LectureNotes14G

8/8/2019 LectureNotes14G

1/12

Math 6455 Nov 1, 20061

Differential Geometry I

Fall 2006, Georgia Tech

Lecture Notes 14

Connections

Suppose that we have a vector field X on a Riemannian manifold M. How can wemeasure how much X is changing at a point p M in the direction Yp TpM?The main problem here is that there exists no canonical way to compare a vectorin some tangent space of a manifold to a vector in another tangent space. Hencewe need to impose a new kind of structure on a manifold. To gain some insight, we

first study the case where M = Rn.

0.1 Differentiation of vector fields in Rn

Since each tangent space TpRn is canonically isomorphic to Rn, any vector field on

Rn may be identified as a mapping X: Rn Rn. Then for any Yp TpRn wedefine the covariant derivative of X with respect to Yp as

YpX :=

Yp(X1), . . . , Y p(X

n)

.

Recall that Yp(Xi) is the directional derivative of Xi at p in the direction of Y, i.e.,

if : (

, )

M is any smooth curve with (0) = p and (0) = Y, then

Yp(Xi) = (Xi )(0) = grad Xi(p), Y.

The last equality is an easy consequence of the chain rule. Now suppose that

Y : Rn Rn is a vector field on Rn, p Y Yp, then we may define a new vec-tor field on Rn by

(YX)p := YpX.

Then the operation (X, Y) XY may be thought of as a mapping : X(Rn)

X(Rn) X(Rn), where X denotes the space of vector fields on Rn.Next note that if X X(Rn) is any vector field and f: M R is a function,

then we may define a new vector field f X

(Rn) by setting (f X)p := f(p)Xp (do

not confuse f X, which is a vector field, with Xf which is a function defined byXf(p) := Xp(f)). Now we observe that the covariant differentiation of vector fieldson Rn satisfies the following properties:

1. Y(X1 + X2) = YX1 + YX21Last revised: November 14, 2006

1


2/12

2. Y(f X) = (Y f)YX+ fYX3. Y1+Y2X = Y1X+ Y2X4. fYX = fYXIt is an easy exercise to check the above properties. Another good exercise to

write down the pointwise versions of the above expressions. For instance note thatitem (2) implies that

Yp(f X) = (Ypf)YpX+ f(p)YpX,

for all p M.

0.2 Definition of connection and Christoffel symbols

Motivated by the Euclidean case, we define a connection on a manifold M as anymapping

: X(M) X(M) X(M)which satisfies the four properties mentioned above. We say that is smooth ifwhenever X and Y are smooth vector fields on M, then YX is a smooth vectorfield as well. Note that any manifold admits the trivial connection 0. In thenext sections we study some nontrivial examples.

Here we describe how to express a connection in local charts. Let Ei be a basisfor the tangent space of M in a neighborhood of a point p. For instance, choose alocal chart (U, ) centered at p and set Ei(q) := d

1(q)(ei) for all q U . Then ifX

and Y are any vector fields on M, we may write X = i XiEi, and Y = i YiEion U. Consequently, if is a connection on M we haveYX = Y

i

XiEi

=i

Y(Xi)Ei + X

iYEi

.

Now note that since (Ej Ei)p TpM, for all p U, then it is a linear combinationof the basis elements of TpM. So we may write

EjEi =k

kjiEk

for some functions kji

on U which are known as the Christoffel symbols. Thus

YX =i

Y(Xi)Ei + X

ij

Yjk

kjiEk

=k

Y(Xk) +

ij

YiXjkij

Ek

2


3/12

Conversely note that, a choice of the functions kij on any local neighborhood of Mdefines a connection on that neighborhood by the above expression. Thus we maydefine a connection on any manifold, by an arbitrary choice of Christoffel symbols

in each local chart of some atlas of M and then using a partition of unity.Next note that for every p U we have:

(YX)p =k

Yp(X

k) +ij

Yi(p)Xj(p)kij(p)

Ek(p). (1)

This immediately shows that

Theorem 0.1. For any point p M, (YX)p depends only on the value of Xat p and the restriction of Y to any curve : (, ) M which belongs to theequivalence class of curves determined by Xp.

Thus if p M, Yp TpM and X is any vector field which is defined on an openneighborhood of p, then we may defineYpX := (YX)p

where Y is any extension of Yp to a vector field in a neighborhood of p. Note thatsuch an extension may always be found: for instance, if Yp =

YipEi(p), where

Ei are some local basis for tangent spaces in a neighborhood U of p, then we mayset Yq :=

YipEi(q) for all q U. By the previous proposition, (YX)p does not

depend on the choice of the local extension Y, so YpX is well defined.

