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mapping Dg corresponds to

U — D0 E ’ V — D0 F
+a)— —¦ g — —¦ ( ’g(a))—
(a, ‚) ’ g(a), ‚ —¦ (
+a) ’ g(a))— .
= g(a), ‚ —¦ (g(

In order to show that this is smooth, its enough to consider the second component
and we compose it with the embedding D0 F ’ W 0 C ∞ (W, R) . The associated
mapping U — D0 E — C ∞ (W, R) ’ R is given by

(a, ‚, f ) ’ ‚ f —¦ (g( +a) ’ g(a)) ,
(1)

28.9
282 Chapter VI. In¬nite dimensional manifolds 28.11

where f —¦ (g( +a) ’ g(a)) is smooth on the open 0-neighborhood Wa := {y ∈ E :
g(y + a) ’ g(a) ∈ W } = g ’1 (g(a) + W ) ’ a in E. Now let a : R ’ U be a smooth
curve and I a bounded interval in R. Then there exists an open neighborhood UI,W
of 0 in E such that UI,W ⊆ Wa(t) for all t ∈ I. Then the mapping (1), composed
with a : I ’ U , factors as

I — D0 E — C ∞ (W, R) ’ C ∞ (UI,W , R) — C ∞ (UI,W , R) ’ R,

given by

(t, ‚, f ) ’ ‚UI,W , f —¦ g( ’ ‚UI,W (f —¦(g(
+a(t))’g(a(t)) +a(t))’g(a(t))))),

which is smooth by cartesian closedness.
(k)
28.10. Let E be a convenient vector space. Recall from (28.2) that Da E is the
space of all operational tangent vectors of order ¤ k. For an open subset U in a
convenient vector space E and k > 0 we consider the disjoint union

Da E ∼ U — D0 E ⊆ E — D0 E.
(k) (k)
D(k) U := (k)
=
a∈U


Lemma. For a smooth mapping f : E ⊃ U ’ V ‚ F the smooth mapping Df :
DU ’ DV from (28.9) induces smooth mappings D(k) f : D(k) U ’ D(k) V .
(k) (k)
Proof. We only have to show that Da f maps Da E into Df (a) F , because smooth-
ness follows then by restriction.
The pullback f — : C ∞ (V, R) ’ C ∞ (U, R) maps functions which are ¬‚at of order k
at f (a) to functions which are ¬‚at of the same order at a. Thus, Da f maps the
(k) (k)
corresponding annihilator Da U into the annihilator Df (a) V .

28.11. Lemma.
(1) The chain rule holds in general: D(f —¦ g) = Df —¦ Dg and D(k) (f —¦ g) =
D(k) f —¦ D(k) g.
(2) If g : E ’ F is a bounded a¬ne mapping then Dx g commutes with the
restriction and the projection to the subspaces of derivations which are ho-
mogeneous of degree k > 1.
(3) If g : E ’ F is a bounded a¬ne mapping with linear part = g ’ g(0) :
[k] [k]
E ’ F then Dx g : Dx E ’ Dg(x) F is induced by the linear mappings
(Lk ( ; R))— : Lk (E, R)— ’ Lk (F, R)— .
sym sym sym
[1]
(4) If g : E ’ R is bounded linear we have Dg.Xx = D(1) g.Xx .

Remark that if g is not a¬ne then in general Dg does not respect the subspaces of
derivations which are homogeneous of degree k > 1:
[k]
In fact let g : E ’ R be a homogeneous polynomial of degree k on which ‚ ∈ D0 E
does not vanish. Then by (4) we have that 0 = ‚(g) = Dg(‚) ∈ R ∼ D0 R = D0 R.
[1]
=

28.11
28.12 28. Tangent vectors 283

Proof. (1) is obvious.
For (2) let Xx ∈ Dx E and f ∈ C ∞ (F, R). Then we have

(Dg.Xx )[k] (f ) = (Dg.Xx )( k! dk f (g(x))( ’g(x))k )
1

k
) ’ g(x))k )
1
= k! Xx (d f (g(x))(g(
[k] [k]
Xx (f —¦ g)
(Dg.Xx )(f ) =
Xx ( k! dk (f —¦ g)(x)( ’x)k )
1
=
k
’x))k ).
1
= k! Xx (d f (g(x))( (

