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

Simple topology question

  1. Feb 21, 2010 #1
    How do you define the inside and the outside of a loop drawn on a closed surface?

    For example, take a sphere. Draw a small circle around the point that is the North pole. Now you can expand the circle by pulling it down and stretching it until it fits around the equator. If you pull it down further, it becomes a small circle encircling the South pole. Does this circle still encircle the North pole?

    Basically, it bothers me that if you have a point on the surface of a sphere, and a circle next to it (but not enclosing it), you can stretch the circle the other way around the sphere until it encloses the point! If you have a basketball and draw a point on it, and get a rubber band and place it next to the point (thereby enclosing an area next to the point), then you can stretch one half of the rubber band around the equator to enclose the point.

    There are some things like indices that one needs to calculate by drawing a loop and counting the singularities inside, but I don't really know if the loop encloses the singularity or not, because stretching the loop without crossing any other singularities allows you to ecnlose any point you want.
     
  2. jcsd
  3. Feb 21, 2010 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi RedX! :smile:
    You can't, for the reasons you've given.
    Are there (on a closed surface)? For example?
     
  4. Feb 21, 2010 #3
    Well, you can put various continuous vector fields (tangent fields) on closed surfaces.

    There's a theorem called Poincare-Hopf's theorem that relates the sum of the indices of a vector field on a surface to the Euler characteristic for that surface.

    In order to prove the theorem, the book does weird things like deforms an arbitrary closed surface (with a vector field attached) into very long submarine sandwich (or hoagie) with some holes in the middle of the sandwich (in case the original surface had holes, to keep the same Euler characteristic after the deformation of the surface). Anyways, the deformation also moves all the singularities of the vector field next to each other, on the right end of the hoagie (where you take a bite from, the cross-section).

    Anyways, that was probably a confusing description. Here's an easier one. Take a closed cylinder. All the singularities of an attached vector field are on the top lid. If I draw a circle around the cylinder near the bottom lid, does it encircle the top lid? That is, if I calculate the winding number of the circle, will I get the sum of the indices of the top lid (which should be equal to 2, the Euler characteristic of a closed cylinder), or the bottom lid (which would be zero, since there are no singularities on the bottom lid).
     
  5. Feb 21, 2010 #4

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Here is a suggested definition:

    If the surface M is oriented, then choose an orientation for the loop. Then at each point p of the loop, let v in TpM be positively oriented wrt the loop. Any vector w in TpM that is transversal to the loop is either such that (w,v) is positively oriented in M or (w,v) is negatively oritented. If w is such that (w,v) is positively oriented, call the interior of the loop the region that w points towards.

    More precisely, by this definition, a point q is in the interior of the loop iff there is a path c starting at p such that c'(0)=w and c(1)=q.
     
  6. Feb 21, 2010 #5
    The equator of the sphere encloses a singularity at either pole. The index of the singularity is independent of the coordinate chart's orientation. This is not a problem.
     
  7. Feb 21, 2010 #6

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    A situation similar to this turns up in evaluating contour integrals by residues. Remember that the complex plane can be turned into a sphere by the addition of a point at infinity. Sometimes there is a pole at this point; sometimes not. In either case, once you wrap the complex plane into a sphere, the definition of which poles are "inside" a loop becomes ambiguous.

    However, in complex integration, the situation automagically resolves itself: simply choose which side of the loop to call the "inside", and make sure you count the residues as positive or negative, depending on which way around the curve you have to go in order to keep the "inside" to the left. You will find that if you carefully keep track of the pole at infinity (when it exists), then you always get the correct answer. That is, one can just as easily add up all the residues outside a curve, but with minus signs, and one should get the same result.

    It seems to me that on the sphere, the reason you put all the poles to one side of the curve is that the curve can then be contracted to a point on the opposite side. The goal is, after all, to find the total deficit angle over the entire sphere (which is [itex]2\pi[/itex] times the Euler number). So what you're really doing is counting all the singularities on the surface, and keeping track of the signs.

    I'm not familiar with the proof you're reading, but maybe the reason for pushing all the singularities into one region is to make it easier to keep track of the signs.
     
  8. Feb 21, 2010 #7
    What if the loop were slightly off the equator? Then does it enclose the pole that belongs to the smaller hemisphere? But aren't you allowed to stretch the sphere in topology, so that the size of each region doesn't really matter?

    You can let the North Pole of a sphere have a dipole field on it with index +2 with no other singularities. For example the picture can be found in this figure on page 32:

    http://books.google.com/books?id=Aq...ift&pg=PA32#v=onepage&q=poincare-hopf&f=false

    I think the problem is that I'm using a topology book that's for high school students, so it can't use calculus or analysis or anything to explain its concepts: just pictures. So I'm just describing things with descriptions of pictures rather than equations, so I'm getting into trouble.

    They attempt to prove the Poincare-Hopf theorem (page 40):

    http://books.google.com/books?id=Aq...gift&pg=PA40#v=onepage&q=poincare-hopf&f=true

    but all other proofs I've found on the internet are not so easy: therefore, I think the book is wrong.

