Connections Chapter5

7/25/2019 Connections Chapter5

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Chapter 5

Curvature on Bundles

Contents5.1 Flat sections and connections . . . . . . . . . . 115

5.2 Integrability and the Frobenius theorem . . . 117

5.3 Curvature on a vector bundle . . . . . . . . . . 122

5.1 Flat sections and connections

A connection on a fiber bundle : E Mallows one to define when asection is constant along smooth paths (t) M; we call such sectionshorizontal, or in the case of a vector bundle, parallel. Since by defini-tion horizontal sections always exist along any given path, the nontrivialimplications of the following question may not be immediately obvious:

Given p M and a sufficiently small neighborhoodp U M, doesEadmit any sections (E|U) for whichs 0?

A section whose covariant derivative vanishes identically is called a flat orcovariantly constant section. It may seem counterintuitive that the an-

swer could possibly be noafter all, one of the most obvious facts aboutsmooth functions is that constant functions exist. This translates easilyinto a statement about sections of trivial bundles. Of course, all bun-dles are locally trivial, thus locally one can always find sections that areconstant with respect to a trivialization. These sections are covariantlyconstant with respect to a connection determined by the trivialization. Aswe will see however, connections of this type are rather special: for genericconnections, flat sections do not exist, even locally!

115


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116 CHAPTER 5. CURVATURE ON BUNDLES

p0v0

Figure 5.1: Parallel transport along a closed path on S2.

Definition 5.1. A connection on the fiber bundle : EMis called flatif for every x M and p Ex, there exists a neighborhood x U Mand a flat section s: U E with s(x) =p.

The flatness of a connection is closely related to the following question:

If EM is a fiber bundle with a connection and : [0, 1] M is asmooth path with(0) = (1), does the parallel transport Pt : E(0) E(t) satisfyP

1 = Id?

In other words, does parallel transport around a closed path bring everyelement of the fiber back to itself? Clearly this is the case if the path (t)lies in an open set U Mon which flat sections s : U Eexist, for thenPt(s((0)) =s((t)).

We can now easily see an example of a connection that is not flat. Let S2

be the unit sphere in R3, with Riemannian metric and corresponding Levi-Civita connection on T S2 S2 inherited from the standard Euclideanmetric on R3. Construct a piecewise smooth closed path as follows:beginning at the equator, travel 90 degrees of longitude along the equator,then upward along a geodesic to the north pole, and down along a differentgeodesic back to the starting point (Figure 5.1). Parallel transport of anyvector along this path has the effect of rotating the original vector by90 degrees. This implies that the triangle bounded by this path does notadmit any covariantly constant vector field.


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5.2. INTEGRABILITY AND THE FROBENIUS THEOREM 117

5.2 Integrability and the Frobenius theorem

Our main goal in this chapter is to formulate precise conditions for identi-

fying whether a connection is flat. Along the way, we can solve a relatedproblem which is of independent interest and has nothing intrinsically todo with bundles: it leads to the theorem of Frobenius on integrable distri-butions.

The word integrabilityhas a variety of meanings in different contexts:generally it refers to questions in which one is given some data of a lin-ear nature, and would like to find some nonlinear data which produce thegiven linear data as a form of derivative. The problem of finding an-tiderivatives of smooth functions on R is the simplest example: it is alwayssolvable (at least in principle) and therefore not very interesting for thepresent discussion. A more interesting example is the generalization ofthis question to higher dimensions, which can be stated as follows:

Given a 1-form on an n-manifold M, under what conditions is locally the differential of a smooth functionf :M R?

Our use of the word locally means that for every p M, we seek aneighborhood p U M and smooth function f : U R such thatdf = |U. The answer to this question is well known: the right conditionis that must be a closed 1-form, d = 0. Indeed, the Poincare lemmageneralizes this result by saying that every closed differentialk-form locallyis the exterior derivative of some (k 1)-form.

Here are two more (closely related) integrability results from the theoryof smooth manifolds which we shall find quite useful.

Theorem 5.2. For any vector fields X, Y Vec(M), the flows sX andtY commute for alls, t R if and only if[X, Y] 0.

Corollary 5.3. Suppose X1, . . . , X n Vec(M) are vector fields such that[Xi, Xj] 0 for all i, j {1, . . . , n}. Then for everyp M there exists aneighborhoodp U Mand a coordinate chartx = (x1, . . . , xn) :U Rn

such that xj

Xj|U for allj.

