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For the converse recall the smooth mapping „ W : P —M P [W, ρ] ’ W from (37.12),
which satis¬es „ W (u, q(u, w)) = w, q(ux , „ W (ux , vx )) = vx , and „ W (ux g, vx ) =
ρ(g ’1 )„ W (ux , vx ).
For ¦ ∈ „¦k (M ; P [W ]) we de¬ne q ¦ ∈ „¦k (P, W ) as follows. For Xi ∈ Tu P we put

(q ¦)u (X1 , . . . , Xk ) := „ W (u, ¦p(u) (Tu p.X1 , . . . , Tu p.Xk )).

Then q ¦ is smooth and horizontal. For g ∈ G we have

((rg )— (q ¦))u (X1 , . . . , Xk ) = (q ¦)ug (Tu (rg ).X1 , . . . , Tu (rg ).Xk )
= „ W (ug, ¦p(ug) (Tug p.Tu (rg ).X1 , . . . , Tug p.Tu (rg ).Xk ))
= ρ(g ’1 )„ W (u, ¦p(u) (Tu p.X1 , . . . , Tu p.Xk ))
= ρ(g ’1 )(q ¦)u (X1 , . . . , Xk ).

Clearly, the two constructions are inverse to each other.

37.31. Let (p : P ’ M, G) be a principal ¬ber bundle with a principal connection
¦ = ζ —¦ ω, and let ρ : G ’ GL(W ) be a representation of the structure group G on
a convenient vector space W , as in (49.1). We consider the associated vector bundle
¯
(p : E := P [W, ρ] ’ M, W ), the induced connection ¦ on it, and the corresponding
covariant derivative.

Theorem. The covariant exterior derivative dω from (37.23) on P and the covari-
ant exterior derivative for P [W ]-valued forms on M are connected by the mapping
q from (37.30), as follows:

q —¦ d = dω —¦ q : „¦(M ; P [W ]) ’ „¦hor (P, W )G .


Proof. Let us consider ¬rst f ∈ „¦0 (P, W )G = C ∞ (P, W )G , then f = q s for
hor

s ∈ C (M ← P [W ]), and we have f (u) = „ W (u, s(p(u))) and s(p(u)) = q(u, f (u))
by (37.30) and (37.16). Therefore, we have T s.T p.Xu = T q(Xu , T f.Xu ), where
T f.Xu = (f (u), df (Xu )) ∈ T W = W — W . If χ : T P ’ HP is the horizontal

37.31
37.32 37. Bundles and connections 403

projection as in (37.23), we have T s.T p.Xu = T s.T p.χ.Xu = T q(χ.Xu , T f.χ.Xu ).
So we get

(q d s)(Xu ) = „ W (u, (d s)(T p.Xu ))
= „ W (u, T p.Xu s) by (37.29.5)
= „ W (u, K.T s.T p.Xu ) by (37.28.1)
= „ W (u, K.T q(χ.Xu , T f.χ.Xu )) from above
= „ W (u, pr2 .vl’1 .¦.T q(χ.Xu , T f.χ.Xu ))
¯ by (37.26)
P [W ]

„ W (u, pr2 .vl’1 ] .T q.(¦ — Id)(χ.Xu , T f.χ.Xu ))
= by (37.24)
P [W

„ W (u, pr2 .vl’1 ] .T q(0u , T f.χ.Xu ))
= since ¦.χ = 0
P [W

= „ W (u, q. pr2 .vl’1 .(0u , T f.χ.Xu )) since q is ¬ber linear
P —W

= „ W (u, q(u, df.χ.Xu )) = (χ— df )(Xu )
= (dω q s)(Xu ).

Now we turn to the general case. Let Yi for i = 0, . . . , k be local vector ¬elds on M ,
and let CYi be their horizontal lifts to P . Then T p.CYi = yi —¦ p, so Yi and CYi are
p-related. Since both q d ¦ and dω q are horizontal, it su¬ces to check that they
coincide on all local vector ¬elds of the form CYi . Since C[Yi , Yj ] = χ[CYi , CYj ],
we get from the special case above and the de¬nition of q :

(’1)i (CYi )(q ¦)(CY0 , . . . , CYi , . . . , CYk )
(dω q ¦)(CY0 , . . . , CYk ) =
0¤i¤k

(’1)i+j (q ¦)([CYi , CYj ], CY0 , . . . CYi . . . CYj . . . , CYk )
+
i<j

(’1)i (CYi )(q (¦(Y0 , . . . , Yi , . . . , Yk )
=
0¤i¤k

(’1)i+j (q ¦)(C[Yi , Yj ], CY0 , . . . CYi . . . CYj . . . , CYk )
+
i<j

(’1)i q (
= Yi (¦(Y0 , . . . , Yi , . . . , Yk )))
0¤i¤k

(’1)i+j q (¦([Yi , Yj ], Y0 , . . . Yi . . . Yj . . . , Yk ))
+
i<j

= q (d ¦(Y0 , . . . , Yk )) = (q d ¦)(CY0 , . . . , CYk ).


