Hyperbolic Hopf Fibrations

Hyperbolic Hopf Fibrations

The Hopf Fibration of S^3 is amazing and beautiful.  Rather than describe it here, I'll point you to a lovely online reference with pictures and videos by Niles Johnson.

nilesjohnson.net/hopf.html

To understand it better (and fibrations in general), I recommend this talk by Niles too.

www.youtube.com/watch?v=QXDQsmL-8Us

It turns out there is an analogue of the Hopf fibration for H^3.  In fact, there is not just one "fiberwise homogenous" fibration in the hyperbolic case.  There is a 2-parameter family of them, plus one additional fibration that does not fit the family.  As with S^3, fibers in the H^3 cases are geodesics.  They are ultraparallel in fibrations from the family, and parallel in the exceptional fibration.  I found the following dissertation by Haggai Nuchi a good intro and resource to help think about all this.

www.math.upenn.edu/grad/dissertations/NuchiThesis.pdf

I'm posting three early pictures of H^3 fibrations below, labeled with the notation of Haggai's dissertation.  I plan to experiment more!  I'd love to see an interactive visualization rather than still renders, and to hold some kind of 3D print in hand.  Fibers don't touch each other, but perhaps an approach similar to Henry Segerman's S^3 models can work for a print (www.youtube.com/watch?v=fUWyHbOiwbI).

I don't yet know if the fibers project down to a 2D surface, say the hyperbolic plane, like they do in the S^3 case.  I bet they do.  I also don't yet know what kinds of surfaces of fibers (like tori in the S^3 case) are natural here. Anyone know about these questions?


Comments

  1. Fig 2.3 in Nuchi's thesis suggests that a vertical H^2 plane coming out of the line on the boundary of H^3 would intersect each fiber once, so the fibration would project to it.

    ReplyDelete
  2. Ah cool, that seems like a clue to the second question too.  Maybe one can get nice surfaces of fibers by drawing curves on that plane (say a circle), then unprojecting them.

    ReplyDelete
  3. For the special fibration (all geodesics meeting a fixed point at infinity) there is a horosphere that meets all of the geodesics perpendicularly.  So that is the quotient.  Also, this one is somewhat better than just "fiberwise homogeneous" - we can take any marked geodesic to any other.  The same is true of the Hopf fibration.  (The thesis calls this pointwise homogeneous.)

    The set of geodesics perpendicular to a given hyperbolic plane is also fiberwise homogeneous, but is not pointwise homogeneous.  

    In the Hopf fibration the quotient is a sphere, S^2.  The natural thing to do is pull back a great circle, and so get a Clifford torus.  In the two examples above, the quotients are E^2 and H^2.  The natural thing to do is pull back a geodesic, and so get... a copy of H^2.  A bit disappointing. 

    I would need to think some more to understand the classification of fiberwise homogeneous fibrations.  Hmm.  Your question makes me wonder about a related topic - what are the "natural" ruled surfaces in H^3?  A helicoid in hyperbolic space would look pretty cool.

    ReplyDelete
  4. Saul Schleimer, I agree getting a copy of H^2 from the pulled back surfaces is disappointing.  

    However, I was thinking about this further tonight, and realized there is more going on.  You get a copy of H^2 in the special case and in one choice from the family (when z=i), that is from the second two images in this post.  In one sense, these are the two least interesting of the lot, because the fibers are not twisted.

    But for the first image, you don't get a copy of H^2.  The surface you get in the upper half space model is an affine transformation of the H^2 hemisphere, or half of an ellipsoid.  One of the three ellipsoid axes remains the radius of the hemisphere, but the other two change, one growing and one shrinking, so it will be a triaxial ellipsoid.  Now I'm trying to understand what that means (the properties of this shape in hyperbolic space), and if there might be a certain choice of the shearing that might be "best".  

    The intersection of the pulled back surface with the boundary will be an ellipse, and it's two axes are those of the ellipsoid.  It'd be nice to be able to calculate them, given a certain amount of shearing.  Looks like there is a construction for this (en.wikipedia.org/wiki/Rytz%27s_construction), but that isn't helping me much so far :)

    ReplyDelete
  5. I'm glad to see that people are interested in this! Some thoughts about the discussion, which have probably already occurred to everyone but which I figured I would say explicitly: in these hyperbolic fibrations, the fibers are not a constant distance apart, so the base space of the fibrations will not have a metric structure.

    The base space does have a transitive group action which it inherits from the fibration, but usually this group action is not the full isometry group of the hyperbolic plane. As Saul Schleimer  pointed out, in two special cases those transitive group actions are those belonging to the hyperbolic plane and to the Euclidean plane. But in the rest, the group is just some subgroup of the hyperbolic group.

    Great pictures and animation!

    ReplyDelete
  6. This worries me a good deal. The base plane of the construction is a euclidean plane, or more correctly, a Möbius plan, as it should be, which means that it's not a H2 plane. It's a curved surface in H3.

    What I see here is that the fibers are simply perpendiculars to a plane, defined by the line in the tunnel, and the point at the other end.

    ReplyDelete
  7. wendy krieger, the base plane is a euclidean plane in only one of the fibrations, the last of the images above.  It is a hororsphere there.  

    For the first two images, the base plane is an H2 plane.  Like Haggai did in his paper, I've been taking it to be the xz plane (both in the ball model and the upper half space models).  

    I wasn't sure what you meant by "tunnel", but In the more general fibrations like the first above, the fibers are not orthogonal to the base plane.  In the second image, they are orthogonal.

    ReplyDelete

Post a Comment

Popular posts from this blog

{4,3,∞} Domains

Stretched Model of the Hyperbolic Plane