Graphics Reference
In-Depth Information
a
a
c
2
c
1
c
2
c
3
c
p-1
c
1
c
0
c
0
c
4
b
b
(a)
(b)
Figure 7.14.
The lens paces L(p,q) and L(5,1).
Definition.
Let p and q be relatively prime positive integers and assume that
0 £ q £ p/2. Define the
lens space
L(p,q) as follows: Let S be the reflection in
R
3
about
the x-y plane and let R be the rotation about the z-axis through an angle of 2pq/p
radians. Then
(
)
=
D
3
Lpq
,
~,
where ~ is the equivalence relation induced by the identification of
x
Œ
S
+
with R(S(
x
)).
In other words, we are identifying the point
x
in the upper hemisphere with the point
in the lower hemisphere obtained by reflecting
x
about the xy-plane and then rotat-
ing by 2pq/p. We are not identifying any points in the interior of the disk
D
3
.
Here is another description of L(p,q). See Figure 7.14(a). Let
=
(
)
c
k
cos
2
p
kp
,sin
2
p
kp
,
0
,
k
=
0 1
, ,...,
p
-
1
.
The points
c
k
divide the equator of the sphere into p equal arcs. Let
a
=
e
3
and
b
=
-
e
3
be the north and south pole of the sphere, respectively. We get the following cell
decomposition of
D
3
:
0-cells:
a
,
b
,
c
k
1-cells: the great arcs from
a
to
c
k
and from
b
to
c
k
and the arcs along the unit
circle from
c
k
to
c
k+1
2-cells: 2p curved triangles, denoted by
vc
k
c
k+1
, bounded by the arcs from
v
to
c
k
,
from
c
k
to
c
k+1
, and from
c
k+1
back to
v
, where
v
is
a
or
b
3-cells:
D
3
(All indices are taken modulo p.) L(p,q) is now the disk
D
3
where we identify the
curved triangle
ac
k
c
k+1
in the upper hemisphere with the curved triangle in the lower
hemisphere
bc
k+q
c
k+q+1
and the vertices are identified in the order listed. Figure 7.14(b)
shows the case of L(5,1) where we have linearized the construction and have replaced
the sphere by a suspension of a five-sided polygon. The shaded triangle
ac
3
c
4
gets
identified with the shaded triangle
bc
4
c
0
.
