued). We have for each i = 1, . . . , n that

ui = cxi (1) and using the above similar triangles, we see that r c = . (2) r − xn+1

1 Hence stereographic projection φ : Sn \{N} → Rn is given by rx1 rxn φ(x1, . . . , xn, xn+1) = ,..., r − xn+1 r − xn+1

The inverse φ−1 of stereographic projection is given by (exercise)

2r2ui xi = , i = 1, . . . , n |u|2 + r2 2r3 xn+1 = r − . |u|2 + r2 and so we see that stereographic projection is a diffeomorphism.

Stereographic projection is conformal

Deﬁnition 16.2. A local coordinate chart (U, φ) on Riemannian manifold (M, g) is conformal if for any p ∈ U and X,Y ∈ TpM,

2 gp(X,Y ) = F (p) hdφp(X), dφp(Y )i where F : U → R is a nowhere vanishing smooth function and on the right-hand side we are using the usual Euclidean inner product.

Exercise 16.3. Prove that this is equivalent to:

1. dφp preserving angles, where the angle between X and Y is deﬁned (up to adding multi- ples of 2π) by g(X,Y ) cos θ = pg(X,X)pg(Y,Y ) and also to

2 2. gij = F δij for some nowhere vanishing smooth function F on U.

Example 16.1 (Continued). We now proceed to show that stereographic projection is a conformal map. Now −1 −1 gij(u) = hdφu (ei), dφu (ej)i and ∂xj ∂ dφ−1(e ) = (u) u i ∂ui ∂xj 2r2 4r2(ui)2 ∂ 4r3ui ∂ = − + |u|2 + r2 (|u|2 + r2)2 ∂xi (|u|2 + r2)2 ∂xn+1 | {z } | {z } A(i) B(i)

2 so

gij = A(i)A(j)δij + B(i)B(j) and simplifying gives 4r4 gij = δij. (3) (|u|2 + r2)2 Pre-composing with the rotation of the sphere is another coordinate chart which includes the point N = (0,..., 0, 1) and since rotation is an isometry, (3) holds for this coordinate chart also and hence we have covered the sphere with an atlas satisfying 4r4 gij = δij. (|u|2 + r2)2

We next use this to compute the sectional curvature of Sn(r). In fact we will do the computation for a general conformal coordinate chart.

Recall Corollary 15.10

Corollary (Constant Sectional Curvature). Let (M, g) be a Riemannian manifold, p ∈ M and

{e1, . . . , en} an orthonormal basis of TpM. Then the following are equivalent:

1. sec(π) = K0 for all 2-planes π ⊂ TpM

2. Rmijkl = K0(δilδjk − δikδjl).

3. Rmijji = −Rmijij = K0 for all i 6= j and Rmijkl = 0 otherwise.

Hence if we have a conformal coordinate chart with

Rmijji Kij = = K0 for all i 6= j giigjj and Rmijkl = 0 otherwise, then our manifold of constant sectional curvature K0.

Proposition 16.4. Suppose (U, φ) is a conformal coordinate chart on the Riemannian manifold (M, g) and 2 gij = F δij Writing f = log F , the sectional curvature on U satisﬁes ! −2 2 2 X 2 Kij = −F ∂i f + ∂j f + (∂kf) . k6=i,j

Furthermore Rmijkl = 0 if all four indices are distinct, and if i, j, k are distinct we have

2 i 2 Rmijki = F Rijk = F ((∂kf)(∂jf) − ∂k(∂j(f))) .

The symmetries of Rm then determine all Rmijkl with precisely 3 distinct indices.

3 Proving this proposition is your 1st assignment question.

Example 16.1 (Continued). For the sphere Sn(r) we have 2r2 F = , f = log(2r2) − log(|u|2 + r2) |u|2 + r2 so l l 2 −2u 2 −2 4(u ) ∂lf = , ∂ f = + |u|2 + r2 l |u|2 + r2 (|u|2 + r2)2 and 4ulum ∂m(∂lf) = . (|u|2 + r2)2 Example 16.1 (Continued). Hence

Kij = 2 ! (|u| + r2)2 4 4((ui)2 + 4(uj)2 X 4(ul)2 − − 4r4 2 2 2 2 2 2 2 2 |u| + r (|u| + r ) l6=i,j (|u| + r ) 4r2 1 = = . 4r4 r2 Furthermore

i Rijk = (∂kf)(∂jf) − ∂k(∂j(f)) 1 = 4ukuj − 4ukuj = 0 (|u|2 + r2)2 Example 16.1 (Continued). Similarly

j Rijk = −(∂if)(∂kf) − ∂k(∂if) = 0. Hence by the corollary from last time about constant sectional curvature we have 1 sec(π) = r2 n n for any 2-plane π ⊂ TpS (r) and any p ∈ S (r). 1 The sphere n(r) of radius r has constant sectional curvature equal to . S r2

Hyperbolic Space Example 16.5. The upper half space model for hyperbolic space is given by

1 n n n M = {(x , . . . , x ) ∈ R | x > 0} together with the Riemannian metric 1 g (x1, . . . , xn) = δ ij (xn)2 ij where we are taking the standard basis vectors ∂i = ei = (0,..., 0, 1, 0,..., 0).

1 n 1 This then is clearly a conformal coordinate system, with F (x , . . . , x ) = xn .

4 Example 16.5 (Continued). Our proposition gives that (for i 6= j) ! −2 2 2 X 2 Kij = −F ∂i f + ∂j f + (∂kf) k6=i,j and then since f(x1, . . . , xn) = − log(xn) we have ( 0, if i 6= n ∂if = −(xn)−1, if i = n

( 2 0, if i 6= n ∂i f = (xn)−2, if i = n so ( 2 −(xn)2 ((−xn)−1) = −1, for i, j, n all distinct Kij = −(xn)2(xn)−2 = −1, for i 6= j = n Example 16.5 (Continued). To conclude that the sectional curvature is −1 for every tangential 2-plane, not just the cooordinate ones, according to the corollary it remains to check that the

Riemann coefﬁcients which are not of the form Rmijji or Rmijij vanish. From Proposition

15.10, Rmijkl = 0 if all four indices are distinct, and for i, j, k distinct we have

n −2 i n −2 Rmijki = (x ) Rijk = (x ) ((∂kf)(∂jf) − ∂k(∂j(f))) = 0. and the other one is similiar. Hence

hyperbolic space has constant sectional curvature −1.