Example 1.5
(open subsets)
.
Any open subset
U
⊂
M
of a topological mani-
fold is also a topological manifold, where the charts are simply restrictions
ϕ
|
U
of charts
ϕ
for
M
. For instance, the real
n
×
n
matrices
Mat(
n,
R
)
form a vector
space isomorphic to
R
n
2
, and contain an open subset
GL
(
n,
R
) =
{
A
∈
Mat(
n,
R
) : det
A
= 0
}
,
(3)
known as the general linear group, which is a topological manifold.
Example 1.6
(Spheres)
.
The
n
-sphere is defined as the subspace of unit vectors
in
R
n
+1
:
S
n
=
{
(
x
0
, . . . , x
n
)
∈
R
n
+1
:
x
2
i
= 1
}
.
Let
N
= (1
,
0
, . . . ,
0)
be the north pole and let
S
= (
−
1
,
0
, . . . ,
0)
be the south
pole in
S
n
. Then we may write
S
n
as the union
S
n
=
U
N
∪
U
S
, where
U
N
=
S
n
\{
S
}
and
U
S
=
S
n
\{
N
}
are equipped with coordinate charts
ϕ
N
, ϕ
S
into
R
n
, given by the “stereographic projections” from the points
S, N
respectively
ϕ
N
: (
x
0
, x
)
→
(1 +
x
0
)
−
1
x,
(4)
ϕ
S
: (
x
0
, x
)
→
(1
−
x
0
)
−
1
x.
(5)
Remark 1.7.
We have endowed the sphere
S
n
with a certain topology, but is
it possible for another topological manifold
˜
S
n
to be homotopy equivalent to
S
n
without
being homeomorphic to it? The answer is no, and this is known as the
topological Poincaré conjecture, and is usually stated as follows: any homotopy
n
-sphere is homeomorphic to the
n
-sphere. It was proven for
n >
4
by Smale,
for
n
= 4
by Freedman, and for
n
= 3
is equivalent to the smooth Poincaré
conjecture which was proved by Hamilton-Perelman. In dimensions
n
= 1
,
2
it
is a consequence of the classification of topological 1- and 2-manifolds.
Example 1.8
(Projective spaces)
.
Let
K
=
R
or
C
. Then
K
P
n
is defined to be
the space of lines through
{
0
}
in
K
n
+1
, and is called the projective space over
K
of dimension
n
.
More precisely, let
X
=
K
n
+1
\{
0
}
and define an equivalence relation on
X
via
x
∼
y
iff
∃
λ
∈
K
∗
=
K
\{
0
}
such that
λx
=
y
, i.e.
x, y
lie on the same line
through the origin. Then
K
P
n
=
X/
∼
,
and it is equipped with the quotient topology.
The projection map
π
:
X
−→
K
P
n
is an
open
map, since if
U
⊂
X
is
open, then
tU
is also open
∀
t
∈
K
∗
, implying that
∪
t
∈
K
∗
tU
=
π
−
1
(
π
(
U
))
is
open, implying
π
(
U
)
is open. This immediately shows, by the way, that
K
P
n
is second countable.
To show
K
P
n
is Hausdorff (which we must do, since Hausdorff is preserved
by subspaces and products, but
not
quotients), we show that the graph of the
equivalence relation is closed in
X
×
X
(this, together with the openness of
π
,
gives us the Hausdorff property for
K
P
n
). This graph is simply
Γ
∼
=
{
(
x, y
)
∈
X
×
X
:
x
∼
y
}
,
2