## MATH 782 Differential Geometry : solutions to homework ...

MATH 782 Differential Geometry : solutions to homework assignment two. 1. Let ∇ be the connection on R2 that has vanishing Christoffel symbols Γk.

MATH 782 Differential Geometry : solutions to homework assignment two 1. Let ∇ be the connection on R2 that has vanishing Christoffel symbols Γkij with respect to Cartesian coordinates (x, y). Calculate the Christoffel symbols with respect to polar coordinates (r, θ). In other words, calculate the coefficients of ∂ ∂ ∂ , ∇∂ , ∇∂ , ∂r ∂r ∂θ ∂r ∂r ∂θ  ∂ in the basis ∂∂r , ∂θ . p Since r = x2 + y 2 and θ = arctan(y/x), we find ∇∂



dr dθ



x x2 +y 2 −y x2 +y 2

=

y

x2 +y 2 x x2 +y 2

∂ , ∂θ ∂θ

∇∂

and

!

dx dy

 .

Dually, the bases of the tangent space are related by the inverse matrix !  ∂  ∂  √x √ y 2 +y 2 2 +y 2 ∂x x x ∂r . = ∂ ∂ −y x ∂y ∂θ We can then calculate     ∂ x ∂ ∂ y ∂ ∂ ∇∂ = p ∇ ∂ −y +x +p ∇ ∂ −y +x ∂r ∂θ ∂x ∂y ∂x ∂y x2 + y 2 ∂x x2 + y 2 ∂y     x ∂x ∂ y ∂(−y) ∂ = p +p ∂y ∂x x2 + y 2 ∂x ∂y x2 + y 2 x −y ∂ ∂ +p = p 2 2 2 2 x + y ∂x x + y ∂y 1∂ = , r ∂θ and with a little more work ∂ ∂ = −r , ∂θ ∂θ ∂r

∇∂

∂ 1∂ = , ∂θ ∂r r ∂θ

∇∂

Note that ∇∂

∂θ

and

∂ = 0. ∂r ∂r

∇∂

∂ ∂ = ∇∂ ∂r ∂r ∂θ

as the connection is torsion-free. 2. Let M ⊂ RN be a submanifold of Euclidean space. We obtain a Riemannian metric on M by restricting the Euclidean metric, and this metric g has an associated Levi-Civita connection ∇. Instead we could take the standard connection on RN and restrict it to a connection ∇0 on M . Specifically, let X and Y be vector fields on M which extend to vector fields X and Y on a neighbourhood of M in RN . Writing Y as (Y 1 , . . . , Y N ), the standard connection on RN is given by ∇X Y = (DX Y 1 , . . . , DX Y N ) ∈ T RN . 1

Then we define ∇0X Y to be the orthogonal projection of ∇X Y to T M ⊂ T RN |M . Prove that ∇0 is the same as the Levi-Civita connection ∇ on M . We first show that ∇0 is a connection on M . The equations ∇0f X+gY Z = f ∇0X Z + g∇0Y Z ∇0X (Y + Z) = ∇0X Y + ∇0X Z follow automatically from the corresponding equations for ∇, and ∇0X f Y

= = = =

proj(∇X f Y ) proj(f ∇X Y + X(f )Y ) f proj(∇X Y ) + X(f )proj(Y ) f ∇0X Y + X(f )Y

as required, proving that ∇0 is indeed a connection. Note that when we project, we are also restricting to M . Moreover f |M = f , X(f )|M = X(f )|M (in general, the value of Z(g) at a point p only depends on the value Zp of the vector field at p), and then X(f )|M = X(f ) (as X is a vector field on M , we only need to differentiate f along M , not transversally; but f |M = f ), etc. Next we show that ∇0 is compatible with the Riemannian metric on M . XhY, Zi = = = = = =

XhY , Zi XhY , Zi h∇X Y , Zi + hY , ∇X Zi h∇X Y , Zi + hY, ∇X Zi hproj(∇X Y ), Zi + hY, proj(∇X Z)i h∇X Y, Zi + hY, ∇X Zi

