Lie Derivatives on Manifolds

William C. SchulzDepartment of Mathematics and Statistics,

Northern Arizona University, Flagstaff, AZ 86011

1. INTRODUCTION

This module gives a brief introduction to Lie derivatives and how they acton various geometric objects. The principal difficulty in taking derivatives ofsections of vector bundles is that the there is no cannonical way of comparingvalues of sections over different points. In general this problem is handled byinstalling a connection on the manifold, but if one has a tangent vector fieldthen its flow provides another method of identifying the points in nearby fibres,and thus provides a method (which of course depends on the vector field) oftaking derivatives of sections in any vector bundle. In this module we developthis theory. 1

2. TANGENT VECTOR FIELDS

A tangent vector field is simply a section of the tangent bundle. For our purposeshere we regard the section as defined on the entire manifold M . We can alwaysarrange this by extending a section defined on an open set U of M by 0, aftersome appropriate smoothing. However, since we are going to be concerned withthe flow generated by the tangent vector field we do need it to be defined on allof M .

The objects in the T (M) are defined to be first order linear operators actingon the sheaf of C∞ functions on the manifold. Xp(f) is a real number (or acomplex number for complex manifolds). At each point p ∈ M we have

Xp(f + g) = Xp(f) + Xp(g)

Xp(fg) = Xp(f)g(p) + f(p)Xp(g)

Xp(f) should be thought of as the directional derivative of F in the direction X

at P . Thus X inputs a function (at p) and outputs a real (or complex) number.We regard f as being defined on some neighborhood of p (which depends on f).The rules above describe how Xp acts on sums and products.

It is possible to show, although we will not do so here, that Tp(M) is ann-dimensional vector space (n = dim M , the dimension of the Manifold M)with a basis in local coordinates { ∂

∂u1 , . . . , ∂∂un }. Hence

Xp = X i(p)∂

∂ui

∣

∣

∣

p

17 Oct 2011

1

and

Xp(f) = X i(p)∂f

∂ui

∣

∣

∣

p

If we change coordinates from ui to uj the the local expression of X changeslike a contravariant tensor:

X i ∂f

∂ui= X i ∂f

∂uj

∂uj

∂ui= Xj ∂f

∂uj

so

Xj = X i ∂uj

∂uiand X i = Xj ∂ui

∂uj

3. THE LIE BRACKET

If we attempt to compose X and Y the results are not encouraging; locally

Y (f) = Y i ∂f

∂ui

X(Y (f)) = Xj ∂

∂uj

(

Y i ∂f

∂ui

)

= Xj ∂Y i

∂uj

∂f

∂ui+ XjY i ∂2f

∂uj∂ui

which shows that XY is not a tangent vector since it contains the second deriva-tive of f and is thus not a first order operator. However, we take heart fromthe observation that the objectionable term is symmetric in i and j. Thus if weform

Y (X(f)) = Y j ∂X i

∂uj

∂f

∂ui+ Y jX i ∂2f

∂uj∂ui

and subtract, the objectionable terms will drop out and we have

X(Y (f)) − Y (X(f)) =

(

Xj ∂Y i

∂uj− Y j ∂X i

∂uj

)

∂f

∂ui

which is a first order operator and hence a tangent vector. We now introducenew notation

[X, Y ] = X(Y (·)) − Y (X(·)) =

(

Xj ∂Y i

∂uj− Y j ∂X i

∂uj

)

∂

∂ui

[X, Y ]i =

(

Xj ∂Y i

∂uj− Y j ∂X i

∂uj

)

For practise the user may wish to verify that [X, Y ]i transforms properly undercoordinate change.

X and Y are said to commute if [X, Y ] = 0. For example ∂∂ui and ∂

∂uj

commute.It is obvious that [X, Y ] is linear in each variable.

2

Next we form

[X, [Y, Z]](f) = X((Y Z − ZY )(f)) − (Y Z − ZY )(X(f))

= X(Y (Z(f))) − X(Z(Y (f))) − Y (Z(X(f))) + Z(Y (X(f)))

Let us abbreviate this by surpressing the f and the parentheses and then per-mute cyclically.

[X, [Y, Z]] = XY Z − XZY − Y ZX + ZY X

[Y, [Z, X ]] = Y ZX − Y XZ − ZXY + XZY

[Z, [X, Y ]] = ZXY − ZY X − XY Z + Y XZ

Now if we add the three equations the terms cancel in pairs and we have theimportant Jacobi Identity

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

Thus the vector fields in T (M) form an (infinite dimensional) Lie Algebra,which is a vector space with a skew symmetric multiplication [X, Y ] linear ineach variable satisfying the previous identity. The Jacobi identity substitutesfor associativity.

4. BACK AND FORTH WITH DIFFEOMOR-

PHISMS

The user may wonder why this area of mathematics tends to emphasize diffeo-morphisms instead of differentiable maps. In fact, the level of generality impliedby φ : M → N is largely spurious; most of the time N = M and so diffeomor-phisms are the natural objects to study. Nevertheless we follow convention inthis regard and formulate the results for φ : M → N

The user will recall that a mapping φ : M → N induces a mapping φ∗ = dφ

from Tp(M) to Tq(N) where q = φ(p). Since a vector in Tq(N) may be identifiedby its action on a function f : V → R, V a neighborhood of q in N , we candefine φ∗ by

(φ∗X)q(f) = Xp(f ◦ φ)

since f ◦ φ : M → R. For notational amusement we define

φ∗f = f ◦ φ

and then we have(φ∗X)(f) = X(φ∗f)

and thusφ∗(X) = X ◦ φ∗

3

for X ∈ Tp(M) and φ∗ : Tp(M) → Tφ(p)(N).In local coordinates we describe φ as follows: p has coodinates u1, . . . , un

and φ(p) has coodinates v1, . . . , vn and thus p → φ(p) is given by

vi(u1, . . . , un) i = 1, . . . , n

Then, if

X = X i ∂

∂uiY = Y i ∂

∂vj

and Yφ(p) = φ∗(Xp) we have, (surpressing some p subscripts)

Yφ(p)(f) = Y j ∂f

∂vj= Y j ∂f ◦ φ

∂ui

∂ui

∂vj

and

Yφ(p)(f) = Xp(f ◦ φ) = X i ∂f ◦ φ

∂ui

from which we see that

X i = Y j ∂ui

∂vj

and symmetrically

Y j = X i ∂vj

∂ui

which resembles the coordinate change rules. This is for the best of reasons;because φ is a diffeomorphism a coordinate patch on N becomes, via φ, acoordinate patch on M . Thus, locally, a diffeomorphism looks like a coordinatechange. This is not particularly helpful in keeping things straight in ones mind,although occasionally technically useful.

