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Question On Homotopy Groups

  1. Feb 7, 2005 #1
    Hi

    I have this question on homotopy groups: Spacial infinity in two dimensional space is a unit circle S1 (topologically). I understand that. Now, in physics one can prove that fields will exhibit an equation (expressed by the map phy --> v =constant) that also represents a unit circle. Now, a map between two unit circles (S1 to S1) is topologically stable because this map cannot be transformed into a map where S1 is mapped onto a point of S1. Can someone explain that last sentence, please.

    Now, a homotopy group counts the number of topologically inequivalent maps. Can some one give me an example of those kinds of maps. Does this mean that you cannot go from one map to the other without breaking it apart ?

    A map phi--->v*exp(imx) (where x is the angle and m an integer) represents a map that wraps one circle around the other m times. Does anyone know what this means and how can i envision this. This is about solitons and instantons

    thanks in advance

    regards
    marlon
     
  2. jcsd
  3. Feb 7, 2005 #2
    PS these questions come from QFT in a Nutshell by A.Zee, page 283

    marlon
     
  4. Feb 7, 2005 #3

    Hurkyl

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    It sounds like they're talking about contractability. Here's an example:

    An example of a map from S1 to R^2 is one that maps S1 homeomorphically to a circle centered at the origin. Let f be one such map.

    We can smoothly transform f into a constant map as follows: define g(x, t) = (1 - t) f(x) where t ranges over [0, 1]. We have that g(x, 0) = f(x) and g(x, 1) = (0, 0). So, we would say that this map is contractible.

    If, however, we mapped into R^2 - (0, 0), then f would not be contractible. (because it "contains" a hole)

    A less trivial example would be a map from S^1 into a circle that loops around a torus. (Either meaning to around is fine: they're inequivalent, I think)


    Disclaimer: I'm not too familiar with homotopy, and I'm borrowing heavily from the other things I know about topology. I may have the terminology all wrong for this. :frown:


    Think of how the functions zm behave on the complex plane: it's essentially the same. (it is the same thing when you embed S^1 as the unit circle)
     
  5. Feb 8, 2005 #4

    matt grime

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    The maps of the circle to some space are the same as the maps from [0,1] to the space with the images of 0 and 1 agreeing. Think of it as inserting a bit of length of elastic into the space and joining up the ends. Two of these loops are topologically the same if they can be deformed continuously into one another. INSIDE the space. Normally we fix the point where 0 and 1 are sent to, it is called the base point.

    In the case when we're mapping into the circle we can loop the elastic once, twice, three times, etc round the circle. All these are different. We can send the elastic to a single point, or we can wrap it round in the opposite direction, ie minus once, -2, or -3 times. Thus these maps are the same as the abstract group of integers (i'm omitting details, obviously).

    If you're used to winding numbers from, say, complex analysis, that is exactly what we're measuring here - the winding number of something round the origin.

    "Now, a map between two unit circles (S1 to S1) is topologically stable because this map cannot be transformed into a map where S1 is mapped onto a point of S1. Can someone explain that last sentence, please."

    I can't explain that since it is wrong. Though I admit they are using some odd terminology (bloody applied mathematicians). I think there must need to be some qualifications about what kind of map they're talking about.

    Anything else?
     
  6. Feb 8, 2005 #5
    Hi matt, thanks for the reply. However, i must admit that i do not understand this. I have read previous explanations like this but i really don't get it. What does that mean : the images of 0 and 1 agreeing... What about this elastic that you insert into the space ??? What is this elastic and what is the space ??? I really don't get it.


    Well the field configuration becomes phy --> |v|² at spatial infinity. This is a unit circle S1 and the spatial infinity itself is also a unit circle S1 (in the case of a two dimensional space) Hence the fact that the field MUST become a S1 circle at infinity leads to the map between S1 and S1...

    marlon
     
  7. Feb 8, 2005 #6

    matt grime

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    You don't follow the maths, and I don't follow the physics, so we're stuck: I've no clue as to what the phrase "spatial infinity" might begin to mean.

    Let's start properly on the maths. Let X be a topological space, and let x be a distinguished point, eg X=S^1 and x=1.

    A path is a continuous map (or function), f, from the interval I=[0,1] to X such f(0)=f(1)=x.

    If you like, imagine a particle moving in the space X and at time t=0 it is at x, and at time t=1 it is back at x, inbetween it traces out a path in X.

    Elastic is a metaphor for a trace of the path. Two paths are the same if there is a way to continuously deform one to the other, in the metaphor, move the two bits of elastic so that one covers the other exactly, in the technical terminology there is a homotopy between them.

    Example X+S^1, x=1, define f(t)=1 for t in I, and g(t) = e(t) for 0<=t<=1/2, and e(1-t) for 1/2<=t<=1, where e(s) = exp{2.s.\pi}


    so f would keep the particle fixed at 1, and g moves it around the upper half circle to -1 and then back again. g can be continuously deformed into f without moving outside the circle. Explicitly, define

    h(t,s) = g(t) for 0<=t<=s/2, then it stays at g(s/2) for s/2<=t<1-s/2, and for 1-s/2<=t< 1 it is g(t) again, where s runs from 0 to 1, h is continuous and h(t,0) = f(t), h(t,1)=g(t). this is a homotopy.

