Confused as to what constitutes a manifold?

[saminator910]

I am having trouble getting a set definition of what constitutes a manifold for example ,

I have the real plane R^2, and the sphere
s = {(x,y,z)|(x,y,z)£R^2, x^2+y^2+z^2=1}
Note £, is meant to be "element of".
And I have a continuous function f mapping the real plane onto s such that
f:R^2-->S

Is S considered a manifold?, please tell why or why not and I really need some examples of manifolds, and how one defines its structure . Oh and I know whole courses are taught on this, but, I appreciate any response

 

[HallsofIvy]

I have no idea how "(x, y, z)" can be in R²! Did you mean R³?
But there is NO continuous mapping of the real plane to a sphere.

However, it is still a manifold because you can find two mappings of the plane to a sphere such that each maps to all except one point, and they 'overlap' smoothly. For example, Set the sphere so that its "south pole" is at the on the plane and draw a line from its "north pole" to a point on the plane, the point where that line passes through the sphere being being the point that point on the plane is mapped to. That maps a point on the plane to every point on the sphere except the north pole. Set the sphere so that its north pole is at the origin and draw lines from the south pole to get the mapping that maps points on the plane to every point on the sphere except the south pole.

An "n dimensional manifold" is defined as a topological space, M, together with a collection of open sets in M, \{U_i\} and a corresponding collection of functions, \{\phi_i(p)\} such that \{U_i\} "covers" M (every point in M is in at least one U_i, every \phi_i maps U_i one to one onto an open subset of Rⁿ, and, if U_i and U_j has non-empty intersection, then \phi_j(\phi_i^{-1}) is a continuous function from Rⁿ to R².

 

[quasar987]

I am having trouble getting a set definition of what constitutes a manifold for example ,

I have the real plane R^2, and the sphere
s = {(x,y,z)|(x,y,z)£R^2, x^2+y^2+z^2=1}
Note £, is meant to be "element of".
And I have a continuous function f mapping the real plane onto s such that
f:R^2-->S

Is S considered a manifold?, please tell why or why not and I really need some examples of manifolds, and how one defines its structure . Oh and I know whole courses are taught on this, but, I appreciate any response

There certainly are continuous surjective mappings R²-->S, but that is not the point of manifolds. To show that S is a (topological) manifold, you must find a collections of open sets U_i such that \bigcup_iU_i=S as well as maps f_i:U_i\rightarrow f(U_i)\subset \mathbb{R}^2 which are
1) continuous
2) surjective (automatic)
3) injective
4) f_i(U_i) open in R²
5) f_i^-1 continuous
In other words, the maps f_i are homeomorphisms.

These maps f_i which map subsets of S bijectively to subsets of R² must be thought of as assigning coordinates on S, so that with respect to one such map f_i, it makes sense to talk about "the point of coordinate (x,y) in S" (assuming (x,y) belongs to f_i(U_i)).

 

[mathwonk]

if you don't move around much, it just looks like an open n- ball in euclidean space.

 

[Studiot]

I'm glad this question was asked because my knowledge here has always been a bit hazy.

My maths dictionary has two definitions, the second of which is the one so far supplied here and comes from differential topology.

My dictionary's first definition is 'the collection of elements of a set' .

This first definition is less restrictive since the second implies that all such manifolds have infinite numbers of members.

 

[A. Bahat]

Wouldn't 'the collection of elements of a set' just be a set? That doesn't sound like it has anything to do with manifolds.

 

[saminator910]

So what has been said is really helpful, if I'm correct it is like mapping sets, such as the real plane, and their topologies , onto a structure like a sphere, and the sphere will now resemble at every point, a point of R^2. Now I realize that manifold wasn't completely correct but can anyone please define another manifold as an example, like one with a different structure, and sets?

 

[Studiot]

I can see the idea is that if you want to do calculus on the set so you want a set, with a distance function, in which every point has a neighbourhood so you can define continuity and derivatives. Loosely you call such a set a (differentiable) manifold.

Other particular property restrictions might define other types of manifold.

