Talk:Seifert fiber space

Torus bundles

Well it is fine someone else is willing to share knowledge.

And, i've a Q for you -R.e.b-

• It is know that the ${MCG}$ determine some surface bundles over a simple loop which they are SFS, coz the use of periodic monodromies.

• You can see that there are only four K-bundles coz MCG(K)=Z_2+Z_2 (only 4 periodic monodromies)

• Also, seven (periodic) T-bundles, five orientable and two non.

• • so the question is: Do you know if all the seven periodic $N_{3}$-bundles are SFS? Which are each symbol of, if any? yes, it is widely accepted that $MCG(N_{3})=GL_{2}(\mathbb {Z} )$, gave it by Birman-Chillingworth.

—The preceding unsigned comment was added by Juan Marquez (talk • contribs) 2006-11-12T21:36:09.

Yes, all surface bundle over a circle of finite order automorphisms of the 2-torus are Seifert fiber spaces. More generally the surface bundle over a circle of a 2-torus automorphism is a Seifert fiber space if and only if some power of the automorphism is unipotent. I have added this to the article. R.e.b. 14:57, 13 November 2006 (UTC)

¨

$N_{3}$-bundles

Let me rephrase the Q: For the genus three non-orientable surface $N_{3}$, do you know if all the seven periodic $N_{3}$-bundles over $S^{1}$ are SFS? Remember that, there is a Scott's definition of a Scott-SFS that said: an S-SFS is a three manifold that can be foliated by circles, which is more general than Seifert one. Greets --kiddo 18:15, 13 November 2006 (UTC)

I dont know offhand, but I see no reason why they should be SFS unless the automorphisms have isolated fixed points. R.e.b. 17:00, 14 November 2006 (UTC)

In fact there are at least one which has a periodic monodromy with a circle of fixed points.

Do you know of somebody who already did?

Do you know someone (human) who can help us?

Do you know of any interested in this stuff?

Thanx --kiddo 19:09, 14 November 2006 (UTC)

The SFS that fiber over the circle have been classified by Orlik and Raymond; you might be able to answer your question by checking their results, which are sumarized in Orlik's book on Seifert manifolds. R.e.b. 20:55, 14 November 2006 (UTC)

Nah, Orlik wasn't awared of the result: $MCG(N_{3})=GL_{2}(\mathbb {Z} )$, so there wasn't any urge in determining $N_{3}$-bundles over the circle. And it is difficult to give the homeomorphism between the circle bundle and the surface bundle by means of a homomorphism of fundamental groups, as Orlik did for torus bundles and Klein bundles, back in 1972 $:)$ --kiddo 21:30, 14 November 2006 (UTC)

Hempel claims on page 121 of his book that any suface bundle is Seifert fibered. However his proof on page 124 only seems to work for homeomorphisms such that no power has 1-dimensional fixed sets. This applies to all orientation preserving homeomorphisms of orientable surfaces, and all homeomorphisms of odd order, but does not quite seem to cover your case. In any case it seems to show that your surface bundles have double covers that are Seifert; I'm not sure if this implies that they are Seifert. R.e.b. 18:04, 15 November 2006 (UTC)

Thanks again friend, i'm gonna check it out and lets keep pushing the subjet.--kiddo 04:02, 16 November 2006 (UTC)

Some are surface bundle

I think that a fibering over the circle with fiber a surface isn't the same as mapping cylinders or

--kiddo 18:30, 13 November 2006 (UTC)

A mapping cylinder suface bundle associated to a surface automorphism always has a natural fibering. In general this is not a Seifert fibration, but it is if the automorphism has finite order and looks locally like a rotation near all fixed points. R.e.b. 19:45, 13 November 2006 (UTC)

Can you tell me how a SF can be seen as a MC, please? If isn't quite annoying friend ...--kiddo 16:12, 14 November 2006 (UTC)

I have just realized that I was misusing the term "mapping cylinder"; I confused it with the term mapping torus. R.e.b. 17:10, 14 November 2006 (UTC)

Mapping torus generalizes the notion of surface bundle -which for 3-manifolds- the non-trivials are (the awesomes) surfaces bundle over the circle.

Allow me here

to give the Seifert-translation of the first 11 (χ>0)

• 1. $(Oo,0|b)\,$: which are
  • • (Oo,0|0)=$S^{2}\times S^{1}$,
    • (Oo,0|1)=$S^{3}\,$.
    • And if b>1; (Oo,0|b)=L(b,1)

• 1. $(Oo,0|b:(a_{1},b_{1}))=L(ba_{1}+b_{1},a_{1}),\,$
  2. $(Oo,0|b:(a_{1},b_{1}),(a_{2},b_{2}))\,$
  3. $(Oo,0|b:(2,1),(2,1),(a_{3},b_{3}))\,$
  4. $(Oo,0|b:(2,1),(3,b_{2}),(3,b_{3}))\,$
  5. $(Oo,0|b:(2,1),(3,b_{2}),(4,b_{3}))\,$
  6. $(Oo,0|b:(2,1),(3,b_{2}),(5,b_{3}))=(Oo,0|-1:(2,1),(3,1),(5,1))\,$
  7. $(NnI,1|b)\,$
  8. $(NnI,1|b:(a_{1},b_{1}))\,$
  9. $(On,1|b:)\,$
  10. $(Oo,0|b:(a_{1},b_{1}))\,$

then, the next 11 (χ=0)

• 1. $(Oo,0|b:(3,b_{1}),(3,b_{2}),(3,b_{3},))\,$
  2. $(Oo,0|b:(2,1),(4,1),(4,b_{3}))\,$
  3. $(Oo,0|b:(2,1),(3,b_{2}),(6,b_{3}))\,$
  4. $(Oo,0|b:(2,1),(2,1)(2,1),(2,1),(2,1))\,$
  5. $(Oo,1|b)\,$
  6. $(No,1|b))\,$
  7. $(NnI,1|0:(2,1),(2,1))\,$
  8. $(NnI,2|b)\,$
  9. $(On,1|b:(2,1),(2,1))\,$
  10. $(On,2|b)\,$
  11. $(NnII,2|b)\,$

The invariants

So I tried to compute the Seifert space {(O,o₁),b;(a₁,b₁),(a₂,b₂)}, and I get something a bit different- the lens space L(a₁b₂+a₂b₁-ba₁a₂,ma₂-lb₂). There's a minus instead of a plus before a₁a₂. Can anyone recheck this or give me a reference for this formula so I can rethink about it? Talipos (talk) 07:28, 11 November 2009 (UTC)