# Computation of homology group

1. Apr 18, 2010

### Sumanta

Hello,

I am reading the book on algebraic topology by Fulton and the geometric intusion that is supposed to be given for homological group is the number of connected components.

I wanted to understand an example.

For a circle ( S1) which has got 2 points say A at ( 1, 0) and another point B at ( - 1, 0) is the calculation of the Homology group in the following way

$$\stackrel{ a + b }{a -b }$$ where a and b are the two generators along the upper half plane and the lower half plane. Then is the result 2a or 2b ie 2*Z.

Thx

2. Apr 20, 2010

### Bacle

Are you interested in H_o(X,G) , the 0-th homological group.?. In that case, the homology
group (working with coefficients from an Abelian group G) is given by kG, where k is the
number of path-connected components, i.e

Ho(X;G)=G(+)G(+)...(+)G , k copies.

The issues as a I see it, is that , in each path-component, a single point generates
the full Ho(X;G), since the only possible boundaries are paths, and there is, by assumption,
a path between any two points.

Expanding: consider a path-connected component X1 , and let x1 be a point
in X1. Then x1 itself generates all the 0-th homology: let x1' =/ x1 in x1. Then there
is a path p(t) between x1 and x1' in X1 .
This means that either x1-x1' or x1'-x1 is the boundary of p(t) (basically, the paths
are the only 1-dimensional objects that are/can be bounded by the points in H_o(X1,G)
.
You can easily see that there can be no other homology classes in X1; given any other
point x1'' in X1, x1-x1'' or x1''-x1 are a boundary of this path. This means that every
point in X1 is homologous to x1.

Same goes for any connected component.

HTH.

3. Apr 20, 2010

### Bacle

I also don't know how Fulton's book lays out the homology groups.

4. Apr 20, 2010

### zhentil

Just from the definitions,

$$H_0(X;\mathbb{Z})$$

is defined as the closed 0-chains (every 0-chain is closed) modulo the boundaries (i.e. beginning and end points of maps of the interval into X). In our case (where I don't draw a distinction between connected and path-connected), two points a and b lie in the same connected component if and only they can be connected by a path

$$\gamma: [0,1] \rightarrow X$$

Hence a and b are homologous if and only if

$$b-a = \partial (\gamma)$$.

It follows therefore that the zeroth homology group is freely generated by the homology class of a point selected from each connected component.

5. Apr 21, 2010

### Bacle

Re: computation of homology group. Meaning of H_n(X,Z)

Just Curious:

It is straightforward to interpret the meaning of either (X an n-manifold, Z=integers.):

H0(X,Z)=kG , or Hn(X,Z)=Z (assume for now G has no torsion

o.wise use Universal Coeff. Thm.) .

First tells us that X has k path components, and last tells us that X is orientable.

How would we interpret , tho, H1(X;Z) =Z (or something else;

obviously, n>1 here.).?. I guess , Z being Abelian, we can also conclude (Hurewicz)

that Pin=Z too.

Otherwise , AFAIK, Hn measures "k-connectedness" . Anyone have