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> where two vectors are equivalent if one is a complex scalar multiple of the other. I can get an intuitive understanding of equivalence classes of real vector

> where two vectors are equivalent if one is a complex scalar multiple of the other. I can get an intuitive understanding of equivalence classes of real vectors being equal if they are real scalar multiples of each others, and understand that this is an equivalent case because of the field and linear space definitions/axioms, but I struggle to visualize it, possibly because of the larger dimensionality and the fact that there are two "kind" of dimensions, 2 (complex plane) x 3 (the linear bases). Are you able to visualize objects like that? Any help?

GolDDranks 3y ago
Ah, I think I was able to think a kind of a visualization myself. You can think of a complex 3-vector as a triplet of labeled points on the complex plane. Scaling this vector by a complex number is equivalent of rotating and scaling this triplet (0 as origin, so no affine transformations). Therefore, two vectors in this space are equivalent up to a scalar if you can find a complex number that scales/rotates one triplet to another. I can visualize a rotation/scaling transformation of a point cloud on the complex plane pretty well so I'm kind of satisfied this visualization.
- GolDDranks 3y ago
  Also, as a neat interpolation between the two vectors, you can think of doing linear interpolation between x := 0 and 1; c1 * c^x where c1 is a complex number component of the first vector and c the complex scalar multiple between the two vectors, rinse and repeat for each component. (Didn't test this in code yet, though) In my mind, this should lead the point cloud of the first vector to spirally slide to form the second vector.
matheist 3y ago
It is harder! Calabi's trick is one attempt at making it easier. Hmmmmm. If you're ok with the real version... well another way to write down the real version is "lines through the origin in real 3-space" (that's what the equivalence relation turns out to be). How do you feel about "lines through the origin in real n-space"? How about "planes through the origin in real n-space"? If we start with "complex lines through the origin in complex 3-space" (the complex projective plane we're trying to understand), and throw away all the complex stuff, then that's "some real 2-planes through the origin in real 6-space", i.e. it's a subset of planes through the origin in real 6-space. It's a proper subset because every complex line is a 2-plane but not every 2-plane in 6-space is a complex line. So maybe that's a start? It's not a perfect way to visualize it because (a) 2-planes in 6-space aren't exactly easy to visualize either, and (b) okay but which planes in 6-space are the special ones that come from the complex structure that we ignored earlier. But maybe it's something.

GolDDranks 3y ago

Ah, I think I was able to think a kind of a visualization myself. You can think of a complex 3-vector as a triplet of labeled points on the complex plane. Scaling this vector by a complex number is equivalent of rotating and scaling this triplet (0 as origin, so no affine transformations). Therefore, two vectors in this space are equivalent up to a scalar if you can find a complex number that scales/rotates one triplet to another. I can visualize a rotation/scaling transformation of a point cloud on the complex plane pretty well so I'm kind of satisfied this visualization.

GolDDranks 3y ago
Also, as a neat interpolation between the two vectors, you can think of doing linear interpolation between x := 0 and 1; c1 * c^x where c1 is a complex number component of the first vector and c the complex scalar multiple between the two vectors, rinse and repeat for each component. (Didn't test this in code yet, though) In my mind, this should lead the point cloud of the first vector to spirally slide to form the second vector.

matheist 3y ago

It is harder! Calabi's trick is one attempt at making it easier. Hmmmmm. If you're ok with the real version... well another way to write down the real version is "lines through the origin in real 3-space" (that's what the equivalence relation turns out to be). How do you feel about "lines through the origin in real n-space"? How about "planes through the origin in real n-space"? If we start with "complex lines through the origin in complex 3-space" (the complex projective plane we're trying to understand), and throw away all the complex stuff, then that's "some real 2-planes through the origin in real 6-space", i.e. it's a subset of planes through the origin in real 6-space. It's a proper subset because every complex line is a 2-plane but not every 2-plane in 6-space is a complex line. So maybe that's a start? It's not a perfect way to visualize it because (a) 2-planes in 6-space aren't exactly easy to visualize either, and (b) okay but which planes in 6-space are the special ones that come from the complex structure that we ignored earlier. But maybe it's something.