Intro to Diff. Geom. PLEASE COMMENT.


by mordechai9
Tags: comment, diff, geom, intro
mordechai9
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#1
Feb7-09, 06:11 PM
P: 205
I am taking an introductory course in differential geometry this semester and I want to try and sketch my knowledge so far. We are using the book "Elementary topics in differential geometry" by Thorpe, which from what I have read, provides a fairly unusual treatment. I am looking for agreements/disaagreements/ or general comments on my introductory, conceptual interpretation. I would especially appreciate any points you think are subtle/extra important. THANKS.

First we consider arbitrary smooth functions. Smooth functions are continous, with continuous derivatives of all orders. We consider these functions as mapping from some open set U to R, in which case it is the typical scalar function. Otherwise, if the function maps from U to G, where G has dimension equal to U, then we have a vector field.

For the vector field, consider the domain and image, with the standard meaning. Then each level set of the domain of the vector field can be parametrized by a smooth curve. If this curve has a velocity vector everywhere equal to the image of the vector field, then this curve is an integral curve. By the existence and uniqueness theorem we can show that there exists a maximal integral curve for each level set of the vector field.

Next we go back to the consideration of smooth scalar fields. Then it can be shown that gradient of the scalar field at some point p on the curve is orthogonal to all vectors tangent to the integral curve at the point p. Conversely we can show that if a vector is tangent to the integral curve at point p, then it is orthogonal to the gradient of the scalar field at that point. So we can establish a tangent space, meaning the subspace at point p of all vectors tangent to the gradient of the scalar field. Note that this subspace, the tangent space, has dimension value that is one less than the dimension of the domain.

Next we consider surfaces as level sets of smooth scalar fields whose gradient at all points on the surface is nonzero, so that each point on the surface can be associated with a unique tangent space.

This is my brief conceptual summary... next chapters are, "Vector fields on surfaces, orientation", "the gauss map", "geodesics", "parallel transport"...
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WWGD
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#2
Feb8-09, 11:04 AM
P: 391
Quote Quote by mordechai9 View Post
I am taking an introductory course in differential geometry this semester and I want to try and sketch my knowledge so far. We are using the book "Elementary topics in differential geometry" by Thorpe, which from what I have read, provides a fairly unusual treatment. I am looking for agreements/disaagreements/ or general comments on my introductory, conceptual interpretation. I would especially appreciate any points you think are subtle/extra important. THANKS.

First we consider arbitrary smooth functions. Smooth functions are continous, with continuous derivatives of all orders. We consider these functions as mapping from some open set U to R, in which case it is the typical scalar function. Otherwise, if the function maps from U to G, where G has dimension equal to U, then we have a vector field.

For the vector field, consider the domain and image, with the standard meaning. Then each level set of the domain of the vector field can be parametrized by a smooth curve. If this curve has a velocity vector everywhere equal to the image of the vector field, then this curve is an integral curve. By the existence and uniqueness theorem we can show that there exists a maximal integral curve for each level set of the vector field.

The concept of vector fields is formalized by using the tangent bundle, which you will
surely be seeing soon --a lot of it. The issue with the maximal integral curve will take
you to 1-parameter groups too.

Next we go back to the consideration of smooth scalar fields. Then it can be shown that gradient of the scalar field at some point p on the curve is orthogonal to all vectors tangent to the integral curve at the point p. Conversely we can show that if a vector is tangent to the integral curve at point p, then it is orthogonal to the gradient of the scalar field at that point. So we can establish a tangent space, meaning the subspace at point p of all vectors tangent to the gradient of the scalar field. Note that this subspace, the tangent space, has dimension value that is one less than the dimension of the domain.

Nicely said. Just a few comments:

I don't know if this is what you said, but dimT_pM =dimM , for any manifold M.

Also this def. of the tangent space works for manifolds that are embedded , tho not for stand-alone manifolds. In the last case, you need to define an abstract tangent space as a space of derivations based at p, which of course will have dimension dimM .

Another important issue to think about is that of an important difference between
R^n and other manifolds: All the tangent spaces T_pR^n are naturally isomorphic
to each other (meaning that there is a linear iso. between them that is independent
of the choice of basis), i.e, T_pR^n ~T_qR^n . This fact implies that differentiation
in R^n is much simpler than on a manifold where this is not the case: in R^n , the
difference f(x+h)-f(x) can be dealt with "naturaly": the vector f(x+h) lives in T_(x+h),
while f(x) lives in T_x . By the natural iso. between these, we can translate /move
vectors from T_x to T_(x+h) . OTOH, when we don't have this iso., we need to use
contraptions like Connections/Parallel transport to be able to make sense of the difference quotient and differentiation.

