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X constant (but arbitrary) we have to consider only the following expression:

’ ω(x), (dY (x)— —¦ ιE )— .X(x)
X(x), ω(x), dY (x) =
E E E

’ ω(x), ι— .dY (x)—— .X(x)
= X(x), ω(x) —¦ dY (x) E E
E
= X(x), dY (x)— .ω(x) ’ ι—— .ω(x), dY (x)—— .X(x)
E E
E
= ιE .ω(x), dY (x)—— .X(x) ’ ι—— .ω(x), dY (x) .X(x)
——
E E
E
’ ι—— ).ω(x), dY (x) .X(x)
——
= (ιE ,
E
E


which is not 0 in general since ker(ιE ’ ι—— ) = ιE (E ) at least for Banach spaces,
E
see [Cigler, Losert, Michor, 1979, 1.15], applied to ιE . So we may assume that
(ιE ’ ι—— ).ω(x) = 0 ∈ E . We choose a non-re¬‚exive Banach space which is
E
isomorphic to its bidual ([James, 1951]) and we choose as dY (x) this isomorphism,
then dY (x)—— is also an isomorphism, and a suitable X(x) makes the expression
nonzero.

Note that this also shows that for general convenient vector spaces E the exterior
(1)
derivative dω is in C ∞ (U, L2 (D0 E, R)) only if ω ∈ C ∞ (M ← T M ). Note that
alt
even for ω : U ’ E a constant 1-form of order 1 we need not have dω = 0.

33.14. Example. There exist c∞ -open subsets U and V in a Banach space E, a
di¬eomorphism f : U ’ V , and a 1-form ω ∈ C ∞ (U, L(E , R)) such that dloc f — ω =
f — dloc ω.

Proof. We start in a more general situation. Let f : U ’ V ‚ F be a smooth
(1)
mapping, and let Xx , Yx ∈ Dx U = E . Then we have

dloc (f — ω)x (Xx , Yx ) = Dx (f — ω( ).Yx ).Xx ’ Dx (f — ω( ).Xx ).Yx
’ ...
= Dx (ω(f ( )).D( ) f.Yx ).Xx

= Xx ω —¦ f, D( ) f.Yx F ’ . . .

ω —¦ f, df ( )—— .Yx F (x)—— .Xx ’ . . . by (32.6)
=d
ω(f ( )), df (x)—— .Yx F (x)—— .Xx +
=d
d ω(f (x)), df ( )—— .Yx F (x)—— .Xx ’ . . . by
+ (32.6)
f — (dloc ω)x (Xx , Yx ) = (dloc ω)f (x) (Dx f.Xx , Dx f.Yx )
) .Dx f.Yx ).Dx f.Xx ’ Df (x) (ω( ) .Dx f.Xx ).Dx f.Yx
= Df (x) (ω(
), df (x)—— .Yx F (f (x))—— .df (x)—— .Xx ’ . . .
= d ω(

——
Recall that for ∈ H = L(H, R) the bidual mapping satis¬es L(H , R) =
ιH ( ) ∈ H . Then for the di¬erence we get

dloc (f — ω)x (Xx , Yx ) ’ f — (dloc ω)x (Xx , Yx )
)—— .Yx (x)—— .Xx ’ d ω(f (x)), df ( )—— .Xx (x)—— .Yx
= d ω(f (x)), df ( F F

) .Yx )(x)—— .Xx ’ d(df (
——
)—— .Xx )(x)—— .Yx
= iF ω(f (x)), d(df ( .
F


33.14
346 Chapter VII. Calculus on in¬nite dimensional manifolds 33.15

This expression does not vanish in general, e.g., when the following choices are
made: We put ω(f (x)) = ιF . = —— for ∈ F , and we have

)—— Yx )(x)—— .Xx = d(d , f )—— Yx )(x)—— .Xx
d(d( —¦ f )( F(

)—— Yx (x)—— .Xx
= d ιF , df ( F
——
) Yx )(x)—— .Xx
——
= ιF , d(df ( ,
F


which is not symmetric in general for —¦ f = ev : G — G ’ R (for a non re¬‚ex-
ive Banach space G) by the argument in (32.7). It remains to show that such a
factorization of ev over a di¬eomorphism f and ∈ (G — G) is possible. Choose
(±, x) ∈ G — G such that ±, x = 1, and consider

f
G — G = G — ker ± — R.x ’ G — ker ± — R.x ’ R
’ ’
(β, y, tx) ’ (β, y, β, y + tx ’ β, y + tx
G .x) G