(You should think carefully about why each equality holds, perhaps writing things in local coordinates to see more clearly. For example, the first equality holds because Y and Z give Y and Z when we restrict to M , so hY , Zi gives hY, Zi when restricted to M . For the fifth equality, Z lies in T M ⊂ T RN , so when we take the inner product with ∇X Y we can ignore the component of ∇X Y which is orthogonal to T M ; equivalently, we can project ∇X Y to T M .) Finally, we show that ∇0 is torsion-free. ∇0X Y − ∇0Y X = proj(∇X Y ) − proj(∇Y X) = proj([X, Y ]) This time we will use local coordinates. Let {x1 , . . . , xn } be local coordinates on M and complete to local coordinates {x1 , . . . , xn , xn+1 , . . . , xN } on RN . Then X = X1

∂ ∂ ∂ ∂ + . . . + Xn + Xn+1 + . . . + XN ∂x1 ∂xn ∂xn+1 ∂xN 2

where Xn+1 , . . . , XN all vanish on M since X|M = X. There is a similar expression for Y . Now !  N  X ∂Xj ∂ ∂Yj − Yi proj([X, Y ]) = proj Xi ∂x ∂xi ∂xj i i,j=1  N X n  X ∂Xj ∂ ∂Yj = − Yi Xi ∂xi ∂xi ∂xj i=1 j=1  n X n  X ∂Yj ∂Xj ∂ = Xi − Yi ∂xi ∂xi ∂xj i=1 j=1 = [X, Y ] where the second equality follows from the fact that projection to T M means that we ∂ keep only the terms involving ∂xj for j = 1, . . . , n, and the third equality follows from the fact that Xi and Yi vanish on M for i = n + 1, . . . , N . It follows that ∇0 is torsion-free. In conclusion, we have shown that ∇0 is a torsion-free connection on M which is compatible with the Riemannian metric. By uniqueness of such connections, ∇0 must be the LeviCivita connection on M . 3. Consider the upper half-plane R2+ := {(x, y) ∈ R2 |y > 0} with the hyperbolic metric 1 (dx2 + dy 2 ). y2 a) Calculate the Christoffel symbols of the hyperbolic metric. Letting (x1 , x2 ) = (x, y), we see that gij = δij /y 2 . Substituting this into 1 Γkij = g kl (gil;j + gjl;i − gij;l ) 2 we find Γ112 = Γ121 =

−1 y

1 y −1 = y

Γ211 = Γ222 and all other Christoffel symbols vanish.

b) Let v0 be the vector (0, 1) ∈ T(0,1) R2+ and let c : R → R2+ be the curve t 7→ (t, 1). Find the parallel transport of v0 along c, i.e., find a formula for the parallel vector field V (t) ∈ T(t,1) R2+ that satisfies V (0) = v0 . 3

 Let V (t) =

v(t) w(t)

 . Then DV (t) = ∇ ∂ V (t) ∂x dt  0    v (t) v(t) i = + Γ1j w0 (t) w(t)  0     0 −1 v (t) v(t) y = + 1 w0 (t) 0 w(t) y  0  v (t) − y1 w(t) = w0 (t) + y1 v(t)  0  v (t) − w(t) = w0 (t) + v(t)

where the last equality is due to the fact that  y always   equals 1 on the curve c. The v(t) 0 unique solution with initial condition = is w(t) 1   sin(t) V (t) = . cos(t) 4. Geodesics on a surface of revolution: Consider the surface of revolution S given by the parameterization (u, v) 7→ (f (v)cosu, f (v)sinu, g(v)) where f and g are smooth functions with f (v) > 0 and (f 0 (v))2 + (g 0 (v))2 > 0 for all v. In Homework 1 we calculated the metric on S induced by the Euclidean metric on R3 :   (f (v))2 0 g= 0 (f 0 (v))2 + (g 0 (v))2 a) Derive the local form of the geodesic equations: d2 u 2f 0 du dv + =0 dt2 f dt dt  2  2 d2 v ff0 du f 0 f 00 + g 0 g 00 dv − 0 2 + 0 2 =0 dt2 (f ) + (g 0 )2 dt (f ) + (g 0 )2 dt Let (x1 , x2 ) = (u, v). As in problem 3, the only non-vanishing Christoffel symbols are Γ112 = Γ121 , Γ211 , and Γ222 , because the metric is diagonal and independent of the first coordinate u. We find 1 d f0 Γ112 = Γ121 = 2 (f 2 ) = 2f dv f Γ211 =