There are some nice posts in this site , on connections. If you have time, try doing
a search on them.

Next we consider surfaces as level sets of smooth scalar fields whose gradient at all points on the surface is nonzero, so that each point on the surface can be associated with a unique tangent space.

This is my brief conceptual summary... next chapters are, "Vector fields on surfaces, orientation", "the gauss map", "geodesics", "parallel transport"...

It would be nice if you could read ahead on bundles; specially the tangent bundle. It
appears everywhere. It is a way of formalizing the def. of vector fields as maps from
a manifold to its tangent space.

Hope it Helped.
WWGD
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#3
Feb8-09, 11:06 AM
P: 391
Sorry, Mordechai9, I don't know how to use the quoting option too well; my post
got mixed up with yours. Let me know if it is difficult to parse.

mordechai9
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#4
Feb9-09, 10:12 AM
P: 205

Intro to Diff. Geom. PLEASE COMMENT.


Thank you very much WWGD for your comments. I would especially appreciate any more comments, especially on point #2. Also note I am still unsure about the formal idea of a manifold, and hence I may be using the term incorrectly.

1.) I will be on the look out for the tangent bundle as a formalization of the vector field.

2.) You stated dimT_pM = dimM, for any manifold M. If I understand your notation correctly, you mean something like, "The dimension of the tangent space at point p of manifold M is equal to the dimension of the manifold M." However, I thought that the tangent space was actually 1 dimension less; i.e., dimT_pM = dimM - 1. I was under this impression because the tangent space is orthogonal to the gradient at that point; hence it seems to have one less basis vector defining the space (i.e., the basis vector in the direction of the gradient). However, now that I think about it, I'm not sure this is true. Could you please comment?

3.) You point out that the idea of the tangent space we mentioned does not hold for stand-alone manifolds. This sounds interesting but I am rather ignorant here because I am still not rigorously clear on what it means to be embedded.

4.) You point out that in R^n, the isomorphism between tangent spaces at different points provides for a simpler differentiation process. This sounds like a good motivation for the parallel transport/connections topics and I will keep this in mind. This also sounds very interesting becaues I am again rather ignorant on this topic. Indeed I had never considered that a manifold might not have this kind of isomorphism. Could you make an example of a nonfriendly manifold of this type, also, could you clarify briefly what you mean by the word manifold?
WWGD
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#5
Feb9-09, 05:47 PM
P: 391
Hi, Mordechai9:
I will get back to you later with other comments; sorry, kind of busy.:

Re:
2.) You stated dimT_pM = dimM, for any manifold M. If I understand your notation correctly, you mean something like, "The dimension of the tangent space at point p of manifold M is equal to the dimension of the manifold M."

Yes, this is what I meant.

However, I thought that the tangent space was actually 1 dimension less; i.e., dimT_pM = dimM - 1. I was under this impression because the tangent space is orthogonal to the gradient at that point; hence it seems to have one less basis vector defining the space (i.e., the basis vector in the direction of the gradient). However, now that I think about it, I'm not sure this is true. Could you please comment?

Well, if I understand you well, the normal vector is not a vector in the tangent space,
which in this case ( embedding in R^n) coincides with the tangent plane. So the
tangent plane contains all vectors that are tangent to the manifold at a given point.
And this is not the case with the normal vector. In an informal way, the tangent plane T_pM
contains all the directions of curves passing thru p, with the derivative of the curve
being tangent to the mfld. at p. The derivative as a vector (V. field, that is) determines
a direction. You will see that all these directions can be spanned
by the "basis directions" given by e_1=(1,0,..,0), e_2=(0,1,0,..,0),..,e_n=(0,..,0,1),
(Where n is the dim. of the manifold) and this is why the dimension of the tangent space is the same as dim. of the mfld.

Will get back to you with the rest.

You'll see
mordechai9
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#6
Feb9-09, 07:12 PM
P: 205
You know I'm a very introductory student here, so I would tend to accept your argument, except that I spoke to the professor about this earlier today and I thought that he was actually agreeing with what I said -- that dimT_pM = dimM - 1.

You say that the tangent plane T_pM contains all the directions of curves passing thru p, with the derivative of the curve being tangent to the mfld. (I agree). We assume the dimension of the embedding space (and manifold) is R^n. Let me try a proof by contradiction. Assume the dimension of the tangent space is n. Then the dimension of the embedding space equals the dimension of the tangent space. Thus any vector in the embedding space can be written in terms of basis vectors of the tangent space. Take a vector from the embedding space, orthogonal to the surface at point p. Then we can rewrite this vector in terms of the tangent space basis, so that we must conclude that this vector also belongs to the tangent space. Contradiction, because we know the tangent space contains no orthogonal vectors.
--> So dimT_pM cannot equal dimM.