 

[A. Bahat]

There are really two ways of defining a manifold, which are easily shown to be equivalent but give useful differing perspectives.

One definition (the more concrete one) is as a topological space which is locally homeomorphic to Euclidean space, and you can put charts on in which are required to overlap continuously/differentiably/smoothly/analytically.

The other (more abstract one) is that a manifold is a locally ringed space with a structure sheaf locally isomorphic to the sheaf of continuous/differentiable/smooth/analytic functions on Euclidean space. That is, we are essentially saying a manifold is a set with certain functions specified on each open set, which we designate as the continuous/differentiable/smooth/analytic functions, and these determine the 'manifoldy' structure. This approach is analogous to how schemes are defined in algebraic geometry.

 

[Studiot]

There are really two ways of defining a manifold,

So what is an affine manifold in 3 space?

 

[A. Bahat]

An affine manifold is one where the transition maps (i.e. change of coordinates) are not only smooth, but are actually affine maps, so a linear map followed by a translation.

 

[mathwonk]

to give examples, consider a smooth mapping from R^n-->R^m with m ≤ n.

and assume the rank of the map is m at every preimage of 0. then the preimage of 0 is a manifold of dimension n-m.its just like the fundamental theorem of linear algebra, wherein the preimage of 0 under a linear map R^n-->R^m of rank m, is a subspace of dimension n-m.

indeed in the setup above the preimage of 0 under the derivative of the map is the tangent space of the manifoldspecifically, let f(x,y,z) = x^2 + y^2 + z^2 - 1. then the preimage of zero is all points of the unit sphere,

and at any such point the differential (2x,2y,2z) is non zero, hence of rank 1. so the nullspace of the linear map (2x,2y,2z).( ) is the tangent plane to the sphere at (x,y,z). i.e. the tangent plane to the sphere at (x,y,z) can be considered as the set of vectors perpendicular to the radius vector (x,y,z).

 

[quasar987]

R, R², R³, ... themselves are manifolds. These are the "trivial examples".

For another 2-dimensional manifold, there is the torus, and more generally, the so-called "n-holed tori". for example, here is a 3-holed torus:

http://en.wikipedia.org/wiki/File:Triple_torus_illustration.png

 

[A. Bahat]

In fact, we know that all of the compact, orientable 2-dimensional manifolds are diffeomorphic to some n-holed torus (with the sphere as the case n=0). This is the celebrated 'classification of closed surfaces'.

[The orientability condition isn't really necessary; we just need to add things like projective planes. But the orientable case is easier to visualize.]

 

[saminator910]

Are there any quick, basic books you can recommend for someone interested more in the subject of differential topology/ manifolds, with a background in general topology/ calculus, but no linear algebra or advanced calculus(I am aware some books on the subject will teach all the linear algebra and advanced calculus needed).

 

[A. Bahat]

What exactly do you mean by 'advanced calculus'? (I ask because different people can mean very different things by this phrase.) Also, you should learn linear algebra ASAP...

 

[saminator910]

I know single variable, and I know the concepts of multivariate plus how to perform the basics, plus a little vector calculus. I think I may want to go into linear algebra next though. Do you think linear algebra is more useful than more advanced calculus is differential topology?

 

[A. Bahat]

Let me put it this way: advanced calculus is basically using linear algebra to study the local properties of functions on Euclidean space, and differential topology is just the generalization of advanced calculus to manifolds. So without those two subjects, I wouldn't recommend trying to learn differential topology yet. (By all means read about it if it interests you! I'm just saying that you'll probably miss some things because you don't know the 'easier version', i.e. advanced calculus.)

 

[quasar987]

Linear algebra is not very difficult to learn imo. I think the most useful results for diff. top. are:
i) a linear transformation is injective iff its kernel is 0
ii) a linear transformation is bijective iff its determinant is non 0
iii) the formalism of internal direct sums (what does it mean that a vector space splits as the direct sum of two of its subspaces?)
iv) the rank-nullity formula relating the dimension (aka rank) of the kernel to that of the image and domain space.
v) a linear transformation is entirely determined by its action on a basis.

