View Full Version : Defining points in 4D space

Frog march

2009-Jul-09, 02:19 PM

If you have two points on a 2-sphere, and draw a line between them, you can define points inside the 2-sphere, in 3D space.

I was just wondering, if the Universe were a 3-sphere, could you define points in 4dimensions, by using two points in our space?

Is this done?

maybe this is trite, but I thought I would post anyway.

Ken G

2009-Jul-09, 04:36 PM

I'm not completely sure what you are asking, but if you can embed the 3-sphere into a 4D vector space (I can't say if you can always do that or not), then two separate points determine points on a line between them, as that holds in any dimensional vector space. Just take the vector that connects from point A to point B, and multiply it by scalars from 0 to 1, and interpret the result as a displacement from point A. The set of all such displacements would generate the points you are talking about. It sounds like you are really asking "can the universe be embedded in a 4D spatial vector space if it is a 3 sphere at any given age", and I don't see why not, though it's not terribly clear what those points would mean because they would not be in our universe.

gzhpcu

2009-Jul-09, 04:46 PM

If you have two points on a 2-sphere, and draw a line between them, you can define points inside the 2-sphere, in 3D space.

I was just wondering, if the Universe were a 3-sphere, could you define points in 4dimensions, by using two points in our space?

Is this done?

maybe this is trite, but I thought I would post anyway.

My understanding: You can warp a 2 dimensional plane into the figure of the surface of a sphere in 3 dimensional space. The same way (theoretically), in the presence of a 4th spatial dimension, a 3-sphere can be warped into another shape. Any point in our 3 dimensional space would also correspond to a point in the fourth dimensional space, just as any point on a 2-sphere is contained in 3 dimensional space.

Frog march

2009-Jul-09, 04:48 PM

well, I wasn't sure what I was asking. :)

I suppose that you could define the 8 corners of a 4D cube, with different length sticks(with the ends being the defining points)....It's going to be my entry for the Turner prize..:D

does a cube even have only 8 corners in 4D?

DrRocket

2009-Jul-09, 06:53 PM

If you have two points on a 2-sphere, and draw a line between them, you can define points inside the 2-sphere, in 3D space.

I was just wondering, if the Universe were a 3-sphere, could you define points in 4dimensions, by using two points in our space?

Is this done?

maybe this is trite, but I thought I would post anyway.

Yes. Yoiu can realize the n-sphere as the surface of an (n+1)-ball. and you can extend the idea of spherical coordinates to higher dimensions.

A 2-sphere is naturally realized as the points of norm 1 in 3-space -- the surface of the unit ball.

The 3-sphere is just the points of norm 1 in 4-space -- again the surface of the unit ball in 4-space.

This idea extends to the n-sphere in (n+1)-space.

While it is not in general possible to embed an arbitrary n-manifold in (n+1)-space, you can always do it with the n-sphere. You can also embed an n-manifold in a Euclidean space of suitably high dimension (Whitney embedding in dimension 2n). You can even do this isometrically for Riemannian manifolds, with a bit more work (Nash embedding). The theorem also extends to metrics with arbitrary signature.

tdvance

2009-Jul-09, 08:46 PM

well, I wasn't sure what I was asking. :)

I suppose that you could define the 8 corners of a 4D cube, with different length sticks(with the ends being the defining points)....It's going to be my entry for the Turner prize..:D

does a cube even have only 8 corners in 4D?

Yes, but a hypercube has 16 corners :)

OT but: Actually, there's a formula for number of r-faces in an n-cube--

r is the dimensionality of the face: 0 for corner, 1 for edge, etc., n the dimensionality of the (solid) cube, 0 for a single point, 1 for a line segment, 2 for a square, 3 for a cube, 4 for a 4-d hypercube, etc.

The number of r-faces in an n-cube is 2^(n-r) * (n choose r)

One way to think of an n-cube is the set of all n-coordinate points

(a,b,c,d,...,x) with each coordinate being between 0 and 1 inclusive.

Then, an r-dimensional face of the n cube is selected by choosing n-r of the coordinates (n choose r ways to do so, since n choose r = n choose n-r), and consider all possible ways of making them 0 or 1 (2^(n-r) ways to do so). Each choice gives a face--and the other points, taking values from 0 to 1, define the r-dimensional coordinates on that face.

DrRocket

2009-Jul-09, 09:07 PM

Yes, but a hypercube has 16 corners :)

OT but: Actually, there's a formula for number of r-faces in an n-cube--

r is the dimensionality of the face: 0 for corner, 1 for edge, etc., n the dimensionality of the (solid) cube, 0 for a single point, 1 for a line segment, 2 for a square, 3 for a cube, 4 for a 4-d hypercube, etc.

The number of r-faces in an n-cube is 2^(n-r) * (n choose r)

One way to think of an n-cube is the set of all n-coordinate points

(a,b,c,d,...,x) with each coordinate being between 0 and 1 inclusive.