Now if φ were just a differentiable mapping then it could not be used tomap vector fields on M to vector fields on N . The obvious way to do this is asfollows: given q ∈ N , select p ∈ M so that φ(p) = q and then let Yq = φ∗Xp.This won’t work for two reasons. First, there might be q ∈ N not in the rangeof φ, and even if q is in the range of φ we might have φ(p1) = φ(p2) = q forp1 6= p2 so that Yq would not be uniquely defined. However, neither of these isa problem for diffeomorphisms so that

Def If X is a vector field on M and φ : M → N is a diffeomormorphism thenwe can define a vector field Y = φ∗(X) on N by

(φ∗X)q = Xφ−1q ◦ φ = φ∗(Xφ−1q)

If f : N → R then (recall φ∗(f) = f ◦ φ)

(φ∗X)q(f) =(

Xφ−1q ◦ φ)

(f) = Xφ−1q(φ∗f)

= Xφ−1q(f ◦ φ)

Expressed slightly differently

if Y = φ∗X

then Yq(f) = Xφ−1q(f ◦ φ)

4

Notice that if φ1 : M1 → M2 and φ2 : M2 → M3 are diffeomorphisms then

φ2∗ ◦ φ1∗ = (φ2 ◦ φ1)∗

This is just an abstract expression of the chain rule, but we can say it reallyfancy: ∗ is a covariant functor from the category of Manifolds and Diffeomor-phisms to the category of vector bundles and isomorphisms.

Now we want to show that the Lie Bracket is preserved under diffeomor-phisms. First we note that if φ : M → N is a diffeomorphism then it can beregarded locally as a coordinate change. Those persons who verified that theLie Bracket was invariant under coordinate change when I requested them todo so need not read the following.

Let φ : M → N be a diffeomorphism and let u1, . . . , un be coordinatesaround p ∈ M and v1, . . . , vn be coordinates around q = φ(p) ∈ N . Then wehave, for f : N → R,

(φ∗X)qf = Xφ−1q(f ◦ φ)

= X iui(vj)

∂

∂ui(f ◦ φ)

= X iui(vj)

∂f

∂vi

∂vj

∂ui= X

j

vk

∂f

∂vj

where we set

Xj

vk = X iuℓ(vk)

∂vj

∂ui

Now we can calculate the Lie Bracket. We surpress the subscripts on X, X, Y, Y

because they are always the same. Then

[X, Y ]i = Xj ∂Y i

∂vj− Y j ∂X i

∂vj

= Xk ∂vj

∂uk

∂

∂uℓ

(

Y m ∂vi

∂um

)∂uℓ

∂vj− Y k ∂vj

∂uk

∂

∂uℓ

(

Xm ∂vi

∂um

)∂uℓ

∂vj

= Xk ∂Y m

∂uℓ

∂uℓ

∂vj

∂vj

∂uk

∂vi

∂um+ XkY m ∂vj

∂uk

∂2vi

∂ui∂um

∂vℓ

∂uj

− Y k ∂Xm

∂uℓ

∂uℓ

∂vj

∂vj

∂uk

∂vi

∂um− Y kXm ∂vj

∂uk

∂2vi

∂ui∂um

∂vℓ

∂uj

= Xk ∂Y m

∂uℓδℓk

∂vi

∂um− Y k ∂Xm

∂uℓδℓk

∂vi

∂um

+ XkY mδℓk

∂2vi

∂uℓ∂um− Y kXmδℓ

k

∂2vi

∂uℓ∂um

=(

Xℓ ∂Y m

∂uℓ− Y ℓ ∂Xm

∂uℓ

) ∂vi

∂um+ XkY m ∂2vi

∂uk∂um− Y kXm ∂2vi

∂uk∂um

= [X, Y ]m∂vi

∂um+ 0

= [X, Y ]i

5

5. FLOWS

A vector field (section of T (M)) gives rise to a flow φt(p) = φ(p, t) : M ×I → M

where the φt are diffeomorphisms and the interval I is (−ǫ, ǫ), where ǫ may, ingeneral, depend on the value of p. Let {u1, . . . , un} be local coordinates and thevector field be given locally by

X = X i ∂

∂ui

The φt also have a local expression where φt(p) is given by {u1(t), . . . , un(t)}(where p corresponds to ui(0) = ui

0). Then φ(t) is determined by differentialequations with initial conditions. We denote the ui(t) with initial conditionsui(0) = ui

0) by ui(uk0 , t). The differential equations are:

duk(ui0, t)

dt= Xk(uj(ui

0, t)) uk(ui0, 0) = uk

0

The ui0 function as parameters in these equations. The flowline begins with the

point p that has these coordinates and flows out of p to points with the coor-dinates uj(ui

0, t). The Picard-Lindelof theorem guarantees that the solutionswill be as differentiable as the input data and depend as differentiably on theparameters as the X i do. Solutions will exist for some interval (−ǫ, ǫ) whereǫ > 0. However, in general the ǫ will depend on the intial point p. This won’tdo us any harm. Uniqueness of the solutions guarantees that

φt1(φt2 (p)) = φt1+t2(p)

as long as t1, t2, t1 + t2 remain within (−ǫ, ǫ). If M is compact we can say muchmore; φt is defined for all t ∈ R. However, for Lie purposes the local existenceis sufficient, so we will not pursue the matter here.

6. LIE DERIVATIVES

Here is the idea of the Lie Derivative. Given a tangent vector field X on M , thederivative in the X direction of an object is the rate of change along the flowlineφt(p) if this makes sense. Unfortunately, it only makes sense for functions. Tofind derivatives of objects that live in bundles we must change the game slightly,since objects in neighboring fibres cannot be directly compared. Let E be abundle over M . To compare an object Wq in the fibre Eq over q to an objectWp in the fibre Ep over p we need an isomorphism from Eq to Ep. This is notgenerally available, but if q is on the flowline out of p determined by the vectorfield X then we may be able to find such an isomorphism and this suffices todefine the Lie Derivative. Indeed if q = φt(p) then p = φ−1

t (q) = φ−t(q) and ifwe can find (φ−t)∗ : π−1[q] → π−1[p] then we can compare (φ−t)∗Wq with Wp.In fact

(φ−t)∗Wφt(p) ∈ π−1[p]

6

is, for small t, a curve in π−1[p] and hence can be differentiated at t = 0, yieldingan object in Tp(π

−1[p]) = TpWp. This is the Lie Derivative for the bundle E.The importance of understanding this construction is easy to underesti-

mate. If we understand it we understand a) where the Lie Derivative lives (inthe tangent space to the fibre of the bundle). If the fibre is a vector space it ispossible to identify the tangent space with the fibre itself, and this is often done.b) Knowing that we are dealing with a curve in the fibre Wp focuses our atten-tion properly when we are doing the technical calculations. c) it prepares us forthe idea of connections in vector bundles and principal fibre bundles where theideas are somewhat similar.