We refer to[Spi99] for the proofs, though it will be important to recallwhy Corollary5.3follows from Theorem5.2: the desired chart x : U Rn

is constructed as the inverse of a map of the form

f(t1, . . . , tn) =t1

X1 . . . t

n

Xn(p),

which works precisely because the flows all commute.The main integrability problem that we wish to address in this chap-

ter concerns distributions: recall that if M is a smooth n-manifold, a k-dimensional distribution T M is a smooth subbundle of rank k in the


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tangent bundle T M M, in other words, it assigns smoothly to eachtangent space TpM a k-dimensional subspace p TpM. We say that avector fieldXVec(M) is tangent to ifX(p) pfor allp M. In this

case Xis also a section of the vector bundle M.

Definition 5.4. Given a distribution T M, a smooth submanifoldNMis called an integral submanifold ofif for all p N, TpNp.

Definition 5.5. A k-dimensional distribution T M is called inte-grableif for every p Mthere exists ak-dimensional integral submanifoldthrough p.1

It follows from the basic theory of ordinary differential equations thatevery 1-dimensional distribution is integrable: locally one can choose a

nonzero vector field that spans the distribution, and the orbits of this vec-tor field trace out integral submanifolds. More generally, any distributionadmits 1-dimensional integral submanifolds, and one can further use thetheory of ordinary differential equations to show that k-dimensional inte-gral submanifolds of a k-dimensional distribution are unique if they exist.Existence is however not so clear whenk 2. For instance, a 2-dimensionaldistribution in R3 may twist in such a way as to make integral subman-ifolds impossible (Figure5.2).

A simple necessary condition for T Mto be integrable comes fromconsidering brackets of vector fields tangent to . Indeed, supposeis inte-

grable, and for every p M, denote by Np Mthe unique k-dimensionalintegral submanifold containing p. The key observation now is that a vec-tor field X Vec(M) is tangent to if and only if it is tangent to allthe integral submanifolds, in which case it has well defined restrictionsX|Np Vec(Np). Thus if X, Y Vec(M) are both tangent to , so is[X, Y], as it must also have well defined restrictions [X, Y]|Np Vec(Np)which match [X|Np, Y|Np ]. We will see from examples that in general, thebracket of two vector fields tangent to is not also tangent to , which isclearly a necessary condition for integrability. The content of Theorem5.16below is that this condition is also sufficient.

We will approach the proof of this via a version that applies specifically

to connections on fiber bundles, and is also highly relevant to the flatnessquestion. Recalling the definition of the covariant derivative on a fiberbundle E M, we see that a section s : U E is flat if and only ifthe submanifold s(U) E is everywhere tangent to the chosen horizontal

1Some authors give a different though equivalent definition for an integrable distri-bution: they reverse the roles of Definition 5.5 and Theorem5.16so that a distributionis said to be integrable if it satisfies the conditions stated in the theorem. Then theirversion of the Frobenius theorem states that this definition is equivalent to ours. Itcomes to the same thing in the end.


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z

x

y

Figure 5.2: A non-integrable 2-dimensional distribution on R3.

subbundleH E T E. This means it is ann-dimensional integral subman-ifold for an n-dimensional distribution on E, and allows us therefore toreformulate the definition of a flat connection as follows.

Proposition 5.6. If E M is a fiber bundle, then a connection is flat

if and only if the corresponding horizontal distributionHET E is inte-grable.

Exercise 5.7. Show that ifEis a fiber bundle over a 1-manifold M, thenevery connection on Eis flat.

Exercise 5.8. Show that the trivial connection on a trivial bundle E =M F is flat.

Let : E M be a fiber bundle with connection HE T E anddenote by

K:T EV E, H :T E H E

the two projections defined by the splitting T E = HE V E. A vectorfield onEis calledhorizontalorverticalif it is everywhere tangent to H Eor V E respectively. For XVec(M), define the horizontal lift ofXto bethe unique horizontal vector fieldXh Vec(E) such thatT Xh= X ,or equivalently,

Xh(p) = Horp(X(x))

for each x M, p Ex.


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Exercise 5.9. Show that for any XVec(M) and fC(M), LXh(f) = (LXf) .

Lemma 5.10. Suppose, Vec(E)are both horizontal and satisfyLfLf for every functionfC

(E) such thatdf|V E0. Then .

Proof. If (p) = (p) for some p E, assume without loss of generalitythat (p) = 0. Since (p) HpE, we can then find a smooth real-valuedfunction f defined near p which is constant in the vertical directions butsatisfiesdf((p)) = 0 and df((p)) = 0, so Lf(p)=Lf(p).

Lemma 5.11. For anyX, Y Vec(M), [X, Y]h= H [Xh, Yh].

