[ Pobierz całość w formacie PDF ]
S1 S1
But, H2(BS1; Z) = Z, H2(S2; Z) = Z, H3(BS1; Z) = 0, so we get
j
0 -�! Z -�! H2(S2S1; Z) -�! Z -�! 0
So, H2(S2S1; Z) = Z �" Z.
For S2, i.e., c1(P ) = +1, we can take the following S1-liftings, a different lifting
for each n " Z:
z � (z1, z2) �! (znz1, zn-1z2).
These are S1-liftings because of the following commutative diagram:
S1\
(z1, z2) - �! z1/z2 " S2 = CP1 z1z1 + z2z2 = 1
---
�� ��
�� ��
�z �z
S1\
z1
(znz1, zn-1z2) - �! z � rotation about the poles of S2
---
z2
For L(m, 1), i.e., c1(P ) = m, we define z1, z2 " L(m, 1) by taking the orbit
space of the diagonal Zm-action on S3 given by
k k k
m m m
e2�i � (z1, z2) �! (e2�i z1, e2�i z2), k = 0, 1, � � � , m - 1.
Thus we get a different S1-lifting for each n " Z:
z � z1, z2 �! znz1, zn-1z2 .
z1
3.4.2 Exercise. If we take the ineffective S1-action on S2 given by z � �! zp z1 ,
z2 z2
then determine the S1-liftings to the Hopf fibering S3 �! S2 up to equivalence.
Which of these S1-liftings will be effective?
What about the product case, i.e., on S1 � S2? We can take
e2�i� � (z, w) - �! (z, e2�i�w)
---
�� ��
�� ��
(e2�i�, w) - �! e2�i�w
---
Then
e2�i� � (z, w) - �! (e2�in�z, e2�i�w)
---
�� ��
�� ��
(e2�i�, w) - �! e2�i�w
---
100 3. B-HATTORI: HATTORI
would give us an infinite number of spaces covering the original? These are all
inequivalent S1-liftings.
3.4.3 Exercise. Points of CPn are represented by [z1 : z2 : � � � : zn+1], the homo-
geneous coordinates which is the S1-orbit of (z1, z2, � � � , zn+1) " S2n+1 under the
Hopf map. We can take an S1-action, say
z � [z1 : z2 : z3 : � � � : zn+1] �! [znz1 : zn-1z2 : z3 : � � � : zn+1]
for definiteness. Describe the S1-liftings to S2n+1.
3.4.4 Exercise. Let Q be a connected Lie group acting properly on W and P be
a principal G-bundle over W with G connected. Let
H = Im(evw : �1(Q, 1) �! �1(W, w)),
"
and K be the kernel of �1(Q, 1) �! H. Corresponding to K, there is a unique
connected covering group QK of Q (whose fundamental group is K). Let E be
the image of evu : �1(G, 1) �! �1(P, u), where u �! w under the bundle projection
"
map. Then the action of G lifts to P , the covering space of P corresponding to
the subgroup E. This is a principal G-bundle over W , the universal covering of W .
Show that the action of Q is liftable to P over W , if and only if, the action of QK
is liftable to P over W .
Hint: Use the following exercise and Lemma ??.
3.4.5 Exercise (cf. Lemma ??). Let G and Q be Lie groups and act on a space
Z. Suppose there exists a continuous homomorphism � : Q �! Aut(G) such that
h(az) = �(a)h(z), "a " G, h " Q.
(Assume that the image of Q is closed). Form the Lie group G Q by defining
(a, h)(b, k) = (a � �(h)b, hk).
Let G Q act on Z by (a, h)�z = a(h(z)). Show that this is well defined and encodes
the actions of G and Q on Z.
3.4.6 Exercise. Let Q be discrete and act properly on W . Let �1(W ) �! Q �!
Q �! 1 be the lifting sequence to W , the universal covering of W . Let P be a
principal G-bundle over W , G connected and E be the image of evu : �1(G, 1) �!
"
�1(P, u). Then G lifts to P , the covering space of P corresponding to the subgroup
E �" �1(P ). Show this is a principal G-bundle over W and that the group Q is
liftable to a group of weak bundle automorphism of P over W , compatible with
the homomorphism � : Q �! Aut(G) if and only if Q is liftable to a group of weak
bundle automorphism Q �! Q �! Aut(G).
3.4.7 Exercise. Formulate Exercise ?? for � : Q �! Aut(G) and principal G-
bundle over W , where G and Q are connected and in terms of Q-liftings to weak
bundle automorphisms compatible with �. Again use ??
3.5 101
These 4 exercises, in some sense, reduces the problems of liftable over W to
simply connected W s.
Question to answer if we can
3.5
In our stuff, if we take Q �! TOP(W ) and it is a very nice subgroup of TOP(W ),
say Isom(W ), then any extension � that we can stick into TOPG(P ) should not
be very weird. In fact, as we vary a possible embedding (or a homomorphism into
TOPG(P )) keeping � � � : Q �! Out(G) � TOP(W ) fixed, we can really only alter
by elements of (G) �ZG M(P, G), along the fibers. So what chance do we have in
taking some embedding of � into TOPG(P ) and conjugating it into a nice subgroup
of TOPG(P )? Of course, we discussed this in the product case several times in the
[ Pobierz całość w formacie PDF ]