Math 6455 Oct 10, Differential Geometry I Fall 2006, Georgia Tech

Transcription

1 Mth 6455 Oct 10, Differentil Geometry I Fll 2006, Georgi Tech Lecture Notes 12 Riemnnin Metrics 0.1 Definition If M is smooth mnifold then by Riemnnin metric g on M we men smooth ssignment of n innerproduct to ech tngent spce of M. This mens tht, for ech p M, g p : T p M T p M R is symmetric, positive definite, biliner mp, nd furthermore the ssignment p g p is smooth, i.e., for ny smooth vector fields X nd Y on M, p g p (X p, Y p ) is smooth function. The pir (M, g) then will be clled Riemnnin mnifold. We sy tht diffeomorphism f : M N between pir of Riemnnin mnifolds (M, g) nd (N, h) is n isometry provided tht for ll p M nd X, Y T p M. g p (X, Y ) = h f(p) (df p (X), df p (Y )) Exercise Show tht the ntipodl reflection : S n S n, (x) := x is n isometry. 0.2 Exmples The Eucliden Spce The simplest exmple of Riemnnin mnifold is R n with its stndrd Eucliden innerproduct, g(x, Y ) := X, Y Submnifolds of Riemnnin mnifold A rich source of exmples re generted by immersions f : N M of ny mnifold N into Riemnnin mnifold M (with metric g); for this induces metric h on N given by h p (X, Y ) := g f(p) (df p (X), df p (Y )). In prticulr ny mnifold my be equipped with Riemnnin metric since every mnifold dmits n embedding into R n. 1 Lst revised: November 23,

2 0.2.3 Quotient of Riemnnin mnifold by group of isometries Note tht the set of isometries f : M M forms group. Another source of exmples of Riemnnin mnifolds re generted by tking the quotient of Riemnnin mnifold (M, g) by subgroup G of its isometries which cts properly discontinuously on M. Recll tht if G cts properly discontinuously, then M/G is indeed mnifold. Then we my define metric h on M/G by setting h [p] := g p. More precisely recll tht the projections π : M M/G, given by π(p) := [p] is locl diffeomorphism, i.e., for ny q [p] there exists n open neighborhood U of p in M nd n open neighborhood V of [p] in M/G such tht π : U V is diffeomorphism. Then we my define h [p] (X, Y ) := g q ((dπ q ) 1 (X), (dπ q ) 1 (Y )). One cn immeditely check tht h does not depend on the choice of q [p] nd is thus well defined.a specific exmple of proper discontinuous ction of isometries is given by trnsltions f z : R n R n given by f z (p) := p + z where z Z n. Recll tht R n /Z n is the torus T n, which my now be equipped with the metric induced by this group ction. Similrly RP n dmits cnonicl metric, since RP n = S n /{±1}, nd reflections of sphere re isometries Conforml trnsformtions As nother set of exmples note tht if (M, g) is ny Riemnnin mnifold, then (M, λg) is lso Riemnnin mnifold where λ: M R + is ny smooth positive function. Note tht this chnge of metric does not effect the ngles between ny pir of vectors in tngent spce of M. Thus (M, λg) is sid to be conforml to (M, g). Exercise Show tht the inversion i: R n {o} R n given by i(x) := x/ x is conforml trnsformtion. Exercise Show tht the stereogrphic projection π : S 2 {(0, 0, 1)} R 2 is conforml trnsformtion The hyperbolic spce Finlly, n importnt exmple is the hyperbolic spce which my be represented by number of models. One model, known s Poincre s hlf spce model, is to tke the open upper hlf spce of R n nd define there metric vi g p (X, Y ) := X, Y (p n ) 2, 2

3 where p n denotes the n th coordinte of p. Another description of the hyperbolic spce my be given by tking the open unit bll if R n nd defining This is known s Poincre s bll model. g p (X, Y ) := X, Y (1 p 2 ) 2. Exercise Show tht the the Poincre hlf-plne nd the hlf-disk re isometric (Hint: identify the Poincre hlf-plne with the region y > 1 in R 2 nd do n inversion). 0.3 Metric in locl coordintes Let (U, φ) be locl chrt for (M, g). Then, recll tht if e 1,... e n denote the stndrd bsis of R n, we obtin bsis for ech T p M, for p U by setting E i (p) := dφ 1 φ(p) (e i). Now if X, Y T p M, then X = n i=1 Xi E i nd Y = n i=1 Y i E i. Further, if we set g ij (p) := g p (E i, E j ), then, since g is biliner we hve g p (X, Y ) = X i Y j g p (E i, E j ) = X i Y j g ij (p). Thus in ny locl coordinte (U, φ) metric is completely determined by the functions g ij which my be regrded s the coefficients of positive definite mtrix. To obtin concrete exmple, note tht if M R n is submnifold, with the induce metric from R n, nd (φ, U) is locl chrt of M, then if we set f := φ 1, f : φ(u) R n is prmetriztion for U, nd d(f)(e i ) = D i f. Consequently, g ij (p) = D i f(f 1 (p)), D j f(f 1 (p)). For instnce, note tht surfce of revolution in R 3 which is given by rotting the curve (r(t), z(t)) in the xz-plne bout the z xis cn be prmetrized by So f(t, θ) = (r(t) cos θ, r(t) sin θ, z(t)). D 1 f(t, θ) = (r (t) cos θ, r (t) sin θ, z (t)) nd D 2 f(t, θ) = ( r(t) sin θ, r(t) cos θ, 0), nd consequently g ij (f(t, θ)) is given by ( (r ) 2 + (z ) r 2 Note tht if we ssume tht the curve in the xz-plne is prmetrized by rclength, then (r ) 2 + (z ) 2 = 1, so the bove mtrix becomes more simple to work with. 3 ).