0.3 Induced connection on submanifolds

As we have already seen M admits a standard connection when M = Rn. To giveother examples of manifolds with a distinguished connection, we use the followingobservation.

Lemma 0.2. LetM be a manifold, M be an embedded submanifold of M, andX bea vector field of M. Then for every point p M there exists an open neighborhoodU of p in M and a vector filed X defined on U such that Xp = Xp for all p M.Proof. Recall that, by the rank theorem, there exists a local chart (U, ) of Mcentered at p such that (U M) = Rnk where k = dim(M) dim(M). Now,note that d(X) is a vector field on Rnk and let Y be an extension of d(X) toRn (any vector field on a subspace of Rn may be extended to all of Rn). Then set

X := d1(Y).

Now if M is a Riemannian manifold with connection , and M is any subman-ifold of M, we may define a connection on M as follows. First note that for anyp M,

TpM = TpM (TpM),

3


4/12

that is any vector X TpM may written as sum of a vector X TpM (which istangent to M and vector X := X X (which is normal to M). So for any vectorfields X and Y on M we define a new vector field on M by setting, for each p

M,

(YX)p := (YX)pwhere Y and X are local extensions ofX and Y to vector fields on a neighborhoodof p in M. Note (YX)p is well-defined, because it is independent of the choice oflocal extensions X and Y by Theorem 0.1.

0.4 Covariant derivative

We now describe how to differentiate a vector field along a curve in a manifold Mwith a connection . Let : I M be a smooth immersion, i.e., dt = 0 for allt

I, where I

R is an open interval. By a vector filed along we mean a mapping

X: I T M such that X(t) T(t)M for all t I. Let X() denote the space ofvector fields along .

For any vector field X X(), we define another vector field DX X(),called the covariant derivative of X along , as follows. First recall that is locallyone-to-one by the inverse function theorem. Thus, by the previous lemma on theexistence of local extensions of vector fields on embedded submanifolds, there existsan open neighborhood U of (t0) and a vector field X defined on U such thatX(t) = X(t) for all t (t0 , t0 + ). Set

DX(t0) := (t0)X.Recall that (t0) := dt0(1)

T(t

0

)M. By Theorem 0.1, DX(t0) is well defined,

i.e., it does not depend on the choice of the local extension X. Thus we obtain amapping D: X() X(). Note that if X, Y X(), then (X+ Y)(t) := X(t) +Y(t) X(). Further, if f: I R is any function then (f X)(t) := f(t)X(t) X(). It is easy to check that

D(X+ Y) = D(X) + D(Y) and D(f X) = f D(X).

Proposition 0.3. If : I Rn, and X X(), then DX = X. In particular,D = .

Proof. Let X be a vector field on an open neighborhood of (t0) such that

X((t)) = X(t),

for all t (t0 , t0 + ). ThenDX(t0) = (t0)X = (X )(t0) = X(t0).

4


5/12

Corollary 0.4. Let M be an immersed submanifold of Rn with the induced con-nection , and corresponding covariant derivative D. Suppose : I M is animmersed curve, and X

XM() is a vector field along in M. Then DX =

(X).

0.5 Geodesics

Note that, by the last exercise, the only curves : I Rn with the property that

D 0

are given by (t) = at + b, which trace straight lines. With this motivation, wedefine a geodesic (which is meant to be a generalization of the concept of lines) asan immersed curve : I M which satisfies the above equality for all t I. A nicesupply of examples of geodesics are provided by the following observation:

Proposition 0.5. Let M Rn be an immersed submanifold, and : I M animmersed curve. Then is a geodesic of M (with respect to the induced connectionfrom Rn) if and only if 0. In particular, if : I M is a geodesic, then = const.Proof. The first claim is an immediate consequence of the last two results. The lastsentence follows from the leibnitz rule for differentiating inner products in Euclideanspace: , = 2, . Thus if 0, then 2 = const.

As an application of the last result, we can show that the geodesics on the sphereS2 are those curves which trace a great circle with constant speed:

Example 0.6 (Geodesics on S2). A C2 immersion : I S2 is a geodesic if andonly if has constant speed and lies on a plane which passes through the center ofthe sphere, i.e., it traces a segment of a great circle.