?
bombadillo
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#7
Feb10-09, 03:44 AM
P: 30
Quote Quote by mordechai9 View Post
You know I'm a very introductory student here, so I would tend to accept your argument, except that I spoke to the professor about this earlier today and I thought that he was actually agreeing with what I said -- that dimT_pM = dimM - 1.
That is completely wrong. The dimension of a (non-singular) manifold is the same as the (algebraic) dimension of the tangent space. Construct a simple example: the surface of the sphere is two-dimensional (in the sense that it can be covered with overlapping coordinate patches. Now look at the tangent plane to the sphere at any point: it is a plane.

Thorpe is a nice book but I prefer the text by O'Neill, and even more, the text by Oprea.

Don't bother with tangent bundles right now: you just want to learn surface theory. Fibre bundles and cohomology can come later.
zhentil
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#8
Feb10-09, 06:40 AM
P: 491
Quote Quote by mordechai9 View Post
You know I'm a very introductory student here, so I would tend to accept your argument, except that I spoke to the professor about this earlier today and I thought that he was actually agreeing with what I said -- that dimT_pM = dimM - 1.

You say that the tangent plane T_pM contains all the directions of curves passing thru p, with the derivative of the curve being tangent to the mfld. (I agree). We assume the dimension of the embedding space (and manifold) is R^n. Let me try a proof by contradiction. Assume the dimension of the tangent space is n. Then the dimension of the embedding space equals the dimension of the tangent space. Thus any vector in the embedding space can be written in terms of basis vectors of the tangent space. Take a vector from the embedding space, orthogonal to the surface at point p. Then we can rewrite this vector in terms of the tangent space basis, so that we must conclude that this vector also belongs to the tangent space. Contradiction, because we know the tangent space contains no orthogonal vectors.
--> So dimT_pM cannot equal dimM.

?
How could the tangent space depend on the embedding space? If you can embed the manifold in R^k, you can also embed it in R^n for any n>k. Your argument would imply that the tangent space can have arbitrary dimension.
mordechai9
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#9
Feb10-09, 03:25 PM
P: 205
Bombadillo - If we are talking about a regular sphere, then I assume you mean the sphere existing from the scalar function f: R^3 -> R^1 defined by f(x1,x2,x3) = x1^2+x2^2+x3^2. In this case, the dimension of the tangent plane is 2, but the dimension of the domain is 3. So do you see what I'm saying? I'm saying that the dimension of the tangent plane is 1 less than the dimension of the original domain. I may have used the incorrect terminology to call the domain a manifold or the embedding space; this may be my problem. Please respond. BTW, thanks a lot for the book recommendations.

Zhentil - You tell me! I still haven't learned the meaning of embedding space OR manifold! How do you expect me to understand this???? Please provide more information to help me.
Hurkyl
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Feb10-09, 03:49 PM
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Quote Quote by mordechai9 View Post
but the dimension of the domain is 3.
You know, you can embed the sphere in a 4-dimensional "domain", or a 5-dimensional "domain" or....

P.S. why do you care about the "domain"?
WWGD
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#11
Feb11-09, 01:08 AM
P: 391
Another way of seeing that Dim T_pM =dim M is this:

we do a correspondence between the directions and the tangent vectors.
Then we show that each tangent vector can be made to correspond to a
directional derivative: the direction d_0 corresponds to a directional deriva
tive along the direction d_0 , like in calculus in R^n. Now, all directional
derivatives are spanned by the e_i's . We just dot-product along any direction
other than the standard directions represented by the e_i's.

As Hurkyl said, the dimension of the underlying space has no effect: by
his comment, if this were the case, then the dimension of the manifold
would not be well-defined , in the sense that it would depend on the
dimension of the space in which M is embedded.

Re the issue of embedding vs. stand-alone: I don't know if you have seen
any point-set topology yet. If you have, then the relation between the two
is that between X being a space itself vs. X being a subspace of Y . If not,
let me know, I will try to think of something.

Re the def. of mflds: As a way of setting-up context, the key fact is that
the spaces that seem to be easier to understand for us (people) , are the
Euclidean spaces. So , whenever possible, we try to compare other non-Euc.
spaces to Euclidean ones.