(β, y, t’β,x .x) ← (β, y, tx).
β,y



33.15. Proposition. Let f : M ’ N be a smooth mapping between smooth
manifolds. Then we have

f — —¦ d = d —¦ f — : C ∞ (Lk (T N, N — R)) ’ C ∞ (Lk+1 (T M, M — R)).
alt alt



Proof. Since by (33.12) the local and global formula for the exterior derivative
coincide on spaces C ∞ (Lk (D± M, M — R)) we shall prove the result with help of
alt
the local formula. So we may assume that f : U ’ V is smooth between c∞ -open
sets in convenient vector spaces E and F , respectively. Note that we may use the
global formula only if f is a local di¬eomorphism, see (33.9).
For ω ∈ C ∞ (V, Lk (F, R)), x ∈ U , and Xi ∈ E we have
alt


(f — ω)(x)(X1 , . . . , Xk ) = ω(f (x))(df (x).X1 , . . . , df (x).Xk ),

so by (33.12.1) we may compute

k
(df — ω)(x)(X0 , . . . , Xk ) = (’1)i d(f — ω)(x)(Xi )(X0 , . . . , Xi , . . . , Xk )
i=0
k
(’1)i (dω(f (x)).df (x).Xi )(df (x).X0 , . . . , i , . . . , df (x).Xk )
=
i=0
k
(’1)i ω(f (x))(df (x).X0 , . . . , d2 f (x).(Xi , Xj ), . . . , i , . . . , df (x).Xk )
+
i=0 j<i
k
(’1)i ω(f (x))(df (x).X0 , . . . , i , . . . , d2 f (x).(Xi , Xj ), . . . , df (x).Xk )
+
i=0 j>i


33.15
33.18 33. Di¬erential forms 347

k
(’1)i dω(f (x))(df (x).X0 , . . . , df (x).Xk )
=
i=0

(’1)i+j ω(f (x))(d2 f (x).(Xi , Xj ) ’ d2 f (x).(Xj , Xi ),
+
j<i
df (x).X0 , . . . , j , . . . , i , . . . , df (x).Xk )
= (f — dω)(x)(X0 , . . . , Xk ) + 0.


33.16. Example. There exists a smooth function

f ∈ C ∞ (E, R) = C ∞ (E, L0 (D(1) E, R))
alt


such that
0 = dloc dloc f ∈ C ∞ (E, L2 (D(1) E, R)).
alt


(1)
Proof. Let f ∈ C ∞ (E, R), Xx , Yx ∈ Dx E = E . Then we have

(dloc f )x (Xx ) = df (x)—— .Xx = ιF .df (x), Xx E

= Xx , df (x) E

(dloc dloc f )x (Xx , Yx ) =
(x)—— .Xx ’ d Xx , df ( (x)—— .Yx
= d Yx , df ( ) )
E E

= ιE .Yx , d(df )(x)—— .Xx ’ ιE .Xx , d(df )(x)—— .Yx
E E

= d(df )(x)—— .Xx , Yx ’ d(df )(x)—— .Yx , Xx ,
E E


which does not vanish in general by the argument in (32.7).

33.17. Lie derivatives. Let D± denote one of T , D, or D[1,∞) . For a vector
¬eld X ∈ C ∞ (M ← D± M ) and ω ∈ Homk,∞ (M,R) (C ∞ (M ← D± M ), C ∞ (M, R))
alt
C
we de¬ne the Lie derivative LX ω of ω along X by

k
(LX ω)|U (Y1 , . . . , Yk ) = X(ω(Y1 , . . . , Yk )) ’ ω|U (Y1 , . . . , [X, Yi ], . . . , Yk ),
i=1

for Y1 , . . . , Yk ∈ C ∞ (U ← D± U ). From (32.5) it follows that

LX ω ∈ Homk,∞ (M,R) (C ∞ (M ← D± M ), C ∞ (M, R)).
alt
C


33.18. Theorem. The following formulas hold for C ∞ (Lk (T M, M — R)) and
alt
k, alt ∞ ∞
for the spaces HomC ∞ (M,R) (C (M ← D M ), C (M, R)) where D± is any of D,
±

D[1,∞) , or T .
(1) iX (• § ψ) = iX • § ψ + (’1)deg • • § iX ψ.
(2) LX (• § ψ) = LX • § ψ + • § LX ψ.
(3) d(• § ψ) = d• § ψ + (’1)deg • • § dψ.

