Dimensions! Clarification needed.

Swapnil

1) Why is it that a sphere is a 2-dimensional object even though we need 3-dimensions to actually *draw* the sphere.

2) If the Cartesian plane is a 2-dimensional *space*, then the number-line should be a 1-dimensional *space*, right? (not sure if space is the appropriate term here)

3) Would a point be considered a 1-dimensional or a 0-dimensional entity?

Thanks.

cesiumfrog

It's the surface of the sphere that is 2D, not the sphere itself. Yes, a line or axis is only one dimension. A point has no dimensions.

disregardthat

But a 2 dimensial space cannot connect to the ends. In 2 dimensions up and down gives no meaning! it's only 2 dimensions, back and forth, and left and right. up and down is preserved to the third dimension. So if you want a ball you must add the third dimension even though you only think if the surface of it. Am I not right?

arildno

No. Look up on non-Euclidean geometries.
Although such geometries can readily be regarded as geometries-on-curved-surfaces i.e, within a 3-D world), their postulates (in particular their alternative to Euclid's fifth postulate) contains no mention of a third dimension.

disregardthat

Well, they have to include some dimension. a 2 dimensional field cant have more information than it's own area... or is that incorrect too?

Swapnil

I see. So would you still call sphere a 3-dimensional object?


Also, what does it mean to say that a particular function lives in a certain dimension. For example, say [tex]f[/tex] lives in [tex]\mathbb{R}^3[/tex]

robphy

Technically, a sphere is a 2D-object [a closed surface] whereas a [tacitly assumed solid] ball is a 3D-object.

To answer your last question, you are asking about the dimensionality of the domain of f.

Swapnil

Do you mean the range instead of the domain?

I mean, you can have a function f that maps every point on the number line to a set of 3 points. Then would say that the function lives in 1-dimension because of its domain? Or would you say that the function lives in 3-dimension because of its range?

robphy

well... clearly "lives" is not a precise enough term to be used seriously.
Function may be somewhat vague as well... hence terms like real-valued function and vector-valued function.

Maybe "lives" should be replaced by a more technical notation like the mapping
[tex]f: M \rightarrow N[/tex]
and discuss the dimensionality [if it exists] of the domain M and of the range N.

Gokul43201

Here's the general approach: ask yourself how many numbers you need to specify to completely define a point on a sphere. That number tells you the dimensionality (crudely, but rule-of-thumbishly speaking - see link) of the sphere(edited from "object" to "sphere").

Read this carefully.

Both statements are true, and space is a correct usage. In particular, the set of Reals is a vector space (defined with the usual addition and mutiplication of reals).

Apply the test described above.

Hurkyl

As robphy stated, the notation is somewhat imprecise. But I would expect it to mean that f is a function into R³.

Informally, people sometimes think of such an f as a vector that varies with time, or with position (depending on how we think of f's domain), which explains why one might think of it as "living" in R³. It also provides a convenient way to be sloppy with notation. But it's probably not best to actually think of f in that way, unless you're willing to swallow a hefty dose of formal logic before doing so.

matt grime

Because that is the definition of dimension we have chosen - it is intrinsic to the object, not any embedding of it in an ambient space (which is clearly not a well defined number anyway). The dimension refers to the 'local dimension' (non-standard term) of that object: locally, i.e. in any small patch about a point, it is homeomprhic to the open disc or R^2. Since this is true for every point on the surface, we say it is 2-dimensional. Some things don't have a well defined dimension because they are not everywhere locally homeomorphic to R^n. There is another definition of dimension. The sphere is defined, in R^3, by the equation f:=x^2+y^2+z^2-1=0. The (algebraic) functions from the sphere to R are then R[x,y,z]/f, and this ring has krull dimension 1. Which is just saying that f is irreducible.


The real line is a 1-dimensional vector space, and a 1-dimensional manifold, yes.

zero dimensional.

matt grime

Yes. You are not right. I don't understand what the first sentence means ('the' ends of what?).

I don't think you're using the correct definition of 'dimension'.

Gib Z

I don't seem to completely understand Gokul's definition...you would need 2 values to define a point on a Cartesian Plane, but I would think a point is 0 dimensional.

matt grime

Again, you're thinking of the point embedded in the plane. The embedding idea is wrong. A point is 0 dimensional. If you must, imagine that some reference point within the object in question is given. The dimension is then roughly the minimal number of parameters required to specify all other points in relation to that reference point. If the reference point is the only point in the space, then nothing else is required. So, for a sphere, we can imagine the globe. Our reference point is 0 deg lat, 0 deg long. Any other point is then uniquely determined by lat and long. But this is a completely unrigorous definition. It is better to use a proper definition of dimension. I already gave two about 4 posts before now.

mathwonk

poincare has a nice little popular essay on the topic in one of his books of essays.

he does it inductively. apoint is called zewro dimensional, as is any finite set.

then a (connected) geometric set is called 1 dimensional if it can be disconnected by removing a zero dimensional set.

thus a circle is 1 dimensional because removing two points, disconnects it into two pieces.

a sphere is 2 dimensional becausew removing the 1 dimensional equator disconnect it nito an upper and lower hemisphere.

etc....


of course then one gets into the realm of analysis of singularities as follows. consider the union of two spheres touching at one point. removing that point renders them disconnected, but they are not one dimensional.

so one must exclude "singualr" points such as the point where the tow spheres meet, orelse one mjust make the definition local.

there is anotherd efinition, the geometric version of matts reference to "krull" dimension.

namely nesated sequences of geometric objects. a circle is 1 diemnsiuonal bvecause in it you can find a subobject namely a point.

a sphere is 2 dimensional because in it you can find not only a point but also a curve through that point.

etc.....

e.g. in linear algebra, the vector dimension of a finite dimensional space equals the number of proper subspaces it contains, all of which are nested.

e.g. the plane contains the origin and a line through the origin, so dim = 2.

mathwonk

dimension also depends on how you measure it, i.e. it is always relative to something else. put otherwise it depends on what you consider to be a point.

in field theory if k, L, M, are nested increasing fields, then M has different dimensions over k and over L.

a polynomial ring over C has a different krull dimension from its vector dimension, etc....


a man in an airp[lane might copnsider the earth to be a space and the people to be points. a woman in a spaceship might consider our galaxy as the space and the earth to be a point.