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Computation of homology group

  1. Apr 18, 2010 #1
    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

    [tex]\stackrel{ a + b }{a -b }[/tex] 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. jcsd
  3. Apr 20, 2010 #2
    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.
     
  4. Apr 20, 2010 #3
    I also don't know how Fulton's book lays out the homology groups.
     
  5. Apr 20, 2010 #4
    Just from the definitions,

    [tex]H_0(X;\mathbb{Z}) [/tex]

    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

    [tex]\gamma: [0,1] \rightarrow X [/tex]

    Hence a and b are homologous if and only if

    [tex]b-a = \partial (\gamma) [/tex].

    It follows therefore that the zeroth homology group is freely generated by the homology class of a point selected from each connected component.
     
  6. Apr 21, 2010 #5
    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

    any comments on this.?
     
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