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(Z(t), 0) := AdG (0, h(t))’1 (U (t), Y (t)) ’ T µ(0,h(t)) .(0, ‚t h(t)) ,


’1
) ’ T (c( , h(t)’1 )) . ‚t h(t) ,
Z(t) = T0 (±h(t)’1 ). U (t) + T (±c(h(t) ,h(t)) ‚


k(t) := EvolK (Z)(t) ∈ K.

38.9. Semidirect products. From theorem (38.6) we see immediately that the
semidirect product of regular Lie groups is regular. Since we shall need explicit
formulas later we specialize the proof of (38.6) to this case.
Let H and K be regular Lie groups with Lie algebras h and k, respectively. Let
± : H — K ’ K be smooth such that ±∨ : H ’ Aut(K) is a group homomorphism.
Then the semidirect product K H is the Lie group K — H with multiplication
(k, h).(k , h ) = (k.±h (k ), h.h ) and inverse (k, h)’1 = (±h’1 (k)’1 , h’1 ). We have
then T(e,e) (µ(k ,h ) ).(U, Y ) = (T (µk ).U + T (±k ).Y, T (µh ).Y ).
Now we consider a curve t ’ X(t) = (U (t), Y (t)) in the Lie algebra k h. Since
s : h ’ (e, h) is a smooth homomorphism of Lie groups, from the proof of (38.6)
we get that
g(t) := EvolK H (U, Y )(t) = (e, h(t)).(k(t), e) = (±h(t) (k(t)), h(t)), where
h(t) := EvolH (Y )(t) ∈ H,
’1
(Z(t), 0) := AdK H (e, h(t) )(U (t), 0) = (Te (±h(t)’1 ).U (t), 0),
k(t) := EvolK (Z)(t) ∈ K.

38.10. Corollary. Let G be a Lie group. Then via right trivialization (κr , πG ) :
T G ’ g — G the tangent group T G is isomorphic to the semidirect product g G,
where G acts by Ad : G ’ Aut(g).
Therefore, if G is a regular Lie group, then T G ∼ g G also is regular, and T evolr
= G
evolr G . ∞ ∞
In particular, for (Y, X) ∈ C (R, g — g) = T C (R, g),
corresponds to T
where X is the footpoint, we have
1
Ad(Evolr (X)(s)’1 ).Y (s) ds, evolr (X)
evolr G (Y, X) Ad(evolr (X))
= G G G
g
0
1
Ad(Evolr (X)(s)’1 ).Y (s) ds,
TX evolr .Y = T (µevolr (X) ).
G G
G
0
t
Ad(Evolr (X)(s)’1 ).Y (s) ds.
TX (Evolr ( )(t)).Y = T (µEvolr (X)(t) ).
G G
G
0

38.10
416 Chapter VIII. In¬nite dimensional di¬erential geometry 38.10

The expression in (38.2) for the derivative of the exponential mapping is a special
case of the expression for T evolG , for constant curves in g. Note that in the semidi-
rect product representation T G ∼ g G the footpoint appears in the right factor
=
G, contrary to our usual convention. We followed this also in T g = g g.

Proof. Via right trivialization the tangent group T G is the semidirect product
g G, where G acts on the Lie algebra g by Ad : G ’ Aut(g), because by (36.2)
we have for g, h ∈ G and X, Y ∈ g, where µ = µG is the multiplication on G:

T(g,h) µ.(RX (g), RY (h)) = T (µh ).RX (g) + T (µg ).RY (h)
= T (µh ).T (µg ).X + T (µg ).T (µh ).Y
= RX (gh) + RAd(g)Y (gh),
’1
Tg ν.RX (g) = ’T (µg ).T (µg’1 ).T (µg ).X
= ’RAd(g’1 )X (g ’1 ),

so that µT G and νT G are given by

(1) µg G ((X, g), (Y, h)) = (X + Ad(g)Y, gh)
= (’Ad(g ’1 )X, g ’1 ).
νg G (X, g)


