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Sudden confusion about definitions of singular homology

  1. Apr 23, 2009 #1
    Hi. I'm currently doing a course on algebraic topology and thought I was getting on great, was usually getting full marks on homework etc.

    But I've just got onto a question which should be fairly easy but its suddenly made me realise how badly I understand singular homology.

    Basically, it says to work out the degree of the map z^n when considered on the Riemann sphere.

    So it would be a good idea to understand the generating element for the 2nd homology group of the 2-sphere. But this is where I got stuck.

    I can't picture how the simplex "wraps" round the sphere in some sense. I mean you could use 2 of them, one for each hemisphere which have the opposite orientation for their boundary so that the boundary cancels out. But then you are left with something that has one orientation on the top and another at the bottom.

    Another question (that I really should have found out) is are homotopic simplexes represented by the same element in homology. And if not, how do we (intuitively) see that two simplexes are the same thing in homology? I feel this is quite essential in understanding the elements of homology. Like when we have 2 simplexes like this, are we to imagine that we have "glued" them together so to speak?

    My inkling is that the identity for the 2nd homology group is mapped to the element n (so z^n has degree n) since you can imagine each circle of points with modulus r being mapped to a circle which winds around n times with modulus r^n. The only thing is then however, that the points 1 and infinity are only mapped to once (by 0 and infinity). So surely if the identity of the homology group was mapped to n, each point should have been covered n times.


    I'm sorry for the extremely basic question and how badly I explained it, but its difficult to explain why you don't understand something! I just need to get this sorted because its so basic and is doing my head in!!! Any help will be very much appreciated :)
     
  2. jcsd
  3. Apr 23, 2009 #2
    Sorry, obviously there is no "nice" way to wrap around the sphere because of the "hairy ball theorem". I think the confusion is coming in in the form of the way I picture the elements. It clearly doesn't need to be a nice continuous sheet all the way around, it can be anything as long as every point is covered right? But clearly orientation still matters. I think my idea of homotopic simplexes leading to the same element in homology was causing the confusion since if this was true then a loop around the circle would be homotopic to a constant map (as the endpoints aren't fixed).

    So I think I'm mainly confused as how to classify element in homology visually. For example, the sinplex I described for z^n. How do we see this in homology? It does go all the way around the sphere, but some points more than others. I mean obviously the way the simplex is mapped has importance and not just the the number of times each point is mapped to. How could this element be associated to n when 0 and infinity are mapped to only once???
     
  4. Apr 23, 2009 #3
    It's easiest to picture it for the circle. Once you have that down, try to visualize the suspension map.

    Another way to think about is that you can cover the sphere with two two-simplices (hemispheres) which overlap slightly on a great circle from north to south pole. The generator for the two-homology of the sphere is the sum of these. When you hit them with z^2, they each suddenly cover the entire sphere (think of what happens when you hit half the circle with z^2), giving you twice the generator for the homology.
     
  5. Apr 23, 2009 #4
    The other thing (thinking about this in degree terms), is that you calculate the degree by counting the number of pre-images of regular values. North and south poles are not regular values for the z^n map.
     
  6. Apr 23, 2009 #5
    Ok, thanks for the good explanation. But I'm not sure what you mean by the regular values. I can clearly see that intuitively, the degree should be n. But I don't see how the induced map can be degree n when the North and South poles only have 1 pre-image. If we think of the corresponding 2-simplex, surely this can't represent an element greater than 1. What should be so special about the North and South poles?

    And also, is just counting the number of pre-images a good idea? I mean (I know this doesn't happen) but we could go one way and then another and get several pre-images but a lower degree map. So would there be a way to prove this without hand-waving?

    Another thing is, if there is an overlap of your 2 hemispheres, surely when you compute the boundaries, you will get something non-zero, so such a combination of 2-simplexes would not be in the homology. I can see that you can take 2 hemispheres which have the same boundary though, but as I said, reversing their orientation to each other.

    And one more thing, do we still get a degree n map for any degree n polynomial?

    Thanks for the help.
     
  7. Apr 23, 2009 #6
    The counting pre-images thing is a differential topological concept, and it goes as follows: take a differential (smooth) map f from S^n to itself. An amazing resulting result (which depends ultimately on transversality) is that if a point in S^n is a regular value of f (which means that the Jacobian has full rank at every point in the point's pre-image), then the number of points in the pre-image is the same for every regular value. More amazing results: 1) This is a homotopy invariant; 2) This is the only homotopy invariant of self-maps of the sphere (if two maps have the same degree, they're homotopic!).
     
