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I took a differential geometry topics course from Gene Calabi about 12 years ago in grad school. I was surprised to see his name on the upcoming course calendar

I took a differential geometry topics course from Gene Calabi about 12 years ago in grad school. I was surprised to see his name on the upcoming course calendar — I'd heard of him ("Calabi-Yau manifold") but hadn't expected he might still be teaching classes. (And ordinarily he wasn't — he was already long-retired at that point.) Turned out the reason he was teaching a course that semester was that he was really excited to share a new idea he'd had about how to visualize the complex projective plane. There were two (or three?) students in attendance. Okay, here's Calabi's visualization of the complex projective plane. The complex projective plane (as everyone knows) is the set of equivalence classes of complex 3-vectors, where two vectors are equivalent if one is a complex scalar multiple of the other. Calabi's visualization: for a given representative, take the real part and imaginary part of that 3-vector separately, now that's a pair of vectors in real 3-space, and those are much easier to visualize. Choose a particular representative for the equivalence class: use up the dilation part of your complex scalar to normalize your 3-vectors, then use up the rotation part to make them orthogonal. Now notice that multiplication by a unit-length complex scalar will rotate the real and imaginary 3-vectors through the plane they span, so we can visualize an element of the complex projective plane as an oriented ellipse in real 3-space. Actually we have some singular ones — if the real and imaginary parts are the same length (and orthogonal) then instead of ellipses we have circles, and if the real or imaginary part is zero then we have line segments. But we have a 1-parameter family of oriented ellipses (the parameter is the eccentricity) where at each parameter we have a quotient of SO(3) (it's almost the special orthogonal group but not quite) and at one end of the family it reduces to a 2-sphere and at the other end to a real projective plane. But the main point is instead of having to visualize 3 or 2 complex dimensions (6 or 4 real dimensions) now we get to visualize ellipses (1 real dimension!) inside real 3-space. Much more accessible to geometric intuition.

matheist 3y ago
A slightly lesser-known fact: the complex projective plane double covers the 4-sphere via the conjugation map. i.e. identify two equivalence classes of complex 3-vectors if one is the complex conjugate of the other. The resulting space is the 4-sphere. Calabi's visualization: take our oriented ellipses in real 3-space (spanned by the real and imaginary part of our complex 3-vector). Complex conjugation means negating the imaginary part, so the result of identifying complex conjugates is throwing out the orientation of the ellipse. So here's a visualization of the 4-sphere: unoriented ellipses in 3-space. There's a 1-parameter family with varying eccentricity; for each eccentricity there's a manifold which is 4-covered by SO(3); the singular sets at each end are a real projective plane.
fiforpg 3y ago
Nice construction. Took me a little while to figure out that by oriented ellipse you really mean a fixed curve in 3d, not just a fixed direction of the normal to the plane of the ellipse, which was my first reading. That way orientation takes 3 (real) dimensions + 1 dimension for eccentricity, which gives the 4 real dimensions in CP^2. How accessible it is to intuition is debatable though :P
xeonmc 3y ago
so would you say that it's basically a panorama of Jones vectors?
pradn 3y ago
Thank you, you write well - being able to transform objects in many contexts is super useful, it unlocks so much creativity.
GolDDranks 3y ago
> 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.

matheist 3y ago

A slightly lesser-known fact: the complex projective plane double covers the 4-sphere via the conjugation map. i.e. identify two equivalence classes of complex 3-vectors if one is the complex conjugate of the other. The resulting space is the 4-sphere. Calabi's visualization: take our oriented ellipses in real 3-space (spanned by the real and imaginary part of our complex 3-vector). Complex conjugation means negating the imaginary part, so the result of identifying complex conjugates is throwing out the orientation of the ellipse. So here's a visualization of the 4-sphere: unoriented ellipses in 3-space. There's a 1-parameter family with varying eccentricity; for each eccentricity there's a manifold which is 4-covered by SO(3); the singular sets at each end are a real projective plane.

fiforpg 3y ago

Nice construction. Took me a little while to figure out that by oriented ellipse you really mean a fixed curve in 3d, not just a fixed direction of the normal to the plane of the ellipse, which was my first reading. That way orientation takes 3 (real) dimensions + 1 dimension for eccentricity, which gives the 4 real dimensions in CP^2. How accessible it is to intuition is debatable though :P

xeonmc 3y ago

so would you say that it's basically a panorama of Jones vectors?

pradn 3y ago

Thank you, you write well - being able to transform objects in many contexts is super useful, it unlocks so much creativity.

GolDDranks 3y ago

> 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.