Is a curved line 1 dimensional?

[tgm1024]

There are models I've come across in physics involving things such as a 4 dimensional surface of a 5 dimensional sphere. The reasons for that are not important, but it has always given me pause.

I can imagine such things readily enough, but I am stuck on one small thing: Is the surface of (say) a 3 dimensional sphere really only 2 dimensions? Even if an [infinitely thin] surface, can it really be 2 dimensions if bent through out 3 dimensional space?

 

[Office_Shredder]

For example, what do you think is the difference between a parabola and the x-axis that would make the parabola 2 dimensional?

Assuming we're talking about manifolds (which I think is the case) the basic idea is that if you zoom in far enough, the set looks like it's a flat subset of R^k for some k. The dimension is just whatever the value of k is.

You could easily take a sphere and embed it into R¹⁰⁰⁰ would this mean that it's 1000 dimensional?

 

[tgm1024]

For example, what do you think is the difference between a parabola and the x-axis that would make the parabola 2 dimensional?

Assuming we're talking about manifolds (which I think is the case) the basic idea is that if you zoom in far enough, the set looks like it's a flat subset of R^k for some k. The dimension is just whatever the value of k is.

You could easily take a sphere and embed it into R¹⁰⁰⁰ would this mean that it's 1000 dimensional?

Manifold terminology here throws a wrench into the works for what I'm visualizing...but only because it's a definition that itself is what I'm tackling visually. (Or less clumsy words to that effect.) :)

What I'm saying is that yes, when you zoom in far enough to the surface of an N-dimensional sphere you are able to see a piece that more and more resembles something (N-1) dimensional, but here's the rub:

When you look at a plain plane :) you have something with 2 degrees of freedom. The surface of a sphere, however confuses me. It is arcing throughout 3 degrees of freedom *regardless* of what it resembles close up, no?

 

[Office_Shredder]

I can embed a three dimensional sphere into R¹⁰⁰⁰, does that mean it has 1000 degrees of freedom?

You shouldn't let the topology of the space that your surface does not occupy affect your judgement of what the surface itself looks like. The sphere itself is in three dimensions, but on the sphere you only have two independent directions really. Similar to how the x-y plane in three dimensional space is "arcing" through three dimensions (very little in one direction of course)

 

[tgm1024]

I can embed a three dimensional sphere into R¹⁰⁰⁰, does that mean it has 1000 degrees of freedom?

You shouldn't let the topology of the space that your surface does not occupy affect your judgement of what the surface itself looks like. The sphere itself is in three dimensions, but on the sphere you only have two independent directions really. Similar to how the x-y plane in three dimensional space is "arcing" through three dimensions (very little in one direction of course)

For the record, I'm not arguing with you (nor are you reacting like I am--thank you), I am just trying to express the concern here. Using your example of R¹⁰⁰⁰: I question why that is so. For example, pairing it down, I cannot embed a plane into / onto a sphere without it cutting into it. It is no longer a plane at that point.

It's the topology itself being considered a dimension less than "it is" (in my book) that I'm having trouble with.

Note: It's not that I cannot visualize this. I'm 45 and since a kid I've been tormenting friends with N dimensional "slices" of N+1 dimensional objects---even with an arbitrary one of them being time. The problem I have is more that a 2 dimensional manifold of a sphere cannot "live" within a flat 2 dimensional space.

It is not expressible in that 2 dimensional Cartesian coordinate system. <---Am I still at odds with geometry here?

 

[quasar987]

When you look at a plain plane :) you have something with 2 degrees of freedom. The surface of a sphere, however confuses me. It is arcing throughout 3 degrees of freedom *regardless* of what it resembles close up, no?

No no, when you are moving on the surface of a sphere, you do not have 3 degrees of freedom, you actually have only 2: you can go from any point to any other by specifying 2 numbers: the longitudinal angle you have to rotate by and the latitudinal angle you have to rotate by.

Another way to see that the sphere is 2-dimensional: around each point, you can take a small patch of sphere and flatten in onto a part of a plane. Now it is even more clear that the points of the sphere can be parametrized by 2 numbers. And this is what it means to be 2-dimensional.