  8. Apr 23, 2009 #7
    For your other questions: yes, reversing the orientation is necessary to get a generator for the homology. I'm not sure why you're troubled by this.

    Secondly, any degree n polynomial extended to the Riemann sphere will have degree n. This follows from Sard's Theorem and the Fundamental Theorem of Algebra.
     
  9. Apr 23, 2009 #8
    Wow, I didn't know that maps having the same degree implied that they were homotopic.

    So what you said before implies that the number of pre-images is the same for all the points (except 0 and infinity since they are not regular points). We already knew this. My problem was that is this justification for saying that the map is therefore a degree n map?

    I was troubled by the orientations being opposite because if you take the suspension map from the circle like you said, we seem to get a map which looks totally different; although I'm probably being naive and in fact both are the same in homology.

    Was what I said about homotopic simplexes being the same element in homology correct?

    And for the general polynomial, we have repeated roots [although I'm guessing that these will correspond to non-regular points, which is important for some reason that I don't understand yet :( ]

    We haven't covered any of this stuff in class, so I'm assuming that we were supposed to do it a different way, but thanks a lot for the help so far, at least I'm not quite as confused as to what the generating element of the homology actually is now!
     
  10. Apr 24, 2009 #9
    Homotopic simplices correspond to the same element in homology; it's one of the Eilenberg-Steenrod axioms. For the polynomial with repeated roots, it will correspond to non-regular points, as follows: expand the polynomial as a Taylor series about the repeated root. Its derivative will vanish there, so it can't be a regular point. But if you look at the equation P(z)=c for arbitrary c, then for almost every c, this will have n distinct pre-images (this is Sard's Theorem).
     
  11. Apr 26, 2009 #10
    Ok, thank you so much for the help. Understand it all now, except just a couple of things. I don't understand how the whole regular points etc. explains the degree, but I'm sure its proven somewhere, I'll have a look. Secondly, I still don't understand how a degree n map can map to something where we can have less than n points in the pre-image for each point; but I guess the answer to the first question will answer the second as so often happens in mathematics :)
     
  12. Apr 27, 2009 #11
    The answer to this (and almost every question) is the inverse function theorem. The inverse image of a regular point will be a submanifold, hence if the manifolds have equal dimensions, a finite number of points. Degree is invariant under homotopy, so we can "move" the regular point to any other regular point and the degree will be unchanged. This breaks down if we look at critical values, since the inverse function theorem doesn't work, and since the homotopy-invariance of degree requires that the map remain transverse throughout the homotopy.
     
  13. Apr 29, 2009 #12
    Ok, cheers. I think the main thing that was troubling me (and still is a bit) is that homotopic maps represent the same element in homology. I then imagine the element as sort of like an onion skin, the element n being n skins of the onion. But then if there are some points missing on a skin, we can squeeze it down to a point by homoptopy geometrically meaning that the element does not represent n.
     
  14. Apr 29, 2009 #13

    Hurkyl

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    The points aren't missing: they are "pinched". They are singularities of the mapping.
     
  15. Apr 30, 2009 #14
    Hmm. I was trying to visualize the actual element as the image on the sphere.

    The problem with the teaching method that I've been subjected to is that if we have to do something non-rigorously (which I assume we are supposed to since we haven't been taught about regular points etc.) then we need to use intuition that we haven't really developed properly yet.

    I hope I am going on about the same kind of degree as everyone else? It is defined as the representative from the integers of the element that the identity of the 2nd homology group of the sphere is mapped to.

    Sorry for my stupidity, but all I see is a canvas around the sphere that doesn't go over each point enough times and therefore can't represent n. Just don't get it and don't understand how these points are singularities either :(
     
  16. May 1, 2009 #15
    Here's an example to make it clear. Think of standing a torus on the xy-plane vertically (so it looks like an annulus in the vertical cross-section). Look at the height function. There are four critical points. If you look at the level sets between the second and third critical points (where the slice goes through the "hole"), you obtain two disjoint circles. As you move towards either critical point, they get closer and closer together, until finally they "pinch" and become a wedge of two circles at either singularity. It seems easier to understand "critical points = singularities" if you look at a different example.
     
  17. May 1, 2009 #16
    I guess the same example might work on the sphere as well. Look at z^2, and look at the pre-images of points moving closer and closer to the south pole. At every point except the south pole, there are two distinct pre-images, but they move closer and closer together, until finally they merge ("pinch") at the south pole itself.
     
  18. May 3, 2009 #17
    Ahha, i get you. The way I was looking at it was too simplistic and I overlooked this. Thanks a lot :)
     
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