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Product space

  1. Jan 20, 2012 #1
    Hi all,

    I would really appreciate if someone can explain to me what is meant by a product of spheres.
    What would for example S1 x S0 look like? The first being a circle and the second being a pair of boundary points... So what kind of "object" is their product?

    And how about S1 x S1 or S1 x S2

    I hope someone can clear this up for me.

    Kind regards,
  2. jcsd
  3. Jan 20, 2012 #2

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    Welcome to PF, Jooolz! :smile:

    S1 x S0 is a circle that is combined with each of the boundary points.
    Consider a circle in the X-Y-plane.
    And let the boundary points be at z=+1 and z=-1.
    What you get is 2 circles.

    S1 x S1 is a circle, and at each point of the circle we have another circle that extends in a dimension.
    That's... a donut!

    S1 x S2 is a bit more difficult, since we can't visualize it in 3D anymore.
    It's a circle with at each point a sphere in a different dimension.
    That's a kind of hyperdonut.
  4. Jan 22, 2012 #3
    thank you very much helping!! :-)

    I dont think I really got it... So S1 x S0 = S1 x {-1, 1} ?
    with -1 and 1 on the x-axis? then draw two circles.. with midpoints (-1, 0) and (1, 0)? Or does it not make sense to talk about -1 and 1 on x-axis cause there actually isn't one...cause -1 and 1 are just some loose points that could've also be named A and B??
    AAARRRRRGGG I am making myself more confused now... :-S

    How about S1 x ℝ?

    kind regards,
  5. Jan 23, 2012 #4

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    This is isomorphic to 2 circles parallel to the X-Y-plane with their midpoints at (0,0,-1) and (0,0,+1).
    Or if you want you could pick midpoints ((0,0), -1) and ((0,0), +1) which matches the cartesian product.

    This is (isomorphic to) a cylinder that is infinitely long.
  6. Jan 23, 2012 #5
    Of course... it is a cylinder :shy:

    Can you tell me if I'm thinking in the right direction with this:

    If we define an equivalence relation on S1 X ℝ by

    (x, t) ~ (-x, -t)

    Do they mean, that if we choose a point x on the cylinder (which is a point on a circle) the t would be automaticaly the midpoint of that circle??? And then that point must be identified with (-x, -t) ? I tried to pictured it... but the -t part doesn't make sense to me...

    first this: is it true that if we have a circle, and the equivalence relation x ~ -x that we would end up with half a circle?

    I so much appreciate that you take time to help me.. Thank you!
    I like Topology, but god it is not easy...

    kind regards,
  7. Jan 23, 2012 #6

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    "t" would not be a midpoint of that circle, (0, t) would be the midpoint of a circle.

    x would be a point on the circle, which would actually be (x,y).
    t would be the z coordinate.

    Your equivalence relation would identify each point of the cylinder with the point on the opposite side of the origin.
    I'm getting a headache trying to visualize that. :eek:

    Hmm, I dunno.
    Never did something like that before.
    Will you let me know if you find out?
  8. Jan 23, 2012 #7
    Nope, you will end up with a very weird space: a projective space. http://en.wikipedia.org/wiki/Projective_space

    This space is extremely important in geometry and algebraic topology.
  9. Jan 23, 2012 #8
    I was thinking of it this way: having the infinite cylinder standig so that the height is the z-axis... and then a point of a circle would be (x, y, z) where z is the midpoint...Cause the height of points on the same circle would be the same... Wrong?

    So Am I... :-s
  10. Jan 23, 2012 #9
    Now you've made me cry...

    In my book they are saying that the one-point-compactification of the quotiënt space (S1 X ℝ)/~ Is in fact homeomorphic to the real projective plane...

    So of course I want to know why that is... pfffff.... first I wanted to know what I am actually dealing with.. And now I dont like it... cause I really want to understand this, but when I read things like : the projective plane, isn't actually a plane... I'm not getting happier...

    Can I use that the one-point-compactification of ℝis homeomorphic to S1?

    I really like the activity on this forum... I am gonna see if I can actually help someone...

    Thanks again guy's!

    still crying though... ;-)
  11. Jan 23, 2012 #10

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    Crying is the prerogative of the supreme onion, the saddener of worlds :cry:
    Perhaps we should try to peel off a few of those projective layers. :wink:
  12. Jan 23, 2012 #11
    What book are you using??

    Did you ever hear of projective planes before??
  13. Jan 23, 2012 #12

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    I learned from the book Topologie by Klaus Jänich, but there's nothing in there about Projektiver Räume. :cry:
  14. Jan 23, 2012 #13
    Yes I've heard about Pn(ℝ), but I always try to avoid them... cause they're creeping me out...

    Well, I use a book/reader that is made by a professor of my university...
    And every now and then I skip through the pages of General Topology by Stephen Willard
    Do you know it?
  15. Jan 29, 2012 #14


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    the way i always thought about the projective plane is this:

    start with a disk (you know, like a pancake, sort of). now if we sewed the edges of the disk "the right way", we'd get a sphere (imagine "closing up a basketball").

    but if we "twisted" the edges, and sewed them up "backwards" (we'd need some extra dimensions to actually DO this, without the disk "intersecting itself"), we'd get the projective plane (it's weird, because it only has one side).

    you can see a nifty animation of a polygonized model here (you have to imagine the self-intersections "aren't there"):

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