Now we shall prove that the following diagram commutes and that the equations
of the corollary follow. The lower triangle commutes by de¬nition.


xw C

=
∞ ∞
T C (R, g) (R, g g)
x
xx
u  evol
x u
evolg
T evolG G
TG

wg
TG G

=
For that we choose X, Y ∈ C ∞ (R, g). Let us ¬rst consider the evolution operator
of the tangent group T G in the picture g G. On (g, +) the evolution mapping is
the de¬nite integral, so going through the prescription (38.9) for evolg G we have
the following data:

(2) evolg G (Y, X) = (h(1), g(1)), where
g(t) := EvolG (X)(t) ∈ G,
Z(t) := Ad(g(t)’1 ).Y (t) ∈ g,
t
Ad(g(u)’1 ).Y (u) du ∈ g,
h0 (t) := Evol(g,+) (Z)(t) =
0
t
Ad(g(u)’1 ).Y (u) du ∈ g.
h(t) := Ad(g(t))h0 (t) = Ad(g(t))
0

This shows the ¬rst equation in the corollary. The di¬erential equation for the
curve (h(t), g(t)), which by lemma (38.3) has a unique solution starting at (0, e),

38.10
38.10 38. Regular Lie groups 417

looks as follows, using (1):
(h(t),g(t))
(h (t), h(t)), g (t) = T(0,e) (µg ). (Y (t), 0), X(t)
G
g(t)
= Y (t) + dAd(X(t)).h(t), 0 + Ad(e).h(t) , T (µG ).X(t) ,
(3) h (t) = Y (t) + ad(X(t))h(t),
g(t)
g (t) = T (µG ).X(t).

For the computation of T evolG we let

g(t, s) := evolG u ’ t(X(tu) + sY (tu)) = EvolG (X + sY )(t),
δ r g(‚t (t, s)) = X(t) + sY (t).
satisfying

Then T evolG (Y, X) = ‚s |0 g(1, s), and the derivative ‚s |0 g(t, s) in T G corresponds
to the element
’1
(T (µg(t,0) ).‚s |0 g(t, s), g(t, 0)) = (δ r g(‚s (t, 0)), g(t, 0)) ∈ g G

via right trivialization. For the right hand side we have g(t, 0) = g(t), so it remains
to show that δ r g(‚s (t, 0)) = h(t). We will show that δ r g(‚s (t, 0)) is the unique
solution of the di¬erential equation (3) for h(t). Using the Maurer Cartan equation
1
dδ r g ’ 2 [δ r g, δ r g]§ = 0 from lemma (38.1) we get

‚t δ r g(‚s ) = ‚s δ r g(‚t ) + d(δ r g)(‚t , ‚s ) + δ r g([‚t , ‚s ])
= ‚s δ r g(‚t ) + [δ r g(‚t ), δ r g(‚s )]g + 0
= ‚s (X(t) + sY (t)) + [X(t) + sY (t), δ r g(‚s )]g ,

so that for s = 0 we get

‚t δ r g(‚s (t, 0)) = Y (t) + [X(t), δ r g(‚s (t, 0))]g
= Y (t) + ad(X(t))δ r g(‚s (t, 0)).

Thus, δ r g(‚s (t, 0)) is a solution of the inhomogeneous linear ordinary di¬erential
equation (3), as required.
It remains to check the last formula. Note that X ’ tX(t ) is a bounded linear
operator. So we have

Evolr (X)(t) = evolr (s ’ tX(ts)),
G G
TX (Evolr ( r
)(t)).Y = TtX(t ) evolG .(tY (t ))
G
1
))(s)’1 .tY (ts) ds
AdG Evolr (tX(t
= T (µevolr (tX(t )) ). G
G
0
1
AdG evolr (stX(st ))’1 .tY (ts) ds
= T (µEvolr (X)(t) ). G
G
0
t
AdG Evolr (X)(s)’1 .Y (s) ds.
= T (µEvolr (X)(t) ). G
G
0




38.10
418 Chapter VIII. In¬nite dimensional di¬erential geometry 38.12

38.11. Current groups. We have another stability result: If G is regular and M
is a ¬nite dimensional manifold then also the space of all smooth mappings M ’ G
is a a regular Lie group, denoted by C∞ (M, G), with evolC∞ (M,G) = C∞ (M, evolG ),
see (42.21) below.