Proof. Observe first that for any Vec(E) and fC(M),

L(f ) =LH(f ) + LK(f ) =LH(f )

sinced(f )|V E0. Then for X, Y Vec(M), using Exercise5.9,

LH[Xh,Yh](f ) =L[Xh,Yh](f )

=LXhLYh(f ) LYhLXh(f )

=LXh((LYf) ) LYh((LXf) )

= (LXLYf) (LYLXf) = (L[X,Y]f) .

Likewise, again applying Exercise 5.9,

L[X,Y]h(f ) = (L[X,Y]f) = LH[Xh,Yh](f ),

so the result follows from Lemma 5.10

Proposition 5.12. The distributionHE T Eis integrable if and only iffor every pair of vector fieldsX, Y Vec(M), [Xh, Yh] is horizontal.

Proof. If HEis integrable, then the lifts Xh, Yh Vec(E) are tangent tothe integral submanifolds, implying that [Xh, Yh] is as well. To prove theconverse, for any x M, pick pointwise linearly independent vector fieldsX1, . . . , X n on a neighborhood x U Msuch that [Xi, Xj] = 0 for each

pair, and denote j := (Xj )h. By assumption [i, j] is horizontal, thus byLemma5.11,

[i, j] =H [i, j] = [Xi, Xj ]h= 0.

Therefore for anyp Ex, we can construct an integral submanifold throughp via the commuting flows ofi: it is parametrized by the map

f(t1, . . . , tn) =t1

1 . . . t

n

n(p) (5.1)

for real numbers t1, . . . , tn sufficiently close to 0.


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Exercise 5.13. Verify that the map (5.1) parametrizes an embedded in-tegral submanifold ofH E.

Exercise 5.14. Show that the bilinear map K : Vec(E) Vec(E) Vec(E) defined by

K(, ) = K([H(), H()]) (5.2)

is C-linear in both variables.

By the result of this exercise, (5.2) defines an antisymmetric bundlemap K :T E T EV E, called the curvature2-formassociated to theconnection. Combining this with Prop.5.12,we find that the vanishing ofthis 2-form characterizes flat connections:

Theorem 5.15. A connection on the fiber bundle : E M is flat if

and only if its curvature2-form vanishes identically.Proof. If HE is integrable then the bracket of the two horizontal vectorfields H() and H() is also horizontal, thus K(, ) = 0. Converselyif this is true for every , Vec(E), then it holds in particular for thehorizontal lifts Xh and Yh of X, Y Vec(M), implying that [Xh, Yh] ishorizontal, and by Prop.5.12,HEis integrable.

We conclude this section with the promised integrability theorem fordistributions in generalthe result has nothing intrinsically to do withfiber bundles, but our proof rests on the fact that locally, both situationsare the same.

Theorem 5.16 (Frobenius). A distribution T M is integrable if andonly if for every pair of vector fieldsX, Y Vec(M) tangent to , [X, Y]is also tangent to .

Proof. The question is fundamentally local, so we can assume withoutloss of generality that M is an open subset U Rn, and arrange thek-dimensional distribution TRn|U so that for each x U, p R

n

is transverse to a fixed subspace Rnk Rn. We can then view as aconnection on a fiber bundle which has Uas its total space and U Rnk

as the vertical subbundle, and the result follows from Theorem 5.15.

Exercise 5.17. Using cylindrical polar coordinates ( ,,z) on R3, definethe 1-form

= f() dz+ g() d, (5.3)

where f and g are smooth real-valued functions such that (f(), g()) =(0, 0) for all, and define a 2-dimensional distribution by = ker TR3,i.e. forp R3,p= ker(p). An example of such a distribution is shown inFigure5.2. In this problem we develop a general scheme for determiningwhether distributions of this type are integrable. Indeed, if is any 1-formon R3 that is nowhere zero, show that the following are equivalent:


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1. d 0.

2. The restriction ofd to a bilinear form on the bundle is everywhere

degenerate, i.e. at every p R3

, there is a vector X p such thatd(X, Y) = 0 for all Y p.

3. is integrable.

Conclude that the distribution defined as the kernel of (5.3) is integrableif and only iff()g() f()g() = 0 for all .

5.3 Curvature on a vector bundle

If E M is a vector bundle and is a linear connection, it is natural

to ask whether covariant derivative operators X and Y in different di-rections commute. Of course this is not even true in general for the LiederivativesLX and LY on C

(M), which one can view as the trivial con-nection on a trivial line bundle. Their lack of commutativity can howeverbe measured via the identity

LXLY LYLX=L[X,Y],

and one might wonder whether it is true more generally that XY YX =[X,Y] as operators on (E). The answer turns out to be no ingeneral, but the failure of this identity can be measured precisely in terms

of curvature.