Then, an r-dimensional face of the n cube is selected by choosing n-r of the coordinates (n choose r ways to do so, since n choose r = n choose n-r), and consider all possible ways of making them 0 or 1 (2^(n-r) ways to do so). Each choice gives a face--and the other points, taking values from 0 to 1, define the r-dimensional coordinates on that face.

You don't even have to stop there. Dick Anderson had a distinguished career studying- the Hilbert cube, an infinite product of the unit interval, and also the infinite product of lines in which it lives.

This can also be handled with the use of the L-infinity norm (the norm of (x1, x2,..., xn) is the max of {|x1|, |x2|,..., |xn|}). Then the cube is the unit ball in this norm and faces constitute the unit sphere . Faces in this context consist of points in which one of the xi is +/- 1. This is one way to quickly see the validity of your expression for the number of faces, and easily extends to r-faces. (You probably know this, but not everyone will.)

a1call

2009-Jul-10, 01:52 PM

If you have two points on a 2-sphere, and draw a line between them, you can define points inside the 2-sphere, in 3D space.

I was just wondering, if the Universe were a 3-sphere, could you define points in 4dimensions, by using two points in our space?

Is this done?

maybe this is trite, but I thought I would post anyway.

I believe that's the concept behind LISA (http://lisa.nasa.gov/).

Supposedly gravity acts across 4D. So LISA will detect it with it's vast 3D span.

Corrections are welcome.

ETA: See this video (http://www.wimp.com/bigtheory/) for reference to LISA concept. :whistle:

DrRocket

2009-Jul-10, 04:48 PM

I believe that's the concept behind LISA (http://lisa.nasa.gov/).

Supposedly gravity acts across 4D. So LISA will detect it with it's vast 3D span.

Corrections are welcome.

ETA: See this video (http://www.wimp.com/bigtheory/) for reference to LISA concept. :whistle:

I don't think so.

The original question had to do with realizations of n-spheres in (n+1)-space. Specifically 2-spheres and 3-spheres in 3-space and 4-space.

When you say "gravity acts across 4D" I think you are referring to the description of gravitation n general relativity as the curvature of a 4-dimensional manifold. That manifold is somewhat different, even locally, from ordinary Euclidean 4-space. What is called the "metric" in GR is a smooth selection of a non-degenerate quadratic form for the formation of "inner products". But the form that is used is not positive-definite, which makes the geometry non-Euclidean, and the metric does not define a "metric" in the usual topological sense (since it is possible for distinct points to have zero "distance" in this metric). This rather scrambles the notion of a sphere, at least geometrically.

In contrast the question was posed in terms of Euclidean spaces, which come equipped with a positive-definite inner product, the ordinary dot product. This provides a simple way to realize a sphere -- just the set of ponts equidistant from a fixed point. Distance defined using a positive-definite quadratic form does meet the usual conditions for a metric in the sense of topology, and the geometry is what you would expect it to be.

nokton

2009-Jul-10, 05:49 PM

If you have two points on a 2-sphere, and draw a line between them, you can define points inside the 2-sphere, in 3D space.

I was just wondering, if the Universe were a 3-sphere, could you define points in 4dimensions, by using two points in our space?

Is this done?

maybe this is trite, but I thought I would post anyway.

Hi Frog, just my point on a 3D gravity well of a black hole, and no,

you cannot. 4 dimensions ( excluding time, that is) are beyond our

current brain power to understand, let alone define.

Perhaps that is why we struggle so much with the dimensions in

string theory. Aw, we think we know it all, the compexities of

the universe are beyond our current comprehension, but, perhaps

grasping that, we may eventualy understand.

Nokton.

DrRocket

2009-Jul-10, 06:01 PM

Hi Frog, just my point on a 3D gravity well of a black hole, and no,

you cannot. 4 dimensions ( excluding time, that is) are beyond our

current brain power to understand, let alone define.

.Nokton.

That would be correct except for one tiny detail -- it is wrong.

See earlier posts.

a1call

2009-Jul-10, 06:25 PM

I don't think so.

When you say "gravity acts across 4D" I think you are referring to the description of gravitation n general relativity as the curvature of a 4-dimensional manifold.

What I meant was a reference to the video where it was stated that dark matter is the result of gravitational pull of the extra-universal matter(my wording). Hence across 4D (a dimension other than our 3D). I understand your concept of time being the 4th dimension. However the video deals with multi-verses and actual Euclidean extra dimensions (not time). Listen to the video again and note the 2D pond analogy.

DrRocket

2009-Jul-10, 07:34 PM

What I meant was a reference to the video where it was stated that dark matter is the result of gravitational pull of the extra-universal matter(my wording). Hence across 4D (a dimension other than our 3D). I understand your concept of time being the 4th dimension. However the video deals with multi-verses and actual Euclidean extra dimensions (not time). Listen to the video again and note the 2D pond analogy.

Nuts.

That's Kaku trying to sell his wildly speculative books. IMO he has discovered that there is more money in hyped-up books than there is in physics.

There seems to be a new paradigm -- if you can't produce a prediction with strings, maybe you can produce a book. This is unfortunate as it destroys credibility of people who are actually trying to do real research.

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