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Topologising RP2 using open sets in R3

  1. May 19, 2012 #1
    I am reading Martin Crossley's book - Essential Topology - basically to get an understanding of Topology and then to build a knowledge of Algebraic Topology! (That is the aim, anyway!)

    On page 27, Example 3.33 (see attachment) Crossley is explaining the toplogising of [itex] \mathbb{R} P^2 [/itex] where, of course, [itex] \mathbb{R} P^2 [/itex] consists of lines through the origin in [itex] \mathbb {R}^3 [/itex].

    We take a subset of [itex] \mathbb{R} P^2 [/itex] i.e. a collection of lines in [itex] \mathbb {R}^3 [/itex], and then take a union of these lines to get a subset of [itex] \mathbb {R}^3 [/itex].

    Crossley then defines a subset of [itex] \mathbb{R} P^2[/itex] to be open if the corresponding subset of [itex] \mathbb {R}^3 [/itex] is open.

    Crossley then argues that there is a special problem with the origin, presumably because the intersection of a number of lines through the origin is the origin itself alone and this is not an open set in [itex] \mathbb {R}^3 [/itex]. (in a toplological space finite intersections of open sets must be open) [Is this reasoning correct?]

    After resolving this problem by omitting the origin from [itex] \mathbb {R}^3 [/itex] in his definition of openness, Crossley then asserts:

    "Unions and intersections of [itex] \mathbb{R} P^2 [/itex] correspond to unions and intersections of [itex] \mathbb {R}^3 [/itex] - {0} ..."

    But I cannot see that this is the case.

    If we consider two lines [itex] l_1 [/itex] and [itex] l_2 [/itex] passing through the origin (see my diagram - topologising RP2 using open sets in R3 - attached) then the union of these is supposed to be an open set in [itex] \mathbb {R}^3 [/itex] - {0} . But surely this would only be the case if we consider a complete cone of lines through the origin. With two lines - take a point x on one of them - then surely there is no open ball around this point in [itex] \mathbb {R}^3 [/itex] - {0} ??? ( again - see my diagram - topologising RP2 using open sets in R3 - attached) So the set is not open in [itex] \mathbb {R}^3 [/itex] - {0}?

    Can someone please clarify this for me?

  2. jcsd
  3. May 19, 2012 #2


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    There is a natural map from 3 space minus the origin onto the projective plane. A point is mapped to the line through the origin that contains it. the topology of the projective plane is just the quotient topology under this map. Inverse images of open sets therefore are open sets in 3 space minus the origin. In fact, a set is open in the plane only if its inverse image is open.

    A basis for the topology of projective space is the projections of open cones of lines through the origin. (The origin is removed from these cones to give an open set in 3 space. )
  4. May 19, 2012 #3
    Thanks for the help

    So that means that I cannot simply take two lines (points) in [itex] \mathbb{R} P^2 [/itex] as an open set because the corresponding set in [itex] \mathbb{R}^3 [/itex] is not open - as in the attached digram.

    Only cones of lines in [itex] \mathbb{R} P^2 [/itex] are open.

    But this seems to contradict what Crossley says - see attachement of Crossley page 27.


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  5. May 20, 2012 #4


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    I do not see a problem with what the book says. It is the same as what we are saying. The author is just pointing out that you have to remove the origin from each line, otherwise the cones will not be open.
  6. May 24, 2012 #5
    Just to add - we need to remove the origin to get an interesting result here. Perhaps an easier situation to visualise (but entirely analogous) is to look at the 2d version, or 1d real projective space (the circle).

    Try and find a subset of the plane formed by straight lines through the origin which is open - you won't be able to unless you take the whole plane since, if you look at the origin, if the set is to be open, there must be a small ball in it containing the origin. But then it contains points at all angles from the origin, so it contains all possible lines. If you take away the origin, you get what you want (the inverse sets look like double cones without boundary with the origin removed).

    Lavinia's way of looking at it is nicer than the author's - the reason for the above definition is precisely that real projective space can be defined as the quotient of Euclidean space minus the origin. Intuitively, an open set containing some line should contain all lines sufficiently "close to it" (i.e. pointing in a similar direction). You may want to define the space as instead collapsing the sphere by identifying antipodal points on it - you should be able to check that this gives you the same result.
  7. May 24, 2012 #6
    By the way, single lines will correspond to points. Since real projective space is Hausdorff, singletons will be closed. Your union of two lines that you talk of therefore will be a closed set, not an open one (real projective space is also connected, so a closed set can't also be open unless it is empty or the whole space).
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