These expressions are equal.

ww y wD E
[k]
Dx E Dx E x



u
u u
Dg Dg
Dg

ww y wD F
[k]
Dg(x) F Dg(x) F x


For (3) we take • ∈ Lk (E; R) which vanishes on all decomposable forms, and let
sym
[k]
k
Xx = ‚• |x ∈ Dx E be the corresponding homogeneous derivation. Then

k k
(Dg.‚• |x )(f ) = ‚• |x (f —¦ g)
= •( k! dk (f —¦ g)(x))
1

= •( k! dk f (g(x)) —¦ k
1
)
= (Lk ( ; R)— •)( k! dk f (g(x)))
1
sym
[k]
|
= ‚Lk (f ).
;R)— • g(x)
sym (



y wL
[k] k
Dx E Sym (E, R)

Lk ( , R)—
u
D[k] g
u Sym


y wL
[k] k
Dg(x) F Sym (F, R)

(4) is a special case of (2).

28.12. The operational and the kinematic tangent bundles. Let M be a

manifold with a smooth atlas (M ⊃ U± ’’ E± )±∈A . We consider the following

equivalence relation on the disjoint union

D(u± (U± )) — {±},
D(u± (U± )) :=
±∈A ±∈A
(‚, ±) ∼ (‚ , β) ⇐’ D(u±β )‚ = ‚.

We denote the quotient set by DM and call it the operational tangent bundle
of M . Let πM : DM ’ M be the obvious foot point projection, let DU± =

28.12
284 Chapter VI. In¬nite dimensional manifolds 28.12

’1
πM (U± ) ‚ DM , and let Du± : DU± ’ D(u± (U± )) be given by Du± ([‚, ±]) = ‚.
So Du± ([‚ , β]) = D(u±β )‚ .
The charts (DU± , Du± ) form a smooth atlas for DM , since the chart changings are
given by
Du± —¦ (Duβ )’1 = D(u±β ) : D(uβ (U±β )) ’ D(u± (U±β )).
This chart changing formula also implies that the smooth structure on DM depends
only on the equivalence class of the smooth atlas for M .
The mapping πM : DM ’ M is obviously smooth. The natural topology is
automatically Hausdor¬: X, Y ∈ DM can be separated by open sets of the form
’1
πM (V ) for V ‚ M , if πM (X) = πM (Y ), since M is Hausdor¬, and by open subsets
of the form (T u± )’1 (E± — W ) for W open in E± , if πM (X) = πM (Y ) ∈ U± .
’1
For x ∈ M the set Dx M := πM (x) is called the operational tangent space at x or
the ¬ber over x of the operational tangent bundle. It carries a canonical convenient
vector space structure induced by Dx (u± ) := Du± |Dx M : Du± (x) E± ∼ D0 (E± ) for
=
some (equivalently any) ± with x ∈ U± .
Let us construct now the kinematic tangent bundle. We consider the following
equivalence relation on the disjoint union

U± — E± — {±},
±∈A
(x, v, ±) ∼ (y, w, β) ⇐’ x = y and d(u±β )(uβ (x))w = v
and denote the quotient set by T M , the kinematic tangent bundle of M . Let
’1
πM : T M ’ M be given by πM ([x, v, ±]) = x, let T U± = πM (U± ) ‚ T M ,
and let T u± : T U± ’ u± (U± ) — E± be given by T u± ([x, v, ±]) = (u± (x), v). So
T u± ([x, w, β]) = (u± (x), d(u±β )(uβ (x))w).
The charts (T U± , T u± ) form a smooth atlas for T M , since the chart changings are
given by

T u± —¦ (T uβ )’1 : uβ (U±β ) — Eβ ’ u± (U±β ) — E± ,
(x, v) ’ (u±β (x), d(u±β )(x)v).
This chart changing formula also implies that the smooth structure on T M depends
only on the equivalence class of the smooth atlas for M .
The mapping πM : T M ’ M is obviously smooth. It is called the (foot point)
projection of M . The natural topology is automatically Hausdor¬; this follows
from the bundle property and the proof is the same as for DM above.
’1
For x ∈ M the set Tx M := πM (x) is called the kinematic tangent space at x or
the ¬ber over x of the tangent bundle. It carries a canonical convenient vector
space structure induced by Tx (u± ) := T u± |Tx M : Tx M ’ {x} — E± ∼ E± for some
=
(equivalently any) ± with x ∈ U± .
Note that the kinematic tangent bundle T M embeds as a subbundle into DM ; also
for each k ∈ N the same construction as above gives us tangent bundles D(k) M
which are subbundles of DM .