37.32. Corollary. In the situation of theorem (37.31), for the Lie algebra valued
curvature form „¦ ∈ „¦2 (P, g) and the curvature RP [W ] ∈ „¦2 (M ; L(P [W ], P [W ]))
hor
we have the relation
qL(P [W ],P [W ]) RP [W ] = ρ —¦ „¦,

where ρ = Te ρ : g ’ L(W, W ) is the derivative of the representation ρ.

37.32
404 Chapter VIII. In¬nite dimensional di¬erential geometry 38.1

Proof. We use the notation of the proof of theorem (37.31), by which we have for
X, Y ∈ Tu P

(dω dω qP [W ] s)u (X, Y ) = (q d d s)u (X, Y )
= (q RP [W ] s)u (X, Y )
= „ W (u, RP [W ] (Tu p.X, Tu p.Y )s(p(u)))
= (qL(P [W ],P [W ]) RP [W ] )u (X, Y )(qP [W ] s)(u).


On the other hand, let g(t) be a smooth curve in G with g(0) = e and g(t) =
‚t 0
„¦u (X, Y ) ∈ g. Then we have by theorem (37.23.8)

(dω dω q s)u (X, Y ) = (χ— iR dq s)u (X, Y )
since R is horizontal
= (dq s)u (R(X, Y ))
= (dq s)(’愦(X,Y ) (u)) by (37.20)
’1
(q s)(rg(t) (u))

= ‚t 0
„ W u.g(t)’1 , s(p(u.g(t)’1 ))

= ‚t 0
„ W (u.g(t)’1 , s(p(u)))

= ‚t 0
ρ(g(t))„ W (u, s(p(u)))

= by (37.12)
‚t 0
= ρ („¦u (X, Y ))(q s)(u).




38. Regular Lie Groups

38.1. The right and left logarithmic derivatives. Let M be a manifold, and
let f : M ’ G be a smooth mapping into a Lie group G with Lie algebra g. We
de¬ne the mapping δ r f : T M ’ g by the formula
’1
δ r f (ξx ) := Tf (x) (µf (x) ).Tx f.ξx = (f — κr )(ξx ) for ξx ∈ Tx M,

where κr is the right Maurer-Cartan form from (36.10). Then δ r f is a g-valued
1-form on M , δ r f ∈ „¦1 (M, g). We call δ r f the right logarithmic derivative of f ,
since for f : R ’ (R+ , ·) we have δ r f (x).1 = f (x) = (log —¦f ) (x).
(x)
f

Similarly, the left logarithmic derivative δ l f ∈ „¦1 (M, g) of a smooth mapping f :
M ’ G is given by

δ l f.ξx := Tf (x) (µf (x)’1 ).Tx f.ξx = (F — κl )(ξx ).

Lemma. Let f, g : M ’ G be smooth. Then we have

δ r (f.g)(x) = δ r f (x) + Ad(f (x)).δ r g(x).

38.1
38.1 38. Regular Lie groups 405

Moreover, the di¬erential form δ r f ∈ „¦1 (M, g) satis¬es the ˜left Maurer-Cartan
equation™ (left because it stems from the left action of G on itself )

dδ r f (ξ, ·) ’ [δ r f (ξ), δ r f (·)]g = 0
1
or dδ r f ’ [δ r f, δ r f ]g = 0, §
2
]g was de¬ned in
where ξ, · ∈ Tx M , and where the graded Lie bracket [ , §
(37.20.1).
The left logarithmic derivative also satis¬es a ˜Leibniz rule™ and the ˜right Maurer
Cartan equation™:

δ l (f g)(x) = δ l g(x) + Ad(g(x)’1 ).δ l f (x),
1
dδ l f + [δ l f, δ l f ]g = 0.
§
2

For ˜regular Lie groups™ we will prove a converse to this statement later in (40.2).

