Lie Bracket

Lemma 7.2 Let U ⊂ M be an open set. If LX f(p) = 0 for allf ∈ C^∞(U) and allp∈U then X|U = 0.

Proof. Working locally in a chart (x, U), letX be the principal represen-tative of X (defined in section 7.1). Suppose that ` : E → R is a continu-ous linear map such that `(X(p)) 6= 0 (Hahn-Banach). Then Dp` = ` and d(x^∗_α`p)(X(p)) =Dp`·X6= 0

Theorem 7.1 For every pair of vector fields X, Y ∈ X(M) there is a unique vector field [X, Y] which for any open set U ⊂ M and f ∈ C^∞(U) we have [X, Y](p)f =Xp(Y f)−Yp(Xf) and such that in a local chart U,x the vector field[X, Y]has principal part given by

DY·X−DX·Y

7.4. LIE BRACKET 109 or more fully if f ∈ C^∞(U)then letting f:=f◦x⁻¹ be the local representative we have

([X,Y]f)(x) =Df·(DY(x)·X(x)−DX(x)·Y(x)) where x∼pand[X,Y]f means the local representative of[X, Y]f.

Proof. We sketch the main points and let the reader fill in the details.

One can prove that the formula f 7→Df(DY·X−DX·Y) defines a derivation locally. Also, DY·X−DX·Y transforms correctly and so gives a vector at each point. Thus by pulling back via charts we get a vector field on each chart domain. But these agree on overlaps because they are all coming from the local representations of a single derivationf 7→Xp(Y f)−Yp(Xf). In the finite dimensional case we can see the result from another point of view. Namely, every derivation of C^∞(M) isLX for some X so we can define [X, Y] as the unique vector field such thatL[X,Y]=LX ◦ LY − LY ◦ LX.

We ought to see what the local formula for the Lie derivative looks like in the finite dimensional case where we may employ classical notation. Suppose we haveX =Pm

i=1X^{i ∂}_∂xi andY =Pm

i=1Y^{i ∂}_∂xi. Then [X, Y] =P

i

P

j∂Yⁱ

∂x^jX^j−^∂X_∂x^jiY^j‘

∂

∂xⁱ

Exercise 7.1 Check this.

Definition 7.7 The vector field[X, Y] from the previous theorem is called the Lie bracket ofX andY.

The following properties of the Lie Bracket are checked by direct calculation.

For anyX, Y, Z∈X(M), 1. [X, Y] =−[Y, X]

2. [X,[Y, Z]]+ [Y,[Z, X]] + [Z,[X, Y]] = 0

Definition 7.8 (Lie Algebra) A vector spaceais called aLie algebraif it is equipped with a bilinear map a×a→a(a multiplication) denotedv, w7→[v, w]

such that

[v, w] =−[w, v]

and such that we have a Jacobi identity

[x,[y, z]] + [y,[z, x]] + [z,[x, y]] = 0 for all x, y, z∈a.

We have seen thatX(M) (orX(U)) is a Lie algebra with the Bracket defined in definition 7.7.

7.5 Localization

It is a fact that if M is finite dimensional then every derivation of C^∞(M) is given by some vector field and in fact the association X 7→ LX is a bijection fromX(M) onto the set of all derivations of the algebra ofC^∞=F(M) functions C^∞(M). Actually, this is true not just for finite dimensional manifolds but also for manifolds modelled on a special class of Banach spaces which include Hilbert spaces. In order to get an isomorphism between the set of all derivations of the algebra C^∞(M) = F(M) we need to be able to construct appropriate bump functions or more specifically cut-off functions.

Definition 7.9 Let K be a closed subset of M contained in an open subset U ⊂ M. A cut-off function for the nested pair K ⊂ U is a C^∞ function β:M →Rsuch that β|K ≡1 and β|M\U ≡0.

Definition 7.10 A manifold M is said to admit cut-off functions if given any pointp∈M and any open neighborhoodU ofp, there is another neighbor-hood V of psuch that V ⊂U and a cut-off function β_{V ,U} for the nested pair V ⊂U.

Lemma 7.3 All finite dimensional smooth manifolds admit cut-off functions.

Proof. Exercise

Definition 7.11 Let E be a Banach space and suppose that the norm on E is smooth (resp.C^r). The we say thatEis asmooth (resp. C^r) Banach space.