−1 d 2 −f f 0 (f ) = 2((f 0 )2 + (g 0 )2 ) dv (f 0 )2 + (g 0 )2 4

Γ222 =

1 d f 0 f 00 + g 0 g 00 0 2 0 2 ((f ) + (g ) ) = 2((f 0 )2 + (g 0 )2 ) dv (f 0 )2 + (g 0 )2

where 0 always denotes derivatives with respect to v. The result follows by substituting these Christoffel symbols into the geodesic equations. b) Fix a geodesic γ. A parallel is a circle given by v = constant (and hence also r2 = x2 + y 2 = constant). Suppose that at γ(t), the geodesic γ meets a parallel of radius r(t) at an angle β(t). Prove that r(t)cosβ(t) is independent of t. This is called Clairaut’s relation. Let us first find a formula for cosβ(t). Suppose the parallel and the geodesic meet at the point (x, y, z) = (f cosu, f sinu, g). The unit tangent to the parallel is given by 1 p (−y, x, 0) = (−sinu, cosu, 0). 2 x + y2 The velocity vector field of the geodesic is given by d dv du dv du dv (f cosu, f sinu, g) = (f 0 cosu − f sinu , f 0 sinu + f cosu , g 0 ). dt dt dt dt dt dt Rescaling t if necessary, we can assume that the geodesic is normalized (parametrized by arc length), so that the velocity vector has length one. The inner product of these two unit tangent vectors then gives du cosβ(t) = f dt p and multiplying by r(t) = x2 + y 2 = f gives r(t)cosβ(t) = f 2

du . dt

Now

d d2 u du dv (r(t)cosβ(t)) = f 2 2 + 2f f 0 dt dt dt dt which vanishes by the first geodesic equation. Therefore r(t)cosβ(t) is constant (independent of t) as required. * c) The paraboloid is the surface of revolution built from a parabola (f (v), g(v)) = (v, v 2 ). Use Clairaut’s relation to show that a geodesic on the paraboloid intersects itself an infinite number of times (unless it is given by u = constant). Firstly, substituting f = v and g = v 2 into the geodesic equations gives d2 u 2 du dv + =0 dt2 v dt dt  2  2 d2 v v du 4v dv − + = 0. 2 2 2 dt 1 + 4v dt 1 + 4v dt 5

We will only need the first equation, which as we saw in part b) is equivalent to Clairaut’s relation. Separating u and v  −1 2 du du 2 dv = dt dt2 v dt and integrating gives log Therefore

du = −2logv + const. dt C du = 2 dt v

for some constant C. Without loss of generality, let’s assume that r(t) = v achieves its minimum value v0 at t = 0. At this point the geodesic is tangent to the parallel and β(0) = 0. Since r(t)cosβ(t) must be constant, we see that for t > 0, β(t) > 0 and r(t) = v is increasing (the geodesic is “climbing”). On the other hand, for t < 0, β(t) < 0, and the geodesic is “climbing” in the reverse direction. These two directions will intersect each other infinitely many times provided the geodesic keeps circling the surface, i.e., provided u → ∞ as t → ∞ (and by symmetry, u → −∞ as t → −∞). Since the z-coordinate on the surface is g = v 2 , we must have (v(t))2 − v02 ≤ t as the increase in the z-coordinate cannot be greater than the distance covered by the geodesic in time t (recall that the geodesic is parametrized by arc length). Therefore C C du = 2 ≥ dt v t + v02 and u(t) ≥ Clog|t + v02 | + const. We conclude that u(t) has the desired asymptotic behaviour as t → ∞. 5. Killing fields: A Killing vector field X ∈ X (M ) is an infinitesimal isometry, meaning that the flow ψt that it generates (locally and for small t) is a local isometry. a) For M = Rn , we can think of a vector field as a map X : Rn → Rn (by identifying the tangent space at every point of Rn with Rn itself). We call X a linear field if this map is given by an n × n matrix. Prove that a linear field is a Killing field if and only if this matrix is anti-symmetric. Suppose X is given by an n × n matrix A, so that the vector field at a point (x1 , . . . , xn ) is given by X ∂ Aij xj . ∂xi ij

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The equation for a flow line (x1 (t), . . . , xn (t)) is then     x1 x1 d  .   .   ..  = A  ..  dt xn xn with solution

   x1 (0) x1 (t)  ..   .   .  = exp(tA)  ..  . xn (t) xn (0)