A manifold ( with a few exceptions, like open subsets of the real line, etc.)
are not globally Euclidean (meaning that an n-mfld. is not, for the most part,
globally homeomorphic with R^n.).
But a manifold is something of a second-best: it is locally-Euclidean, in the
sense that at each point p in the manifold, there is a small 'hood ( 'hood =
neighborhood) that is Euclidean. One standard example is that of the surface
of the earth: it is not globally Euclidean (e.g., because it is compact, and R^n is
not. You also see the lack of flatness by walking long-enough and seeing how
objects change in size.)

but around each point it is Euclidean. This is formalized by saying that
each point p in M has a 'hood that is diffeomorphic with an open subset of R^n
(an open n-ball, actually. ), so for each p in M, there is an open U_p containing
p, and a function f_p:U_p --> O_n , where O_n is open in R^n. The pair
(U_p,f_p) is called a 'chart' (just like in a regular map.)
You also have a compatibility requirement , in that if two points p,q , are in
different charts , you want the charts to 'glue well' , so that f_pof_q^-1
(o:= composition) , and f_q o f_p^-1 are at least continuous. If this last
is the case, then your manifold is a topological manifold. If the overlap is
differentiable, then your manifold is differentiable, etc.



Re the issue of the bundle, I just meant that it formalizes vector fields
as maps between points in the manifold and vectors in the tangent space.
The bundle itself , the disjoint union of all the tangent spaces, can itself
be made into a manifold -- with dimension twice that of the manifold.
Considering the T.Bundle as a manifold allows you to talk about smoothness,
etc.
mordechai9
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#12
Feb11-09, 10:32 AM
P: 205
OK, I am understanding things better now. Thanks a lot. I was just interested in the domain/tangent plane relation because it gave some intuition about how the tangent plane is "smaller" (i.e. more restricted) than the domain. Even if you embed a sphere in R^4, R^5, or whatever, you still notice that the tangent plane dimension is always smaller than the dimension of the domain. I don't think this is absolutely useless, because it provides some intuition, like I said.

Anyways, things are clearer now... I will follow up after learning a bit more in class.
Hurkyl
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#13
Feb11-09, 03:39 PM
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If you embed a ball in R^3 (i.e. as the set of points (x, y, z) with x^2 + y^2 + z^2 < 1), how does the dimension of its tangent space compare with the dimension of the ambient space?



By the way, an important fact to note is that if you differentiably embed a manifold M inside R^n in a way that M is the solution to a set of scalar equations
[itex]f_1(v) = f_2(v) = \cdots = f_k(v) = 0[/itex]
then the cotangent space of M is the solution set to
[itex]d f_1(v) = d f_2(v) = \cdots = d f_k(v) = 0[/itex]
wofsy
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#14
Feb12-09, 07:34 AM
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Quote Quote by mordechai9 View Post
You know I'm a very introductory student here, so I would tend to accept your argument, except that I spoke to the professor about this earlier today and I thought that he was actually agreeing with what I said -- that dimT_pM = dimM - 1.

You say that the tangent plane T_pM contains all the directions of curves passing thru p, with the derivative of the curve being tangent to the mfld. (I agree). We assume the dimension of the embedding space (and manifold) is R^n. Let me try a proof by contradiction. Assume the dimension of the tangent space is n. Then the dimension of the embedding space equals the dimension of the tangent space. Thus any vector in the embedding space can be written in terms of basis vectors of the tangent space. Take a vector from the embedding space, orthogonal to the surface at point p. Then we can rewrite this vector in terms of the tangent space basis, so that we must conclude that this vector also belongs to the tangent space. Contradiction, because we know the tangent space contains no orthogonal vectors.
--> So dimT_pM cannot equal dimM.



?
A good way to see tangent spaces is to look at paramterizations of surfaces.
For instance a paramterized hemisphere sphere has the equation

(x,y) -> (sinxcosy,sinxsiny,cosx)

the RHS is in three dimenational space but is in the image of a two dimensional disk. Intuitively this disk is warped into the shape of a hemisphere and is obviously 2 dimensional. But also it is clear that the the possible directions on the hemisphere come from directions on the disk. At any point these directions form the tangent space.

The tangent vectors can be explicitly written down using the Jacobian of the parameterization. That is a good excercise.

Generally, manifolds are thought of as spaces that can be locally parametrized with disks in Euclidean space. These parameterizations provide directions in which functions can be differentiated. These directions are your tangent space.

The trick in manifolds is how to compare two different parametrizations of the same region. But this is not so bad. It is the same as a change of coordinates just like using say regular and polar coordinates in Euclidean space.

wofsy


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