Naturally for different bundles (φ−t)∗ will have a different form, but this isjust technical stuff; the basic idea is given above.

First we will deal with the Lie Derivative of a function. In this case (φ−t)∗ =Identity and (φ−t)∗f = f . Hence

£X(f) =d

dt(f ◦ φt)

∣

∣

t=0

=d

dt(f(ui(t)))

∣

∣

t=0

=∂f

∂ui

dui

dt

∣

∣

∣

t=0

=∂f

∂uiX i

= X(f)

Our next project is the Lie Derivative of a Tangent Vector Field Y ∈ T (M).X is still the vector field with flow φt(p) with d

dtφt|t=0 = X . Recall that

φ−t = φ−1t . The isomorphism is

(φ−t)∗ : Tφt(p) → Tp(M)

where p = φ0(p) and q = φt(p). The rest is mere intricate calculation. Let p

have coordinates u1(0), . . . , un(0) and q have coordinates u1(t), . . . , un(t). Forease of presentation we will use vi = ui(t). Then (φt)∗ : Tp(M) → Tq(M) isgiven in coordinates, with Yq = (φt)∗Zp, by

Y iq =

∂vi

∂ujZj

p

and then (φ−t)∗ = (φ−1t )∗ is given by

(

∂vi

∂uj

)−1

Y 1q

...Y n

q

=

Z1p

...Zn

p

∈ Tp(M)

Notice that(

∂vi

∂uj

)

∣

∣

∣

t=0=

(

∂ui(t)

∂uj

)

∣

∣

∣

t=0= Id

7

We are going to need

d

dt

(

∂vi

∂uj

)−1∣

∣

∣

t=0= (−1)

(

∂vi

∂uj

)−2d

dt

(

∂vi

∂uj

)

∣

∣

∣

t=0

= (−1)

(

∂vi

∂uj

)−2(∂

∂uj

d

dtui(t)

)

∣

∣

∣

t=0

= (−1)(Id)

(

∂

∂ujX i

)

= −

(

∂X i

∂uj

)

Now we have, for Yq ∈ Tq(M), or more precisely

Yφt(p) ∈ Tφt(p)(M)

Z1p

...Zn

p

=

(

∂vi

∂uj

)−1

Y 1φt(p)

...Y n

φt(p)

a curve in the fibre Tp(M) whose derivative at t = 0 is the Lie Derivative£X(Y ) ∈ Tp,0(Tp(M)). We now compute

£X(Y ) =d

dt

Z1p

...Zn

p

∣

∣

∣

∣

∣

∣

∣

t=0

=

d

dt

(

(

∂vi

∂uj

)−1)

Y 1φt(p)

...Y n

φt(p)

+

(

∂vi

∂uj

)−1d

dt

Y 1φt(p)

...Y n

φt(p)

t=0

= −

(

∂X i

∂uj

)

Y 1p

...Y n

p

+

(

∂vi

∂uj

)−1∂

∂vj

Y 1φt(p)

...Y n

φt(p)

dvj

dt

t=0

= −

(

∂X i

∂uj

)

Y 1p

...Y n

p

+ Id

∂

∂uj

Y 1p

...Y n

p

Xj

If the coordinates of £X(Y ) in the local basis are denoted by £X(Y )i this gives

£X(Y )i = Xj ∂

∂ujY i − Y j ∂

∂ujX i = [X, Y ]i

and we have proved£X(Y ) = [X, Y ]

8

We note that £X(Y ) has two terms because it is the derivative of (φ−t)∗Yφt(p)

both factors of which depend on t, and the origin of the negative sign lies in theinverse: φ−t = φ−1

t . We also note that the above derivation is highly dependenton the continuity of the second derivatives.

Next we want to derive a formula for the Lie Derivative of the Lie Bracketof two Vector Fields. This is immediate from the Jacobi Identity. Indeed recall

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

from which we get

[X, [Y, Z]] = [Y, [X, Z]] − [Z, [X, Y ]]

£X([Y, Z]) = £Y ([X, Z]) − £Z([X, Y ])

£X([Y, Z]) = £Y £X(Z) − £Z£X(Y )

In contrast to the preceding the calculation of the Lie Derivative of a 1-formis somewhat easier as the formulation of (φ−t)∗ is a bit easier. It is simply thefamiliar pullback of differential forms. If φt(p) = q then

φ∗t : T ∗

q (M) → T ∗p (M)

is the pullback and we set, for this bundle, (φ−t)∗ = φ∗t . (Remember, to get a

Lie Derivative on a bundle we must have an isomorphism (φ−t)∗ from π−1[q] toπ−1[p]. We hope to find such an isomorphism which makes sense for the bundle.In the sequel we will also have to worry about whether the Lie Derivatives forthe various bundles fit together well. For T ∗(M) the above turns out to bea good choice, as well as the only obvious choice.) Let’s now recall how φ∗

works. If φ : M → N and u1, . . . , un are coordinates on M and v1, . . . , vn arecoordinates on N and φ is given locally by vi(u1, . . . , un) and a section ω isgiven locally on N by ωidvi then ω = φ∗(ω) is given locally on M by

ωi(uj) dui = ωj(v

k(uj))∂vj

∂uidui

In our case φt is expressed by vi = vi(uj), and we know that

dvk

dt

∣

∣

∣

t=0= Xk

sincedφt

dt

∣

∣

∣

t=0= X

Thus

£X(ω) =d

dt[(φ−t)∗ω(φt(p))]t=0

=d

dt[(φt)

