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

Projective spaces

  1. Feb 28, 2007 #1
    I'm trying to get a kind of geometric understanding of [tex]RP^{n}[/tex] and [tex]CP^{n}[/tex], unfortunately various books and websites I've read seem to give different descriptions, at least sufficently different to be confusing me.

    The projective space seems to be defined via the equivalence relation x~y is x=ky for k not 0 and [tex]RP^{n} = \mathbb{R}^{n+1} - \{ 0 \} / \sim[/tex].
    Therefore you can rescale any point in [tex]\mathbb{R}^{n+1}[/tex] down onto the surface of a unit sphere [tex]S^{n}[/tex] and you also have the fact antipodal points are equivalent, so you end up with [tex]RP^{n}[/tex] defined as the hemispherical surface of one half of [tex]S^{n}[/tex], but with the entire 'equator' (ie boundary of the hemisphere) included.

    Why is the entire equator included, when you only need half of it? Is it to make sure tne boundary is entirely closed and allows for this hemisphere to be identitied with [tex]D^{n}[/tex], the 'circular' disk?

    That all seems fairly okay to me, but then when it comes to the complex version, [tex]CP^{n}[/tex], the extension doesn't seem to be "It's the same, but in complex space", a whole new description seems to be done (which varies from place to place I've read) and even my supervisor wasn't sure about it (not that this stuff is her thing though). Is [tex]CP^{n}[/tex] just the collection of lines through the origin in [tex]\mathbb{C}^{n+1}[/tex] or is it more subtle than that now you're in complex space?

    Any help would be appreciated. It's not something I desperately need to know, but it's mentioned enough times in various books I'd like to get my head around it.
  2. jcsd
  3. Feb 28, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    in real space the equator is included but opposite points of the equator are identified.

    in complex space, yes it is "lines" through the origin but they are of course complex lines.

    perhaos more intuitively, ikt is the compactification of affine n space obtained by adding a projective n-1 space at infinity. so real projective plane is the disjoint union of A^2 + A^1 + one point.
  4. Mar 1, 2007 #3


    User Avatar
    Science Advisor
    Homework Helper

    imagine real 3 space, and a real plane at height z =1. then every line through the origin that is not horizontal hits that plane in one point.

    so the points of that afine plane represent a large portion of the points of the projective plane. but the hoprizontal lines, i.e. the lines through the oprigin of the real x,y plane are not included.

    that is how the projetive plane augments the affine plane by a projective line, i.e. by the horizontal lineas through the origin.

    those lines again mostly meet the affine line in the x,y plane with y = 1, but one line escapes, the line parallel to the x axis, i.e. the x axis itself.

    that is the point.

    but as you can see, there is great latitude in choosing these esceptionalplanes lines and point, indeed any plane through the origin can contain the extra lines, so the projective plane is quite homogeneous, and the complement of any projective line is an affine plane.

    it is actually a lovely gadget.
  5. Mar 1, 2007 #4
    Cheers Mathwonk. I've got to admit, I didn't follow the A^2+A+1 description initially, but after your second post I've got what you mean by that and it's a pretty nifty way of thinking about it! Thanks :smile:
  6. Mar 9, 2007 #5


    User Avatar
    Science Advisor
    Homework Helper

    thank you my friend.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook