DIFFERENTIAL GEOMETRY HW 7

DIFFERENTIAL GEOMETRY HW 7 CLAY SHONKWILER 1 Show that within a local coordinate system x 1,..., x n ) on M with coordinate vector fields X 1 / x 1,..., X n / x n, if we pick n 3 smooth real-valued functions Γ k ij at random, and define V W [v i X i w k ) + v i w j Γ k ij X k, then will satisfy the definition of an affine connection within the range of that coordinate system. Proof. First, we show linearity in the first vector field; let f 1, f 2 be smooth functions on the range of the coordinate system and let V 1, V 2, W be smooth vector fields on the range of the coordinate system. Then f1 V 1 +f 2 V 2 W [f 1 v1 i + f 2 v2)x i i w k ) + f 1 v1 i + f 2 v2)w i j Γ k ij X k [f 1 v1x i i w k ) + f 1 v1w i j Γ k ij X k + [ f 2 v2x i i w k ) + f 2 v2w i j Γ k ij f 1 [v i 1X i w k ) + v i 1w j Γ k ij f 1 V1 W + f 2 V2 W. X k + f 2 [ v i 2X i w k ) + v i 2w j Γ k ij Next, we show additivity in the second coordinate; let V, W 1, W 2 be smooth vector fields on the range of the coordinate system. Then V W 1 + W 2 ) [v i X i w1 k + w2) k + v i w j 1 + wj 2 )Γk ij X k [v ) i X i w1) k + X i w2) k + v i w j 1 + wj 2 )Γk ij [v i X i w k 1) + v i w j 1 Γk ij V W 1 + V W 2. X k X k + [ v i X i w k 2) + v i w j 2 Γk ij Finally, we show that the Leibniz rule is satisfied; let V, W be smooth vector fields on the range of the coordinate system and let g be a smooth function 1 X k X k X k

2 CLAY SHONKWILER on the range of the coordinate system. Then V gw ) [v i X i gw k ) + v i gw j Γ k ij X k [v ) i X i g)w k + gx i w k ) + v i gw j Γ k ij X k v i X i g)w k X k + [v i gx i w k ) + v i gw j Γ k ij V g)w k X k + g [v i X i w k ) + v i w j Γ k ij V g)w + g V W. X k X k Having shown linearity in the first coordinate, additivity in the second coordinate and satisfaction of the Leibniz rule, we conclude that is an affine connection on the range of the coordinate system. 3 Show that an affine connection is compatible with a Riemannian metric on M if and only if, for any vector fields V and W along a smooth curve c : I M, we have d V, W, W + V, DW. Proof. Suppose that, for any vector fields V and W along a smooth curve c : I M, we have d V, W, W + V, DW. Let c : I M be a smooth curve and let V and W be parallel vector fields along c. Then d V, W, W + V, DW 0, W + V, 0 0, so we see that V, W is constant along c, meaning that is compatible with the Riemannian metric on M. On the other hand, suppose is compatible with the Riemannian metric on M. Let c : I M be a smooth curve and let V and W be vector fields along c. Let U 1,..., U n be an orthonormal basis of T c0) M and let U i t) be the parallel transport of U i along c. Then, since is compatible with the Riemannian metric on M, U i t), U j t) U i, U j δ ij, so U 1 t),..., U n t) forms an orthonormal basis of T ct) M for all t I. Hence, at each ct), V f i t)u i t)

DIFFERENTIAL GEOMETRY HW 7 3 and W g j t)u j t) j1 for smooth f 1,..., f n, g 1,..., g n. Now, V, W f i t)u i t), g j t)u j t) j1 i,j i,j f i t)g j t) U i t), U j t) f i t)g j t)δ ij so 1) d V, W d fi t)g i t) On the other hand, [ D f i t)u i t) and DW D g j t)u j t) j1 f i t)g i t), [ f i t)g i t) + f i t)g it). D [f it)u i t) j1 D [g jt)u j t) Hence, W + V, DW f it)u i t), 2) f it)g i t) + [ f it)u i t) + f i t) DU it) j1 f it)u i t) [ g jt)u j t) + g j t) DU jt) g j t)u j t) + f i t)u i t), j1 f i t)g it) [ f i t)g i t) + f i t)g it). Equating??) and??), we see that d V, W, W + V, DW. g jt)u j t). j1 g jt)u j t) j1