∗ω(φt(p))]t=0

9

=d

dt

[

ωi

∂vi

∂ujduj

]

t=0

=

[

∂ωi

∂uk

duk

dt

∂vi

∂ujduj + ωi

d

dt

∂vi

∂ujduj

]

t=0

=∂ωi

∂ukXkδi

j duj +

[

ωi

∂

∂uj

dvi

dtduj

]

t=0

=∂ωi

∂ukXk dui + ωi

∂

∂ujX i duj

=

[

∂ωi

∂ujXj + ωj

∂Xj

∂ui

]

dui

= X(ωi) dui + ωi dX i

This last is an interesting formula and we will find use for it from time to time.The following calculations are not strictly necessary; they can be derived

abstractly from the fact that (φt)∗ and d commute. However, the formulas may

come in handy and and there is (for me) a certain amusement in the formulas.We first derive

£X(df) = d£X(f)

by computing both sides.

df =∂f

∂ujduj

£X(df) =

[

∂

∂ui

∂f

∂ujXj +

∂f

∂uj

∂Xj

∂ui

]

dui

=

[

∂2f

∂uj∂uiXj +

∂f

∂uj

∂Xj

∂ui

]

dui

d£X(f) = d(Xf) = d

(

∂f

∂ujXj

)

=

[

∂2f

∂ui∂ujXj +

∂f

∂uj

∂Xj

∂ui

]

dui

Next I want

d£X(uk) = d(X(uk)) = d

(

Xj ∂uk

∂uj

)

= d(Xjδkj ) = dXk

and then

£X(duk) = d£X(uk)

= dXk

Now we can verify that Leibniz’ rule works here; for ω ∈ Λ1

£X(ω) = £X(ωi dui)

10

= £X(ωi) dui + ωi£X(dui)

= X(ωi) dui + ωi dX i

which coincides with our previous formula. We can read this in two ways; if wehave Leibniz’ formula for products then this says the general formula for £X(ω)is derivable from the formulas for £X(f) and £X(duk), or we can view it as ajustification for Leibniz’s formula given our previous work.

We can sweat a little more out of the formula

£X(ω) = X(ωi) dui + ωi dX i

by applying it to a vector field Y . We get

£X(ω)(Y ) = X(ωi) dui(Y ) + ωi dX i(Y )

= X(ωi)Yi + ωiY (X i)

= X(ωiYi) − ωiX(Y i) + ωiY (X i)

= X(ω(Y )) − ωi[X, Y ]i

= X(ω(Y )) − ω([X, Y ])

= £X(ω(Y )) − ω([X, Y ])

This is often written as

£X(ω(Y )) − £X(ω)(Y ) = ω([X, Y ])

This is a handy formula and we will meet it again in another context.

7. THE dω(X0, . . . , Xr) FORMULA

In this section we wish to derive a formula which shows how dω acts on T (M).Specifically, for ω in Λr(M), we want the formula

dω(X0, X1, · · ·Xr) =r∑

i=0

(−1)iX i(ω(X0, · · · , Xi−1, Xi+1, · · · , Xr)

+∑

0≤i<j≤r

(−1)i+jω([Xi, Xj ], · · · , Xi−1, Xi+1, · · · , Xj−1, Xj+1, · · · , Xr)

This way of expressing the formula is a little cumbersome so we will make ofuse of the convention that a hat on a letter means the letter is omitted . Theformula then reads

dω(X0, X1, · · ·Xr) =

r∑

i=0

(−1)iX i(ω(X0, · · · , Xi, · · · , Xr)

+∑

0≤i<j≤r

(−1)i+jω([Xi, Xj ], · · · , Xi, · · · , Xj, · · · , Xr)

11

This interesting formula explicitly gives the value of dω evaluated on vec-tors. It is even possible to use it to define dω, as Godbillion[2], does instead ofdetermining dω by the three rules

df =∂f

∂uidui

d(ω1 ∧ ω2) = dω1 ∧ ω2 + (−1)deg(ω1)ω1 ∧ dω2

ddω = 0

However, we will follow the classical route which requires us to prove this for-mula.

It is worth doing the special case of a one-form first to get the feel for thesituation. We have ω ∈ Λ1(M) and X0 and X1 are two vector fields on M .Then locally

ω = ωi dui

dω(X0, X1) =∂ωi

∂ujduj ∧ dui(X0, X1)

=∂ωi

∂uj

(

duj(X0)dui(X1) − dui(X0)duj(X1))

=∂ωi

∂uj

(

Xj0X i

1 − X i0X

j1

)

= Xj0

∂ωi

∂ujX i

1 − Xj1

∂ωi

∂ujX i

0

= Xj0

∂

∂uj

(

ωiXi1

)

− Xj0ωi

∂X i1

∂uj

−Xj1

∂

∂uj

(

ωiXi0

)

+ Xj1ωi

∂X i0

∂uj

= X0(ω(X1) − X1(ω(X0) − ωi

(

Xj0

∂X i1

∂uj− X

j1

∂X i0

∂uj

)

= X0(ω(X1) − X1(ω(X0) − ωi

(

[X0, X1]i)

= X0(ω(X1) − X1(ω(X0) − ω(

[X0, X1])

This verifies the formula for the case r = 1. Unfortunately for r > 1 the situationis not so simple. As Helgason[3] remarks, this formula is essentially trivial butit is a bit tricky to push through in detail. In fact, it’s not too easy to find asource where this is done. The steps are, in the big picture, the same as thosefor r = 1 but there are so many things to keep straight that the proof is a bitunwieldy.

Before starting we remind the reader of the formula for diffentiating adeterminant. It is, for r = 3, simply

∂

∂u

∣

∣

∣

∣

∣

∣

f g h

i j k

l m n

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

∂f∂u

g h∂i∂u

j k∂l∂u

m n

∣

∣

∣

∣

∣

∣

+

∣

∣

∣

∣

∣

∣

f ∂g∂u

h

i ∂j∂u

k

l ∂m∂u

n

∣

∣

∣

∣

∣

∣

+

∣

∣

∣

∣

∣

∣

f g ∂h∂u

i j ∂k∂u

l m ∂n∂u

∣

∣

∣

∣

∣

∣

12

as is immediate from the formula∣

∣

∣

∣

∣

∣

a11 a1

2 a13

a21 a2

2 a23

a31 a3

2 a33

∣

∣

∣

∣

∣

∣

=∑

π∈S3

sgn(π) aπ(1)1 a

π(2)2 a

π(3)3

By linearity the preceding formula gives, for a vector field X ,

X

∣

∣

∣

∣

∣

∣

f g h

i j k

l m n

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

X(f) g h

X(i) j k

X(l) m n

∣

∣

∣

∣

∣

∣

+

∣

∣

∣

∣

∣

∣

f X(g) h

i X(j) k

l X(m) n

∣

∣

∣

∣

∣

∣

+

∣

∣

∣

∣

∣

∣

f g X(h)i j X(k)l m X(n)