Proof. We treat only the right logarithmic derivative, the proof for the left one is
similar.
’1
.f (x)’1
δ r (f.g)(x) = T (µg(x) ).Tx (f.g)
’1 ’1
= T (µf (x) ).T (µg(x) ).T(f (x),g(x)) µ.(Tx f, Tx g)
’1 ’1
= T (µf (x) ).T (µg(x) ). T (µg(x) ).Tx f + T (µf (x) ).Tx g
= δ r f (x) + Ad(f (x)).δ r g(x).

We shall now use principal bundle geometry from section (37). We consider the
trivial principal bundle pr1 : M — G ’ M with right principal action. Then
the submanifolds {(x, f (x).g) : x ∈ M } for g ∈ G form a foliation of M — G,
whose tangent distribution is complementary to the vertical bundle M — T G ‚
T (M —G) and is invariant under the principal right G-action. So it is the horizontal
distribution of a principal connection on M — G ’ M . For a tangent vector
(ξx , Yg ) ∈ Tx M — Tg G the horizontal part is the right translate to the foot point
(x, g) of (ξx , Tx f.ξx ). The decomposition in horizontal and vertical parts according
to this distribution is
’1 ’1
(ξx , Yg ) = (ξx , T (µg ).T (µf (x) ).Tx f.ξx ) + (0x , Yg ’ T (µg ).T (µf (x) ).Tx f.ξx ).

Since the fundamental vector ¬elds for the right action on G are the left invariant
vector ¬elds, the corresponding connection form is given by
’1
ω r (ξx , Yg ) = T (µg’1 ).(Yg ’ T (µg ).T (µf (x) ).Tx f.ξx ),
ω(x,g) = T (µg’1 ) ’ Ad(g ’1 ).δ r fx ,
r


ω r = κl ’ (Ad —¦ ν).δ r f,
(1)

38.1
406 Chapter VIII. In¬nite dimensional di¬erential geometry 38.1

where κl : T G ’ g is the left Maurer-Cartan form on G (the left trivialization),
given by κl = T (µg’1 ). Note that κl is the principal connection form for the
g
(unique) principal connection p : G ’ {point} with right principal action, which is
¬‚at so that the right (from right action) Maurer-Cartan equation holds in the form

dκl + 1 [κl , κl ]§ = 0.
(2) 2

The principal connection ω r is ¬‚at since we got it via the horizontal leaves, so the
principal curvature form vanishes:

0 = dω r + 1 [ω r , ω r ]§
(3) 2
= dκl + 1 [κl , κl ]§ ’ d(Ad —¦ ν) § δ r f ’ (Ad —¦ ν).dδ r f
2
’ [κl , (Ad —¦ ν).δ r f ]§ + 2 [(Ad —¦ ν).δ r f, (Ad —¦ ν).δ r f ]§
1

= ’(Ad —¦ ν).(dδ r f ’ 1 [δ r f, δ r f ]§ ),
2

where we used (2) and the fact that for ξ ∈ g and a smooth curve c : R ’ G with
c(0) = e and c (0) = ξ we have

Ad(c(t)’1 .g ’1 ) = ’ad(ξ)Ad(g ’1 )

d(Ad —¦ ν)(T (µg )ξ) = ‚t 0

= ’ad κl (T (µg )ξ) (Ad —¦ ν)(g),
d(Ad —¦ ν) = ’(ad —¦ κl ).(Ad —¦ ν).
(4)

So we have dδ r f ’ 1 [δ r f, δ r f ]§ as asserted.
2
For the left logarithmic derivative δ l f the proof is similar, and we discuss only the
essential deviations. First note that on the trivial principal bundle pr1 : M —G ’ M
with left principal action of G the fundamental vector ¬elds are the right invariant
vector ¬elds on G, and that for a principal connection form ω l the curvature form
1
is given by dω l ’ 2 [ω l , ω l ]§ . Look at the proof of theorem (37.20) to see this. The
connection form is then given by

ω l = κr ’ Ad.δ l f,
(1™)
’1
where the right Maurer-Cartan form (κr )g = T (µg ) : Tg G ’ g satis¬es the left
Maurer-Cartan equation
1
dκr ’ [κr , κr ]§ = 0.
(2™)
2
Flatness of ω l now leads to the computation

0 = dω l ’ 1 [ω l , ω l ]§
(3™) 2
= dκr ’ 1 [κr , κr ]§ ’ dAd § δ l f ’ Ad.dδ l f
2
+ [κr , Ad.δ l f ]§ ’ 1 [Ad.δ l f, Ad.δ l f ]§
2
= ’Ad.(dδ l f + 1 [δ l f, δ l f ]§ ),
2

where we have used dAd = (ad —¦ κr )Ad from (36.10.3) directly.