28.12
28.15 28. Tangent vectors 285

28.13. Let us now give an obvious description of T M as the space of all velocity
vectors of curves, which explains the name ˜kinematic tangent bundle™: We put
on C ∞ (R, M ) the equivalence relation : c ∼ e if and only if c(0) = e(0) and in
d
one (equivalently each) chart (U, u) with c(0) = e(0) ∈ U we have dt |0 (u —¦ c)(t) =
d
dt |0 (u —¦ e)(t). We have the following diagram


u e C (R, M )
C ∞ (R, M )/ ∼ ∞

δ ee
ee
∼δ
uh
e u
ev
= 0


wM
TM π M

where to c ∈ C ∞ (R, M ) we associate the tangent vector δ(c) := [c(0), ‚t 0 (u± —¦


c)(t), ±]. It factors to a bijection C ∞ (R, M )/ ∼’ T M , whose inverse associates to
[x, v, ±] the equivalence class of t ’ u’1 (u± (x) + h(t)v) for h a small function with
±
h(t) = t near 0.
Since the c∞ -topology on R — E± is the product topology by corollary (4.15), we
can choose h uniformly for (x, v) in a piece of a smooth curve. Thus, a mapping g :
T M ’ N into another manifold is smooth if and only if g—¦δ : C ∞ (R, M ) ’ N maps
˜smooth curves™ to smooth curves, by which we mean C ∞ (R2 , M ) to C ∞ (R, N ).

28.14. Lemma. If a smooth manifold M and the squares of its model spaces
are smoothly paracompact, then also the kinematic tangent bundle T M is smoothly
paracompact.
If a smooth manifold M and V — D0 V for any of its model spaces V are smoothly
paracompact, then also the operational tangent bundle DM is smoothly paracom-
pact.

Proof. This is a particular case of (29.7) below.

28.15. Tangent mappings. Let f : M ’ N be a smooth mapping between
manifolds. Then f induces a linear mapping Dx f : Dx M ’ Df (x) N for each
x ∈ M by (Dx f.‚x )(h) = ‚x (h —¦ f ) for h ∈ C ∞ (N ⊇ {f (x)}, R). These give a
mapping Df : DM ’ DN . If (U, u) is a chart around x and (V, v) is one around
f (x), then Dv —¦ Df —¦ (Du)’1 = D(v —¦ f —¦ u’1 ) is smooth by lemma (28.9). So
Df : DM ’ DN is smooth.
By lemma (28.10), Df restricts to smooth mappings D(k) f : D(k) M ’ D(k) N
and to T f : T M ’ T N . We check the last statement for open subsets M and
N of convenient vector spaces. (Df.Xa )(g) = Xa (g —¦ f ) = d(g —¦ f )(a)(X) =
dg(f (a))df (a)X = (df (a)X)f (a) (g).
If f ∈ C ∞ (M, E) for a convenient vector space E, then Df : DM ’ DE =
E — D0 E. We then de¬ne the di¬erential of f by df := pr2 —¦ Df : DM ’ D0 E. It
(k)
restricts to smooth ¬berwise linear mappings D(k) M ’ D0 E and df : T M ’ E.
If f ∈ C ∞ (M, R), then df : DM ’ R. Let Id denote the identity function on R,
then (T f.‚x )(Id) = ‚x (Id —¦f ) = ‚x (f ), so we have df (‚x ) = ‚x (f ).