∣

∣

∣

∣

∣

∣

We also remind the reader that the hat on an expression means that it is omitted,so that for example

∣

∣

∣

∣

∣

∣

f g h i

j k l m

n o p q

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

f g i

j k m

n o q

∣

∣

∣

∣

∣

∣

We are now going to prove the general formula

dω(X0, X1, · · ·Xr) =

r∑

i=0

(−1)iX i(ω(X0, · · · , Xi, · · · , Xr)

+∑

0≤i<j≤r

(−1)i+jω([Xi, Xj ], X0, · · · , Xi, · · · , Xj , · · · , Xr)

By linearity, it suffices to prove the formula for

ω = f dui1 ∧ . . . ∧ duir

We recall that, with Xk = Xjk

∂∂uj , that

dui1 ∧ . . . ∧ duir (X1, . . . , Xr) = det |duij (Xk)|

Thus we are trying to prove that

dω(X0, . . . , Xr) =

r∑

i=0

(−1)iXi

(

fdui1 ∧ . . . ∧ duir (X0, . . . , Xi, . . . , Xr))

+∑

0≤i<j≤r

(−1)i+jf dui1 ∧ . . . ∧ duir ([Xi, Xj ], X0, . . . , Xi, . . . , Xj , . . . , Xr)

This is in some sense straightforward, and there is only one substantive step;everything else is cosmetic rearrangement. I will point out the substantive stepwhen we get there. For ease of reading I will leave out the wedges in thecalculation. Here we go....

dω(X0, . . . , Xr) =

r∑

j=1

∂f

∂ujdujdui1 . . . duir (X0, X1, . . . , Xr)

13

=∑

j

∂f

∂uj

∣

∣

∣

∣

∣

∣

∣

∣

duj(X0) duj(X1) . . . duj(Xr)dui1(X0) dui1(X1) . . . dui1(Xr)

· · · · · · · · · · · ·duir (X0) duir (X1) . . . duir (Xr)

∣

∣

∣

∣

∣

∣

∣

∣

=∑

j

∂f

∂uj

∣

∣

∣

∣

∣

∣

∣

∣

Xj0 X

j1 . . . Xj

r

X i10 X i1

1 . . . X i1r

· · · · · · · · · · · ·

X ir

0 X ir

1 . . . X irr

∣

∣

∣

∣

∣

∣

∣

∣

Now we do an expansion off the ith column to get

=∑

j

r∑

i=0

(−1)i ∂f

∂ujX

ji

∣

∣

∣

∣

∣

∣

∣

∣

X i10 . . . X i1

i . . . X i1r

X i20 . . . X i2

i . . . X i2r

· · · · · · · · · · · ·

X ir

0 . . . X ir

i . . . X irr

∣

∣

∣

∣

∣

∣

∣

∣

=r∑

i=0

(−1)iXi(f)

∣

∣

∣

∣

∣

∣

∣

∣

X i10 . . . X i1

i . . . X i1r

X i20 . . . X i2

i . . . X i2r

· · · · · · · · · · · ·

X ir

0 . . . X ir

i . . . X irr

∣

∣

∣

∣

∣

∣

∣

∣

Now comes the one substantive step; we use X(f)g = X(fg)− fX(g):

=

r∑

i=0

(−1)iXi

f

∣

∣

∣

∣

∣

∣

∣

∣

X i10 . . . X i1

i . . . X i1r

X i20 . . . X i2

i . . . X i2r

· · · · · · · · · · · · · · ·

X ir

0 . . . X ir

i . . . X irr

∣

∣

∣

∣

∣

∣

∣

∣

−

r∑

i=0

(−1)i fXi

∣

∣

∣

∣

∣

∣

∣

∣

X i10 . . . X i1

i . . . X i1r

X i20 . . . X i2

i . . . X i2r

· · · · · · · · · · · · · · ·

X ir

0 . . . X ir

i . . . X irr

∣

∣

∣

∣

∣

∣

∣

∣

=

r∑

i=0

(−1)iXi

f

∣

∣

∣

∣

∣

∣

∣

∣

∣

dui1(X0) . . . dui1(Xi) . . . dui1(Xr)

dui2(X0) . . . dui2(Xi) . . . dui2(Xr)· · · · · · · · · · · · · · ·

duir (X0) . . . duir (Xi) . . . duir (Xr)

∣

∣

∣

∣

∣

∣

∣

∣

∣

−

r∑

i=0

∑

k 6=i

(−1)i f

∣

∣

∣

∣

∣

∣

∣

∣

X i10 . . . Xi(X

i1k ) · · · X i1

i . . . X i1r

X i20 . . . Xi(X

i2k ) · · · X i2

i . . . X i2r

· · · · · · · · · · · · · · · · · · · · ·

X ir

0 . . . Xi(Xir

k ) · · · X ir

i . . . X irr

∣

∣

∣

∣

∣

∣

∣

∣

=

r∑

i=0

(−1)iXi

(

fdui1 · · · duir (X0, . . . , Xi, . . .Xr))

14

−r∑

i=0

∑

k<i

(−1)i f

∣

∣

∣

∣

∣

∣

∣

∣

X i10 . . . Xi(X

i1k ) · · · X i1

i . . . X i1r

X i20 . . . Xi(X

i2k ) · · · X i2

i . . . X i2r

· · · · · · · · · · · · · · · · · · · · ·

X ir

0 . . . Xi(Xir

k ) · · · X ir

i . . . X irr

∣

∣

∣

∣

∣

∣

∣

∣

−

r∑

i=0

∑

k>i

(−1)i f

∣

∣

∣

∣

∣

∣

∣

∣

X i10 . . . X i1

i . . . Xi(Xi1k ) . . . X i1

r

X i20 . . . X i2

i . . . Xi(Xi2k ) . . . X i2

r

· · · · · · · · · · · ·

X ir

0 . . . X ir

i . . . Xi(Xir

k ) . . . X irr

∣

∣

∣

∣

∣

∣

∣

∣

=

r∑

i=0

(−1)iXi(ω(X0, . . . , Xi, . . . , Xr)