    You need to draw pictures and understand what is going on!

    The map e(t), for 0<=t<=1 is not homotopic to f, it cannot be continuosly deformed to be f.
     
  8. Feb 8, 2005 #7
    this is clear now, thanks matt.


    Howdo you come to this ? You say that the h(t,s)=g(t) for 1-s/2<t<1. But let us take s=1 then we have that 1/2<t<1 and then the h is NOT equal to g(t) because this is only the case for 0<t<1/2, given the definition of h(t,s)

    What am i doing wrong here ?


    Sorry, i don't see how ?

    regards
    marlon and thanks again for helping me out
     
  9. Feb 8, 2005 #8

    matt grime

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    Put s=1, then

    h(t,1) = g(t) for 0<=t<=1/2

    h(1/2,1) = g(1/2) for 1/2<=t<=1/2

    h(t,1) =g(t) for 1/2<=t<=1

    so h(t,1)= g(t) for all t in I.

    Nothing wrong there.

    The picture you should have in you mind is that at time 0 we stay put, then at time s we go an s'th of the way round the upper half circle, stay at the s'th of the way round and then come back, and so on until we reach at time s=1 the loop g given by going half way round and then turning around.



    The proof that looping once round the circle is not homotopic (continuously deformable) to the constant map is complicated, but the result is obvious: you have a strip of elastic looped round the circle fixed at either end to the point one. How are you going to shrink it so that it is concentrated at 1 WITHOUT breaking the elastic or moving it outside the circle (which isn't allowed). If you think about it, it is the same as taking the punctured plane, ie one missing the origin. Wrap a loop round the (missing) origin. Sliding and stretching the loop in the plane only, try and shrink it down without passing over the hole at the origin. Cannot be done. The actual mathematical proof is long, about 600 words, or a page in a densely written text book. You don't want to see it here, honestly. Go buy any decent book on algebraic topology, and it will prove it directly (the long way) or indirectly by creating a lot of machinery. However, if the loop were contractible, then it would follow that the integral of 1/z round the unit circle would be zero, but it isn't, is it?
     
    Last edited: Feb 8, 2005
  10. Feb 8, 2005 #9
    Thanks matt

    again a crystal clear explanation

    regards
    marlon
     
  11. Feb 8, 2005 #10

    mathwonk

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    the fact that a map from a circle to a circle, whose winding numer, i.e. homotopy class, is non zero, cannot be contracted to a point, is just the fact that if you take a piece of string, wind it around a newspaper some non zero number of times, and then tie it tight, it won't just fall off. I.e. it would have to either pass through ther newspaper, or slide off the end to come off.

    neither of those operations is allowed in your homotopy.
     
  12. Feb 8, 2005 #11

    mathwonk

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    by the way here is the first topic in homotopy theory: why the identity map of the circle to itslef in the plane, is not homotopic to a constant map, as a map into the opunvtured plane, i.e. the unit circle cannot be contracted to a point without passing through the origin.

    Problem: prove that a continuous map from the unit disc to the plane, that sends the unit circle identically onto itself, must send some point of the disc to the origin.

    Hint: use polar cordinates. This can also be proved using calculus, if you give yourself a smooth, i.e. continuously differentiable map. Then it all follows from the existence of the angle form, "dtheta", and integration i.e. polar coordinates again.

    this generalizes to 2 dimensions, the old intermediate value theorem, that a continuous map from the closed unit interval to the line, which sends the end points 0 and 1, to numbers of opposite sign, must send some point of the open unit interval to zero.
     
  13. Feb 8, 2005 #12

    mathwonk

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    as to winding number or homotopy, think of your belt. if you wrap it one or two or three times around your waist then hook it, it does not just come off. But if you wrap it around one way then back the opposite way the same number of timnes, and then hook it, it could slide off through the belt loops, since the total winding number around your waist is zero.
     
  14. Feb 8, 2005 #13
    nice mathwonk, thanks

    marlon
     
  15. Feb 8, 2005 #14

    Hurkyl

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    I think he means the "points at infinity" in a compactification of the plane. In particular, the one where equivalence classes of parallel rays form points. (Or, alternatively, the one analogous to adding the boundary to an open disk)
     
  16. Feb 8, 2005 #15

    mathwonk

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    I. here is a short argument that a unit circle cannot be pulled away from the origin without passing through the origin. the unit circle has winding number 1 around the origin, and any small enough deformation of the circle will not change that. hence the concept of winding number, which is defined for any deformation of the circle not passing through the origin, is a continuous function.

    but if you pulled the circle completely away from the origin in a continuous motion, the winding number would have become zero. that is impossible. a continuous integer valued function on a time interval cannot take values zero and one.