First suppose that : I S2 has constant speed, i.e. = const., and that traces a part of a great circe, i.e., , u = 0 for some fixed vector u S2 (whichis the vector orthogonal to the plane in which lies). Since , = 2 isconstant, it follows from the Leibnitz rule for differentiating the innerproduct that, = 0. Furthermore, differentiating , u = 0 yields that , u = 0. So, lies in the plane of , and is orthogonal to . So, since traces a circle, must beparallel to . This in turn implies that must be orthogonal to TS

2, since isorthogonal to TS

2. So we conclude that () = 0.Conversely, suppose that () = 0. Then is parallel to . So ifu := ,

then u = + = 0+ 0 = 0. So u is constant. But is orthogonal to u, so lies in the plane which passes through the origin and is orthogonal to u. Finally, has constant speed by the last proposition.

5


6/12

0.6 Ordinary differential equations

In order to prove an existence and uniqueness result for geodesic in the next section

we need to develop first a basic result about differential equations:

Theorem 0.7. Let U Rn be an open set and F: U Rn be C1, then for everyx0 U, there exists an > 0 such that for every 0 < < there exists a uniquecurve x : (, ) U with x(0) = x0 and x(t) = F(x(t)).

Note that, from the geometric point of view the above theorem states that therepasses an integral curve through every point of a vector field. To prove this resultwe need a number of preliminary results. Let I R be an interval, (X, d) be acompact metric space, and (I, X) be the space of maps : I X. For every pairof curves 1, (I, X) set

(1, 2) := suptI d1(t), 2(t).It is easy to check that (, ) is a metric space. Now let C(I, X) (I, X) be thesubspace of consisting of continuous curves.

Lemma 0.8. (C, ) is a complete metric space.

Proof. Let i C be a Cauchy sequence. Then, for every t I, i(t) is a Cauchysequence in X. So i(t) converges to a point (t) X (since every compact metricspace is complete). Thus we obtain a mapping : I X. We claim that iscontinuous which would complete the proof. By the triangular inequality,

d((s), (t)) d((s), i(s)) + d(i(s), i(t)) + d(i(t), (t)) 2(, i) + d(i(s), i(t)).

So, since i is continuous,

limts

d((s), (t)) 2(, i).

All we need then is to check that lim i (, i) = 0: Given > 0, choose isufficiently large so that (i, j) < for all j i. Then, for all t I, d(i(t), j(t)) , which in turn yields that d(i(t), (t)) . So (i, ) .

Now we are ready to prove the main result of this section:

Proof of Theorem 0.7. Let B = Bnr (x0) denote a ball of radius r centered at x0.Choose r > 0 so small that that B U. For any continuous curve C((, ), B)we may define another continuous curve s() ((, ), Rn) by

s()(t) := x0 +

t0

F(a(u))du.

6


7/12

We claim that if is small enough, then s() C((, ), B). To see this note that

s()(t) x0 = t

0 F((u))du t

0 F((u))du supB F.So setting r/ supB F, we may then assume that

s : C((, ), B) C((, ), B).

Next note that for every , C((, ), B) , we have

(s(), s()) = supt

t0

F((u)) F((u))du sup

t

t0

F((u)) F((u))du.

Further recall that, since F is C1, by the mean value theorem there is a constant K

such that F(x) F(y) Kx y,for all x, y B (in particular recall that we may set K := n supB |DjFi|). Thust

0F((u)) F((u))du K

t0

(u) (u)du K(, ).

So we conclude that(s(), s()) K(, ).

Now assume that < 1/K(in addition to the earlier assumption that r/ supB F),then, s must have a unique fixed point since it is a contraction mapping. So for

every 0 < < where

:= min

r

supB F,

1n supB |DjFi|

there exists a unique curve x : (, ) B such that x(0) = s(x)(0) = x0, andx(t) = s(x)(t) = F(x(t)).

It only remains to show that x : (, ) U is also the unique curve withx(0) = x0 and x(t) = F(x(t)), i.e., we have to show that if y : (, ) U is anycurve with y(0) = x0 and y(t) = F(y(t)), then y = x (so far we have proved thisonly for y : (, ) B). To see this recall that r/ supB F where r is theradius of B. Thus

y(t) x0 t0

y(u)du =t0

F(y(u))du supB

F r.

So the image of y lies in B, and therefore we must have y = x.

7


8/12

0.7 Existence and uniqueness of geodesics

Note that for every point p Rn and and vector X TpRn Rn, we may finda geodesic through p and with velocity vector X at p, which is given simply by(t) = p + Xt. Here we show that all manifolds with a connection share thisproperty:

Theorem 0.9. LetM be a manifold with a connection. Then for every p M andX TpM there exists an > 0 such that for every 0 < < there is a uniquegeodesic : (, ) M with (0) = p and (0) = X.