−

r∑

i=0

i−1∑

k=0

(−1)i f (−1)k

∣

∣

∣

∣

∣

∣

∣

∣

Xi(Xi1k ) X i1

0 · · · X i1k · · · X i1

i · · · X i1r

Xi(Xi2k ) X i2

0 · · · X i2k · · · X i2

i · · · X i2r

· · · · · · · · · · · · · · · · · · · · · · · ·

Xi(Xir

k ) X ir

0 · · · X ir

k · · · X ir

i · · · X irr

∣

∣

∣

∣

∣

∣

∣

∣

−

r∑

i=0

r∑

k=i+1

(−1)i f (−1)k−1

∣

∣

∣

∣

∣

∣

∣

∣

Xi(Xi1k ) X i1

0 · · · X i1i · · · X i1

k · · · X i1r

Xi(Xi2k ) X i2

0 · · · X i2i · · · X i2

k · · · X i2r

· · · · · · · · · · · · · · · · · · · · · · · ·

Xi(Xir

k ) X ir

0 · · · X ir

i · · · X ir

k · · · X irr

∣

∣

∣

∣

∣

∣

∣

∣

=r∑

i=0

(−1)iXi(ω(X0, . . . , Xi, . . . , Xr)

−∑

0≤i<k≤r

(−1)i+k f

∣

∣

∣

∣

∣

∣

∣

∣

Xi(Xi1k ) − Xk(X i1

i ) X i10 · · · X i1

i · · · X i1k · · · X i1

r

Xi(Xi2k ) − Xk(X i2

i ) X i20 · · · X i2

i · · · X i2k · · · X i2

r

· · · · · · · · · · · · · · · · · · · · · · · ·

Xi(Xir

k ) − Xk(X ir

i ) X ir

0 · · · X ir

i · · · X ir

k · · · X irr

∣

∣

∣

∣

∣

∣

∣

∣

=r∑

i=0

(−1)iXi(ω(X0, . . . , Xi, . . . , Xr)

−∑

0≤i<k≤r

(−1)i+k f

∣

∣

∣

∣

∣

∣

∣

∣

∣

dui1([Xi, Xk]) dui1(X0) · · · dui1(Xi) · · · dui1(Xk) · · · dui1(Xr)

dui2([Xi, Xk]) dui2(X0) · · · dui2(Xi) · · · dui2(Xk) · · · dui2(Xr)· · · · · · · · · · · · · · · · · · · · · · · ·

duir([Xi, Xk]) duir (X0) · · · duir (Xi) · · · duir (Xk) · · · duir (Xr)

∣

∣

∣

∣

∣

∣

∣

∣

∣

=

r∑

i=0

(−1)iXi(ω(X0, . . . , Xi, . . . , Xr)

−∑

0≤i<k≤r

(−1)i+k f dui1dui2 · · ·duir ([Xi, Xk], X0, . . . , Xi, . . . , Xk . . . , Xr)

=

r∑

i=0

(−1)iXi(ω(X0, . . . , Xi, . . . , Xr)

15

−∑

0≤i<k≤r

(−1)i+k ω([Xi, Xk], X0, . . . , Xi, . . . , Xk . . . , Xr)

and the proof is complete.

8. SOME ALGEBRAIC IDEAS

The calculation in the previous section might be regarded by some people asunpleasant. In the next section we will develop what we will call the FrenchMethod for proving identities. This will give us a less computational mode ofattack. However, to do this efficiently we want to bring on board some algebraicideas which will clarify the structure of the theory.

We suppose that we have an Algebra A (i.e. a vector space with an associa-tive multiplication respecting the vector space operations) which is GRADED

in the sense that

A =

∞⊕

k=0

Ak

Ai · Aj ⊆ Ai+j

Elements of Ai are called homogeneous of degree i. A0 and⊕∞

k=0 A2k formnatural subalgebras of the graded algebra A. A paradigmatic example of thisconcept is the Exterior Algebra Λ(M) =

⊕∞k=0 Λk(M)

In recent years the term communtative has come to be applied to gradedalgebras that satisify the equation

xixj = (−1)ij xjxi xi ∈ Ai, xj ∈ Aj

However, to prevent confusion we will refer to this as a Graded Commuatitive

Algebra. There is a commuatator that goes with this situation given by

[xi, xj ] = xixj − (−1)ij xjxi

and of course the algebra is graded communtative if and only if the commutatorvanishes.

The case of interest to us is when the graded algebra A is generated bythe elements of degrees 0 and 1. In the case of A = Λ(M) this would be thefunctions and 1−forms. Abstractly we say

A is generated by A0 ⊕ A1

Next we introduce the concept of a Graded Derivation of degree p. We willtemporarily use the letters a and b for graded derivations. A function a : A → A

is a graded derivation if for each r it satisfies

a : Ar → Ar+p

a(xy) = a(x)y + (−1)rp xa(y) for x ∈ Ar

16

We also define a bracket [a, b] of two graded derivations of degrees p and q

respectively by[a, b] = ab − (−1)pq ba

The theorem is then

Theorem If a and b are graded derivations of degrees p and q respectively then[a, b] is a graded derivation of degree p + q.

Proof The proof is straightforward if a trifle intricate. Recall that

a : Ar → Ar+p

b : Ar → Ar+q

a(xy) = a(x)y + (−1)rpxa(y) x ∈ Ar

b(xy) = b(x)y + (−1)rqxb(y) x ∈ Ar

By the definition of bracket for graded derivation we have

[a, b] = ab − (−1)pqba

Hence

[a, b](xy) = (ab)(xy) − (−1)pq(ba)(xy)

= a[

(bx)y + (−1)rqx(by)]

− (−1)pqb[

(ax)y + (−1)rpx(ay)]

= (abx)y + (−1)p(r+q)(bx)(ay) + (−1)rq(ax)(by) + (−1)rpx(aby)

−(−1)pq[

(bax)y + (−1)q(r+p)(ax)(by) + (−1)rp(bx)(ay) + (−1)rp(−1)rqx(bay)]

=(

(ab − (−1)pqba)x)

y + (−1)r(p+q)x(

(ab − (−1)pqba)y)

+(

(−1)p(r+q) − (−1)pq+rp)

(bx)(ay) +(

(−1)rq − (−1)pq+qr+qp)

(ax)(by)

=(

([a, b](x))

y + (−1)(p+q)rx[a, b](y)

because the 3rd and 4th terms in the penultimate equation are 0 due to cancel-lation of the powers of -1 in the coefficients.

