[PDF] THE FROBENIUS THEOREM 1. Distributions Suppose M is an n

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LECTURE 11: THE FROBENIUS THEOREM

1.Distributions

SupposeMis ann-dimensional smooth manifold. We have seen that any smooth vector eld XonMcan be integrated locally near any point to an integral curve. Moreover, IfXp= 0, then the corresponding integral curve is the constant curve p(t)p.

IfXp6= 0, then the corresponding integral curve

p(t) is a 1-dimensional curve nearp. In what follows we would like to some higher dimensional analogue of this fact. Denition 1.1.Ak-dimensionaldistributionVonMis a map which assigns to every point p2Mak-dimensional vector subspaceVpofTpM.Vis calledsmoothif for everyp2M, there is a neighborhoodUofpand smooth vector eldsX1;;XkonUsuch that for everyq2U, X

1(q);;Xk(q) are a basis ofVq. (In particular,Xi(q)6= 0 for all 1ik.)

Remarks.(1) In what follows, all distributions will be smooth. (2) We say a vector eldXbelongs to a distributionVifXp2 Vpfor allp2M. (3) By denition, ak-dimensional distribution is rankksub-bundle ofTM. Denition 1.2.SupposeVis ak-dimensional distributions onM. An immersed submanifold NMis called anintegral manifoldforVif for everyp2N, the image ofdN:TpN!TpM isVp. We say the distributionVisintegrableif through each point ofMthere exists an integral manifold ofV. Example.Any non-vanishing vector eldXis a 1-dimensional distribution. The image of any integral curve ofXis an integral manifold.

Example.The vector elds@@x

1;;@@x

kspan ak-dimensional distributionVinRn. The integral manifolds ofVare planes that are dened by the system of equations x i=ci(k+ 1in): Remark.An integral manifold need not to be an embedded submanifold ofM. For example, considerM=S1S1R2xR2y. Fix any irrational numbera, the integral manifold of the non-vanishing vector eld X a= (x2@@x

1x1@@x

2) +a(y2@@y

1y1@@y

2) is a dense \curve" inM. (However, it is an immersed submanifold.) Although any 1-dimensional distribution is integrable, a higher dimensional distribution need not to be integrable. 1

2 LECTURE 11: THE FROBENIUS THEOREM

Example.Consider the smooth distributionVonR3spanned by two vector elds X 1=@@x

1+x2@@x

3; X2=@@x

2: I claim that there is no integral manifold through the origin. In fact, ifVis integrable, then the integrable manifoldNofVcontaining the origin must also contain the integrable curve ofX1 passing the origin, which is a piece of thex1-axis, i.e.Ncontains all points of the form (t;0;0) forjtj< ". AlsoNmust contain the integral curves of the vector eldX2passing all these points (t;0;0). It follows that for eachjtj< ",Ncontains a small piece of line segment parallel to the x

2-axis, i.e.Ncontains for eachjtj< "all points of the form (t;s;0);jsj< t. In other words,N

contains a piece of thex1-x2plane that contains the origin. This is a contradiction, because the vector @@x

1is a tangent vectors of this piece of plane but is not inVpfor anyp6= 0.

We are interested in the conditions to make a distribution integrable. A necessary condition is easy to nd. In fact, we have Theorem 1.3.If a distributionVis integrable, then for any two vector eldsXandYbelonging toV, their Lie bracket[X;Y]belongs toValso. Proof.Fix anyp2M, suppose:N ,!Mis an integrable manifold ofV. SinceNis an immersed submanifold ofM, one can \shrink"Nso that(N) is in fact an embedded submanifold ofM. Now supposeX;Yare vector elds belonging toV, then the restrictionsXjN;YjNtoNare vector elds that are tangent to the submanifoldN. By denition,XNisrelated toXandYjNis-related toY. It follows that [XjN;YjN] is-related to [X;Y]. In other words, [X;Y]p=dp([XjN;YjN]p) is tangent toNalso. It follows that for anyp2M, [X;Y]p2 Vp. So [X;Y] belongs toValso. Denition 1.4.A distributionVisinvolutiveif it satises the followingFrobenius condition: If

X;Y21(TM) belong toV, so is [X;Y].

Example.Any 1 dimensional distribution is involutive since [fX;gX] is a multiple ofX. Example.Thek-dimensional distribution spanned by@@x

1;;@@x

kis involutive.

Example.The distributionVspanned by

X 1=@@x

1+x2@@x

3; X2=@@x

2 is not involutive, since [X1;X2] =@@x 3 is not inV. Example.Letf:M!Nbe a submersion. Then the distributionVwithVp= Ker(dfp) is involutive. In fact, ifX;Yare vector elds belonging toV, thendfp(Xp) =dfp(Yp) = 0, i.e. both XandYaref-related to the zero vector eld onN. It follows thatdf([X;Y]p) = 0. It is easy to see thatVis also integrable. In fact, the integrable manifold passingp2Mis the submanifold f

1(f(p)).