Lemma 7.4 If E is a smooth (resp. C^r) Banach space and Br ⊂BR nested open balls then there is a smooth (resp. C^r) functionβ defined on all ofEwhich is identically equal to 1on the closure Br and zero outside ofBR.

Proof. We assume with out loss of generality that the balls are centered at the origin 0∈E. Let

φ1(s) = Rs

−∞g(t)dt R_∞

−∞g(t)dt where

g(t) =





exp(−1/(1− |t|²) if |t|<1

0 otherwise

.

This is a smooth function and is zero ifs <−1 and 1 ifs >1 (verify). Now let β(x) =g(2− |x|). Check that this does the job using the fact that x 7→ |x| is assumed to be smooth (resp. C^r).

Corollary 7.1 If a manifold M is modelled on a smooth (resp. C^r) Banach spaceE, (in particular, ifM is ann-manifold), then for everyαp∈T^∗M, there is a (global) smooth (resp. C^r) function f such that Df|p=αp.

7.5. LOCALIZATION 111 Proof. Let x0 =ψ(p)∈Rⁿ for some chartψ, U. Then the local represen-tative ¯αx0 = (ψ⁻¹)^∗αpcan be considered a linear function onRⁿ since we have the canonical identification Rⁿ∼={x0}×Rⁿ=Tx0 Rⁿ.Thus we can define

ϕ(x) =

š β(x)¯αx0(x) for x∈BR(x0)

0 otherwise

and now making sure that R is small enough that BR(x0) ⊂ ψ(U) we can transfer this function back toM viaψ⁻¹ and extend to zero outside ofU getf.

Now the differential ofϕatx0is ¯αx0 and so we have forv∈TpM df(p)·v=d(ψ^∗ϕ)(p)·v

= (ψ^∗dϕ)(p)v dϕ(Tpψ·v)

= ¯αx0(Tpψ·v) = (ψ⁻¹)^∗αp(Tpψ·v)

=αp(T ψ⁻¹Tpψ·v) =αp(v) so df(p) =αp

Lemma 7.5 The map from X(M) to the vector space of derivations Der(M) given by X 7→ L^X is a module monomorphism if M is modelled on a C^∞ Banach space.

Proof. The fact that the map is a module map is straightforward. We just need to get the injectivity. For that, suppose LX f = 0 for all f ∈ C^∞(M).

Then Df|pXp = 0 for all p∈ M. Thus by corollary 7.1 αp(Xp) = 0 for all αp∈T_p^∗M. By the Hahn Banach theorem this means thatXp= 0. Sincepwas arbitrary we concluded thatX = 0.

Theorem 7.2 LetL:X(M)→ C^∞(M)be aC^∞(M)−linear function on vector fields. IfM admits cut off functions then L(X)(p)depends only on the germ of X atp.

If M is finite dimensional then L(X)(p)depends only on the value of X at p.

Proof. Suppose X = 0 in a neighborhood U and let p∈ U be arbitrary.

Let O be a smaller open set containing with closure inside U. Then letting β be a function that is 1 on a neighborhood of pcontained in O and identically zero outside of Othen (1−β)X =X. Thus we have

L(X)(p) =L((1−β)X)(p)

= (1−β(p))L(X)(p) = 0×L(X)(p)

= 0.

Applying this toX−Y we see that if two fieldsX andY agree in an open set thenL(X) =L(Y) on the same open set. The result follows from this.

Now suppose that M is finite dimensional and suppose that X(p) = 0.

WriteX =X^{i ∂}_∂xi in some chart domain containingpwith smooth functionXⁱ satisfyingXⁱ(p) = 0. Letting β be as above we have

β²L(X) =βXⁱL(β ∂

∂xⁱ) which evaluated atpgives

L(X)(p) = 0

sinceβ(p) = 1. Applying this toX−Y we see that if two fieldsX andY agree atpthenL(X)(p) =L(Y)(p).

Corollary 7.2 If M is finite dimensional and L : X(M) → C^∞(M) is a C^∞(M)−linear function on vector fields then there exists an elementα∈X^∗(M) such that α(X) =L(X)for allX ∈X(M).

Remark 7.1 (Convention) Even though a great deal of what we do does not depend on the existence of cut off functions there are several places where we would like to be able to localize.