Definition 5.18. Given a linear connection on a vector bundleEM,the curvature tensor2 is the unique multilinear bundle map

R: T M T M E E: (X , Y , v)R(X, Y)v

such that for all X, Y Vec(M) and v (E),

R(X, Y)v= (XY YX [X,Y])v.

Exercise 5.19. Show that R(X, Y)v is C-linear with respect to each of

the three variables.

Exercise 5.20. Choosing coordinates x = (x1, . . . , xn) : U Rn anda frame (e(1), . . . , e(m)) for E over some open subset U M, define the

components ofR by Rijk so that (R(X, Y)v)i =Rijk X

jYkv. Show that

Rijk = j ik k

ij+

ijm

mk

ikm

mj .

2This choice of terminology foreshadows Theorem5.21, which relates the curvaturetensor to the curvature 2-form for fiber bundles described in the previous section.


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5.3. CURVATURE ON A VECTOR BUNDLE 123

It may be surprising at first sight that R(X, Y)vdoesnt depend on anyderivatives of v: indeed, it seems to tell us less about v than about theconnection itself. Our main goal in this section is to establish a precise

relationship between the new curvature tensor R and the curvature 2-formKdefined in the previous section. In particular, this next result impliesthat covariant mixed partials commute if and only if the connection isflat.

Theorem 5.21. For any vector bundle E M with connection , thecurvature tensorR satisfies

R(X, Y)v= K([Xh, Yh](v))

for any vector fieldsX, Y Vec(M) andv E.

Note thatKon the right hand side of this formula is not quite the sameprojection as in the previous section: as is standard for linear connections,K is now the connection map K : T E E obtained from the verticalprojection via the identificationVvE=Ep forv Ep. In light of this, it isnatural to give a slightly new (but equivalent) definition of the curvature2-form when the connection is linear. We define an antisymmetric bilinearbundle map K :T M T MEnd(E) by the formula

K(X, Y)v= K([Xh, Yh](v)), (5.4)

for any p M, X, Y TpM and v Ep, where on the right hand sidewe choose arbitrary extensions of X and Y to vector fields near p. Itsstraightforward to check that this expression is C-linear in both X andY; whats less obvious is that it is also linear with respect to v. This istrue because the connection map K:T E EsatisfiesK T m= m K(see Definition 3.9), where m :E E is the map v v for any scalar F. Indeed, since m is a diffeomorphism on E whenever = 0, wehave (m)[, ] [(m), (m)] for any , Vec(E), thus

K(X, Y)(v) =K([Xh, Yh](v)) = K([Xh, Yh] m(v))

=K(T m [Xh, Yh](v)) =m K([Xh, Yh](v))

= K(X, Y)v.

It follows from this and Lemma 3.10 that vK(X, Y)v is linear.With our new definition of the curvature 2-form, Theorem5.21can be

restated succinctly as

R(X, Y)v= K(X, Y)v.

It follows that XYv YXv [X,Y]v 0 for all vector fields X, Yand sectionsv if and only if the connection is flat. As an important special


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case, if (s, t) M is a smooth map parametrized by two real variablesand v(s, t) E(s,t) defines a smooth section ofEalong , we have

stv tsv

if and only ifis flat; more generally

stv tsv= R(s, t)v.

We shall prove Theorem 5.21 by relating the bracket to an exteriorderivative using a generalization of the standard formula

d(X, Y) =LX((Y)) LY((X)) ([X, Y])

for 1-forms 1

(M). In particular, the definitions of K, K and Rcan all be expressed in terms of bundle-valued differential forms. For anyvector bundle : EM, define

k(M, E)

to be the vector space of smooth real multilinear bundle maps

: T M . . . T M k

E

which are antisymmetric in the k variables. By this definition,

k

(M) issimply k(M, M R), i.e. the space ofk-forms taking values in the trivialreal line bundle. Similarly, what we referred to in3.3.3 as k(M, g) (thespace of g-valued k-forms for a Lie algebra g) is actually k(M, M g),though well preserve the old notation when theres no danger of confusion.From this perspective, we have

K1(E, E) and K2(M, End(E)).