4 Exercise Compute the metric of S 2 in terms of sphericl coordintes θ nd φ. Exercise Compute the metric of the surfce given by the grph of function f : Ω R 2 R. 0.4 Length of Curves In Riemnnin mnifold (M, g), the length of ny piecewise smooth curves c: [, b] M with c() = p nd c(b) = q is defined s where Length[c] := g c(t) (c (t), c (t)) dt, c (t) := dc t (1). Note tht the definition for the length of curves here is generliztion of the Eucliden cse where we integrte the speed of the curve. Indeed the lst formul bove coincides with the regulr notion of derivtive when M is just R n. To see this, recll tht dc t (1) = (c γ) (0) where γ : (ɛ, ɛ) [, b] is curve with γ(0) = t nd γ (0) = 1, e.g., γ(u) = t+u. Thus by the chin rule (c γ) (0) = c (γ(0))γ (0) = c (t). Exercise Compute the length of the rdius of the Poincre-disk (with respect to the Poincre metric). 0.5 The clssicl nottion for metric For ny curve c: [, b] R n we my write c(t) = (x 1 (t),..., x n (t)). Consequently, if we define g ij (p) := g p (e i, e j ) where e 1,..., e n is the stndrd bsis for R n, then bilinerity of g yields tht g c(t) (c (t), c (t)) = g c(t) (e i, e j )x i(t)x j(t) = g ij (c(t))x i(t)x j(t). Thus we my write Length[c] := g ij (c(t)) dx i dt dx j dt dt. Indeed clssiclly metrics were specified by n expression of the form ds 2 = g ij dx i dx j. 4

5 nd then length of curve ws defined s the integrl of ds, which ws clled the element of rclength, long tht curve: Length[c] = ds. In prticulr note tht, in the clssicl nottion, the stndrd Eucliden metric in the plne is given by ds 2 = n i=1 dx2 i. Further, in the Poincre s hlf-disk model, ds 2 = n i=1 dx2 i /x2 n. 0.6 Distnce For ny pirs of points p, q M, let C(p, q) denote the spce piecewise smooth curves c: [, b] M with c() = p nd c(b) = q. Then, if M is connected, we my define the distnce between p nd q s d g (p, q) := inf{length[c] c C(p, q)}. So the distnce between pir of points is defined s the gretest lower bound of the lengths of curves which connect those points. First we show tht this is generliztion of the stndrd notion of distnce in R n. Lemm For ll continuous mps f : (, b) R n f(t)dt f(t) dt. Proof. By the Cuchy-Schwrts inequlity, for ny unit vector u S n 1, f(t)dt, u = f(t), u dt f(t) dt. In prticulr we my let u := f(t)dt/ f(t)dt, ssuming tht f(t)dt 0 (otherwise the lemm is obviously true). Corollry If (M, g) = (R n, ) then d g (p, q) = p q. Proof. First note tht if we set c(t) := (1 t)p + tq, then Length[c] := 1 0 c p q dt = p q. So d g (p, q) p q. It remins then to show tht d g (p, q) p q. The lter inequlity holds becuse for ll curves c: [, b] R n c b (t) dt c (t)dt = c(b) c(). 5

6 The previous result shows tht (M, d g ) is metric spce when M is the Eucliden spce R n nd g, which induces d, is the stndrd innerproduct. Next we show tht this is the cse for ll Riemnnin mnifolds. To this end we first need locl lemm: Lemm Let (B, g) be Riemnnin mnifold, where B := B n r (o) R n. Then there exists m > 0 such tht for ny piecewise C 1 curve c: [, b] B with c() = o nd c(b) B we hve Length[c] > m. Proof. Define f : S n 1 B R by f(u, p) := g p (u, u). Note tht, since g is positive definite, f > 0. Thus since f is continuous nd S n 1 B is compct f λ 2 > 0. Consequently, bilinerity of g yields tht g p (v, v) λ 2 v 2. The bove inequlity is obvious when v = 0, nd when v 0, observe tht g p (v, v) = g p (v/ v, v/ v ) v 2. Next note tht Length[c] = g c(t) (c (t), c (t)) dt λ c (t) dt. But c (t) is just the length of c with respect to the stndrd metric on R n. Thus, by the previous proposition, So setting m := λr finishes the proof. c (t) dt c(b) c() = r. The proof of the next observtion is immedite: Lemm If f : M N is n isometry, then Length[c] = Length[f c] for ny piecewise C 1 curve c: [, b] M. Note tht if (M, g) is Riemnnin mnifold nd f : M N is diffeomorphism between M nd ny smooth mnifold N, then we my push forwrd the metric of M by defining df(g) p (X, Y ) := g f 1 (p)(df 1 (X), df 1 (Y )). Then f is n isometry between (M, g) nd (N, df(g)). In prticulr we my ssume tht ny locl chrts (U, φ) on Riemnnin mnifold (M, g) is n isometry, with respect to the push forwrd metric dφ(g) on φ(u). This observtion, together with the previous lemm esily yields tht: Proposition If (M, g) is ny Riemnnin mnifold then (M, d g ) is metric spce. 6

