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pp31
Feb7-11, 02:53 PM
Determine the homeomorphism classes of compact 3-manifolds obtained from D^3 by identifying finitely many pairs of disjoint disks in the boundary?


I just started reading some low dimensional topology on my own and I came across this question. I have realized that based on how the identification is done gives us various manifolds for instance if two disks are identified with identity map would give handlebody with unknotted handle. However I am having trouble determining the homeomorphism classes.

Any help would be appreciated.
Thanks

quasar987
Feb7-11, 03:24 PM
Maybe I'm missing something but if you identify one pair of disk, you get a solid torus. If you identify 2 pairs, you get a solid torus with two holes, etc.

the homeormorphism class is independant of the way you embed the resulting manifold in R³ (i.e. with knotted or unknotted handles). (Any two knots, no matter how knotted\twisted they are, are homeomorphic to each other. (They are not however, isotopic to each other.))

lavinia
Feb8-11, 08:57 AM
Maybe I'm missing something but if you identify one pair of disk, you get a solid torus. If you identify 2 pairs, you get a solid torus with two holes, etc.

the homeormorphism class is independant of the way you embed the resulting manifold in R³ (i.e. with knotted or unknotted handles). (Any two knots, no matter how knotted\twisted they are, are homeomorphic to each other. (They are not however, isotopic to each other.))

I think if the identification is orientation preserving you are right. But if the identification is orientation reversing you do not get a multihandled solid torus.

If you glue the polar ice caps by a reflection around a great circle through the poles I think you get a solid Klein bottle.