Thus the flow ψt is given by exp(tA) (which is defined using the power series for exp). Now exp(tA) preserves the metric on Rn if and only if exp(tA) is in the orthogonal group O(n), if and only if A is anti-symmetric. b) If a Killing field X vanishes at a point p ∈ M , show that X is tangent to the geodesic spheres around p (equivalently, the geodesic spheres are preserved by the flow ψt ). If X vanishes at p, then p will be a fixed point of the flow ψt generated by X. Now for small r, the geodesic sphere of radius r centred at p is precisely the locus of points q which are distance r from p. If X is a Killing field then ψt must be an isometry, and thus d(p, ψt (q)) = d(ψt (p), ψt (q)) = d(p, q) = r. Therefore ψt (q) also lies on the geodesic sphere of radius r centred at p. In particular, the curve t 7→ ψt (q) lies on the geodesic sphere and differentiating shows that X is tangent to the sphere. * c) The Killing equation: Show that X is a Killing vector field if and only if h∇Y X, Zi + h∇Z X, Y i = 0 for all vector fields Y and Z ∈ X (M ). A hint for part c) can be found in the textbook, page 82. If X vanishes identically then there is nothing to prove, so choose a point p at which X(p) 6= 0. We will prove the equation holds at p, and therefore by continuity it will hold everywhere. Since X(p) 6= 0 we can find a hypersurface (codimension one submanifold) S ⊂ M which is transversal to X at p, and also in a small neighbourhood of p. Let (x1 , . . . , xn−1 ) be local coordinates on S. There is a map S × (−δ, δ) → M which takes a point (q, t) to ψt (q) where ψt is the flow generated by X. This map is a local diffeomorphism in a neighbourhood of p, and this yields local coordinates (x1 , . . . , xn−1 , t) on M in a neighbourhood of p such that X = ∂∂t . 7

Now for 1 ≤ i, j ≤ n − 1 we calculate ∂ ∂ ∂ ∂ , i + h∇ ∂ , i ∂xi ∂t ∂xj ∂xj ∂t ∂xi

h∇ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ , i + h[ , ], i + h∇ ∂ , i + h[ , ], i ∂t ∂x ∂t ∂x ∂xi ∂t ∂xj ∂xj ∂t ∂xi i ∂xj j ∂xi ∂ ∂ ∂ ∂ = h∇ ∂ , i + h∇ ∂ , i ∂t ∂x ∂t ∂x i ∂xj j ∂xi ∂ ∂ ∂ = h , i ∂t ∂xi ∂xj = h∇ ∂

where the first equality follows from the fact that ∇ is torsion-free. Now in local coordinates the flow ψt of X is given by (x1 , . . . , xn−1 , 0) 7→ (x1 , . . . , xn−1 , t) so d(ψt )(x1 ,...,xn−1 ,0)

∂ ∂ = . ∂xi ∂xi

Since ψt is an isometry, we must have h

∂ ∂ ∂ ∂ , i(x1 ,...,xn−1 ,0) = hd(ψt )(x1 ,...,xn−1 ,0) , d(ψt )(x1 ,...,xn−1 ,0) i(x ,...,x ,t) ∂xi ∂xj ∂xi ∂xj 1 n−1 ∂ ∂ , i(x ,...,x ,t) = h ∂xi ∂xj 1 n−1 ∂

∂ , i(x1 ,...,xn−1 ,t) is independent of t and Therefore h ∂x i ∂xj

∂ ∂ ∂ h , i = 0. ∂t ∂xi ∂xj We have shown that h∇Y X, Zi + h∇Z X, Y i = 0 ∂ for basis vectors Y and Z (the case when Y or Z is ∂t can be proved similarly). Since ∞ this equation is C (M )-linear in both Y and Z, it must hold for all vector fields Y and Z ∈ X (M ).

For the converse, one essentially just reverses the above argument. Thus one again chooses local coordinates (x1 , . . . , xn−1 , t) around a point p where X(p) 6= 0 such that X = ∂∂t . ∂ As above, the equation yields ∂∂t h ∂ ∂xi , ∂xj i = 0 for i, j = 1, . . . , n − 1. In fact, the same ∂

∂ argument shows that ∂∂t h ∂∂xi , ∂xj i = 0 even when i and/or j is n (where ∂∂xn = ∂t ). This means that the flow generated by X preserves the metric, i.e., X is a Killing vector field.

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