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Your last paragraph had me intrigued. I'm now wondering what other observations we have that are simply special cases due to low dimensionality.

by hashhar 10y ago

rmidthun 10y ago
Rotations can be composed in up to 3 dimensions only. That means that for any sequence of rotations, you can come up with a single rotation that is equal to that rotation. In 2d, this is just adding scalars and rotation has no meaning in 1 or 0 dimensions. In 4 dimensions, this is not true. A rotation in XY cannot compose with one in ZW so at least two rotations are required.
- pandaman 10y ago
  Could you explain more about 4D rotation? In my understanding, a rotation is a transform by an orthogonal matrix with determinant equal to 1. For the 4D rotations to be not composable the product of two 4x4 orthogonal matrices should not be an orthogonal matrix. Is this really true?
  - abecedarius 10y ago
    By 'rotation' is meant rotating a plane that goes through the origin, carrying along the other points that are off the plane according to their projection onto it. In 4-d if you rotate around the y-z plane, then around the w-x plane, there's no plane that can express a single composed rotation with the same effect. But as you say, it's composable as a unitary matrix.
    - pandaman 10y ago
      Hmmm, I am still confused. Rotating "around plane" makes sense only in 3D, where a 2D plane is a hyperplane i.e. has a single normal and you can rotate around that normal (which you implied by mentioning projections along that normal). In 4D a hyperplane is a 3D subspace, orthogonal to some normal. So, while it may be true that you cannot find a 2D plane describing a composite rotation in 4D it does not seem to have any significance since "rotation around y-z" plane does not describe one either :) because y-z plane has an infinite number of normals in 4D and any rotation around any of them will be technically a rotation "around y-z plane" while a different rotation in 4D. I am sure you can find a 3D hyperplane in 4D for any composite rotation. Am I wrong?
      - abecedarius 10y ago
        You're quite right that there's no unique normal to a 2-d plane in 4-d, and so it's meaningless to "rotate around the normal vector". That's why I tried to phrase the geometry in terms of planes instead of axes. Here's an example that's easiest to sort-of-visualize: 1. Rotate the y-z plane of 4-d space by half a turn (180 degrees). That is, transform every point in 4-space according to w->w, x->x, y->-y, z->-z. 2. Rotate the w-x plane, also a half-turn. So w->-w, x->-x, y->y, z->z. Composing these two transformations yields w->-w, x->-x, y->-y, z->-z. That composed transformation is an inversion, not a rotation. Is it clearer now? (It was late when I wrote last night, sorry. Let me try and rephrase how I defined a rotation: a linear transformation that carries one unit vector to a second unit vector, and leaves unchanged any vectors that are orthogonal to both. In my previous comment the plane I was talking about was the plane that contains both given vectors. Of course, the null rotation doesn't pick out any particular plane, though it's not otherwise special in these terms. IANAM, but would appreciate correction if I'm goofing.)
        pandaman 10y ago
        I think I see what you mean. In an even dimensions space a -x is a rotation ( determinant = 1), while it is not in an odd dimensions. So what you described is a rotation in 4D but, indeed, not in 3D. Also, I feel, from this follows, that you can only have a rotation axis (eigenvector for value 1) in an odd dimensions space so I was wrong, assuming there is an axis for any 4D rotation. Same as in 2D there may be none. I figure 5D will act as same as 3D in this regard.
        abecedarius 10y ago
        I'm sorry, my 'rotation' was wrong -- as you point out, it would admit reflections. I'm glad to uncover this embarrassing bug, but on the other hand, oops. https://en.wikipedia.org/wiki/Rotation_(mathematics) https://en.wikipedia.org/wiki/Rotation_(mathematics) has more lore to check against after I've done enough mental debugging.
zellyn 10y ago
https://google.com/search?q=high+dimensions+spiky https://google.com/search?q=high+dimensions+spiky gives some discussions.
jerf 10y ago
Well, here's another fun one for you then: Regular polyhedra/polytopes. In zero dimensions the special case is that it's a nonsense question, which is pretty common special case for 0D. In 1D you can't help but have a line be equal to itself, so still a special case of it being nonsense. In 2D you have the special case of being able to form an arbitrary number of regular polygons. In 3D you get 5 regular polyhedra, the five Platonic solids, which also turns out to be a special case. However, in four dimensions, you have six regular polytopes: video [1] webpage [2] I only found this out relatively recently; while this is no secret it only seems to be recently getting around in the math general interest video channels and such. After that, the regular case takes over and you only have 3 forever more; the tetrahedron analogue, the cube analog, and the octahedron analog. So here's a case where 0 through 4 dimensions is a special case, and 5 is the first that fits the general pattern. [1]: https://www.youtube.com/watch?v=oJ7uOj2LRso https://www.youtube.com/watch?v=oJ7uOj2LRso [2]: http://math.ucr.edu/home/baez/platonic.html http://math.ucr.edu/home/baez/platonic.html
- hashhar 10y ago
  Wow. This entire thread is amazing. Thanks a lot for the links to help me go deeper into the rabbit hole.
rmidthun 10y ago
Remembered another one: gravitational orbits can't occur in 4D space or higher. In 3d space, we have inverse-square laws. Think of a light source, the area that is covers at a given distance is the area of a sphere, so it makes sense that it would decrease with that area, or 4piR*R. Gravity is also inverse-square in 3D space (though relativity makes this much more complicated, I expect). Implication is that in a 4D space, it would be an inverse-cubed relation. It turns out that stable orbits can only happen with inverse squared laws. https://en.wikipedia.org/wiki/Bertrand%27s_theorem https://en.wikipedia.org/wiki/Bertrand%27s_theorem

Your last paragraph had me intrigued. I'm now wondering what other observations we have that are simply special cases due to low dimensionality.

by hashhar 10y ago

Your last paragraph had me intrigued. I'm now wondering what other observations we have that are simply special cases due to low dimensionality.

rmidthun 10y ago

Rotations can be composed in up to 3 dimensions only. That means that for any sequence of rotations, you can come up with a single rotation that is equal to that rotation. In 2d, this is just adding scalars and rotation has no meaning in 1 or 0 dimensions. In 4 dimensions, this is not true. A rotation in XY cannot compose with one in ZW so at least two rotations are required.

pandaman 10y ago
Could you explain more about 4D rotation? In my understanding, a rotation is a transform by an orthogonal matrix with determinant equal to 1. For the 4D rotations to be not composable the product of two 4x4 orthogonal matrices should not be an orthogonal matrix. Is this really true?
- abecedarius 10y ago
  By 'rotation' is meant rotating a plane that goes through the origin, carrying along the other points that are off the plane according to their projection onto it. In 4-d if you rotate around the y-z plane, then around the w-x plane, there's no plane that can express a single composed rotation with the same effect. But as you say, it's composable as a unitary matrix.
  - pandaman 10y ago
    Hmmm, I am still confused. Rotating "around plane" makes sense only in 3D, where a 2D plane is a hyperplane i.e. has a single normal and you can rotate around that normal (which you implied by mentioning projections along that normal). In 4D a hyperplane is a 3D subspace, orthogonal to some normal. So, while it may be true that you cannot find a 2D plane describing a composite rotation in 4D it does not seem to have any significance since "rotation around y-z" plane does not describe one either :) because y-z plane has an infinite number of normals in 4D and any rotation around any of them will be technically a rotation "around y-z plane" while a different rotation in 4D. I am sure you can find a 3D hyperplane in 4D for any composite rotation. Am I wrong?
    - abecedarius 10y ago
      You're quite right that there's no unique normal to a 2-d plane in 4-d, and so it's meaningless to "rotate around the normal vector". That's why I tried to phrase the geometry in terms of planes instead of axes. Here's an example that's easiest to sort-of-visualize: 1. Rotate the y-z plane of 4-d space by half a turn (180 degrees). That is, transform every point in 4-space according to w->w, x->x, y->-y, z->-z. 2. Rotate the w-x plane, also a half-turn. So w->-w, x->-x, y->y, z->z. Composing these two transformations yields w->-w, x->-x, y->-y, z->-z. That composed transformation is an inversion, not a rotation. Is it clearer now? (It was late when I wrote last night, sorry. Let me try and rephrase how I defined a rotation: a linear transformation that carries one unit vector to a second unit vector, and leaves unchanged any vectors that are orthogonal to both. In my previous comment the plane I was talking about was the plane that contains both given vectors. Of course, the null rotation doesn't pick out any particular plane, though it's not otherwise special in these terms. IANAM, but would appreciate correction if I'm goofing.)
      - pandaman 10y ago
        I think I see what you mean. In an even dimensions space a -x is a rotation ( determinant = 1), while it is not in an odd dimensions. So what you described is a rotation in 4D but, indeed, not in 3D. Also, I feel, from this follows, that you can only have a rotation axis (eigenvector for value 1) in an odd dimensions space so I was wrong, assuming there is an axis for any 4D rotation. Same as in 2D there may be none. I figure 5D will act as same as 3D in this regard.
        abecedarius 10y ago
        I'm sorry, my 'rotation' was wrong -- as you point out, it would admit reflections. I'm glad to uncover this embarrassing bug, but on the other hand, oops. https://en.wikipedia.org/wiki/Rotation_(mathematics) https://en.wikipedia.org/wiki/Rotation_(mathematics) has more lore to check against after I've done enough mental debugging.

zellyn 10y ago

https://google.com/search?q=high+dimensions+spiky https://google.com/search?q=high+dimensions+spiky gives some discussions.

jerf 10y ago

Well, here's another fun one for you then: Regular polyhedra/polytopes. In zero dimensions the special case is that it's a nonsense question, which is pretty common special case for 0D. In 1D you can't help but have a line be equal to itself, so still a special case of it being nonsense. In 2D you have the special case of being able to form an arbitrary number of regular polygons. In 3D you get 5 regular polyhedra, the five Platonic solids, which also turns out to be a special case. However, in four dimensions, you have six regular polytopes: video [1] webpage [2] I only found this out relatively recently; while this is no secret it only seems to be recently getting around in the math general interest video channels and such. After that, the regular case takes over and you only have 3 forever more; the tetrahedron analogue, the cube analog, and the octahedron analog. So here's a case where 0 through 4 dimensions is a special case, and 5 is the first that fits the general pattern. [1]: https://www.youtube.com/watch?v=oJ7uOj2LRso https://www.youtube.com/watch?v=oJ7uOj2LRso [2]: http://math.ucr.edu/home/baez/platonic.html http://math.ucr.edu/home/baez/platonic.html

hashhar 10y ago
Wow. This entire thread is amazing. Thanks a lot for the links to help me go deeper into the rabbit hole.

rmidthun 10y ago

Remembered another one: gravitational orbits can't occur in 4D space or higher. In 3d space, we have inverse-square laws. Think of a light source, the area that is covers at a given distance is the area of a sphere, so it makes sense that it would decrease with that area, or 4piR*R. Gravity is also inverse-square in 3D space (though relativity makes this much more complicated, I expect). Implication is that in a 4D space, it would be an inverse-cubed relation. It turns out that stable orbits can only happen with inverse squared laws. https://en.wikipedia.org/wiki/Bertrand%27s_theorem https://en.wikipedia.org/wiki/Bertrand%27s_theorem