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Bounding Pairs of Dehn Twists are Trivial

  1. Jul 18, 2011 #1


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    Hi Again:

    Let S_2, be the genus-2 connected, orientable surface, i.e.,

    the connected sum of 2 tori (e.g., tori spelling and some other tori)

    Consider a pair {C1,C2} of homologous, non-trivial curves, i.e.,

    neither C1 nor C2 bounds, but C1-C2 is a (sub)surface in S_2.

    Consider now a pair {D1,D2) of Dehn twists about each of C1,C2 ,

    but in opposite directions; these Dehn twists about curves like

    {C1,C2} are called bounding pairs. Still, for genus g=2, these

    maps are trivial. Why is this so?

    We clearly have only two classes of nonbounding curves , but I don't see

    why , nor in what sense this map is trivial.

    Any Ideas?

  2. jcsd
  3. Jul 19, 2011 #2


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    A Dehn twist is a diffeomorphism T-->T, so it induces an automorphism in homology. I guess what they mean by trivial here is that if you compose two twists as described in the problem, they amount to the identity in homology. This is obviously the case for the torus and with C1, C2 two meridian or two longitudinal curves.

    Also, I don't know what you meant by "We clearly have only two classes of nonbounding curves" but note that H_1(T#T)=Z^4, not Z_2.
  4. Jul 19, 2011 #3
    But Quasar, I think you need something stronger than that, since there are many maps
    diffeomorphisms, which induce the identity in homology; the collections of all those maps is actually a group called the Torelli group.
  5. Jul 19, 2011 #4


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    Stronger than what?
  6. Jul 19, 2011 #5
    I mean that the condition that, if I understood you well, you meant that the BP map
    mentioned by WWGD was referred-to as being trivial, because it induced the identity
    in homology. The condition that map induce the identity in homology seems too broad to consider that map to be trivial, in the sense that the collection of all maps that induce the identity form a group, so that all these elements are considered different from each other, and, in particular, non-trivial.
  7. Jul 19, 2011 #6


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    Then maybe it means that thay are homotopically trivial.. i.e. the composition of the two twists is homotopic to the identity. Again, on T² for the twists relative 2 meridian or longitudinal curves, this is clearly the case.
  8. Jul 20, 2011 #7
    I think I got it, Quasar, you were on the right track:

    If you take a bounding pair d1,d2 in a genus-g surface, then after you

    cut the surface along d1,d2, you end up with 2 subsurfaces S1,S2, with

    respective genus g1,g2, and with g1+g2=g. Then, for the case of g=2,

    we end up with two subsurfaces S1,S2, and we have two cases:

    1)g1=2, g2=0 .

    Then d1,d2 bound a cyclinder (g2=0 ), so that d1,d2 are homotopic, so the

    two induce the same Dehn twist.

    2)g1= g2=1

    Then after the cuts , we get two tori with two boundary components. But if we

    glue together two tori along a boundary component, we get a genus-3 surface

    ( a sort-of connected sum of 3-tori)

    I think this works.
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