Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homology of a connected sum

  1. Apr 10, 2009 #1

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Question: Let M, N be orientable closed n-manifolds. Knowing that the nth homology group of an orientable closed n-manifolds is isomorphic to Z, compute the homology of M#N (the connected sum of M and N).

    I was thinking of using the Mayer-Vietoris sequence with A=M u (a little bit of N), B=N u (a little bit of M) such that

    (1) A n B deformation retracts onto the "circle" S^{n-1} where M and N are glued together,
    (2) A deformation retracts onto M\B^n,
    (3) B deformation retracts onto N\B^n,

    with B^n (something homeomorphic to) an open n-ball.

    This way, the M-V sequence reads

    [tex]\ldots\rightarrow \widetilde{H}_{i+1}(M\# N)\rightarrow \widetilde{H}_i(\mathbb{S}^{n-1})\rightarrow \widetilde{H}_i(M\backslash \mathbb{B}^n)\oplus \widetilde{H}_i(N\backslash \mathbb{B}^n)\rightarrow \widetilde{H}_i(M\# N)\rightarrow \widetilde{H}_{i-1}(\mathbb{S}^{n-1})\rightarrow\ldots[/tex]

    So [tex]\widetilde{H}_i(M\# N)\cong \widetilde{H}_i(M\backslash \mathbb{B}^n)\oplus \widetilde{H}_i(N\backslash \mathbb{B}^n)[/tex] as soon as [itex]i\neq n-1, n[/itex]. But M#N is itself a closed n-manifold so we know that [tex]\widetilde{H}_n(M\# N)\cong\mathbb{Z}[/tex].

    Now if I could just figure out what the [tex]\widetilde{H}_i(M\backslash \mathbb{B}^n)[/tex] and [tex] \widetilde{H}_i(N\backslash \mathbb{B}^n)[/tex] are...

    The M-V sequence for the obvious decomposition of M into M\B^n and D^n with D^n an n-disk covering B^n gives

    [tex]\ldots\rightarrow \widetilde{H}_{i+1}(M)\rightarrow \widetilde{H}_i(\mathbb{S}^{n-1})\rightarrow \widetilde{H}_i(M\backslash \mathbb{B}^n)\oplus 0\rightarrow \widetilde{H}_i(M)\rightarrow \widetilde{H}_{i-1}(\mathbb{S}^{n-1})\rightarrow\ldots[/tex]

    So [tex]\widetilde{H}_i(M\backslash \mathbb{B}^n)\cong \widetilde{H}_i(M)[/tex] as soon as [itex]i\neq n-1, n[/itex], and similarly for N\B^n.

    Hence, [tex]\widetilde{H}_i(M\# N)\cong \widetilde{H}_i(M)\oplus \widetilde{H}_i (N)[/tex] for all [itex]i\neq n-1,n[/itex] and [tex]\widetilde{H}_n(M\# N)\cong \mathbb{Z}[/tex].

    But what about [tex]\widetilde{H}_{n-1}(M\# N)[/tex] ??
     
  2. jcsd
  3. Apr 14, 2009 #2
    Quasar

    I think the Meyer-Vietoris sequence in dimension n-1 looks like

    0 -> H_n-1(M-p) + H_n-1(N-p) -> H_n-1(M#N) ->0.

    Here I am removing a point rather than a ball but this is the same by excision.

    The left and right end points of the sequence are zero because the right end is H_n-2(S^n-1)
    and the left hand end comes from

    H_n(M-p) + H_n(N-p) -> H_n(M#N) -> H_n-1(S^n-1) -> H_n-1(M-p) + H_n-1(N-p)

    but H_n(M-p) + H_n(N-p) is zero because these retract onto manifolds with boundary so you get

    0 -> H_n(M#N) -> H_n-1(S^n-1) -> H_n-1(M-p) + H_n-1(N-p) which is

    0 -> Z -> Z -> H_n-1(M-p) + H_n-1(N-p) so the map of H_n-1(S^n-1) into H_n-1(M-p) + H_n-1(N-p) is zero.

