
#1
Feb709, 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"... 



#2
Feb809, 11:04 AM

P: 391

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. 



#3
Feb809, 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. 



#4
Feb909, 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 standalone 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? 



#5
Feb909, 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 



#6
Feb909, 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. ? 



#7
Feb1009, 03:44 AM

P: 30

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. 



#8
Feb1009, 06:40 AM

P: 491





#9
Feb1009, 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. 



#10
Feb1009, 03:49 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

P.S. why do you care about the "domain"? 



#11
Feb1109, 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 dotproduct 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 welldefined , in the sense that it would depend on the dimension of the space in which M is embedded. Re the issue of embedding vs. standalone: I don't know if you have seen any pointset 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 settingup 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 nonEuc. 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 nmfld. is not, for the most part, globally homeomorphic with R^n.). But a manifold is something of a secondbest: it is locallyEuclidean, 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 longenough 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 nball, 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. 



#12
Feb1109, 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. 



#13
Feb1109, 03:39 PM

Emeritus
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PF Gold
P: 16,101

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] 



#14
Feb1209, 07:34 AM

P: 707

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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