To prove this theorem, we need to record some preliminary observations. LetM and M be manifolds with connections and respectively. We say that adiffeomorphism f: M M is connection preserving provided that

(

YX)p = df(Y)df(X)f(p)for all p M and all vector fields X, Y X(M). It is an immediate consequenceof the definitions that

Lemma 0.10. Let f: M M be a connection preserving diffeomorphism. Then: I M is a geodesic if and only of f is a geodesic.

Note that if f: M M is a diffeomorphism, and M has a connection , thenf induces a connection on M by

(eY X)ep := df1( eX)df1(Y)f1(ep).It is clear that then f: M M will be connection preserving. So we may concludethat

Lemma 0.11. Let(U, ) be a local chart of M, then : I U is a geodesic if andonly of is a geodesic with respect to the connection induced on Rn by .

Now we are ready to prove the main result of this section:

Proof of Theorem 0.9. Let (U, ) be a local chart of M centered at p and let bethe connection which is induced on (U) = Rn by . We will show that there existsan > 0 such that for every 0 < < there is a unique geodesic c : (, ) Rn,with respect to the induced connection, which satisfies the initial conditions

c(0) = (p) and c(0) = dp(X).

Then, by a previous lemma, := 1 c : (, ) M will be a geodesic on Mwith (0) = p and (0) = X. Furthermore, will be unique. To see this supposethat : (, ) M is another geodesic with (0) = p and (0) = X. Let be the supremum of t [0, ] such that (t, t) U, and set c := |(,).

8


9/12

Then, by Theorem 0.7, c = c on (, ), because < . So it follows that = on(, ), and we are done if (, ) = (, ). This is indeed the case, for otherwise,(

, + )

(

, ), for some > 0. Further (

) = (

)

U. So if is

sufficiently small, then ( , + ) U, which contradicts the definition of .So all we need is to establish the existence and uniqueness of the geodesic

c : (, ) Rn mentioned above. For c to be a geodesic we must haveDcc

0.We will show that this may be written as a system of ordinary differential equations.To see this first recall that

Dcc(t) = c(t)cwhere c is a vector filed in a neighborhood of c(t) which is a local extension of c,i.e.,

c(c(t)) = c(t).

By (1) we have

c(t)c =k

c(t)(c

k) +

ij

ci(t)cj(t)kij(c(t))

ek,

where ei are the standard basis of Rn and kij(p) = (eiej)p, ek. But

c(t)(ck

) = (ck c)(t) = (ck)(t) = ck(t).

So Dcc 0 if and only ifck(t) +

ij

ci(t)cj(t)kij(c(t)) = 0

for all t I and all k. This is a system of n second order ordinary differentialequations (ODEs), which we may rewrite as a system of 2n first order ODEs, viasubstitution c = v. Then we have

ck(t) = vk(t)

vk(t) = ij

vi(t)vj(t)kij(c(t)).

Now let (t) := (c(t), v(t)), and define F: R2n R2n, F = (F1, . . . , F 2n) byF(x, y) = y, and F

+n(x, y) = ij

yiyjij(x)

for = 1, . . . , n. Then the system of 2n ODEs mentioned above may be rewrittenas

(t) = F((t)),

which has a unique solution with initial conditions (0) = ((p), d(X)).

9


10/12

0.8 Parallel translation

Let M be a manifold with a connection, and : I M be an immersed curve.Then we say that a vector field X X() is parallel along if

DX 0.

Thus, in this terminology, is a geodesic if its velocity vector field is parallel. Furthernote that if M is a submanifold of Rn, the, by the earlier results in this section, Xis parallel along if and only (X) 0.Example 0.12. Let M be a two dimensional manifold immersed in Rn, : I Mbe a geodesic of M, and X XM() be a vector field along in M. Then X isparallel along if and only if X has constant length and the angle between X(t)and (t) is constant as well. To see this note that () 0 since is a geodesic;therefore, X, = X, + X, = X, .So, if (X) = 0, then it follows that X, is constant which since and Xhave both constant lengths, implies that the angle between X and is constant.Conversely, suppose that X has constant length and makes a constant angle with .Then X, is constant, and the displayed expression above implies that X, = 0is constant. Furthermore, 0 = X, X = 2X, X. So X(t) is orthogonal to bothX(t) and (t). IfX(t) and (t) are linearly dependent, then this implies that X(t)is orthogonal to T(t)M, i.e., (X

) 0. If X(t) and (t) are linearly dependent,then (X) = D(X) = D(f ) = f D() 0.

Lemma 0.13. Let I R and U Rn

be open subsets and F: I U Rn

, beC1. Then for every t0 I and x0 U there exists an > such that for every0 < < there is a unique curve x : (t0 , t0 + ) Rn with x(t0) = x0 andx(t) = F(t, x(t)).