4 CLAY SHONKWILER 4 Conclude that an affine connection is compatible with a Riemannian metric on M if and only if Proof. Suppose U V, W U V, W + V, U W. U V, W U V, W + V, U W. Now, let V, W be vector fields on M, let c : I M be a smooth curve and let U dc. Then d V, W U V, W UV, W + V, U W dc/ V, W + V, dc/ W, W + V, DW Therefore, by problem 3 above, is compatible with the Riemannian metric on M. On the other hand, suppose is compatible with the Riemannian metric on M. Let U, V, W be vector fields on M, let p M and let c : I M be a smooth curve such that c0) p and c 0) Up). Then U V p) p) and U W p) DW p). Hence, using our result from problem 3, U V, W p d V, W p, W V, DW U V, W p + V, U W p. + p Since our choice of p M was arbitrary, we see that U V, W U V, W + V, U W p. on all of M. 5 Looking back over this proof, show that if U, V and W are three vector fields on M which do not necessarily commute with one another, then 2 U V, W U V, W +V W, U W U, V U, [V, W + V, [W, U + W, [U, V. Proof. Let U, V and W be vector fields on M. By problem 4 above, U V, W u V, W + V, U W V W, U V W, U + W, V U W U, V W U, V + U, W V.

DIFFERENTIAL GEOMETRY HW 7 5 Hence, U V, W + V W, U W U, V u V, W + V, U W ) + V W, U + W, V U ) W U, V + U, W V ) W, U V + V U + U, [V, W V, [W, U W, 2 U V [U, V + U, [V, W V, [W, U 2 U V, W W, [U, V + U, [V, W V, [W, U, since [U, V U V V U. Therefore, solving for 2 U V, W, we see that 2 U V, W U V, W +V W, U W U, V U, [V, W + V, [W, U + W, [U, V. Define a map F : S 3 S 3 SO4) by 6 F x, y)z) xzy 1, where x and y are unit quaternions, z is any quaternion, and quaternion multiplication is used on the right hand side. Show that F is a double covering. Proof. In fact, we can show more than this. We can show that F is a Lie group homomorphism which is surjective with kernel {±1, 1)}, meaning that it induces a diffeomorphism S 3 S 3 /{x, y) x, y)}. To that end, note that, if q, r), p, s) S 3 S 3, then F x, y)z, w)) F xz, yw) [v xzvyw) 1 [v xzvw 1 y 1 [v xvy 1 [v zvw 1 F x, y) F z, w), so F is a group homomorphism.

6 CLAY SHONKWILER Now, If x, y S 3 such that x a + bi + cj + dk and y e + fi + gj + hk, then y 1 ȳ, so: x1)y 1 a + bi + cj + dk)e fi gj hk) ae + bf + cg + dh) + ibe af ch + dg) +jce ag df + bh) + kde ah bg + cf) xi)y 1 a + bi + cj + dk)ie fi gj hk) a + bi + cj + dk)f + ei + hj gk) af be ch + dg) + iae + bf cg dh) +jah + cf + de + bg) + kdf ag + bh ce) xj)y 1 a + bi + cj + dk)je fi gj hk) a + bi + cj + dk)g hi + ej + fk) ag + bh ce df) + ibf ah + cf de) +jae + cg dh bf) + kaf + dg + be + ch) xk)y 1 a + bi + cj + dk)ke fi gj hk) a + bi + cj + dk)h + gi fj + ek) ah bg + cf de) + iag + bh + ce + df) +jch af be + dg) + kae + dh bf cg). Hence, F x, y) is given by the matrix ae + bf + cg + dh af be ch + dg ag + bh ce df ah bg + cf de be af ch + dg ae + bf cg dh bf ah + cf de ag + bh + ce + df ce ag df + bh ah + cf + de + bg ae + cg dh bf ch af be + dg. de ah bg + cf df ag + bh ce af + dg + be + ch ae + dh bf cg We can extend this to a map from S 3 S 3 to R 16. The coordinate functions of this map are clearly continuous in fact, smooth); since restricting the range doesn t affect continuity, we see that F is continuous. Now, if x, y) as above in terms of a, b,...) and x, y) ker F, then, looking at the diagonal entries of F x, y) and recalling that x x yȳ 1, we see that ae + bf + cg + dh 1 ae + bf cg dh 1 ae + cg dh bf 1 ae + dh bf cg 1 a 2 + b 2 + c 2 + d 2 1 e 2 + f 2 + g 2 + h 2 1. Thus, it must be the case that ae 1 and b c d f g h 0, so we see that that is, ker F {±1, 1)}. Hence, x, y) ±1, 1); SO4) S 3 S 3 )/ker F S 3 S 3 )/{x, y) x, y)}. Clearly, this implies that F is a double covering.