38.1
38.2 38. Regular Lie groups 407

Remark. The second half of the proof of lemma (38.1) can be shortened consid-
erably. Namely, as soon as we know that κr satis¬es the Maurer-Cartan equation
dκr ’ 2 [κr , κr ]§ we get it also for the right logarithmic derivative δ r f = f — κr .
1

But the computations in this proof will be used again in the proof of the converse,
theorem (40.2) below.

38.2. Theorem. [Grabowski, 1993] Let G be a Lie group with exponential mapping
exp : g ’ G. Then for all X, Y ∈ g we have

1
TX exp .Y = Te µexp X . Ad(exp(’tX))Y dt
0
1
exp X
= Te µ . Ad(exp(tX))Y dt.
0



Remark. If G is a Banach Lie group then we have from (36.8.4) and (36.9) the
i

series Ad(exp(tX)) = i=0 t ad(X)i , so that we get the usual formula
i!


TX exp = Te µexp X . i
1
(i+1)! ad(X) .
i=0


Proof. We consider the smooth mapping

f : R2 ’ G, f (s, t) := exp(s(X + tY )). exp(’sX).

Then f (s, 0) = e and

‚0 f (s, 0) = sTexp(sX) µexp(’sX) .TsX exp .Y,
TX exp .Y = Te µexp(X) .‚t f (1, 0).
(1)

Moreover we get
’1
δ r f (s, t).‚s = Tf (s,t) µf (s,t) T µexp(’sX) .‚s exp(s(X + tY ))
+ T µexp(s(X+tY )) .‚s exp(’sX)
’1
= Tf (s,t) µf (s,t) T µexp(’sX) .RX+tY (exp(s(X + tY )))
’ T µexp(s(X+tY )) .LX exp(’sX)
= X + tY ’ Ad(f (s, t))X.
‚t |0 δ r f (s, t).‚s = Y ’ ad(‚t f (s, 0)).X
(2)

Now we use (38.1) to get

0 = d(δ r f )(‚s , ‚t ) ’ [δ r f (‚s ), δ r f (‚t )]
= ‚s (δ r f )(‚t ) ’ ‚t (δ r f )(‚s ) ’ (δ r f )([‚s , ‚t ]) ’ [δ r f (‚s ), δ r f (‚t )].

38.2
408 Chapter VIII. In¬nite dimensional di¬erential geometry 38.3

Since (δ r f )(‚s )|t=0 = 0 we get ‚s (δ r f )(s, 0)(‚t ) = ‚t (δ r f )(s, 0)(‚s ), and from (2)
we then conclude that the curve

c(s) = (δ r f )(s, 0)(‚t ) = ‚t |0 f (s, 0) = sδ r exp(sX).Y ∈ g
(3)

is a solution of the ordinary di¬erential equation

(4) c (s) = Y + [X, c(s)] = Y + ad(X).c(s), c(0) = 0.

The unique solution for the homogeneous equation with c(0) = c0 is

c(s) = Ad(exp(sX)).c0 , since
c (s) = ‚t |t=0 Ad(exp(tX))Ad(exp(sX))c0 = [X, c(s)],
‚s (Ad(exp(’sX))C(s)) = ’Ad(exp(’sX)).ad(X).C(s)+
+ Ad(exp(’sX))[X, C(s)] = 0

for every other solution C(t). Using the variation of constant ansatz we get the
solution s
c(s) = Ad(exp(sX)) Ad(exp(’tX))Y dt
0

of the inhomogeneous equation (4), which is unique for c(0) = 0 since 0 is the
unique solution of the homogeneous equation with initial value 0. Finally, we have
from (1)

TX exp .Y = Te µexp(X) .c(1)
1
exp(X)
= Te µ .Ad(exp(X)) Ad(exp(’tX))Y dt
0
1
= Te µexp(X) . Ad(exp(’tX))Y dt
0
1
exp(X)
TX exp .Y = Te µ .Ad(exp(X)) Ad(exp(’tX))Y dt
0
1
exp(X)
Ad(exp((1 ’ t)X))Y dt
= Te µ
0
1
exp(X)
= Te µ Ad(exp(rX))Y dr.
0


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