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Complex Projective Space

  1. Mar 5, 2009 #1
    I've been thinking...and am starting to think that I don't understand complex projective space...So, it's defined as ( Cn+1 \{0,0} / C\{0} ). Now, I think this is just the set of planes in 4 space that pass through the origin... and one can consider how they would all intersect a 3 sphere and think of it as S3/U(1) where U(1) is the circle group... and the hopf function will take all these circles and map them to the 2 sphere isomorphically... but the problem I have is... just pick any 3 of the 4 basis vectors in C^2 and span two planes with them...essential you can just look at R^3 for this... and think of the plane spanned by XY and XZ....well they intersect at the whole X axis...which means there are elements that belong to both planes...but in the case of ( Cn+1 \{0,0} / C\{0} ), these planes are supposed to be equivalence classes...meaning it should divide the space into disjoint sets...and thus, you can't have an element in 2 equivalence classes...Unless, both these planes are actually in the same equivalence class, which is just mind blowing since you can find a bunch more planes that will intercet XY and XZ and before you know it, all of R^3 will belong to the same equivalence class...So, clearly something is wrong with this way of thinking of it...Anyone?
     
  2. jcsd
  3. Mar 6, 2009 #2
    The set of 2-dimensional subspaces in R^4 is by definition the Grassmannian G(2,4). It has real dimension 2(4-2)=4, so it can not be the same as CP1, which has 2 real dimensions.
     
  4. Mar 6, 2009 #3

    HallsofIvy

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    Heck, I've thought that for years!
     
  5. Mar 6, 2009 #4

    Ben Niehoff

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    In four dimensions, two planes can intersect in a single point, not a line. Take the x-y and z-w planes, for example, which intersect only at the origin.
     
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