7 Proof. It is immedite tht d is symmetric nd stisfies the tringle inequlity. Furthermore it is cler tht d is lwys nonnegtive. Showing tht d is positive definite, however, requires more work. Specificlly, we need to show tht when p q, then d(p, q) > 0. Suppose p q. Then, since M is Husdorff, there exists n open neighborhood V of p such tht q V. Let (U, φ) be locl chrt centered t p. Choose r so smll tht B r (o) φ(v U), nd set W := φ 1 (B r (o)). Then φ: W B r (o) is diffeomorphism, nd we my equip B r (o) with the push forwrd metric dφ(g) which will turn φ into n isometry. Now let c: [, b] M be ny piecewise C 1 curve with c() = p nd c(b) = q. Then there exist b b such tht c[, b ] W nd c(b ) W (to find b let W := φ 1 (B r (o)) be the interior of W, then c 1 ( W ) is n open subset of [, b] which contins, nd we my let b be the upperbound of the component of c 1 ( W ) which contins.) Let c: [, b ] W be the restriction of c. Then obviously Length[c] Length[c]. But Length[c] = Length[φ c] since φ is n isometry, nd by the previous lemm then length of ny curve in (B n r (o), dφ(g)) which begins t the center of the bll nd ends t its boundry is bounded below by positive constnt. Now recll tht ny metric spce hs nturl topology. In prticulr (M, d g ) is topologicl spce. Next we show tht this topologicl spce is identicl to the originl M. Lemm Let (M, g 1 ), (M, g 2 ) be Riemnnin mnifolds, nd suppose M is compct. Then there exist constnt λ > 0 such tht for ny p, q M we hve d g 1(p, q) λ d g 2(p, q). Proof. Define f : S n 1 M R by f(u, p) := g 1 p(u, u)/g 2 p(u, u). Note tht, since g is positive definite, f > 0. Thus since f is continuous nd S n 1 M is compct f λ 2 > 0. Consequently, bilinerity of g yields tht g 1 p(v, v) λ 2 g 2 p(v, v), for ll v R n. Next note tht the bove inequlity yields Length g1 [c] = g 1 c(t) (c (t), c (t)) dt λ g 2 c(t) (c (t), c (t)) dt = λ Length g2 [c]. for ny curve c: [, b] M. In prticulr the bove inequlities hold for ll curves c: [, b] M with c() = p nd c(b) = q. Proposition The metric spce (M, d g ), endowed with its metric topology, is homeomorphic to M with its stndrd topology. 7

8 Proof. There re two prts to this rgument: Prt I: We hve to show tht every open neighborhood U of M is open in its metric topology, i.e., for every p U there exists n r > 0 such tht B g r (p) U, where B g r (p) := {q M d g (p, q) < r}. To see this first note tht, s we showed in the proof of the previous proposition, there exists n open neighborhood V of p with V U such tht there exists homeomorphism φ: V B n 1 (o). Now, much s in the proof of the previous proposition, if we endow B n 1 (o) with the push forwrd metric induced by φ then (B n 1 (o), dφ(g)) becomes isometric to (V, g). But recll tht, s we showed in the erlier proposition, the distnce of ny point in the boundry B n 1 (o) = Sn of B n 1 (o) from the origin o ws bigger thn some constnt, sy λ. Thus the sme is true of the distnce of V from p. In prticulr, if we choose r < λ, then B g r (p) V U. Prt II: We hve to show tht every metric bll B g r (p) is open in M, i.e., t every q B g r (p) we cn find open neighborhood U of q in M such tht U B g r (p). To see this let V be n open neighborhood of p such tht there exists homeomorphism ψ : V B n 1 (o), nd endow B n 1 (o) with the push forwrd metric dψ(g). Then the distnce of ψ(v B g r (p)) from o is equl to r, with respect to the metric dψ(g). So, by the previous proposition, this distnce, with respect to the Eucliden metric on B n 1 (o) must be t lest λr > 0. Thus if we choose r < λ r, then the Eucliden bll B n r (o) ψ(v ). Consequently, U := ψ 1 (B n r (o)) V, nd U is open in M, since B n r (o) is open. 8