# Tangent vectors to a manifold

1. Feb 29, 2008

### Goldbeetle

Dear all,
I can formally understand one of the many definitions of tangent vectors to a manifold, but what are they in reality? It should depend on the nature of the points of the manifold, for example, if M={set of events of general relativity}, then vectors are velocities. Other examples??

Kindest regards.
Goldbeetle

2. Mar 1, 2008

### Marco_84

All the CM problems involve manifold---> Configuration space (Lagrangian formalism)
Phase space(hamiltonian formalism)

regards
marco

3. Mar 1, 2008

### CompuChip

I think in general one could say that tangent vectors are the "velocity" vectors of curves on the manifold and the question is rather: what do curves on the manifold (physically) mean and is there a meaning to their "velocity vector".

For example, if the manifold is some Euclidean space, we can always view the curves as trajectories of a particle moving in that space, and then the tangent vectors are really velocity vectors in the intuitive sense. But you might define any manifold (for example, outcomes of an experiment, or a space of functions) and then the first question to ask is: what does a curve on such a manifold actually represent?

4. Mar 1, 2008

### HallsofIvy

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.

5. Mar 1, 2008

### Marco_84

completely right... the best wuld be to know perfectly all the fiber bundle..... :)

6. Mar 2, 2008

### Hurkyl

Staff Emeritus
More accurately, the bundle of (certain) differential operators is a tangent bundle. There are many isomorphic 'realizations' of the tangent bundle, each giving a different insight into their use.

Spivak gives three concrete definitions of the tangent bundle:
(1) Tangent vectors at P are defined by differential operators defined near P.
(2) Tangent vectors at P are defined by differentiable curves passing through P.
(3) Tangent vectors at P are defined by specifying its coordinates relative to all bases. (i.e. the classic 'array of numbers that has the right transformation properties)

He also gives an abstract definition by showing that for a particular kind of construction, there is a unique one (up to isomorphism) that, when applied to R^n, yields the usual Euclidean tangent bundle. Then
(4) A tangent bundle is the application of such a construction to a differentiable manifold

Another definition that can be made using nonstandard analysis is to observe that you can make sense of the phrase 'infintiessimally close'. So, you choose an infinitessimal h, and then define:
(5) Tangent vectors at P are defined by points infinitessimally close to *P in the nonstandard version of your manifold. (And two points define the same tangent vector iff they are h-infinitessimally close)

Knowing that each of these are tangent bundles gives us lots of insight. Well, I find it illuminating, anyways.

7. Mar 3, 2008

### Goldbeetle

True! Do you have any other example, where the manifolds points are not "General relativity spacetime events" but whose nature is something else, then what are curves in these cases? And what tanget vectors?

Thanks! I'm aware of these isomorphisms. But these results are illuminting only in the structural sense! In other words: all finite n-dimensional vector spaces on the same field are isomorphic but when it comes to applications the nature of the vector space elements matters: derivatives, polinomials, functional spaces.
In my case, I think the problem is: if somebody gives me this abstract definition of smooth manifold and tangent vectors, once I know what the point of the manifold are, how do I concretely make concrete sense of the objects defined on the manifold? For example, the tangent vectors...A good starting point is "a curve of manifold points". But apart from Euclidean space, I would like to have some other examples...

Thanks to all!
Goldbeetle

8. Mar 3, 2008

### Hurkyl

Staff Emeritus
That is true -- but more interesting things are also at work here!

Observe that we aren't merely considering vector spaces; we're looking at a vector bundle! Each individual point of our manifold has its own vector space attached to it (called a fiber), and a topology that 'sews' the fibers together. (i.e. we have a notion of a 'continuous vector field') An isomorphism of vector bundles has to be:

(a) an invertible linear transformation on each fiber
(b) a homeomorphism of topological spaces
(c) the identity map on the underlying manifold

The structure of a vector bundle turns out to be a richer structure; you can have many nonisomorphic vector bundles of the same dimension.

The classic example is that of the cylinder versus the Möbius strip. These can both be viewed as one-dimensional vector bundles over the circle -- but they are obviously not homoeomorphic as spaces, and therefore cannot be isomorphic as vector bundles!

If you can't visualize them topologically, there is an algebraic description: consider angular coordinates on the circle. A continuous vector field in the cylinder is a periodic function that satisfies the identity $f(x) = f(x + 2\pi)$. However, a continuous vector field in the Möbius strip is an 'anti-periodic' function -- one satisfying the identity $f(x) = - f(x + 2 \pi)$.

(For the Möbius strip, the coordinates $(r, z)$ and $(r + 2 \pi, -z)$ refer to the same point)

You can distinguish them algebraically by the fact that every continuous vector field in the Möbius strip has a zero. (Apply the intermediate value theorem!) But that's not true for the cylinder.

So, we have shown that there are two nonisomorphic one-dimensional vector bundles on the circle.

There are other interesting things too -- for example these isomorphisms are natural in a certain technical sense. And they are specific -- the space of differential operators near P and curves through P are not merely isomorphic, but there is a specifically chosen isomorphism that says which operator is supposed to pair with which curve.

9. Mar 3, 2008

### mathwonk

to me a tangent vector is a small piece of a curve. i.e. a map from spec(k[X]/(X^2)) to your space.

10. Mar 3, 2008

### Cincinnatus

Care to elaborate mathwonk? What does that mean?

11. Mar 4, 2008

### Goldbeetle

For Hurkyl: my question was actually much simpler. I think that given a concrete manifold (where the nature of its points is known) one has to figure out what a curve on that manifold is and what a class of equivalent curves is (having the same directional derivative at a point). Unless one is really interested in studing derivative operators (which are anyway easier to work with from a mathematical point of view)!

For Hurkyl again: you state that the tangent spaces of the tangent bundle are glued together by the fact that the manifold has a topology and that continuous fields can be defined on the tangent bundle. On that respect, what I've always wondered about the definition of continous vector fields is that it is stated in terms of their coordinates with respect with the chart coordinate basis vectors...but are we sure that latter ones vary continously? Or that they vary in such a way that they make the coordinates of the fields become continuous?? (I know I'm wicked mind...)
Thanks.
Goldbeetle

Last edited: Mar 4, 2008
12. Mar 4, 2008

### Hurkyl

Staff Emeritus
A curve is a continuous function R --> M. Sometimes, you might use a subinterval of R instead of the whole thing.

This, of course, only makes sense for differentiable manifolds and differentiable curves. The most straightforward way is to invoke the "locally Euclidean" property of manifolds; choose your favorite coordinates for the manifold, and apply the calculus of R^n. (If your manifold is differentiably embedded in an ambient space, ambient coordinates also work)

The general definition of a vector bundle is akin to that of a manifold. You have the 'standard' n-dimensional vector bundles $\mathbb{R}^m \times \mathbb{R}^n$ over m-dimensional affine space $\mathbb{R}^m$. Just like you build a manifold by 'sewing' together affine spaces, you build a tangent bundle by 'sewing' together these standard vector bundles. (and here, the 'sewing' must respect the topology, the algebra, and the projection down to the base affine space)

Said differently, each point P of your manifold has a neighborhood U for which the subspace of the tangent bundle lying over U is isomorphic to $U \times \mathbb{R}^n$.

13. Apr 8, 2008

### wofsy

a tangent vector is a geometric object. Though it may be interpreted in many ways these interpretations do not tell you what it really is.

Your question is like asking, "what is the real meaning of a triangle? I suppose it depends on the physical situation you are talking about."

14. Apr 9, 2008

### pmb_phy

The position vector is the prototype of a vector. In Cartesian 3-space the position vector represents a spatial displacement from a point chosen as the origin to an point in that space. Something similar holds for the position 4-vector in spacetime. Force, momentum, acceleration, electric field, magnetic field, gravitational force, etc. are all vectors. Things that aren't vectors are things like tidal force which is correctly described by a second rank Cartesian vector.
Vectors in a curved spacetime can also be defined as maps from 1-forms, which are defined at a particular point, to real numbers. Another definition would be according to how the components of the vector transform from one system of coordinates to another. This too has a definite meaning in a curved space and whose value also changes from point to point, in general that is.

Pete

15. Apr 9, 2008

### Goldbeetle

My point is exactly that I would like to have some more interpretations of the that concept. Regarding "what is the real meaning of a triangle", let me add, that I agree, I should rephrase my question by asking for "interpretations" and not for "they are in reality". If my son asks me what a triangle is I can very quickly draw one on a plane on a sphere etc

16. Apr 9, 2008

### wofsy

For a curve in a plane the tangent space at a point is the line which just kisses the curve without crossing it. The tangent vectors are all vectors in this line which begin at the kiss point.

If the curve is allowed to wander in 3 space then there are many lines which kiss it at a point but only one line which is tangent. The tangent line is the one which mosts parallels the curve near that point.

The exact same idea works for surfaces and higher dimensional manifolds.

It turns out that tangent space can be described using calculus as the set of all possible velocity vectors of curves on the surface that pass through the point.
Because of this, all of the Physics interpretations of vectors apply to tangent vectors,e.g.velocities, forces, gradients etc.

The advantage of the calculus definition is that it allows tangent spaces to be defined intrinsically through coordinate charts on the manifold rather than referring to vectors in an ambient Euclidean space. In Physics this method a crucial because we can not stand outside of the Universe in some high dimension Euclidean space an look down upon it. We need a way to find its geometry without examining its shape inside another space.

Historically, one of the first reasons for studying tangent spaces was the study of constrained motion. This is motion where a particle is allowed to move freely except that it is constrained to move on a surface. Other than this constraint no force acts upon the particle so it is like a type of constrained intertial motion. Imagine a magnetic ball bearing rolling on an iron sphere. It was realized that the constraint forces the motion to be "along the surface" in some sense and this means that its velocity vector lies tangent to the surface.

Last edited: Apr 9, 2008
17. Sep 23, 2008

### fisico30

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

18. Sep 24, 2008

### CompuChip

You can look at it this way: a scalar field is a field that assigns to every point of the manifold (e.g.: spacetime), a number. For example, the function $\vec x \mapsto T(x)$ that assigns to every point the temperature at that point is a scalar field. A vector field assigns three numbers to each point, i.e. a vector. For example, the function that assigns to every point the (local) wind direction is an example.

Note that the vectors of a vector field are not on the manifold. For example, if the manifold is a sphere, then you should visualise the vectors as being tangent to the sphere (cf this image) rather than going along the surface of the sphere.

19. Sep 24, 2008

### fisico30

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

20. Sep 25, 2008

### Marco_84

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