The popular terminology for derivations is as follows.If p is even a graded derivation is called a derivation

If p is odd a graded derivation is called an antiderivation

Let a1 and a2 be antiderivations with odd degrees p1 and p2

Let b1 and b2 be derivations with even degrees q1 and q2

Then our result about brackets of graded derivations tell us that

[a1, a2] = a1a2 + a2a1is a derivation of degree p1 + p2

[a1, a1] = 2a21is a derivation of degree 2p1

[a1, b1] = a1b1 − b1a1is an antiderivation of degree p1 + q1

[b1, b2] = b1b2 − b2b1is a derivation of degree q1 + q2

17

The following theorem is almost obvious.

Theorem Suppose that the graded algebra A =⊕

Ai is generated by A0 ⊕A1

as an algebra and suppose a and b are graded derivations of degree p and supposethat they agree on A0 ⊕A1 (that is, on scalars and ”vectors”). Then they agreeon the whole graded algebra

⊕

Ai.

Proof Since the elements of Ai are generated by products of vectors from A1

and scalars from A0, the value of a on an element of Ai is determined by thevalues on A0 and A1. Since a and b coincide on A0 and A1, they coincide onthe entire algebra.

Now the obvious next question is whether one can create a graded derivationon an algebra generated by A0 ⊕ A1 by specifying it it on the elements of A0

and A1 and requiring it to satisfy the graded derivation law. The answer, ingeneral, is NO because an element of Ai may be a product of elements of A0

and A1 in many different ways, and the different ways may lead to inconsistentvalues of the graded derivation on that element. Algebraically this means thatthe graded algebra has relations .

But suppose the algebra A is a FREE graded Algebra, which means thereare no relations between the elements except those that are forced by beinga graded algebra. Then indeed the construction of the derivation is possible.An example of such a free graded algebra is the Exterior Algebra of FormsΛ(M) on a manifold M . It is for precisely this reason that the operator d maybe uniquely determined by its value on functions f and one forms ω and thelaw d(ωη) = d(ω)η + (−1)deg ωωd(η) which says precisely that d is a gradedderivation of degree 1. This bit of trickery works because Λ(M) is free.

9. INTERACTION OF THE LIE DERIVATIVE

WITH d AND INNER PRODUCT

In this section we will use the ideas in the previous section to develop whatwe will call the French Method for proving identities. This will give us a lesscomputational mode of attack.

First however we show that £X commutes with d. Recall that if X ∈Γ(T (M)) (i.e. a vector field) and φt is the corresponding flow then for ω ∈Λr(T ∗(M)) = Λr(M)

£Xω = limt→0

φ∗t (ω) − ω

t

Recall that φ∗t (ω1 ∧ ω2) = φ∗

t (ω1) ∧ φ∗t (ω2). Then

φ∗t (ω1 ∧ ω2) − ω1 ∧ ω2 = φ∗

t (ω1) ∧ φ∗t (ω2)ω1 ∧ ω2

= φ∗t (ω1) ∧ φ∗

t (ω2) − ω1 ∧ φ∗t (ω2)

+ ω1 ∧ φ∗t (ω2) − ω1 ∧ ω2

£X(ω1 ∧ ω2) = limt→0

φ∗t (ω1 ∧ ω2) − ω1 ∧ ω2

t

18

= limt→0

(φ∗t (ω1) − ω1) ∧ φ∗

t (ω2) + ω1 ∧ (φ∗t (ω2) − ω2)

t

= £X(ω1) ∧ ω2 + ω1 ∧ £X(ω2)

as we would expect for a reasonable product.In a similar way, since φ∗

t (dω) = dφ∗t (ω) we have

£X(dω) = limt→0

φ∗t (dω) − dω

t= lim

t→0

dφ∗t (ω) − dω

t= d£X(ω)

Note that t is not one of the variables involved in d. We previously proved theformula above for 1−forms but this proof is valid for r−forms. A different proofof this is possible using methods we are about to introduce.

Next notice that £X is a graded derivation of degree 0 and d is a gradedderivation of degree 1. Hence by the theorem in the previous section,

[d,£X ] = d ◦ £X − (−1)−1·0£X ◦ d = d ◦ £X − £X ◦ d

is a graded derivation of degree 1. We have just shown it is identically 0. How-ever, note that this fact could be proved by showing that [d,£X ] is identically0 on functions and 1−forms. It would then follow that it is 0 on all forms, bythe discussion at the end of the previous section. We will call this the FrenchMethod.

Clearly there is no fun to be had here with d and £X because they commute.For real fun we need another graded derivation which does NOT commute withd or £X .

Our first example is simply to use two Lie Derivatives, £X and £Y . Wewill show that

[£X ,£Y ] = £[X,Y ]

Since these are graded derivations of degree 0, it suffices to show that theycoincide on functions and one forms. Note that we have

[£X ,£Y ] = £X ◦ £Y − (−1)0·0£Y ◦ £X = £X ◦ £Y − £Y ◦£X

On functions this gives

[£X ,£Y ](f) = £X ◦ £Y (f) − £Y ◦ £X(f) = £X(Y (f)) − £Y (X(f))

= X(Y (f)) − Y (X(f)) = (XY − Y X)(f) = [X, Y ](f)

= £[X,Y ](f)

For one-forms it is perfectly possible to perform a direct attack and fill a pagewith computations that establish that the two derivations coincide on one forms.However, we would like to do this in a smarter way. The miminal one-form wecan work with is dui and we can use the commutativity of £X and d effectivelyhere; we don’t even need to look at the formula for £X on one-forms.

(£X£Y − £Y £X)(dui) = d(£X£Y − £Y £X)(ui)

= d£[X,Y ](ui)

= £[X,Y ](dui)

19

Since the two derivations £X£Y −£Y £X and £[X,Y ] agree on functions f andthe special one forms dui, they will agree on all one forms ω = fidui, as required.

Another very important example is the inner product of a vector field anda differential form. This is defined byDef

X | f = 0 for f ∈ Λ0(M)

X |ω = ω(X) for ω ∈ Λ1(M)

X | is a graded derivation of degree − 1

This completely defines X | as we discussed previously, and no further formulaeare theoretically necessary. But it is nice to know that for ω ∈ Λr(M)

X |ω(Y2, Y3, . . . , Yr) = ω(X, Y2, Y3, . . . , Yr)

However, in order not to break up the derivation we will prove the generalformula in appendix 1 of this section. The case r = 2 we need below.