33.18
348 Chapter VII. Calculus on in¬nite dimensional manifolds 33.18

1
d2 = d —¦ d = 2 [d, d] = 0.
(4)
[LX , d] = LX —¦ d ’ d —¦ LX = 0.
(5)
[iX , d] = iX —¦ d + d —¦ iX = LX .
(6)
[LX , LY ] = LX —¦ LY ’ LY —¦ LX = L[X,Y ] .
(7)
[LX , iY ] = LX iY ’ iY LX = i[X,Y ] .
(8)
(9) [iX , iY ] = iX iY + iY iX = 0.
Lf.X • = f.LX • + df § iX •.
(10)

Remark. In this theorem we used the graded commutator for graded derivations
[D1 , D2 ] := D1 —¦ D2 ’ (’1)deg(D1 ) deg(D2 ) D2 —¦ D1 . We will elaborate this notion in
(35.1) below.
The left hand side of (6) maps the subspace C ∞ (Lk (T M, M — R)) of the space of
alt
k, alt ∞
modular di¬erential forms HomC ∞ (M,R) (X(M ), C (M, R)) into itself, thus the Lie
derivative LX also does. We do not know whether this is true for the other spaces
on the left hand side of the diagram in (33.12).

Proof. All results will be proved in Homk,∞ (M,R) (C ∞ (M ← D± M ), C ∞ (M, R)),
alt
C

so they also hold in the subspace C (M ← Lk (T M, M — R)).
alt

(9) is obvious and (1) was shown in (33.11).
(8) Take the di¬erence of the following two expressions:

k
(LX iY ω)(Z1 , . . . , Zk ) = X((iY ω)(Z1 , . . . , Zk )) ’ (iY ω)(Z1 , . . . , [X, Zi ], . . . , Zk )
i=1
k
= X(ω(Y, Z1 , . . . , Zk )) ’ ω(Y, Z1 , . . . , [X, Zi ], . . . , Zk )
i=1
(iY LX ω)(Z1 , . . . , Zk ) = LX ω(Y, Z1 , . . . , Zk )
= X(ω(Y, Z1 , . . . , Zk )) ’ ω([X, Y ], Z1 , . . . , Zk )’
k
’ ω(Y, Z1 , . . . , [X, Zi ], . . . , Zk ).
i=1



(2) Let • be of degree p and ψ of degree q. We prove the result by induction on
p + q. Suppose that (2) is true for p + q < k. Then for X we have by part (8), by
1, and by induction

(iY LX )(• § ψ) = (LX iY )(• § ψ) ’ i[X,Y ] (• § ψ)
= LX (iY • § ψ + (’1)p • § iY ψ) ’ i[X,Y ] • § ψ ’ (’1)p • § i[X,Y ] ψ
= LX iY • § ψ + iY • § LX ψ + (’1)p LX • § iY ψ+
+ (’1)p • § LX iY ψ ’ i[X,Y ] • § ψ ’ (’1)p • § i[X,Y ] ψ
iY (LX • § ψ + • § LX ψ) = iY LX • § ψ + (’1)p LX • § iY ψ+
+ iY • § LX ψ + (’1)p • § iY LX ψ.

33.18
33.18 33. Di¬erential forms 349

Using again (8), we get the result since the iY for all local vector ¬elds Y together act
point separating on each space of di¬erential forms, in both cases of the convention
(33.2).
(6) follows by summing up the following parts.

