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Going crazy trying to understand projective CP1 and CPn

  1. Aug 20, 2010 #1
    I am trying to understand all details of the complex projective space CPn. Since surely CP1 must be the most simply to understand, I started out with it and even there I cannot gain full understanding. I would be eminently thankful for any help :D

    Nearly all texts trying to describe CPn get very general about it, using mostly language and not so much explicitness. Perhaps they assume some foreknowledge that I as a physicist lack :)

    I was wondering if someone could shed some light on this.

    1) Firstly, neither on Wikipedia nor on Wolfram Mathworld can I find a clear, thorough and detailed definition of what a complex line is. Is it a collection of points of dimension 2? How does one parametrize it in the same way that a real line (x,y) in the plane has a parametrization (x(t),y(t)) for some interval of t.

    2) Furthermore, since complex lines are fundamentally different to real lines, I have no idea what a complex line "through the origin" is supposed to mean.

    3) Do we identify ANY two points in C^(n+1) that differ by ANY multiplicative factor lambda? Because from what I understand, any complex point z can be moved to any other complex point w if a suitable lambda is found, just use

    lambda = z / w

    This would mean that all points in C^(n+1) are identified.

    I hope that when I spell it out this way, you guys can see where I have gone wrong.

    4) To specify a "complex line" in CPn, what is the minimum amount of information you will need to uniquely determine 1 exact "line" ? (points on the manifold need to be unique and fully defined).
     
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  3. Aug 20, 2010 #2

    lavinia

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    A complex line is a one dimensional complex linear subspace of a complex vector space.

    It is false that any point in a complex vector space can be reached from any other by scalar multiplication.
     
  4. Aug 20, 2010 #3
    Many thanks for your reply!

    After looking at Wikipedia and Wolfram Mathworld I still don't know what these complex linear subspaces are. Are they copies of the complex plane?

    If I want to reach z= |z|e^(ia) with w=|w|e^(ib) why can't I multiply w by v = |z|/|w| e^(i(a-b))

    wv = |z|/|w| e^(i(a-b)) |w|e^(ib) = |z|e^(ia) = z

    This way any non-zero complex number can be moved to any other right?
     
  5. Aug 20, 2010 #4

    quasar987

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    It is good to know RP^n first, because RP^n is easier to understand, and CP^n is defined analogously. But to answer your question:

    (1) & (2) In R^n, a (real) line through the origin is what? It is a set of the form {ax: a in R} for some nonzero x in R^n. This actually is a line "passing through" 0 and x. By analogy, a "complex line" through the origin in C^n is a set of the form {wz: w in C} for some nonzero z in C^n. If you write w as rexp(iθ), then you can "parametrize" it by r>0 and θ in [0,2π), but you cannot parametrize it by some interval of the real line (not in a 1-1 manner at least) because the complex line, despite the name "line", is a 2-dimensional object.

    (3) In C^1, you are right, we can get from any (non zero) point to any point by multiplication by a suitable λ. Thus, CP^0=[0], the silly 1-point space. But in C^2 a point is actually a pair of points of C. Surely, given two such pairs (z,z'), (w,w'), you realize that there may not exist a λ such that (w,w')=(λz,λz'). For this to hold, λ would have to be equal to w/z and to w'/z' simultaneously. So as soon as these ratios differ, (z,z') and (w,w') do not belong to the same complex line. The same argument goes for complex lines in C^n for all n>1.

    (4) There are no complex lines in CP^n. The elements/points of CP^n are the complex lines in C^{n+1}. Just nitpicking on your use of words.

    Actually, there is a technical detail. The points of CP^n are the complex lines through 0 in C^{n+1} to which the point 0 has been removed.

    An element of CP^n is usually expressed by giving a point by which the line passes. If (z_1,...,z_{n+1}) is a point of C^{n+1}-0, then the line by which it passes, i.e. the set {λ(z_1,...,z_{n+1}):λ in C-0}, is denoted [z_1: ... : z_{n+1}]. So here a point of CP^n is specified by specifying n+1 complex numbers, or, 2(n+1) real numbers. It is possible to specify the same point with only n complex numbers, or 2n real numbers. Indeed, we know that (z_1,...,z_{n+1}) is nonzero, so there is at least one of the z_i wich is nonzero. Suppose z_i≠0. Then since z_i is a nonzero complex number, the point (1/z_i)*(z_1,...,z_{n+1}) = (z_1/z_i,...,1,...,z_n/z_i) belongs to the same complex line as (z_1,...,z_{n+1}). That is to say, [z_1: ... : z_{n+1}] = [z_1/z_i : ... : 1 : ... : z_n/z_i]. But this last way of expressing the same complex line uses only n complex numbers, or 2n real ones.
    This number of "information" (2n real numbers) needed to specify a point of CP^n is minimal, in the sense that CP^n is a manifold of (real) dimension 2n. So you will not be able to parametrize a piece of it in a bijective and bicontinuous manner with less than 2n real numbers.
     