    Is this wrong?

    By the way how do you get the mathematical symbols?
     
  4. Apr 14, 2009 #3

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You're stuck because you're faced with the exact sequence

    [tex]
    0
    \to \widetilde{H}_n(M\backslash \mathbb{B}^n)\oplus \widetilde{H}_n(N\backslash \mathbb{B}^n)
    \to \widetilde{H}_{n}(M\# N)
    \to \mathbb{Z}
    \to \widetilde{H}_{n-1}(M\backslash \mathbb{B}^n)\oplus \widetilde{H}_{n-1}(N\backslash \mathbb{B}^n)
    \to \widetilde{H}_{n-1}(M\# N)
    \to 0
    [/tex]

    correct? Well, can you compute any of those maps, so that you can split it into two shorter sequences?


    Oh, I guess you're stuck with
    [tex]
    0 \to \widetilde{H}_{n-1}(M\backslash \mathbb{B}^n)
    \to \widetilde{H}_{n-1}(M)\rightarrow \mathbb{Z}
    \to \widetilde{H}_{n-2}(M\backslash \mathbb{B}^n)
    \to \widetilde{H}_{n-2}(M)\rightarrow 0
    [/tex]

    Same advice
     
    Last edited: Apr 14, 2009
  5. Apr 14, 2009 #4

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    P.S. the sequence

    [tex]0 \to \mathbb{Z} \to \mathbb{Z} \xrightarrow{f} G[/tex]

    only implies that f maps onto a finite, cyclic subgroup of G....


    If you click on any of the images, you will see the code that was entered to produce them.
     
  6. Apr 14, 2009 #5
    right. my mistake
     
  7. Apr 14, 2009 #6
    Yeah but the manifolds are orientable so the bounding sphere is homologous to zero.
     
  8. Apr 14, 2009 #7

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    So you're saying that if L is a n-manifold with boundary, then H_n(L)=0? Is there an easy to see this?

    To type the math symbols, see this thread: https://www.physicsforums.com/showthread.php?t=8997. And as Hurkyl said, you can click on the math symbols to see what is the code for them.

    Well, I know every maps in this sequence since it is the M-V sequence. I don't know what you're hinting at since even if I could "split" this sequence in two short exact ones or something like that, I don't know most of the groups explicitly in there so how would I be more advanced?
     
  9. Apr 14, 2009 #8
    thanks quasar for the symbols.

    I think the theorem is that a compact n-manifold that is orientable contains a unique cycle in the top dimension. It is the sum of the (appropriately oriented) n-simplices in a triangulation. (I'm not sure about manifolds that have no triangulation.)

    If you remove one of these n-simplices then you no longer have a cycle so the top dimensional homology is zero. But removing an n-simplex is the same as removing an n-ball as you have been doing.

    Over Z an unorientable manifold has zero n-th homology and the unique cycle exists only over Z/2Z as in the case of the real projective plane.

    I also think that because the manifolds you are using are orientable that the bounding sphere is homologous to zero. This is really for the same reason. The fundamental cycle with an n-simplex removed will no longer be a cycle but will have boundary equal to the boundary of the missing n-simplex i.e the bounding sphere.

    Again if the manifold is not orientable this is not true. For instance, if you view real projective space as the connected sum of itself and a small sphere then the bounding circle is not homologous to zero (over Z) in the plane minus the disc. It is homologous to zero over Z/2Z.
     
    Last edited: Apr 14, 2009
  10. Apr 15, 2009 #9

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I meant more concretely than "the induced map from inclusion" or "the boundary map from homology".

    The map I most suspected you can say something about is

    [tex]H_n(M \# N) \to H_{n-1}(S^{n-1})[/tex]

    Since I would imagine any theorem that says "The n-th homology of a closed orientable n-manifold is Z" would come equipped with additional information about what happens to homology under the induced map of a map between manifolds, as well as what happens when you apply the boundary map down to a closed (n-1)-submanifold.

    I could be wrong, though -- that is just what I suspect exists.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook