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Tangent vectors to a manifold

by Goldbeetle
Tags: manifold, tangent, vectors
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fisico30
#19
Sep24-08, 01:03 PM
P: 374
Thanks Compuchip/

Would then a scalar field be a chart? Isn't a chart a map, function that assigns to a point on the manifold (whatever that point means, a spatial point, a configuration point, etc....).....

A manifold is coverered by compatible charts. If we talk about traditional space, the chart give a coordinate to the point. If we talk about a scalar field, like temperature, the scalar field gives a number to the point on the manifold which represents, in an abstract sense, the concept of temperature.... right or wrong?
Can u elaborate a bit on the idea of tangent to the sphere vs on the sphere?
That seems subtle....

thanks again
Marco_84
#20
Sep25-08, 07:05 AM
P: 173
Quote Quote by fisico30 View Post
Hello Marco 84,

I have a question for you.
Take the electric field at a point in space. WE can define a vector field as a function that extracts out of a vector space associated with a point in a manifold one vector, the vector that is there.

I am not sure in what sense a vector space is associated to a point in the manifold....
What does that mean?
I guess my question is: can you explain in simple terms how a vector field is defined on a manifold?
why is a vector field not a manifold?

thanks a lot
OK...

imagine that over a sphere, at each point you put an orthonormal basis (inthis exemple 2 versors) ther you ahave a vector space( sometimes called linear space).

nOw you have taht for every point x in M you have a mapping phi(x): M-----> R^2

those are the charts!!!!

a scalar/vector, or more generally a tensor field is a mapping from the manifold, wich in applied physics is usaually the space of configuration of your system (how can it move), the degree of freedom.

In other words if we take a sphere in a what ever you want electric/gravitational field... we can define charts on that spehere to have a Manifold... the we can define a vector fiield (the electric field) only on the surface of the sphere..... and with the help of the charts we can easly solve the problems in R^2....

i hope this exemple helped to you.....

obviously... the spher is not the case....

it is better to change coordinate at the begginning of the problem cartesina---_>spherical....


But if you think how different and difficullt are the geometries in practical.... having this tools it is easier..... you skip always yo cartesian coordinates......

i think that's normal... our minds is cartesian!!! :)

regards

marco
fisico30
#21
Sep25-08, 08:24 AM
P: 374
Thanks Marco!
by the way my name is Marco too. Are you from Italy?
thanks again for the infos. I need to be refreshed on these matters.
I will respond soon with a summary of what I have learned, maybe you can make sure I truly get it!

marco
mathwonk
#22
Sep25-08, 03:33 PM
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you study a manifold M in two ways, 1) by looking at functions from R into the manifold,
2) by looking at functions from the manifold into R.

these two dual points of view lead, with suitable equivalence relations, to the concepts of vector and covector.

if you take functions from M into R as fundamental, then a vector is something dual to these functions, i.e. a differential operator.

if you think functions from R into M are fundamental, then you think of vectors as certain equivalence classes of such curves.
Marco_84
#23
Sep26-08, 01:52 AM
P: 173
Quote Quote by fisico30 View Post
Thanks Marco!
by the way my name is Marco too. Are you from Italy?
thanks again for the infos. I need to be refreshed on these matters.
I will respond soon with a summary of what I have learned, maybe you can make sure I truly get it!

marco
Yep, I'm italian...from Milano.... i'm sorry if im not so clear with english!!!

I'm taking my Master thesis in theoretical physics, But i took a camputational branch. Meanwhile i'm working in a company as an analyst. I do optical simulations for automotive lighting....

Not so interesting as particle physics, but i keep playing with physical laws.

let's ear each others.

marco
Marco_84
#24
Sep26-08, 01:57 AM
P: 173
Quote Quote by mathwonk View Post
you study a manifold M in two ways, 1) by looking at functions from R into the manifold,
2) by looking at functions from the manifold into R.

these two dual points of view lead, with suitable equivalence relations, to the concepts of vector and covector.

if you take functions from M into R as fundamental, then a vector is something dual to these functions, i.e. a differential operator.

if you think functions from R into M are fundamental, then you think of vectors as certain equivalence classes of such curves.
too mathematical for waht fisico30 was asking, but i like it is completely right :)

and this gives you the meaning of what a "fibrato tangente" and a "fibrato cotangente " is...

i think i can translate in english as co/tangent fiber??

usually is T(g)M and T*(g)M..... where g are the poins of M (manifold)....

bye marco
wofsy
#25
Sep27-08, 02:01 AM
P: 707
A few things

vector fields are not only three dimensional but are defined on a manifold of any dimension. For instance in space time they assign a four vector to each point.

not all vector fields need to be tangent to the manifold. For instance when measuring the flux of a vector field one looks at its normal component to a surface that bounds a volume.

generally speaking if one can continuously assign a vector space to each point of a manifold then a vector field is a choice of a vector from each of these vector spaces at each point. Not all of these vector space choices derive from tangents spaces or normal spaces and do not need to be the same dimension as the manifold. for instance, one can have 10 dimensional vector fields on a surface.

also, a vector field is a manifold if it is chosen smoothly or continuously. This manifold lives inside the manifold of vector spaces at each point.
FredericGos
#26
Oct10-08, 03:58 AM
P: 61
Quote Quote by HallsofIvy View Post
Tangent vectors are derivatives. When I was young and foolish (I'm not young anymore), I spent a lot of time worrying about vectors on a curved surface such as a sphere. Did the vectors "curve" to stay on the surface, or did they go through the sphere from endpoint to tip? Neither, of course, vectors lie in the "tangent plane" to the sphere.
I'm still young and foolish and I recently conviced myself that I have to clear myself of this the 'tangent plane' mind image. To me, this implies that the manifold is embeded in a higher dimensional space for this plane to make sense. What if it's not? What would be the tangent vectors of the space time manifold? I'm trying to get an image i my head then, and was thinking that maybe the tangent vector just follow the curvature... Guess that's not the right way. What's my problem? ;)

/Frederic
CompuChip
#27
Oct11-08, 03:58 AM
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Hi Frederic. Your problem is probably that it is impossible for you to imagine any manifold which is not embedded in some euclidean space. Probably you are so used to the three dimensions we live in that you might even not be aware that you are doing this. That's not your shortcoming by the way, I don't think anyone can imagine a two-sphere and it's tangent space at some point, for example, without actually putting it in three-dimensional space. I also don't really think that there is anything wrong with the mental image of 'tangent planes', as most concepts in differential geometry are founded on such intuitive ideas. Whenever you see something new, you can try to put it in the context of something you can imagine to understand whether the concept makes sense. Luckily there is some theorem, that says that many manifolds can indeed be embedded in some Euclidean space, though one could argue that is not much use (does it help you to know that this specific manifold you are staring at can be drawn in a 20-dimensional space?)

If you really want to get rid of the geometric image, you might want to consider the other definitions of tangent vector (for example, as a directional derivative of functions on the manifold).
wofsy
#28
Oct13-08, 02:30 PM
P: 707
a tangent plane to a surface is exactly those vectors that start on a point of the surface and which point along a direction in the surface i.e. they are the derivatives of curves that lie entirely on the surface.

One does not need a picture to know what a directional derivative is. Just write down a curve and differentiate along it. This idea works in any dimension and needs no idea of shape or geometry! One just writes down the curve in some coordinate system and differentiates. That's it.
wofsy
#29
Oct13-08, 05:39 PM
P: 707
One does not need a higher dimensional space in order to define a tangent plane. One only needs that idea of directional derivative.

On the other hand it is true that any manifold can be realized inside a higher dimensional Euclidean space (the Whitney Embedding Theorem) and the tangent planes then are the vectors that kiss the manifold.

the space of tangent planes on a manifold is also a manifold. One can think of the original manifold as a surface within it and then the tangent plane at a point is the set of points in the manifold that are attached to the point. This manifold of tangent planes is called the tangent bundle.

There is a key technicality though that makes this picture more complicated. If one has overlapping coordinates, coordinates that overlap smoothly (differentiably), then the transformation law for directional derivatives should be the Chain Rule. The tangent bundle may be defined using the Chain Rule to compare vectors in different coordinates. However there are manifolds which can not be covered with coordinate neighborhoods that all overlap smoothly. In these manifolds there really is no tangent bundle and the idea of tangent spaces doesn't apply

More strange, on some manifolds there may be more than one way to cover it with coordinate charts that leads to more than one idea of tangent bundle. I do not understand exactly how the geometric realizations would differ.
jdstokes
#30
Oct15-08, 02:08 AM
P: 527
Personally, I find it extremely satisfying that all manifolds can be embedded in a flat (Euclidean) space of suitably high dimension. This guarantees that the abstract definition in terms of smooth derivations corroborates with your intuition of tangent vectors represented by lines which kiss the surface.

This is a situation which parallels other brances of mathematics (e.g. group theory):
1. Start by axiomatizing a concrete object (regular 2D surface).
2. Inspired by pedagogical examples which meet the above criteria, generalise the definition.
3. Find that all abstract examples have some concrete realization.
wofsy
#31
Oct17-08, 08:20 AM
P: 707
Hi JD

One can look at your idea the other way around as well. Euclidean geometry is an abstraction of measurement on the Earth's surface, a curved manifold with irregular geometry, but which is approximately flat in small regions. From this point of view the curved manifold is the natural and concrete and the flat space is the axiomatically described abstraction.

The same is true of the Universe. It is a curved 4 dimensional surface that is approximately, in regions of small gravitational fields and velocities, a three dimensional Euclidean space that moves through absolute time.

regards

wofsy
quasar987
#32
Oct17-08, 11:30 AM
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Talking about the embedding of Whitney's theorem, it is written in Gallot, Lafontaine, Hullin, "But such an embedding is not canonical: the study of abstract manifolds cannot be reduced to the study of submanifolds of numerical space!"

What did they mean by that? What would it mean for two manifolds to be "naturally diffeomorphic" anyway?
wofsy
#33
Oct17-08, 12:12 PM
P: 707
hi Quasar

The Whitney Embedding Theorem does not tell you how to construct an embedding of a manifold in Euclidean space but only that an embedding exists. The arguements involve approximation theorems and can not be directly applied to any specific example. In general there is no obvious way to embed an abstract manifold. each case must be taken separately.
That what it means to be non-canonical.


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