  6. Aug 21, 2010 #5
    Thank you so much! I am understanding much more now! :D

    I understand this, but I have a problem with the way you chose your representative for the equivalence class. You write [0] but the point z = 0 cannot be brought to the other points in the complex plane.

    Shouldn't we be using any representative BUT zero ? For example, CP^0=[1] ?

    Didn't get the part in bold here :(. (Lambda can assume the value zero according to my book if thats what you mean)

    How can one prove that one point in C^{n+1}-0 corresponds to one, unique complex line? This is what I was worried about, that we are not creating an equivalence relationship.

    This is somehow obvious once one has heard it, but didn't think of this until you wrote it! Thank you :D. I was a bit worried about the sets forming a proper "cover" of CPn.

    The rest I understand very well, thank you again! May a thousand virgins descend upon thee!
     
    Last edited: Aug 21, 2010
  7. Aug 21, 2010 #6

    quasar987

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    There is no difficulty in showing that (z_1,...,z_{n+1})~(z'_1,...,z'_{n+1}) iff (z'_1,...,z'_{n+1})=λ(z_1,...,z_{n+1}) for some λ in C defines an equivalence relation on C^{n+1}-0. Just try it.

    Notice that we must take C^{n+1}-0 as the domain of that equivalence relation and not C^{n+1} otherwise every point would be equivalent to 0 (indeed, every complex line passes through 0). And so defining CP^n:=C^n/~ would yield CP^n=[0], the silly 1-point space. In other words, CP^n is defined as the set of complex lines to which 0 has been removed. This is what I means in the above post.

    Just tell them I'm ready when they are.
     
  8. Aug 21, 2010 #7
    Like you recommended I went back to RPn, to understand this I must first know what in fact a line in R^(n+1) is.

    I got stuck in trying to create a general definition of an origin-line in (n+1)-dimensions.

    For R^(n+1) I get, using set builder notation, the locus L (for any lines that DONT lie in the x_2...x_(n+1)-plane)

    L_{x_1 x_2 .... x_(n+1)} = {(a_1,a_2 , ... ,a_(n+1) ) | such that a_2 / a_1 = x_2/x_1 , a_3/a_1 = x_3/x_1, ..... a_(n+1)/a_1 = x_(n+1)/x_n }

    The label for L comes from the fact that we have inserted sufficient and minimal information into the recipe L, so as to produce a unique line on which we all can agree on.


    Is this wrong? Am I focusing on the wrong thing? I cant see immediately that the object/locus I have described is necessarily one-dimensional. Shouldn't one be able to do this, or write it in a way from which this is clear? I would like to be able to see this through a parametrization of one variable, as I am used to being able to identify a one-dimensional curve by spotting that it depends on one variable, confere L_x = { x(t) | t in some real interval } for some given function x.

    I find that my difficulties arise from not having a good idea of what "points" on these projective manifolds look like, all the authors just say "lines" and their lack of clarity annoys me.
     
    Last edited: Aug 21, 2010
  9. Aug 21, 2010 #8

    quasar987

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    In post #4, I said precisely what a real line is and my definitiong meets all the criteria you're after:

    In R^2 or R^3, seeing x as a vector if it helps you, it is clear I think that {ax: a in R} represents the line passing through 0 and x, and it is parametrized nicely by the parameter a in R.
     
  10. Aug 26, 2010 #9

    lavinia

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    One small point. One does not need to worry about the origin - which lies on all lines - when defining projective space. The lines themselves are distinct and project to distinct points in projective space. Their intersection at the origin is irrelevant. The only thing that matters is the definition of open neighborhoods which can be done by intersecting the lines with the unit sphere then taking the quotient topology after identifying each intersection set to a point.

    In the case of real projective space open sets of lines are projections of polar open sets on the sphere. In complex projective space they look like projections of regions of circles on the unit sphere.
     
  11. Aug 26, 2010 #10
    Couldn't you just define the topology of CPn by taking the quotient space of Cn+1 \ {0} by identifying two points iff they lie on the same line (i.e. one is a scalar multiple of the other)?
     
  12. Aug 26, 2010 #11

    lavinia

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    yes - that works
     
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