    Does anyone know of an easy proof of the Poincare-Hopf theorem, that doesn't require one to know topology?
     
  9. Feb 21, 2010 #8
    I don't recall ever turning the complex plane into a sphere to do a contour integral. In order to do this, does that mean all points at infinity in the complex plane are projected to the same point on the sphere? Why would you do this: are you worried about poles at infinity? I've never done a calculation where my contour enclosed a pole at infinity before: can you give me an example of an integral that has this?
     
  10. Feb 21, 2010 #9
    The size of the loop doesn't matter and the location of the pole doesn't matter. The key is that once you choose an interior region then the index is well defined at any singularity in that region.

    I think of it this way. Maybe it will help you. Choose an interior region that the circle bounds and think of all of the tangent vectors of length one tangent to this region.

    At each point these vectors form a circle and across the entire region they form a solid torus (D^2 x S^1). One can think of the circle above the singularity (isolated of course) as the central meridian of this solid torus. Then the vector field restricted to the boundary circle ( Divide the vector field by its length to get a unit vector) winds around this central meridian a certain number of times. This number is the index of the vector field. It is just a winding number.

    To prove that the sum of the indexes across a manifold is the Euler characteristic is a little hard because it is a global theorem. If one agrees that the number is the same for any vector field with isolated singularities then you can triangulate the manifold and take the vector field that is dual to the first barycentric subdivision of the triangulation. For this vector field the theorem is obvious.
     
  11. Feb 21, 2010 #10
    Thanks all. I think I see it now.

    Is there an obvious reason why the number would be the same for any vector field with isolated singularities?
     
  12. Feb 21, 2010 #11
    I don't think it is obvious but here is a complicated picture.

    Imagine the manifold embedded in some Euclidean space - for an orientable surface this would just be R^3 - and imagine a solid tube surrounding the manifold. For instance around the sphere the tube would be a spherical shell.

    The boundary of the tube is n-1 dimensional and so has a well defined unit normal vector. you think of this normal as a mapping of the boundary into the unit sphere by translating each unit normal to the origin. This mapping is known as the Gauss mapping.

    Now suppose that the manifold has a vector field with isolated singularities. Extend the vector field across the whole tube by turning it outward slowly so that by the time it reaches the boundary it is pointing normal to the surface. It is not hard to show that the singularities of this new vector field on the entire tube have the same index as the original singularities of the original vector field on the manifold.

    Now cut out a little spherical ball around each singularity. This creates a new solid whose boundary is the boundary of the tube plus the bounding spheres of each of the excised balls.
    The vector field on these spheres also maps into the unit sphere by translation to the origin and the degree of this map is just the index of the vector field - by definition.

    This degree must equal the degree of the Gauss map on the boundary of the tube because the vector field is extended across the entire solid. Thus all vector fields have the same index sum.
     
  13. Feb 22, 2010 #12
    Thanks! I've got some quick questions.
    By solid tube, do you mean a closed surface that surrounds the manifold? Or does solid mean the inside of the tube is filled (a 3-dimensional tube).
    What is n? If n is the dimension of the tube, then the tube has an (n-1)-dimensional boundary if the tube is an open surface. But for example a spherical shell (the tube for sphere in your example) has no boundary. Or is the boundary the tube itself? Or is n the dimension of the embedding space (3?).
    When you say turn the vector field outward slowly, do you mean deform the manifold that has the tangent vector field on it, until it touches the boundary of the tube? And you twisted the manifold in such a way that the tangent vector field on the manifold is normal to the surface of the tube that it touches? How can you define a tangential vector field on a tube like this, since this vector field will be normal to the tube and not tangential?
     
  14. Feb 22, 2010 #13
    Once the manifold is embedded in a n-dimensional Euclidean space it has a whole set of normal directions. If you fill out the solid tube along these normal directions you will end up with an n-dimensional solid with an n-1 dimensional boundary. For instance a circle in R^3 has a plane of normal directions at each point. If you fill out a tube along these normal planes you fill out a solid torus - which is 3 dimensional - and its boundary is a torus which is 2 dimensional. This will always work if the manifold is compact. The circle embedded in a 1000 dimensional Euclidean space will fill out to a solid tube with a 999 dimensional boundary. And of course it can be any compact manifold of any dimension.

    If the manifold has a vector field on it then what you want to do is extend this vector field to the whole solid. Imagine that it is a flow of a fluid that stays on the manifold without flowing off of it but near by it starts to flow away outwards towards the boundary.

    In a slice of the tube above a singularity the vector field will always point directly towards the boundary and will increase in speed from zero at the singularity to its maximum at the boundary.

    In a slice of the tube not above a singularity the vector field will start tangent to the manifold then turn increasingly upwards.

    By the time the field reaches the boundary it will be pointing out i.e. the fluid will be flowing out of the surface. It may not be exactly normal but that is a technicality that can be handled without changing the picture that you are after.

    Draw a picture of this construction starting with a circle in R^3 and use the polar vector field that starts at the north pole of the circle and flows to the south pole.
     
    Last edited: Feb 23, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple topology question
  1. Topology question? (Replies: 6)

  2. Question in topology (Replies: 2)

Loading...