DIFFERENTIAL GEOMETRY HW 7 7 7 Show that the orbits of the vector fields U, V and W are great circles on S 3, and that in each case the collection of orbits forms a Hopf fibration of S 3 by parallel great circles. Draw such a fibration. Proof. First, suppose that a + bi S 3 ; then Ua + bi) a + bi)i b + ai. Hence, for any x S 3 in the x 0 x 1 -plane, Ux) is also in the x 0 x 1 - plane. Specifically, the orbit of U through 1 lies in the x 0 x 1 -plane. Since the intersection of the x 0 x 1 -plane with S 3 is a great circle, we see that the orbit of U through 1 is a great circle. Similarly, V a + bj) b + aj and W a + bk) b + ak, so the orbits of V and W through 1 are also great circles. Now, for any x S 3, L x 1) x 1 x. Since U, V and W are leftinvariant, L x ) maps the orbits of U, V and W through 1 to the orbits of U, V and W, respectively, through x. Since L x is an isometry this is an easy computation, which I ve done, but which is a huge pain to type up, so I don t reproduce the calculation here) and isometries take geodesics to geodesics, we see that the orbits of U, V and W through x are also great circles. Since our choice of x S 3 was arbitrary, we see that all the orbits of U, V and W are great circles on S 3. Now, suppose x 1 i + x 2 j + x 3 k S 3. Then Ux 1 i + x 2 j + x 3 k) x 1 i + x 2 j + x 3 k)i x 1 + x 3 j x 2 k, so the orbit of U passing through x 1 i + x 2 j + x 3 k moves out of the purely imaginary quaternions, meaning the orbit does not lie in the purely imaginary quaternions. A similar argument shows that none of the orbits of V or W lie entirely in the purely imaginary quaternions. On the other hand, the purely imaginary quaternions form a surface on S 3, while the orbits of U, V and W are great circles on S 3, so each orbit intersects the purely imaginary quaternions in a single point. Since {x 0 + x 1 i + x 2 j + x 3 k S 3 : x 0 0} S 2, we can define the map f U : S 3 S 2 where f U x) is the unique purely imaginary quaternion lying on the orbit of U passing through x. A similar definition gives maps f V : S 3 S 2 and f W : S 3 S 2. If p S 2, we can represent p by a purely imaginary unit quaternion x 1 i + x 2 j + x 3 k. Now, f 1 Y x 1i + x 2 j + x 3 k) is the orbit of Y on S 3 passing through x 1 i + x 2 j + x 3 k for Y U, V, W. Since these orbits are great circles, we see that the fibers of f Y : S 3 S 2 are S 1 for Y U, V, W. In other words, f U, f V, f W are S 1 fibrations of S 2 ; that is, Hopf fibrations.

8 CLAY SHONKWILER Here s a crude picture of a Hopf fibration: a): Justify the use of symmetry above. Proof. We ve already seen that U V W. Now, 8 ) V W V x 3 + x 2 x 1 + x 0 V x 3 ) + V x 2 ) V x 1 ) + V x 0 ) x 1 + x 0 + x 3 x 2 U and ) W U W x 1 + x 0 + x 3 x 2 W x 1 ) + W x 0 ) + W x 3 ) W x 2 ) x 2 x 3 + x 0 + x 1 V. b): Show that V U W, W V U, U W V.

DIFFERENTIAL GEOMETRY HW 7 9 Proof. ) V U V x 1 + x 0 + x 3 x 2 V x 1 ) + V x 0 ) + V x 3 ) V x 2 ) x 3 x 2 + x 1 x 0 W, ) W V W x 2 x 3 + x 0 + x 1 W x 2 ) W x 3 ) + W x 0 ) + W x 1 ) x 1 x 0 x 3 + x 2 U and ) U W U x 3 + x 2 x 1 + x 0 Ux 3 ) + Ux 2 ) Ux 1 ) + Ux 0 ) x 2 + x 3 x 0 x 1 V. 2 Let X and Y be differentiable vector fields on a Riemannian manifold M. Let p M and let c : I M be an integral curve of X through p; i.e. ct 0 ) p and dc Xct)). Prove that the Riemannian connection of M is X Y )p) d P 1 c,t 0,t Y ct )))) tt0, where P c,t0,t : T ct0 )M T ct) M is the parallel transport along c, from t 0 to t this shows how the connection can be reobtained from the concept of parallelism). Proof. Let X 1,..., X n be an orthonormal basis for T p M; then we can write Y p) y i X i. Let P i t) P c,t0,tx i ). Then, for all t I, Y ct)) y i P i t).

