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Homology of torus and Klein's bottle

  1. Jun 16, 2011 #1

    ddo

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    1. The problem statement, all variables and given/known data
    I'm trying to calculate singular homology groups of the torus and Klein's bottle using the Mayer-Vietoris sequence.

    3. The attempt at a solution

    I represent both spaces as a rectangle with identified edges. Then I take the sets:
    U=rectangle without the boundary
    V=rectangle without the middlepoint
    so U is contractible thus H_n(U)=0 for n>0, H_0(U)=Z
    V=S1vS1 so H_1(V)=ZxZ, H_n(V)=0
    and their intersection = S1, H_n(S1)=0, H_1(S1)=Z, H_0(S1)=Z
    Now from the M-V sequence for n>2 we get an exact sequence
    0->0x0->H_n(T)->0, so H_n(T)=0.
    But I don't know what to do for smaller n...

    Please help!
     
  2. jcsd
  3. Jun 16, 2011 #2

    micromass

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    Hi ddo! :smile:

    Do you need to derive everything from the Mayer-Vietoris sequence? That looks quite hard.

    The zero'th homology group is very easy since it is the number of path connected components.
    The first homology group is also easy by calculating the fundamental group of [itex]S^1\times S^2[/itex]...

    The second homology group can be derived from Mayer Vietoris then.
     
  4. Jun 16, 2011 #3

    ddo

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    Thank for your reply!
    I suppose the Mayer-Vietoris hint was there to make the task easier :)
    So H_0 is Z because there is only one connected component, H_1 is the abelianization of the fundamental group, both torus and Klein's bottle have abelian fundamental groups so for torus it's [itex] Z \times Z [/itex], for Klein's bottle [itex] Z \times Z_2 [/itex].
    Now M-V gives:
    [itex]0 \to H_2(T) \to Z \to 0 \times (Z \times Z) \to H_1(T) \to Z [/itex]
    And I still don't know how to derive H_2(T)...
     
  5. Jun 16, 2011 #4

    micromass

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    Write out the full Mayer-Vietoris. I'll do it for the torus

    [tex]0\rightarrow H_2(X)\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z} \rightarrow 0[/tex]

    Now, it is easy to see that [itex]H_2(X)[/itex] is either 0 or [itex]\mathbb{Z}[/itex].

    Now, reason from left to right as follows:
    The end of the sequence is

    [tex]\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z} \rightarrow 0[/tex]

    So the the map is a surjection, and the kernel of the map is isomorphic to [itex]\mathbb{Z}[/itex].

    Consider the next part of the sequence:

    [tex]\mathbb{Z}\rightarrow \mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}[/tex]

    The kernel of the right map equals the image of the left map. Thus the image of the left map is isomorphic to [itex]\mathbb{Z}[/itex]. Thus the kernel of the left map is necessarily 0.

    Now consider the next part of the sequence, etc.

    Try to complete this argument.
     
  6. Jun 16, 2011 #5

    ddo

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    [tex] \mathbb{Z}\times\mathbb{Z} \rightarrow\mathbb{Z}\rightarrow \mathbb{Z}\times\mathbb{Z}[/tex]
    The kernel of the right map is 0, so the image of the left map is 0, so the kernel of the left map is [itex]\mathbb{Z}\times\mathbb{Z}[/itex]
    Next:
    [tex]\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}[/tex]
    The kernel of the right map is [itex]\mathbb{Z}\times\mathbb{Z}[/itex], so the image of the left map is [itex]\mathbb{Z}\times\mathbb{Z}[/itex], so the kernel of the left map is 0.
    [tex]H_2(X)\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}\times\mathbb{Z}[/tex]
    The kernel of the right map is 0, so the kernel of the middle map is [itex]\mathbb{Z}[/itex], so the image of the left map is also [itex]\mathbb{Z}[/itex], so [itex]H_2(X)[/itex] can't be 0.
    Is that correct?
     
  7. Jun 16, 2011 #6

    micromass

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    Sounds good! So the second homology group is [itex]\mathbb{Z}[/itex]!

    The thing with the Klein Bottle should be analogous!
     
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