Proof. Define F: I U Rn+1 by F(t, x) := (1, F(t, x)). Then, by Theorem 0.7,there exists an > 0 and a unique curve x : (t0, t0+) Rn+1, for every 0 < < ,such that x(t0) = (1, x0) and x(t) = F(x(t)). It follows then that x(t) = (t, x(t)),for some unique curve x : (t0 , t0 + ) Rn. Thus F(x(t)) = (1, F(t, x(t))), andit follows that x(t) = F(t, x(t)).

Lemma 0.14. LetA(t), t I, be a C1 one-parameter family of matrices. Then forevery x0 R

n

and t0 I, there exists a unique curve x : I Rn

with x(t0) = x0such that x(t) = A(t) x(t).Proof. Define F: I Rn Rn by Ft(x) = A(t) x. By the previous lemma,there exists a unique curve x : (t0 , t0 + ) Rn with x(t0) = x0 such thatFt(x(t)) = x(t) for all t (t0 , t0 + ).

10


11/12

Now let J I be the union of all open intervals in I which contains t0 and suchthat x(t) = F(x(t)) for all t in those intervals. Then J is open in I and nonempty.All we need then is to show that J is closed, for then it would follow that J = I.

Suppose that t is a limit point of J in I. Just as we argued in the first paragraph,there exists a curve y : (t , t + ) Rn such that y(t) = F(y(t)) and y(t) = 0.Thus we may assume that y = 0 on (t , t + ), after replacing by a smallernumber. In particular y(t) = 0 for some t (t , t + ) J, and there exists amatrix B such that B y(t) = x(t).

Now let y(t) := B y(t). Since F(y(t)) = y(t), we have F(y(t)) = y(t). Further,by construction y(t) = x(t), so by uniqueness part of the previous result we musthave y = x on (t , t + ) J. Thus x is defined on J (t , t + ). But J wasassumed to be maximal. So (t , t + ) J. In particular t J, which completesthe proof that J is closed in I.

Theorem 0.15. Let X: I M be a C1

immersion. For every t0 I and X0 T(t0)M, there exists a unique parallel vector field X X() such that X(t0) = X0.Proof. First suppose that there exists a local chart (U, ) such that : I U is anembedding. Let X be a vector field on U and set X(t) := X((t)). By (1),

D(X)(t) = (t)X =k

(t)(X

k) +

ij

i(t)Xj(t)kij((t))

Ek((t)).

Further note that(t)X = (X )(t) = X(t).

So, in order for X to be parallel along we need to have

Xk +ij

i(t)kij((t))Xj(t) = 0,

for k = 1, . . . , n. This is a linear system of ODEs in terms of Xi, and therefore bythe previous lemma it has a unique solution on I satisfying the initial conditionsXi(t0) = X

i0.

Now let J I be a compact interval which contains t0. There exists a finitenumber of local charts ofM which cover (J). Consequently there exist subintervalsJ1, . . . , J n of J such that embeds each Ji into a local chart of M. Suppose thatt0 J, then, by the previous paragraph, we may extend X0 to a parallel vectorfield defined on J. Take an element of this extension which lies in a subinterval J

intersecting J and apply the previous paragraph to J . Repeating this procedure,we obtain a parallel vector field on each Ji. By the uniqueness of each local extensionmentioned above, these vector fields coincide on the overlaps of Ji. Thus we obtaina well-defined vector filed X on J which is a parallel extension of X0. Note thatif J is any other compact subinterval of I which contains t0, and X is the parallelextension of X0 on J, then X and X coincide on J J, by the uniqueness of local

11


12/12

parallel extensions. Thus, since each point ofI is contained in a compact subintervalcontaining t0, we may consistently define X on all of I.

Finally let X be another parallel extension ofX0 defined on I. Let A

I be the

set of points where X = X. Then A is closed, by continuity of X and X. FurtherA is open by the uniqueness of local extensions. Furthermore, A is nonempty sincet0 A. So A = I and we conclude that X is unique.

Using the previous result we now define, for every X0 T(t0)M,

P,t0,t(X0) := X(t)

as the parallel transport of X0 along to T(t)M. Thus we obtain a mapping fromT(t0)M to T(t)M.

Exercise 0.16. Show that P,t0,t : T(t0)M T(t)M is an isomorphism (Hint: Usethe fact hat D: X() X() is linear). Also show that P,t0,t depends on thechoice of .

Exercise 0.17. Show that

(t0)X = limtt0X(t0) P1,t0,t(X(t))

t.

(Hint: Use a parallel frame along .)

12

Documents

LectureNotes14G