(LX0 •)(X1 , . . . , Xk ) = X0 (•(X1 , . . . , Xk ))+
k
(’1)0+j •([X0 , Xj ], X1 , . . . , Xj , . . . , Xk )
+
j=1
(iX0 d•)(X1 , . . . , Xk ) = d•(X0 , . . . , Xk )
k
(’1)i Xi (•(X0 , . . . , Xi , . . . , Xk )) +
=
i=0

(’1)i+j •([Xi , Xj ], X0 , . . . , Xi , . . . , Xj , . . . , Xk ).
+
0¤i<j
k
(’1)i’1 Xi ((iX0 •)(X1 , . . . , Xi , . . . , Xk )) +
(diX0 •)(X1 , . . . , Xk ) =
i=1

(’1)i+j’2 (iX0 •)([Xi , Xj ], X1 , . . . , Xi , . . . , Xj , . . . , Xk )
+
1¤i<j
k
(’1)i Xi (•(X0 , X1 , . . . , Xi , . . . , Xk )) ’
=’
i=1

(’1)i+j •([Xi , Xj ], X0 , X1 , . . . , Xi , . . . , Xj , . . . , Xk ).

1¤i<j


(3) We prove the result again by induction on p + q. Suppose that (3) is true for
p+q < k. Then for each local vector ¬eld X we have by (6), (2), 1, and by induction

iX d(• § ψ) = LX (• § ψ) ’ d iX (• § ψ)
= LX • § ψ + • § LX ψ ’ d(iX • § ψ + (’1)p • § iX ψ)
= iX d• § ψ + diX • § ψ + • § iX dψ + • § diX ψ ’ diX • § ψ
’ (’1)p’1 iX • § dψ ’ (’1)p d• § iX ψ ’ • § diX ψ
= iX (d• § ψ + (’1)p • § dψ).

Since X is arbitrary, the result follows.
(4) This follows by a long but straightforward computation directly from the the
global formula (33.12.3), using only the de¬nition of the Lie bracket as a commu-
tator, the Jacobi identity, and cancellation.
(5) dLX = d iX d + ddiX = diX d + iX dd = LX d.
(7) By the (graded) Jacobi identity, by (5), by (6), and by (8) we have [LX , LY ] =
[LX , [iY , d]] = [[LX , iY ], d] + [iY , [LX , d]] = [i[X,Y ] , d] + 0 = L[X,Y ] .
(10) Lf.X • = [if.X , d]• = [f.iX , d]• = f iX d• + d(f iX •) = f iX d• + df § iX • +
f diX • = f LX • + df § iX •.


33.18
350 Chapter VII. Calculus on in¬nite dimensional manifolds 33.20

33.19. Lemma. Let X ∈ X(M ) be a kinematic vector ¬eld which has a local ¬‚ow
FlX . Or more generally, let us suppose that • : R — M ⊃ U ’ M is a smooth
t
mapping such that (t, x) ’ (t, •(t, x) = •t (x)) is a di¬eomorphism U ’ V , where
U and V are open neighborhoods of {0} — M in R — M , and such that •0 = IdM

and ‚t 0 •t = X ∈ X(M ).
k, alt
Then for ω any k-form in HomC ∞ (M,R) (C ∞ (M ← DM ), C ∞ (M, R)) we have

|0 (•t )— ω

= LX ω,
‚t
|0 (FlX )— ω

= LX ω,
t
‚t
X—
= (FlX )— LX ω = LX (FlX )— ω.

‚t (Flt ) ω t t


In particular, for a vector ¬eld X with a local ¬‚ow the Lie derivative LX maps the
spaces C ∞ (Lk (D± M, M — R)) into themselves, for D± = T , D, and D(p) .
alt

Proof. For Yi ∈ C ∞ (M ← D± M ) we have

|0 (ω((•’1 )— Y1 , . . . , (•’1 )— Yk )
( ‚t |0 (•t )— ω)(Y1 , . . . , Yk ) =
‚ ‚
—¦ •t )
t t
‚t
k
ω(Y1 , . . . , ‚t |0 (•’1 )— Yi , . . . , Yk ) + —
‚ ‚
‚t |0 (•t ) (ω(Y1 , . . . , Yp ))
= t
i=1
k
= X(ω(Y1 , . . . , Yk )) ’ ω|U (Y1 , . . . , [X, Yi ], . . . , Yk ),
i=1

where at the end we used (32.15). This proves the ¬rst two assertions.
For the third assertion we proceed as follows

(FlX )— ω |0 (FlX )— (FlX )— ω
d d
=
t t s
dt ds
(FlX )— ds |0 (FlX )— ω
d

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