10 CLAY SHONKWILER Then, for all t I, Pc,t 1 1 0,tY ct))) Pc,t 0,t yi P i t)) y i X i. Hence, 3) d [ P 1 c,t 0,t Y ct ))) tt0 d y i X i ) tt0 dyi X i. tt0 On the other hand, so X Y DY X Y )p) dyk [ dy i + Γk ijy j dx i P k t), P k t 0 ) dyk tt0 X k. tt0 This is equal to the right hand side of??), so we conclude that X Y )p) d P 1 c,t 0,t Y ct )))) tt0. 6 Let M be a Riemannian manifold and let p be a point of M. Consider a constant curve f : I M given by ft) p, for all t I. Let V be a vector field along f that is, V is a differentiable mapping of I into T p M). Show that dv, that is to say, the covariant derivative coincides with the usual derivative of V : I T p M. Proof. Recall that, in local coordinates x 1,..., x n ), [ dv k + Γk ijv j dx i x k, using the Einstein summation convention, where V v k x k. However, since ft) p for all t I, the x i are constant along all of f; that is, they are the chosen coordinate vectors in T p M. Hence, dx i 0 for all i. Therefore, dvk dv x k, by definition.

Consider the upper half-plane DIFFERENTIAL GEOMETRY HW 7 11 8 R 2 + {x, y) R 2 ; y > 0} with the metric given by g 11 g 22 1 y 2, g 12 0 metric of Lobatchevski s non-euclidean geometry). a): Show that the Christoffel symbols of the Riemannian connection are: Γ 1 11 Γ2 12 Γ1 22 0, Γ2 11 1 y, Γ1 12 Γ2 22 1 y. Proof. Recall that Γ m ij 1 2 k [ x i g jk + x j g ki g ij g km, x k where g km ) is the inverse of the matrix g km ). Now, ) 1 0 y g km ) 2 1, 0 y 2 so g km ) y 4 1 y 2 0 0 1 y 2 ) ) y 2 0 0 y 2. Hence, Γ 1 11 1 [ 0 + 0 0) y 2 + 0 + 0 2 ) 2 y 3 0 0 Γ 2 11 1 [ 0 + 0 0) 0 + 0 + 0 2 ) 2 y 3 y 2 1 y Γ 1 12 1 [ 0 + 2 ) 2 y 3 0 y 2 + 0 + 0 0) 0 1 y Γ 2 12 1 [ 0 + 2 ) 2 y 3 0 0 + 0 + 0 0) y 2 0 Γ 1 22 1 [ 2 0 + 0 0) y 2 + 2 y 3 + 2 y 3 2 ) y 2 0 0 Γ 2 22 1 [ 2 0 + 0 0) 0 + 2 y 3 + 2 y 3 2 ) y 2 0 1 y b): Let v 0 0, 1) be a tangent vector at point 1, 0) of R 2+ v 0 is a unit vector on the y-axis with origin at 0, 1)). Let vt) be the parallel transport v 0 along the curve x t, y 1. Show that vt) makes an angle t with the direction of the y-axis, measured in the clockwise sense.

12 CLAY SHONKWILER Proof. Since vt) v 1 t), v 2 t)) is a parallel vector field, it satisfies the equations 0 dvk + i,j Γ k ijv j dx i for k 1, 2. Using the values for Γ k ij these equations reduce to { da computed in part a) above, + Γ1 12 b 0 db + Γ2 11 a 0 Hence, solutions of this system of linear first-order ODEs are of the form a cos θt), b sin θt). Along the given curve, y 1, so Γ 1 12 1 and Γ2 11 1, so 0 da b sin θt)dθ sin θt) 0 db + a cos θt)dθ + cos θt) meaning that dθ 1. Hence, θt) C t for some constant C. Now, since v0) v 0 0, 1), so 0 cosc 0) cos C, so C π 2. Therefore, t π 2 θt); since θ measures the angle vt) makes with the direction of the x-axis measured counter-clockwise, this means that t measures the angle vt) makes with the direction of the y-axis measured in the clockwise sense. DRL 3E3A, University of Pennsylvania E-mail address: shonkwil@math.upenn.edu