28.15
286 Chapter VI. In¬nite dimensional manifolds 28.16

The mapping f ’ df is bounded linear C ∞ (M, R) ’ C ∞ (DM, R). That it is linear
and has values in this space is obvious. So by the smooth uniform boundedness
principle (5.26) it is enough to show that f ’ df.Xx = Xx (f ) is bounded for all
Xx ∈ DM , which is true by de¬nition of DM .
28.16. Remark. From the construction of the tangent bundle in (28.12) it is
immediately clear that
T (pr ) T (pr )
T M ’ ’ 1 T (M — N ) ’ ’ 2 T N
’’ ’ ’’ ’
is also a product, so that T (M — N ) = T M — T N in a canonical way.
We investigate D0 (E — F ) for convenient vector spaces. Since D0 is a functor for 0
preserving maps, we obtain linear sections D0 (injk ) : D0 (Ek ) ’ D0 (E1 — E2 ) and
hence a section D0 (inj1 )+D0 (inj2 ) : D0 (E1 )•D0 (E2 ) ’ D0 (E1 •E2 ). The comple-
ment of the image is given by the kernel of the linear mapping (D0 (pr1 ), D0 (pr2 )) :
D0 (E1 • E2 ) ’ D0 (E1 ) • D0 (E2 ).


w D (E ) )
Id
D0 (E1 )&
&
xˆ )
x€ (pr
x
0 1
&prD (E
&D0 (injE
x
injD0 (E1 ) )
)
&
(
&
x
x
0 1
1
D E1 )
0

w D (E w D (E ) • D (E )
• E2 )&
D0 (E0 ) • D0 (E1 )
)
&
x
x€ (inj
x
0 1 0 0 0 1
& &D0 (prE
xpr
)
injD (E ) &
7 (
&
x x
2
D0 (E2 )
D E2 )
0
0 2


w D (E )
D0 (E2 ) 0 2
2
Lemma. In the case E1 = = E2 this mapping is not injective.
Proof. The space L2 (E1 — E2 , E1 — E2 ; R) can be viewed as L2 (E1 , E1 ; R) —
L2 (E1 , E2 ; R) — L2 (E2 , E1 ; R) — L2 (E2 , E2 ; R) and the subspace formed by those
forms whose (2,1) and (1,2) components with respect to this decomposition are
compact considered as operators in L( 2 , 2 ) ∼ L2 ( 2 , 2 ; R) is a closed subspace.
=
So, by Hahn-Banach, there is a non-trivial continuous linear functional : L2 ( 2 —
22
, — 2 ; R) ’ R vanishing on this subspace. We claim that the linear mapping
‚ : C ∞ ( 2 — 2 , R) f ’ (f (0, 0)) ∈ R is an operational tangent vector of 2 — 2
but not a direct sum of two operational tangent vectors on 2 . In fact, the second
derivative of a product h of two functions f and g is given by
d2 h(0, 0)(w1 , w2 ) = d2 f (0, 0)(w1 , w2 ) g(0, 0)
+ df (0, 0)(w1 ) dg(0, 0)(w2 )
+ df (0, 0)(w2 ) dg(0, 0)(w1 )
+ f (0, 0) d2 g(0, 0)(w1 , w2 ).
Thus ‚ is a derivation since the middle terms give ¬nite dimensional operators in
L2 ( 2 , 2 ; R). It is not a direct sum of two operational tangent vectors on 2 since
functions f depending only on the j-th factor have as second derivative forms with
nonzero (j,j) entry only. Hence D0 (prj )(‚)(f ) = ‚(f —¦ prj ) = ((f —¦ prj ) (0)) = 0,
but ‚ = 0.

28.16
287

29. Vector Bundles

29.1. Vector bundles. Let p : E ’ M be a smooth mapping between manifolds.
By a vector bundle chart on (E, p, M ) we mean a pair (U, ψ), where U is an open
subset in M , and where ψ is a ¬ber respecting di¬eomorphism as in the following
diagram:
w
‘ ψ
E|U := p’1 (U )
  
U —V
‘“
‘   pr
£
 
p 1

U.
Here V is a ¬xed convenient vector space, called the standard ¬ber or the typical
¬ber, real for the moment.
’1
Two vector bundle charts (U1 , ψ1 ) and (U2 , ψ2 ) are called compatible, if ψ1 —¦ ψ2 is
’1
a ¬ber linear isomorphism, i.e., (ψ1 —¦ ψ2 )(x, v) = (x, ψ1,2 (x)v) for some mapping
ψ1,2 : U1,2 := U1 © U2 ’ GL(V ). The mapping ψ1,2 is then unique and smooth

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