 

[quasar987]

The problem I have is more that a 2 dimensional manifold of a sphere cannot "live" within a flat 2 dimensional space.

True, you cannot put the sphere in a plane... but you can do so locally as explained above. And this is what it means for a manifold to be 2-dimensional. The fact that it can or cannot be put in some n-dimensional plane has nothing to do with it. In fact. there are 2-manifolds that can be put in R^2 (a plane) . Others that can't be put in R^2 put can be put in R^3 (a sphere, a torus), and others that can't be put in R^2 or R^3 but can be put in R^4 (the projective plane).

 

[Dickfore]

There are models I've come across in physics involving things such as a 4 dimensional surface of a 5 dimensional sphere. The reasons for that are not important, but it has always given me pause.

I can imagine such things readily enough, but I am stuck on one small thing: Is the surface of (say) a 3 dimensional sphere really only 2 dimensions? Even if an [infinitely thin] surface, can it really be 2 dimensions if bent through out 3 dimensional space?

Imagine d-dimensional hypercubes with infinitesimal sides $\epsilon$. Let $N(\epsilon)$ denote the number of such hypercubes necessary to 'cover' the whole curve. Find the following limit:

$$D = -\lim_{\epsilon \rightarrow 0}{\frac{\log{N(\epsilon)}}{\log{\epsilon}}}$$

This is the dimension of the curve. I suspect this limit is exactly equal to 1.

 

[tgm1024]

Imagine d-dimensional hypercubes with infinitesimal sides $\epsilon$. Let $N(\epsilon)$ denote the number of such hypercubes necessary to 'cover' the whole curve. Find the following limit:

$$D = -\lim_{\epsilon \rightarrow 0}{\frac{\log{N(\epsilon)}}{\log{\epsilon}}}$$

This is the dimension of the curve. I suspect this limit is exactly equal to 1.

Now *that* I'm not able to visualize yet.

 

[tgm1024]

Imagine d-dimensional hypercubes with infinitesimal sides $\epsilon$. Let $N(\epsilon)$ denote the number of such hypercubes necessary to 'cover' the whole curve. Find the following limit:

$$D = -\lim_{\epsilon \rightarrow 0}{\frac{\log{N(\epsilon)}}{\log{\epsilon}}}$$

This is the dimension of the curve. I suspect this limit is exactly equal to 1.

No no, when you are moving on the surface of a sphere, you do not have 3 degrees of freedom, you actually have only 2: you can go from any point to any other by specifying 2 numbers: the longitudinal angle you have to rotate by and the latitudinal angle you have to rotate by.

I understand the point, thanks. But I seem to remain recalcitrant. Hmmmm...

There is a difference between a flat plane and a 2 dimensional manifold over a sphere, yes? If all of space where merely 2-D, a flat manifold would fit nicely and a curved one would merely intersect it, right?

Perhaps I was poisoned by "Flatland" as a kid.

 

[Dickfore]

Please define what you mean by 'dimension' as this is an ambiguous term.

 

[Landau]

Similar to this thread.

 

[tgm1024]

Please define what you mean by 'dimension' as this is an ambiguous term.

Interesting. Basically N dimensions means that I need N pieces of information (cartesian coordinates, polar coordinates, or whatever) to find something within it. No?

...by which you reason that only 2 coordinates are needed with a 2 dimensional manifold regardless of whatever N dimensional thing it's "on". But that's only true if you've *already* defined the manifold as a piece of the N dimensional thing. Thus you need N dimensions I would argue.

On earth, for example, it doesn't tell me how far off the ground I am with just longitude and latitude.

I'll think more on the nature of my objection. I feel like it's teetering on semantic.

 

[Office_Shredder]

Longitude and latitude doesn't tell you how far off the ground you are because once you're off the ground you're no longer on the earth. Assuming that you're on the earth, longitude and latitude tell you exactly where you are

 

[tgm1024]

Longitude and latitude doesn't tell you how far off the ground you are because once you're off the ground you're no longer on the earth. Assuming that you're on the earth, longitude and latitude tell you exactly where you are

Yes. Which makes it 2D apparently. I guess I feel more comfortable with 2D being "flat" when viewed from my 3, 4, etc., dimensions...