 

[saminator910]

So would you recommend first a study of linear algebra, and completion of my study of multivariate calculus before differential manifolds? Is a study of advanced calculus absolutely needed for differential topology? is vector calculus sufficient?

 

[A. Bahat]

Yes, I would recommend studying linear algebra (this need not be too in depth for the purpose of differential topology, but the more linear algebra you know, the better) and multivariable/vector calculus, and then differential topology.

 

[saminator910]

Alright I am already in possession of a family member's multivariate calculus book, I will probably get a Dover book on linear algebra since they're usually quite basic, cheap, and get the job done. by any chance do you know of any books like this for beginners in linear algebra, a basic look at the topic. I am in high school so I don't want an entire coursebook because I will take a course in college, just the basic functional knowledge.

 

[mathwonk]

shilov, linear algebra. dover.

https://www.amazon.com/dp/048663518X/?tag=pfamazon01-20

 

[saminator910]

Thanks Mathwonk, once I'm done with vector Calculus I'll take a look at that book. Pushing this thread to it's original topic , would you consider a parametric surface to be a manifold because for equations x(u,v) y(u,v) z(u,v) u ∈ U, v ∈ V, it is kind of like mapping U x V onto the surface. And could you define equations on U x V like say u = v for a torus would produce a spiral around the surface. For example,

I take the function of a trefoil knot just with t

In this case I have defined parametric equations with variables u,t

x(t) = (2 + cos1.5t)cos(t), y(t) = (2 + cos1.5t)cos(t)
z(t) = sin1.5t

I can find a surface for which the trefoil knot is u = t by subbing in u for some of the variables In this case the surface could be

x(u,t) = (2 + cos1.5u)cos(t)
y(u,t) = (2 + cos1.5u)cos(t)
z(u,t) = sin1.5u

or

x(u,t) = (2+cos(1.5*t))*cos(u)
y(u,t) = (2+cos(1.5*t))*sin(t)
z(u,t) = sin(1.5*t)

or any other combination of U's and T's

then if you want the equation for u = t you just sub in t for u and it yields the trefoil knot again.

the first produces a torus and when both the trefoil knot and it are graphed at the same time, the knot is wrapped around it, the second produces a strange self intersecting surface with the knot wrapped around it, but the point is they both have the trefoil knot as u = t for U x T. When I figured this out it just seemed awesome to me.

This makes sense to me do you get what I am saying? is a parametric surface a manifold?

 

[Bacle2]

Thanks Mathwonk, once I'm done with vector Calculus I'll take a look at that book. Pushing this thread to it's original topic , would you consider a parametric surface to be a manifold because for equations x(u,v) y(u,v) z(u,v) u ∈ U, v ∈ V, it is kind of like mapping U x V onto the surface. And could you define equations on U x V like say u = v for a torus would produce a spiral around the surface. For example,

I take the function of a trefoil knot just with t

In this case I have defined parametric equations with variables u,t

x(t) = (2 + cos1.5t)cos(t), y(t) = (2 + cos1.5t)cos(t)
z(t) = sin1.5t

I can find a surface for which the trefoil knot is u = t by subbing in u for some of the variables In this case the surface could be

x(u,t) = (2 + cos1.5u)cos(t)
y(u,t) = (2 + cos1.5u)cos(t)
z(u,t) = sin1.5u

or

x(u,t) = (2+cos(1.5*t))*cos(u)
y(u,t) = (2+cos(1.5*t))*sin(t)
z(u,t) = sin(1.5*t)

or any other combination of U's and T's

then if you want the equation for u = t you just sub in t for u and it yields the trefoil knot again.

the first produces a torus and when both the trefoil knot and it are graphed at the same time, the knot is wrapped around it, the second produces a strange self intersecting surface with the knot wrapped around it, but the point is they both have the trefoil knot as u = t for U x T. When I figured this out it just seemed awesome to me.

This makes sense to me do you get what I am saying? is a parametric surface a manifold?

Consider using the Inverse Function Theorem on the parametrization; you want to

avoid self-intersections, corners, etc.