38.12. Theorem. For a regular Lie group G we have

evolr (X).evolr (Y ) = evolr t ’ X(t) + AdG (Evolr (X)(t)).Y (t) ,

evolr (X)’1 = evolr t ’ ’AdG (Evolr (X)(t)’1 ).X(t) ,

so that evolr : C ∞ (R, g) ’ G is a surjective smooth homomorphism of Lie groups,
where on C ∞ (R, g) we consider the operations

(X — Y )(t) = X(t) + AdG (Evolr (X)(t)).Y (t),
X ’1 (t) = ’AdG (Evolr (X)(t)’1 ).X(t).

With this operations and with 0 as unit element (C ∞ (R, g), —) becomes a regular
Lie group. Its Lie algebra is C ∞ (R, g) with bracket
t t
[X, Y ]C ∞ (R,g) (t) = X(s) ds, Y (t) + X(t), Y (s) ds
g g
0 0
t t

= X(s) ds, Y (s) ds .
‚t
g
0 0

Its evolution operator is given by
1
AdG (EvolG (Y s )(v)’1 ).X(v)(s) dv,
s
evol(C ∞ (R,g),—) (X) := AdG (evolG (Y )).
0
s
Y s (t) := X(t)(u)du.
0



Proof. For X, Y ∈ C ∞ (R, g) we compute

Evolr (X)(t).Evolr (Y )(t) =

‚t
r r r
= T (µEvol (Y )(t)
).T (µEvol (X)(t)
).X(t) + T (µEvolr (X)(t) ).T (µEvol (Y )(t)
).Y (t)
r
(X)(t).Evolr (Y )(t)
).(X(t) + AdG (Evolr (X)(t))Y (t)),
= T (µEvol

which implies also

Evolr (X)’1 = Evolr (X ’1 ).
Evolr (X).Evolr (Y ) = Evolr (X — Y ),

Thus, Evolr : C ∞ (R, g) ’ C ∞ (R, G) is a group isomorphism onto the subgroup
{c ∈ C ∞ (R, G) : c(0) = e} of C ∞ (R, G) with the pointwise product, which, how-
ever, is only a Fr¨licher space, see (23.1) Nevertheless, it follows that the product
o

38.12
38.12 38. Regular Lie groups 419

on C ∞ (R, g) is associative. It is clear that these operations are smooth, hence the
convenient vector space C ∞ (R, g) becomes a Lie group and C ∞ (R, G) becomes a
manifold.
Now we aim for the Lie bracket. We have

(X — Y —X ’1 )(t) = — ’Ad(Evolr (X)’1 ).X
X + Ad(Evolr (X)).Y (t)

= X(t) + Ad(Evolr (X)(t)).Y (t)’
’ Ad Evolr (X — Y )(t) .Ad Evolr (X)(t)’1 .X(t)

= X(t) + Ad Evolr (X)(t) .Y (t)’

’ Ad Evolr (X)(t) .Ad Evolr (Y )(t) .Ad Evolr (X)(t)’1 .X(t).

