Results 1 to 5 of 5

Math Help - Homology group of S

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    70

    Homology group of S

    Hi,

    I've question about homology group of S.

    How can I show that:

    H_{n}\left(S\right)\cong\mathbb{Z} for n=0,1 ?

    Thanks for any help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    May 2010
    Posts
    95
    Quote Originally Posted by Arczi1984 View Post
    Hi,

    I've question about homology group of S.

    How can I show that:

    H_{n}\left(S\right)\cong\mathbb{Z} for n=0,1 ?

    Thanks for any help.
    There are several ways to solve this problem. I mean, you can use a simplicial, singular or cellular homolgy theory to solve this problem, resulting the same homolgy groups in this case.

    The simplest one might be a simplicial homology using \Delta-complexes.

    A circle S is described as one vertex v (0-simplex) and one edge (1-simplex) in the \Delta-complex structure. Since the boundary map of 1-simplex e is \partial e=v - v = 0, we have H_0(S)\cong \frac{\text{Ker } \partial_0}{\text{Im} \partial_1} = \frac{\mathbb{Z}}{\{e\}}=\mathbb{Z}, where \text{Ker } \partial_0 = \mathbb{Z} since we have a single 0-simplex (vertex v) whose boundary map is trivially zero.

    In a similar manner, we have H_1(S) \cong \frac{\text{Ker } \partial_1}{\text{Im} \partial_2} = \frac{\mathbb{Z}}{\{e\}}=\mathbb{Z} since there is no 2-simplex in our \Delta-complex structure and we have a single 1-simplex (edge e) whose boundary map is zero.

    It is trivial to show that all the remaining higher homology groups of S are zero.

    This is kind of a standard textbook question and you can find this in most algebraic topology books like Hatcher's Algebraic topology p 106.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2009
    Posts
    70
    Thanks for help.

    I've one more question, maybe two.

    Is it possible to show this by exact sequences?
    Could You show me how can I solve this problem via singular homology?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Sampras's Avatar
    Joined
    May 2009
    Posts
    301
    Quote Originally Posted by Arczi1984 View Post
    Hi,

    I've question about homology group of S.

    How can I show that:

    H_{n}\left(S\right)\cong\mathbb{Z} for n=0,1 ?

    Thanks for any help.
    Or one can note that since  S is connected,  H_{0}(S) \cong \mathbb{Z} .

    Also,  H_{1}(S) := \ \text{the number of} \ 1- \text{cycles that are not} \ 1 \ \text{boundaries} . Now  S itself is the only 1-cycle which is not a 1-boundary, since there are no higher dimensional objects. Thus  H_{1}(S) \cong \mathbb{Z} .
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2010
    Posts
    95
    Quote Originally Posted by Arczi1984 View Post
    Thanks for help.

    I've one more question, maybe two.

    Is it possible to show this by exact sequences?
    Could You show me how can I solve this problem via singular homology?
    In a simplicial homology theory, we concern about the triangulations of the given space. In a singular homology, n-simplexes are replaced by their continuous images in the given topological space, where the continuous images do not have to look like simplexs.

    A singular n-simplex is thus a continuous map \sigma:\Delta^n \rightarrow X rather than \Delta^n itself.

    The n-chains, denoted as C_n are finite sums \sum_in_i\sigma_i for n_i \in \mathbb{Z} and \sigma_i:\Delta^n \rightarrow X.

    If you define a homomorphism \epsilon:C_0(X) \rightarrow \mathbb{Z} by \epsilon(\sum_in_i\sigma_i)=\sum_in_i , a little bit of computation shows that \epsilon induces an isomorphism and H_0(X) \cong \mathbb{Z}, where X is nonempty path-connected space (verify that \text{Im }\partial_1=\text{Ker } \epsilon).

    More generally,

    Lemma: If X is a space and \{X_i:i \in I \} are the path components of X, then H_k(X) \cong \sum_{i \in I}H_k(X_i).

    To compute H_1(S^1), you might think of an intuitive way. For instance, a circle can be thought of three intervals (or more) gluing together at their endpoints and their boundary map corresponds to zero and we know that there is no nontrivial singular 2-simplex is available to a topological space of a circle.

    Or you can use an exact sequence and the above lemma and find its homology group (see here).

    Note that H_0(X)=\tilde{H}_0(X) \oplus \mathbb{Z}.
    Last edited by TheArtofSymmetry; May 28th 2010 at 08:58 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Suspension homology
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: December 17th 2011, 12:22 PM
  2. Homology Groups
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: December 13th 2010, 02:33 PM
  3. Homology and Cohomology
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: April 13th 2010, 01:02 AM
  4. Homology Groups of T3
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: August 6th 2009, 08:36 AM
  5. Homology of the 3-Torus
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: March 30th 2009, 03:32 AM

Search Tags


/mathhelpforum @mathhelpforum