THANKS ALL!

 

[Tac-Tics]

The dimension of a surface (more technically, a manifold) is essentially the minimum number of coordinates required to describe a point on it.

For a curve, imagine a curved race path (say, for a marathon). You can describe each point on the path by how far the runners have to run to get to that point. (Alternatively, you could describe each point by the percentage of the race course a runner has run, with the start being 0% and the end being 100%).

For a sphere, we can use traditional longitude, latitude coordinates.

For the surface of a donut, we can use a slightly modified long, lat coordinate system.

Some spaces are more abstract. A rotation in three dimensions is itself a three dimensional space. The traditional coordinates are yaw, pitch, and roll.

The space of invertible 2x2 matrices is a 4 dimensional space. However, it's different from R^4 in the same way the surface of a sphere is different from a plane. The space of 2x2 matrices has "gaps" it in (where non-invertible matrices would be, but we don't include them).

The word "dimension" in physics usage is often used in a ******** way. A "10 dimensional universe" just means that points are described with 10 numbers... the x, y, z spatial coordinates, t for time, and 6 other parameters for whatever the theory does. It's like saying a student taking 5 courses in college is a "5 dimensional student" because he can get different grades in different classes, each of which is independent of the others.

 

[tgm1024]

For example, what do you think is the difference between a parabola and the x-axis that would make the parabola 2 dimensional?

Assuming we're talking about manifolds (which I think is the case) the basic idea is that if you zoom in far enough, the set looks like it's a flat subset of R^k for some k. The dimension is just whatever the value of k is.

You could easily take a sphere and embed it into R¹⁰⁰⁰ would this mean that it's 1000 dimensional?

[...returning...]

A parabola would be 2 dimensional because it cannot be described (shape and position) without both dimensions.

But in a 3 dimensional space it would have to be 3 dimensions to be able to fully describe the shape or position of it. A line, even along the face of a cube is 3 dimensional or it would not be coexisting with the cube.

I guess I need a shift in terminology. Is there a term outside of manifold and dimension that specifically defines the number of dimensions required to describe an object in the space it exists? Or more so, the equator of a 3D sphere is a 1 dimensional manifold. But is there something that allows me to properly label that line as a 3 dimensions because it exists in the 3 dimensional space?

I don't see this as quite a semantic argument.

 

[quasar987]

You might say that the line is "sitting inside R^3".

 

[tgm1024]

You might say that the line is "sitting inside R^3".

Now you've nailed the semantic(?) problem. WHAT line, WHERE?

The one from (x₁) to (x₂)?

Or the one from (x₁,y₁) to (x₂,y₂)?

Or the one from (x₁,y₁,z₁) to (x₂,y₂,z₂)?

The first two don't tell me enough if in 3D space. (?)

If I'm frustrating anyone, please tell me. I'm still not sure of how to regard this.

 

[Landau]

There is only one line, call it L. It is a topological space (or even manifold) in itself. We can embed L in several kinds of spaces, meaning there is an embedding
$$i:L\to X$$
for several spaces X, such as X=R^1, X=R^2, X=R^3, etc.

 

[Etherian]

I think I see where tgm1024 is getting confused.

[...returning...]
A parabola would be 2 dimensional because it cannot be described (shape and position) without both dimensions.

Yes, to define the shape and position of a parabola itself in R^2, it takes two values. To find a point on a parabola relative to R^2 it would take two values. But to find how far along the line you are it only takes one value. It does not matter what shape the line has or what space the line is embedded in. If you call one point on the line zero, another point three units away will stay three units away whether it is in R^2, R^3, or bent onto a number line.
The same is true of a sphere. You can say a point (pi/4, 2pi/3) on a sphere maps to a point (4,7,6) in R^3 or (17/5, 2pi, 15, 3) in R^4. To the Flatland character running around on it, there are no points above or below him, inside or outside of the sphere, only those in front, behind, and beside him. All points he can see can be described as a direction and distance from him.

 

[lavinia]

dimensionality may be thought of as what the world looks like over very small distances. Over a small distance a sphere looks flat - so it is 2 dimensional.