We shall need

T0 AdG (Evolr ( )(t)) .Y = Te AdG .T0 (Evolr ( )(t)).Y
t
= adg Y (s) ds , by (38.10).
0

Using this, we can di¬erentiate the conjugation,

) — X ’1 ).Y )(t)
(AdC ∞ (R,g) (X).Y )(t) = (T0 (X — (
= 0 + Ad(Evolr (X)(t)).Y (t)’
)(t))).Y .Ad(Evolr (X)(t)’1 ).X(t)
’ Ad(Evolr (X)(t)). T0 (Ad(Evolr (
= Ad(Evolr (X)(t)).Y (t)’
t
Y (s) ds .Ad(Evolr (X)(t)’1 ).X(t)
r
’ Ad(Evol (X)(t)).adg
0
t
r r
= Ad(Evol (X)(t)).Y (t) ’ adg . Ad(Evol (X)(t)). Y (s) ds .X(t).
0

Now we can compute the Lie bracket

[X,Y ]C ∞ (R,g) (t) = T0 (AdC ∞ (R,g) ( ).Y ).X (t)
t
r r
)(t)) .X .Y (t) ’ 0 ’ Ad(Evol (0)(t)).
= T0 Ad(Evol ( Y (s) ds, X(t)
g
0
t t

= X(s) ds, Y (t) Y (s) ds, X(t)
g g
0 0
t t
= X(s) ds, Y (t) + X(t), Y (s) ds
g g
0 0
t t

= X(s) ds, Y (s) ds .
‚t
g
0 0

38.12
420 Chapter VIII. In¬nite dimensional di¬erential geometry 38.13

We show that the Lie group (C ∞ (R, g), —) is regular. Let X ∨ ∈ C ∞ (R, C ∞ (R, g))
correspond to X ∈ C ∞ (R2 , g). We look for g ∈ C ∞ (R2 , g) which satis¬es the
equation (38.4.1):

µg(t, )
(Y )(s) = (Y — g(t, ))(s) = Y (s) + AdG (EvolG (Y )(s)).g(t, s)
= T0 (µg(t, )

‚t g(t, s) ).X(t, ) (s)

= X(t, s) + T0 AdG (EvolG ( )(s)) .X(t, ) .g(t, s)
s
= X(t, s) + adg X(t, u)du .g(t, s)
0
s
= X(t, s) + X(t, u)du, g(t, s) .
g
0

This is the di¬erential equation (38.10.3) for h(t), depending smoothly on a further
parameter s, which has the following unique solution given by (38.10.2)
t
AdG (EvolG (Y s )(v)’1 ).X(v, s) dv
s
g(t, s) := AdG (EvolG (Y )(t)).
0
s
Y s (t) := X(t, u)du.
0

Since this solution is obviously smooth in X, the Lie group C ∞ (R, g) is regular.
For convenience (yours, not ours) we show now (once more) that this, in fact, is a
s
solution. Putting Y s (t) := 0 X(t, u)du we have by (36.10.3)

‚t g(t, s) =
t
Ad(Evol(Y s )(v)’1 ).X(v, s) dv
dAd( ‚t Evol(Y s )(t)).

=
0
+ Ad(Evol(Y s )(t)).Ad(Evol(Y s )(t)’1 ).X(t, s)
t
Evol(Y s )(t)
Ad(Evol(Y s )(v)’1 ).X(v, s) dv
r s
= ((ad —¦ κ ).Ad) T (µ ).Y (t) .
0
+ X(t, s)
t
Ad(Evol(Y s )(v)’1 ).X(v, s) dv + X(t, s)
s s
= ad(Y (t)).Ad(Evol(Y )(t)).
0
s
= X(t, u)du, g(t, s) + X(t, s).
g
0


38.13. Corollary. Let G be a regular Lie group. Then as Fr¨licher spaces and
o
groups we have the following isomorphisms

G ∼ {f ∈ C ∞ (R, G) : f (0) = e} G ∼ C ∞ (R, G),
(C ∞ (R, g), —) = =

where g ∈ G acts on f by (±g (f ))(t) = g.f (t).g ’1 , and on X ∈ C ∞ (R, g) by
±g (X)(t) = AdG (g)(X(t)). The leftmost space is a smooth manifold, thus all spaces
are regular Lie groups.

38.13
38.14 38. Regular Lie groups 421

For the Lie algebras we have an isomorphism

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