LECTURE 11: THE FROBENIUS THEOREM 3

2.The Frobenius Theorem

It turns out that the Frobenius condition is not only necessary but also sucient for a distri- bution to be integrable. Theorem 2.1(Global Frobenius Theorem).LetVbe an involutivek-dimensional distribution. Then through every pointp2M, there is a unique maximal connected integral manifold ofV.

Example.Consider the distributionVonR3spanned by

X

1=x1@@x

2x2@@x

1; X2=@@x

3 onM=R3 fx1=x2= 0g. Since [X1;X2] = 0,Vis involutive. What is its integral manifold? Well, let's rst compute the integral curves ofX1andX2. Through any point (x1;x2;x3), the integral curves ofX1are circles in thex3-plane with origin the center, and the integral curves of X

2are the lines that are parallel to thex3-axis. Note that the integral manifold passing (x1;x2;x3)

of the distribution should contains all points of the form'X1t('X2s(x1;x2;x3)) for allt;s. In our case, this is the cylinders centering at thex3-axis. We rst prove the following local version: Any involutive distribution is integrable, i.e. locally near each point one can nd an integrable manifold. Theorem 2.2(Local Frobenius Theorem).LetVbe an involutivek-dimensional distribution. Then for everyp2M, there exists a coordinate patch(U;x1;;xn)centered atpsuch that for allq2U,Vq= spanf@@x

1(q);;@@x

k(q)g: We need the following lemma whose proof is left as an exercise. Lemma 2.3.LetXbe a smooth vector eld onM. Ifp2Msuch thatXp6= 0, then there exists a local chart(U;x1;;xn)nearpsuch thatX=@@x 1onU. Proof of the Local Frobenius Theorem:By the lemma, this is true fork= 1. Suppose the theorem holds fork1 dimensional distributions. LetVbe ankdimensional distribution spanned by X

1;X2;;Xk. SupposeVis involutive, i.e.

[Xi;Xj]0 mod (X1;;Xk);1i;jk: Use the previous lemma, there exits a local chart (U;y1;;yn) nearpsuch thatXk=@@y k. For

1ik1 let

X

0i=XiXi(yk)Xk;

thenX0i(yk) = 0 for 1ik1, andXk(yk) = 1. Note that the vector eldsX01;;X0k1;Xk still spanV. Moreover, if we denote [X0i;X0j] =aijXkmod (X01;;X0k1);1i;jk1; then applying both sides to the functionyk, we seeaij= 0 for all 1i;jk1. In other words, thek1 dimensional distribution V

0= spanfX01;;X0k1g

4 LECTURE 11: THE FROBENIUS THEOREM

is involutive. So there is a local chart (U;z1;;zn) nearpsuch thatV0is spanned byf@@z

1;;@@z

k1g.

Since each

@@z iis a linear combination ofX0jfor 1i;jk1, we conclude@@z i(yk) = 0.

Now denote

[@@z i;Xk] =biXkmod (@@z

1;;@@z

m1): Apply both sides to the functionyk, we seebi= 0 for alli. So we can write @@z i;Xk] =k1X j=1C ij@@z j:

SupposeXk=Pn

j=1j@@z j. Insert this into the previous formula, we see j@z i= 0;1ik1;kjn:

In other words, forjk,j=j(zk;;zn). Let

X 0k=nX j=k j@@z j:

Thenf@@z

1;;@@z

k1;X0kgstill spanV. Finally according to the pervious lemma again, there is a local coordinate change from (z1;;zk;;zn) to (x1;;xk;;xn) withxi=zifor

1ik1, such thatX0k=@@x

k. This completes the proof. Sketch of proof of the Global Frobenius theorem:For anyp2M, let N p=fq2Mj 9a piecewise smooth integral curve inVjointingptoqg: We claim thatNpis the maximal connected integral manifold ofVcontainingp. The manifold structure is dened as follows: for anyq2Np, there is a coordinate patch (U;x1;;xn) centering atqsuch thatV= spanf@@x

1;;@@x

kginU. For each small", let W "=fw2Uj(x1)2(w) ++ (xk)2(w)";xk+1(w) ==xn(w) = 0g: Then any pointw2W"can be joint topby the integral curve (t) =t(x1(w);;xk(w);0;;0):

SoW"Np. Let

':W"!Bk(")Rk; w7!(x1(w);;xk(w)): Now we dene the topology onNpby giving it the weakest topology such that all theseW"'s are open. The atlas onNpis dened to be the set of charts (';W;Bk(")). One can check thatNpis a manifold with this given atlas. For more details, c.f. Warner, pg.48-49.quotesdbs_dbs24.pdfusesText_30

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