We now want to use all our equipment to prove Cartan’s Homotopy formulaTheorem For X a vector field on M and ω ∈ Λr(M).

£Xω = d(X |ω) + X | (dω)

Since d is a graded derivation of degree 1 and X | is a graded derivation ofdegree -1 we know that

[d, X | ] = d ◦ X | + X | ◦ d

will be a graded derivation of degree 0. Since the only graded derivation ofdegree 0 that we know is £X the result is not surprising. Moreover, as wediscussed above we can prove the formula by verifying it for f ∈ Λ0(M) and

ω ∈ Λ1(M). Since £X and [d, X | ] coincide on Λ0(M) and Λ1(M), they mustcoincide on all Λr(M). We proceed to verify the two cases. for f ∈ Λ0(M)

d(X | f) + X | (d f) = X | (d f)

= d f(X) = X(f) = £X(f)

and for ω ∈ Λ1(M), using the formula (X |ω)(Y ) = ω(X, Y ),

(d(X |ω) + X | (dω))(Y ) = d(ω(X))Y + dω(X, Y )

= Y (ω(X)) + X(ω(Y )) − Y (ω(X)) − ω([X, Y ])

= X(ω(Y )) − ω([X, Y ])

= £X(ω)(Y )

where we used the formula £X(ω(Y )) − £X(ω)(Y ) = ω([X, Y ])

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Since we have three graded derivations, d, X | and £X , we should be ableto find formulas using any two of them. We just did the first two, and the paird, £X is dull since the two commute, so the remaining possibility is £X andX | . The commutator of these two should give us a graded derivation of degree-1. Let us see if we can turn it up and simultaneously prove the formula. Forf ∈ Λ0(M)

£X(Y | f) − Y |£X(f) = 0

No help there. For ω ∈ Λ1(M)

£X(Y |ω) − Y |£X(ω) = £X(ω(Y )) − (£X(ω))(Y )

= ω([X, Y ])

= [X, Y ] |ω

AHA!. We now see that

[£X , Y | ] = £X ◦ Y | − Y | ◦ £X = [X, Y ] |

which is the formula we were seeking. The formula is true on functions and1−forms, and since it is a graded derivation is true on all r−forms.

We have now exhasted our supply of graded derivations. However, I wishto present one more formula which has a slightly different feel to it and is notproved in the same way. For this formula we must define the Outer Product orExterior Product.

Def Let ω ∈ Λ1(M). Then the Outer Product ǫ(ω) is defined by

ǫ(ω)(η) = ω ∧ η for η ∈ Λr(M)

Note that ǫ(ω) is not a derivation. However, it does have an interesting formula

which looks a little like a derivation formula. First, recall that X | is a gradedderivation of degree -1 so we have for ω ∈ Λ1(M)

X | (ω ∧ η) = (X |ω) ∧ η − ω ∧ (X | η)

Rewriting this with the outer product we have

X | (ǫ(ω)(η)) = (ω(X))η − ǫ(ω)(X | η)

X | (ǫ(ω)(η)) + ǫ(ω)(X | η) = (ω(X))η

and so we haveX | ◦ ǫ(ω) + ǫ(ω) ◦ X | = ω(X)

where the right hand side is to be interpreted as the operator of scalar multi-plication by ω(X).

Appendix 1 Here we prove the formula for the inner product

(X |ω)(Y2, . . . , Yr) = ω(X, Y2, . . . , Yr)

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for ω ∈ Λr(M). We begin in a leisurely manner by proving the formula forr = 2. For this we have

X | (ω1 ∧ ω2) = (X |ω1)ω2 − ω1(X |ω2)

= ω1(X)ω2 − ω2(X)ω1

(X | (ω1 ∧ ω2))(Y ) = ω1(X)ω2(Y ) − ω2(X)ω1(Y )

= (ω1 ∧ ω2)(X, Y )

For the general case we remind the reader of some formulae. First

ω1 ∧ ω2 ∧ . . . ∧ ωr(Y1, . . . , Yr) = det

ω1(Y1) ω1(Y2) · · · ω1(Yr)ω2(Y1) ω2(Y2) · · · ω2(Yr)· · · · · · · · · · · ·

ωr(Y1) ωr(Y2) · · · ωr(Yr)

ω1∧ω2∧ . . .∧ωr(Y1, . . . , Yr) =

r∑

i=1

(−1)i−1ω1(Yi)ω2∧ . . .∧ωr(Y1, . . . , Yi, . . . , Yr)

Thus for ω ∈ Λr−1

ω1 ∧ ω(Y1, . . . , Yr) =r∑

i=1

(−1)i−1ω1(Yi)ω(Y1, . . . , Yi, . . . , Yr)

Now we begin our proof by induction. Assume the formula is true for r − 1.Then for ω1 ∈ Λ1(M) and ω ∈ Λr(M)

Y1| (ω1 ∧ ω)(Y2, . . . , Yr)

= (Y1|ω1)ω(Y2, . . . , Yr) − (ω1 ∧ (Y1

|ω))(Y2, . . . , Yr)

= ω1(Y1)ω(Y2, . . . , Yr) −

r∑

i=2

(−1)iω1(Yi)(Y1|ω)(Y2, . . . , Yi, . . . , Yr)

Now using the induction assumption we have

Y1| (ω1 ∧ ω)(Y2, . . . , Yr)

= ω1(Y1)ω(Y2, . . . , Yr) −

r∑

i=2

(−1)iω1(Yi)ω(Y1, . . . , Yi, . . . , Yr)

=

r∑

i=1

(−1)i−1ω1(Yi)ω(Y1, . . . , Yi, . . . , Yr)

= (ω1 ∧ ω)(Y1, . . . , Yr)

This completes the proof.

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References

[1] Berline,N; Getzler, E. & Vergne, M. Heat Kernals and Dirac Operators

Springer Verlag, Berlin, 1992

[2] Godbillon, Claude ; Geometrie Differentielle et Mechanique Analytique

Hermann Paris 1969

[3] Helgason, Sigurdur; Differential Geometry and Symmetric Spaces Ameri-can Mathematical Society, Tel Aviv, 2000

[4] Loomis, Lynn H.; Advanced Calculus 2nd edition, Jones and Bartlett,Boston, 1990

[5] Morrey, Charles B.; Multiple Integrals in the Calculus of Variations,

Springer Verlag, Berlin, 1966

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