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

From a vector space to the projective space

  1. Oct 17, 2012 #1


    User Avatar

    Dear all,

    I am not very experienced in this field, so, I have a rather simple question :smile:

    -Consider a linear vector space V of dimension 4.
    -Prescribe that, if two vectors in V differ by a nonvanishing constant, they belong to the same equivalence class.
    -Put together all these equivalence classes, and obtain a 3-dimensional vector space P(V).

    My question is: is the space P(V) obtained as outlined above the "projective space" one encounters in projective geometry -- the one defined using axioms, see e.g. the book by Coxeter.

    Additional question: if the initial vector space V is defined over the field of real numbers, is P(V) the same as the "real projective space" (obtained from ℝ^4\{0} by projection)?

    Any help is very much appreciated,
  2. jcsd
  3. Oct 17, 2012 #2


    User Avatar
    Science Advisor

    So you are defining an equivalence relation "u~ v if and only if u= av for a some non-zero scalar.

    -Put together all these equivalence classes, and obtain a 3-dimensional vector space P(V).

    Okay, what are those axioms?

    ?? If and only if V has dimension 4 to begin with! You didn't say anything about the dimension of V! Every vector space, of dimension n, is isomorphic to Rn and you can use that isomorphism to get an isomorphism between the two projective spaces.

  4. Oct 18, 2012 #3


    User Avatar

    Thanks for your reply HallsofIvy.

    Let me formulate the question more precisely, as per your request.

    In what follows, the vector space V is always 4-dimensional. I would like to proceed like this:
    -Define an equivalence relation: two vectors in V are equivalent if they differ by a non-zero scalar.
    -Obtain the corresponding equivalence classes.
    -Put them together, so as to obtain a 3-dimensional vector space P(V)

    Now, I would like to know if P(V) coincides with the "projective space" one encounters in more traditional projective geometry. Such "projective space" is defined by the following axioms (see e.g. the famous book by Coxeter "Projective Geometry"):

    1. There exists a point and a line that are not incident.
    2. Every line is incident with at least three distinct points.
    3. Any two distinct points are incident with just one line.
    4. If A,B,C,D are four distinct points such that the line AB meets CD, then the line AC meets BD. This is called Veblen's axiom.
    5. If ABC is a plane, there is at least one point not in the plane ABC.
    6. Any two distinct planes have at least two common points.
    7. The three diagonal points of a complete quadrangle are never collinear.
    8. If a projectivity leaves invariant each of three distinct points on a line, it leaves invariant every point on the line.

    My question: is the P(V) derived in the first part of the post from the 4-dimensional vector space V the same as the projective space specified by the axioms above?

    Additional question: Assume that the answer to the first question is "Yes". If V is a 4-dimensional vector space defined over the field of real numbers, is then P(V) the real projective space (obtained from ℝ^4\{0} by projection)?

  5. Oct 19, 2012 #4


    User Avatar

    I, the generator of this post, think that it would perhaps fit better in the category "Geometry and Topology". Can anyone suggest how to relocate a post?

    The relocation would also help the post get some more attention...

    Best Wishes,
  6. Oct 19, 2012 #5


    User Avatar
    Science Advisor
    Homework Helper

    you need to throw out the zero vector, and then yes it is the 3 diml projective space.

    over R yes.
  7. Oct 22, 2012 #6


    User Avatar

    Ok, thank you!
  8. Oct 22, 2012 #7


    User Avatar
    Science Advisor
    Homework Helper

    3 dimlm projective space is by defn the sets of lines through the origin of R^4. there i a map from R^4 - {0} onto this set by sending each non zero point p to the line 0p.

    this map has the same value at each point p on the same line L through 0, hence defines a one one correspondence between the nonzero points, modulo the equivalence relation of being on the same line, and the set of all lines.
  9. Oct 22, 2012 #8

    According to the Veblen-Young theorem, any projective space of dimension at least 3 is equivalent to the projectivization of a vector space over a division ring (not necessarily a field, e.g. quaternions). See http://en.wikipedia.org/wiki/Veblen–Young_theorem

    Obviously there are more axioms in your list than what is listed there. Three of your axioms serve to specify that the dimension is exactly 3 (axioms 1,5,6). Then Axioms 7 and 8 are extra. Axiom 7 guarantees that the space satisfies Desargues' Theorem, which I think rules out vector spaces over the field [itex]\mathbb{F}_2[/itex] (and perhaps all fields of characteristic 2, but I don't know). Regarding axiom 8, I have a feeling that this serves to guarantee that your scalars belong to a field and not just a division ring. However, I don't know enough about abstract projectivities to answer that.

    Consider P(V). Then the "projectivities" are the projective linear transformations, PGL(V). Basically these are just the invertible linear transformations GL(V), but you think of them as acting on the 1D subspaces of V. So in your last axiom, the hypothesis is that three distinct points in a line are preserved. A "point" is a 1D subspace of V and a "line" is a 2D subspace of V. So this means that in a 2D subspace of V, the transformation has three distinct eigenvectors. In otherwords, there are three linearly dependent vectors, u,v,w which are all eigenvectors.

    If your scalars belong to a field, then that automatically implies that u,v,w all have the same eigenvalue AND every other vector in the 2D space that they span is also an eigenvector. This is the conclusion of Axiom 8 once you reword it in projective terms.

    However, if the scalars belong to a noncommutative division ring (i.e. not a field), then you can only conclude that the eigenvalues are conjugates of each other. That is, if Tu=au, Tv=bv, Tw=cw, then, for example, [itex]b = rar^{-1}[/itex] for some scalar r. Further, if some linear combination xu+yv is to be an eigenvector then the coefficients have to satisfy some condition like [itex]a(x^{-1}y)=(x^{-1}y)b[/itex] (if my algebra is right). So apparently there could be three or more eigenvectors, without everything in that 2D space being an eigenvector.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook