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maximum distance of point on $k$ -sphere
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How can we get $k^2$ points on $k$-dimensional hypersphere such that each two points have max euclidean distance ?
geometry
asked Jul 30 at 17:16
Fatemeh
356
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show 6 more comments
up vote
0
down vote
favorite
How can we get $k^2$ points on $k$-dimensional hypersphere such that each two points have max euclidean distance ?
geometry
asked Jul 30 at 17:16
Fatemeh
356
Start simple and work to more complex. If this problem were asking about the 2-sphere, what would the answer be? Then consider higher dimensional spheres.
– InterstellarProbe
Jul 30 at 17:33
in 2- dimensional (circle ) is simple because use the point (1,0), (-1,0),(0,1),(0,-1), but in 3- dimensional I dont know how to get ninth point on sphere
– Fatemeh
Jul 31 at 4:45
A $k$-sphere exists in $k+1$ dimensions. So, $S^0$ is a 0-sphere, which is two points on the line ($pm 1$). It exists in one dimension. $S^1$ is a 1-sphere, which is the standard two-dimensional circle. This is an important distinction because $S^1$, to place one point, you merely choose one point. That is it. Any point will work, because you want $1^2$ points, which will be 1. For the 2-sphere ($S^2$), it is the sphere in three dimensions. Choose any point. Now, you will want three additional points. Basically, you are inscribing a tetrahedron inside the 2-sphere with maximal volume.
– InterstellarProbe
Jul 31 at 13:46
Similarly for the $k$-sphere, my intuition says that you will be inscribing the $k+1$-simplex of maximal volume. So, you choose a base point and then every other point you choose will be fixed up to rotations.
– InterstellarProbe
Jul 31 at 13:50
I have mistake, i like to find $k^2$ points on $k-1$ -dimensional hypersphere for example in 3-dimensional on sphere i like to find 9 points with this property
– Fatemeh
Aug 1 at 2:41
 |Â
show 6 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How can we get $k^2$ points on $k$-dimensional hypersphere such that each two points have max euclidean distance ?
geometry
asked Jul 30 at 17:16
Fatemeh
356
How can we get $k^2$ points on $k$-dimensional hypersphere such that each two points have max euclidean distance ?
geometry
asked Jul 30 at 17:16
Fatemeh
356
asked Jul 30 at 17:16
Fatemeh
356
asked Jul 30 at 17:16
Fatemeh
356
asked Jul 30 at 17:16
Fatemeh
356
356
Start simple and work to more complex. If this problem were asking about the 2-sphere, what would the answer be? Then consider higher dimensional spheres.
– InterstellarProbe
Jul 30 at 17:33
in 2- dimensional (circle ) is simple because use the point (1,0), (-1,0),(0,1),(0,-1), but in 3- dimensional I dont know how to get ninth point on sphere
– Fatemeh
Jul 31 at 4:45
A $k$-sphere exists in $k+1$ dimensions. So, $S^0$ is a 0-sphere, which is two points on the line ($pm 1$). It exists in one dimension. $S^1$ is a 1-sphere, which is the standard two-dimensional circle. This is an important distinction because $S^1$, to place one point, you merely choose one point. That is it. Any point will work, because you want $1^2$ points, which will be 1. For the 2-sphere ($S^2$), it is the sphere in three dimensions. Choose any point. Now, you will want three additional points. Basically, you are inscribing a tetrahedron inside the 2-sphere with maximal volume.
– InterstellarProbe
Jul 31 at 13:46
Similarly for the $k$-sphere, my intuition says that you will be inscribing the $k+1$-simplex of maximal volume. So, you choose a base point and then every other point you choose will be fixed up to rotations.
– InterstellarProbe
Jul 31 at 13:50
I have mistake, i like to find $k^2$ points on $k-1$ -dimensional hypersphere for example in 3-dimensional on sphere i like to find 9 points with this property
– Fatemeh
Aug 1 at 2:41
 |Â
show 6 more comments
Start simple and work to more complex. If this problem were asking about the 2-sphere, what would the answer be? Then consider higher dimensional spheres.
– InterstellarProbe
Jul 30 at 17:33
in 2- dimensional (circle ) is simple because use the point (1,0), (-1,0),(0,1),(0,-1), but in 3- dimensional I dont know how to get ninth point on sphere
– Fatemeh
Jul 31 at 4:45
A $k$-sphere exists in $k+1$ dimensions. So, $S^0$ is a 0-sphere, which is two points on the line ($pm 1$). It exists in one dimension. $S^1$ is a 1-sphere, which is the standard two-dimensional circle. This is an important distinction because $S^1$, to place one point, you merely choose one point. That is it. Any point will work, because you want $1^2$ points, which will be 1. For the 2-sphere ($S^2$), it is the sphere in three dimensions. Choose any point. Now, you will want three additional points. Basically, you are inscribing a tetrahedron inside the 2-sphere with maximal volume.
– InterstellarProbe
Jul 31 at 13:46
Similarly for the $k$-sphere, my intuition says that you will be inscribing the $k+1$-simplex of maximal volume. So, you choose a base point and then every other point you choose will be fixed up to rotations.
– InterstellarProbe
Jul 31 at 13:50
I have mistake, i like to find $k^2$ points on $k-1$ -dimensional hypersphere for example in 3-dimensional on sphere i like to find 9 points with this property
– Fatemeh
Aug 1 at 2:41
Start simple and work to more complex. If this problem were asking about the 2-sphere, what would the answer be? Then consider higher dimensional spheres.
– InterstellarProbe
Jul 30 at 17:33
Start simple and work to more complex. If this problem were asking about the 2-sphere, what would the answer be? Then consider higher dimensional spheres.
– InterstellarProbe
Jul 30 at 17:33
in 2- dimensional (circle ) is simple because use the point (1,0), (-1,0),(0,1),(0,-1), but in 3- dimensional I dont know how to get ninth point on sphere
– Fatemeh
Jul 31 at 4:45
in 2- dimensional (circle ) is simple because use the point (1,0), (-1,0),(0,1),(0,-1), but in 3- dimensional I dont know how to get ninth point on sphere
– Fatemeh
Jul 31 at 4:45
A $k$-sphere exists in $k+1$ dimensions. So, $S^0$ is a 0-sphere, which is two points on the line ($pm 1$). It exists in one dimension. $S^1$ is a 1-sphere, which is the standard two-dimensional circle. This is an important distinction because $S^1$, to place one point, you merely choose one point. That is it. Any point will work, because you want $1^2$ points, which will be 1. For the 2-sphere ($S^2$), it is the sphere in three dimensions. Choose any point. Now, you will want three additional points. Basically, you are inscribing a tetrahedron inside the 2-sphere with maximal volume.
– InterstellarProbe
Jul 31 at 13:46
A $k$-sphere exists in $k+1$ dimensions. So, $S^0$ is a 0-sphere, which is two points on the line ($pm 1$). It exists in one dimension. $S^1$ is a 1-sphere, which is the standard two-dimensional circle. This is an important distinction because $S^1$, to place one point, you merely choose one point. That is it. Any point will work, because you want $1^2$ points, which will be 1. For the 2-sphere ($S^2$), it is the sphere in three dimensions. Choose any point. Now, you will want three additional points. Basically, you are inscribing a tetrahedron inside the 2-sphere with maximal volume.
– InterstellarProbe
Jul 31 at 13:46
Similarly for the $k$-sphere, my intuition says that you will be inscribing the $k+1$-simplex of maximal volume. So, you choose a base point and then every other point you choose will be fixed up to rotations.
– InterstellarProbe
Jul 31 at 13:50
Similarly for the $k$-sphere, my intuition says that you will be inscribing the $k+1$-simplex of maximal volume. So, you choose a base point and then every other point you choose will be fixed up to rotations.
– InterstellarProbe
Jul 31 at 13:50
I have mistake, i like to find $k^2$ points on $k-1$ -dimensional hypersphere for example in 3-dimensional on sphere i like to find 9 points with this property
– Fatemeh
Aug 1 at 2:41
I have mistake, i like to find $k^2$ points on $k-1$ -dimensional hypersphere for example in 3-dimensional on sphere i like to find 9 points with this property
– Fatemeh
Aug 1 at 2:41
 |Â
show 6 more comments
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Start simple and work to more complex. If this problem were asking about the 2-sphere, what would the answer be? Then consider higher dimensional spheres.
– InterstellarProbe
Jul 30 at 17:33
in 2- dimensional (circle ) is simple because use the point (1,0), (-1,0),(0,1),(0,-1), but in 3- dimensional I dont know how to get ninth point on sphere
– Fatemeh
Jul 31 at 4:45
A $k$-sphere exists in $k+1$ dimensions. So, $S^0$ is a 0-sphere, which is two points on the line ($pm 1$). It exists in one dimension. $S^1$ is a 1-sphere, which is the standard two-dimensional circle. This is an important distinction because $S^1$, to place one point, you merely choose one point. That is it. Any point will work, because you want $1^2$ points, which will be 1. For the 2-sphere ($S^2$), it is the sphere in three dimensions. Choose any point. Now, you will want three additional points. Basically, you are inscribing a tetrahedron inside the 2-sphere with maximal volume.
– InterstellarProbe
Jul 31 at 13:46
Similarly for the $k$-sphere, my intuition says that you will be inscribing the $k+1$-simplex of maximal volume. So, you choose a base point and then every other point you choose will be fixed up to rotations.
– InterstellarProbe
Jul 31 at 13:50
I have mistake, i like to find $k^2$ points on $k-1$ -dimensional hypersphere for example in 3-dimensional on sphere i like to find 9 points with this property
– Fatemeh
Aug 1 at 2:41