Defining 0(M, E) := (E), the covariant derivative gives a linear map: 0(M, E) 1(M, E) = (HomR(T M , E )), and by analogy with the

differential d: 0

(M) 1

(M), its natural to extend this to a covariantexterior derivative

d: k(M, E)k+1(M, E),

defined as follows. Every k(M, E) can be expressed in local coordi-natesx= (x1, . . . , xn) : U Rn as

=

1ii


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for some component sections i1...ik (E|U). Then d is defined locallyas

d= 1ii


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TvE respectively. Then using (5.5) and the fact that Kvanishes on hori-zontal vectors,

dK(Xh(v), Yh(v)) =Xh(v)(K(Yh)) Yh(v)(K(Xh)) K([Xh, Yh](v))= K(X, Y)v.

We now show that R(X, Y)vcan also be expressed in this way. Choose asmooth map(s, t) Mfor (s, t) R2 near (0, 0) such that s(0, 0) =Xand t(0, 0) = Y, and extend v Ep to a section v(s, t) E(s,t) along such thatsv(0, 0) = tv(0, 0) = 0. Then expressing covariant derivativesvia the connection map (e.g. sv= K(sv)) and applying (5.5) once more,we find

R(X, Y)v= stv(0, 0) tsv(0, 0)

= s(K(sv(s, t))) t (K(tv(s, t)))|(s,t)=(0,0)

=dK(sv, tv) =dK(Horv(X), Horv(Y)),

where in the last step we used the assumption that v(s, t) has vanishingcovariant derivatives at (0, 0).

We close the discussion of curvature on general vector bundles by ex-hibiting two further ways that it can be framed in terms of exterior deriva-tives. The first of these follows immediately from Equation (5.5): replacing with v for any section v (E), we have

d2v(X, Y) =R(X, Y)v. (5.6)

This elegant (though admittedly somewhat mysterious) expression showsthat the covariant exterior derivative does not satisfy d2 = 0 in general.In fact:

Exercise 5.24. Show that d2= 0 on k(M, E) for all k if and only if the

connection is flat.

Finally, if : E M has structure group G, we can also express

curvature in terms of the local connection 1-formA

1

(U, g

) associatedto a G-compatible trivialization : E|U U Fm. Recall that this is

defined so that

(Xv)= dv(X) + A(X)v(p)

for X TpM and v (E), where v : U Fm expresses v|U with

respect to the trivialization. We define a local curvature 2-form F 2(U, g) by

F(X, Y) =dA(X, Y) + [A(X), A(Y)].


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Exercise 5.25. Show that for any p U, v Ep and X, Y TpM,

(R(X, Y)v)= F(X, Y)v.

Exercise 5.26. If :E|U U Fm is a second trivialization relatedto by the transition map g= g : U UG, show that

F(X, Y) =gF(X, Y)g1.

In general the g-valued curvature 2-formF is only defined locally anddepends on the choice of trivialization, though Exercise5.26brings to lighta certain important case in which there is no dependence: ifG is abelian,thenFFfor any two trivializations and , wherever they overlap.It follows that in this case one can define a global g-valued curvature 2-form

F 2(M, g)

such thatF(X, Y)|U =F(X, Y) for any choice of trivialization on U.Exercise 5.27. Show that ifG is abelian, there is a natural G-action oneach of the fibers ofE, and therefore also a g-action. In this case, F(X, Y)is simply K(X, Y) reexpressed in terms of this action.

Here is an example that will be especially important in the next chap-ter: if (M, g) is an oriented Riemannian 2-manifold, then T M M hasstructure group SO(2), which is abelian, and thus there is a globally de-fined curvature 2-form F 2(M, so(2)). Observe now that so(2) is the1-dimensional vector space of real antisymmetric 2-by-2 matrices, thus allof them are multiples of

J0 := 0 11 0 ,and there is a real-valued 2-form 2(M) such that F(X, Y) =(X, Y)J0.We can simplify things still further by recalling that the metric g defines anatural volume form dA 2(M) (not necessarily an exact form, despitethe notation), such that

dA(X, Y) = 1

whenever (X, Y) is a positively oriented orthonormal basis. Since dim 2TM=1, there is then a unique smooth function

K :M R

such that for any p MandX, Y TpM,F(X, Y) =K(p)dA(X, Y)J0.We see from this that all information about curvature on a 2-manifold canbe encoded in this one smooth function: e.g. the curvature tensor can bereconstructed by

R(X, Y)Z=K(p) dA(X, Y)J0Z.

We call K the Gaussian curvature of (M, g). We will have more to say inthe next chapter on the meaning of this function, which plays a starringrole in the Gauss-Bonnet theorem.


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References

[Spi99] M. Spivak,A comprehensive introduction to differential geometry, 3rd ed., Vol. 1,

Publish or Perish Inc., Houston, TX, 1999.

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Connections Chapter5