7Simplicial homology
II Algebraic Topology
7.7 Homology of surfaces
We want to study compact surfaces and their homology groups. To work with
the simplicial homology, we need to assume they are triangulable. We will not
prove this fact, and just assume it to be true (it really is).
Recall we have classified compact surfaces, and have found the following
orientable surfaces Σ
g
.
We also have non-compact versions of these, known as
F
g
, where we take the
above ones and cut out a hole:
We want to compute the homology groups of these Σ
g
. One way to do so is
to come up with specific triangulations of the space, and then compute the
homology groups directly. However, there is a better way to do it. Given a Σ
g
,
we can slice it apart along a circle and write it as
Σ
g
= F
g−1
∪ F
1
.
Then, we need to compute
H
∗
(
F
g
). Fortunately, this is not too hard, since it
turns out
F
g
is homotopic to a relatively simple space. Recall that we produced
Σ
g
by starting with a 4g-gon and gluing edges:
a
1
b
1
a
1
b
1
a
2
b
2
a
2
b
2
Now what is
F
g
?
F
g
is Σ
g
with a hole cut out of it. We’ll cut the hole out at
the center, and be left with
a
1
b
1
a
1
b
1
a
2
b
2
a
2
b
2
We can now expand the hole to the boundary, and get a deformation retraction
from F
g
to its boundary:
a
1
b
1
a
1
b
1
a
2
b
2
a
2
b
2
Gluing the edges together, we obtain the rose with 2
g
petals, which we shall call
X
2g
:
Using the Mayer-Vietoris sequence (exercise), we see that
H
n
(F
g
)
∼
=
H
n
(X
2g
)
∼
=
Z n = 0
Z
2g
n = 1
0 n > 1
We now compute the homology groups of Σ
g
. The Mayer-Vietoris sequence gives
the following
0 H
2
(S
1
) H
2
(F
g−1
) ⊕ H
2
(F
1
) H
2
(Σ
g
)
H
1
(S
1
) H
1
(F
g−1
) ⊕ H
1
(F
1
) H
1
(Σ
g
)
H
0
(S
1
) H
0
(F
g−1
) ⊕ H
0
(F
1
) H
0
(Σ
g
) 0
We can put in the terms we already know, and get
0 H
2
(Σ
g
) Z Z
2g
H
1
(Σ
g
) Z Z
2
Z 0
By exactness, we know
H
2
(Σ
g
) =
ker{Z → Z
2g
}
. We now note that this map
is indeed the zero map, by direct computation — this map sends
S
1
to the
boundary of the hole of
F
g−1
and
F
1
. If we look at the picture, after the
deformation retraction, this loop passes through each one-cell twice, once with
each orientation. So these cancel, and give 0. So H
2
(Σ
g
) = Z.
To compute
H
1
(Σ
g
), for convenience, we break it up into a short exact
sequence, noting that the function Z → Z
2g
is zero:
0 Z
2g
H
1
(Σ
g
) ker(Z → Z
2
) 0.
We now claim that
Z → Z
2
is injective — this is obvious, since it sends 1
7→
(1
,
1).
So the kernel is zero. So
H
1
(Σ
g
)
∼
=
Z
2g
.
This is a typical application of the Mayer-Vietoris sequence. We write down
the long exact sequence, and put in the terms we already know. This does not
immediately give us the answer we want — we will need to understand one or
two of the maps, and then we can figure out all the groups we want.