    II. to prove it by stokes theorem, note that the winding number, times 2pi, is merely the integral of the form dtheta, over the closed curve. now stokes theorem says that if you deform the circle, without passing through the origin, the integral of dtheta over the deformed circle stays the same. I.e. the deformation sweeps out an annulus in the space missing the origin, and by stokes theorem, the integral of d(dtheta) over the annulus, equals the difference of the integrals of dtheta over the two boundary curves of the annulus. but d(dtheta) = 0, so we are done.

    the same stokes theorem argument says we cannot deform the circle to a point without passing therough the origin, i.e. that deformation would sweep out a disc, a shrinking family of circles, in the plane, missing the origin. Then by stokes, the integral of dtheta over the boundary of the disc, equals the integral of d(dtheta) = 0 over the disc. that is a contradiction since the original position of the boundary circle was around the unit circle, and polar coordinates gives that integral as 2pi.

    as matt said one could also integrate dz/z and get 2pi i times winding number.


    III. To prove it merely using polar coordinates, consider the polar coordinate map from the right half plane to the punctured plane, taking (r,t) to
    (rcos(t), rsin(t)).

    This takes the vertical segment at r = 1, and from t = 0 to t=2pi, onto the unit circle. i.e. as matt grime said, it takes the two end points to the same point of the plane, thus making a closed loop. now the whole point is to prove, using a little uniform continuity, that any deformation of the circle results fom a deformation of the segment.

    i.e. given any continuous motion of the unit circle, not passing through the origin, there is a correponding motion of the segment described above, such that the moving segments map onto the moving closed curves.

    then since the endpoints of the segment must always map to the same point of the plane, to give a closed image curve, the end points of the segments can never come together. i.e they always stay separated by a distance of 2pi vertically in the (r,t) plane.

    since the segments can never be moved to become one point, the image curve can never become one point. so the unit circle cannot be contracted to a point, without passing through the origin.

    so that's three "proofs" in only about a million words.

    by the way since winding numbers are computed by path integrals, and path integrals are computed by parametrization, e.g. by polar coordinates, all three proofs are the same.

    so here is a different proof due to mo hirsch. if there is a map from the disk to the punctured plane which restricts to the identity on the boiundary circle, then since the punctured plane retracts radially onto the unit circle, we can assume we have a map from the unit disc onto its own boundary circle, which restricts to the identity map from the boundary cirfcle to itself.

    now consider a typical point of the circle and ask what its inverse image is under the map from the disc. since the circle is one dimensionala nd the disc two diemnsional, a general point of the circle will have inverse image a one dimensional object, i.e. a union of segments and circles.

    but there is some segment since that point we are looking at does map to itself. but where is the other end of that segment? i.e. no other point of the circle maps to that point, since they all map to themselves.

    \
    but the other endpoint p cannot be inside the disc. i.e. locally, by the inverse function theorem the map looks like as projection near p, so there must be a whole curve through p collapsing to the given point on the circle.

    this is a contradiction. thats 4.
     
    Last edited: Feb 8, 2005
  17. Feb 8, 2005 #16

    mathwonk

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    when i was young and energetic, i taught this stuff in calculus of several variables, as an illustration of the power of stokes theorem.

    an easy corollary is the fundamental theorem of algebra. another is the fact that there is no smooth family of non zero vectors on a 2 - sphere.
     
  18. Feb 10, 2005 #17

    mathwonk

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    I have noticed that when I answer a question too thoroughly I get silence. But I never learn.
     
  19. Feb 10, 2005 #18

    Hurkyl

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    What's wrong with that? Either you've answered all of the audience's questions, or they don't understand one word of it. :smile:
     
  20. Feb 10, 2005 #19

    mathwonk

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    its like singing your best song into a canyon, and getting no echo.

    and experience teaches that anything understood provokes questioons, so the troubling suspicion is they haven't understood a word of it.

    but as I once argued, that's ok too, since one has provided food for future thought.
     
    Last edited: Feb 10, 2005
  21. Feb 11, 2005 #20

    mathwonk

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    here some exciting news on homotopy groups: poincare's conjecture is proved! I just heard this today, is this old news?

    i.e. suppose we say any space where every point has a neighborhood looking like a disc, is a 2 manifold. and we say it is connected if every pair of points can be joined by an arc in the space. and we say it is simply connecte if every closed loop in the space contracts to a point in the space. then the only object with all three proeprties is a sphere.

    I.e. the only connected, simply connecetd, 2 manifold, is a sphere.


    now ask the same question in one dimension up. i.e. a 3 manmifold is a space where evry point has a neighborhood looking like an open ball, and connected and simply connected are as before. then poincare asked in 1904 whetehr every such space is a "3 -sphere", i.,e. homeomoirphioc to the set in 4 space satisfying the equation

    x^2 + y^2 + z^2 + w^2 = 1.


    the answer was unknown for almost 100 years, and now is said to be yes!

    and the proof uses riemannian geometry and curvature, (although the question does not.)
     
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