Talk:Tangent space/Archive 1

Latest comment: 16 years ago by Lambiam in topic Is this definition right?
Archive 1

Meaning of derivative/differential

The linear map (df)p is to be interpreted as the derivative of f at p.

dfp is usually referred to as the differential of f at p, hence the notation. Isn't that a different interpretation than the derivative?

The derivative at a point in calculus is just a number. In multivariable calculus, the derivative at a point is not a number but a matrix or linear map. When dealing with manifolds, it's actually a linear map between the respective tangent spaces. I don't think there's any other notion of derivative at a point than (df)p. If the map f is differentiable at every point of M, then its derivative is a map df : TM → TN between the tangent bundles which comes from glueing together all the maps (df)p. AxelBoldt 17:56, 4 December 2002 (UTC)

Ok, let me be more particular. I was under the impression that the space of maps of said form was considered to be the dual of the space of derivatives, so that a derivative would be a map back the other way between the cotangent spaces. I forget the reason for this, but it had something to do with there being a natural identification with the standard notion of derivatives in the case that N is the real numbers. I will double check when I get the time.

I think you're right. I replaced derivative with differential. AxelBoldt 22:48, 4 December 2002 (UTC)

p or x?

Do we prefer p or x for a point in M? I don't mind, but it would be nice to be consistent between this article and the tangent bundle article. Geometry guy 23:28, 10 February 2007 (UTC)

In most of the other related articles (admittedly, partly because of my edits), x is used, so I have changed p to x here. Geometry guy 23:32, 11 February 2007 (UTC)

Jacobian?

As I understand it, if I have a function, f(x) defined on a manifold, the gradient of f at a point, x is in the tangent space of x. But what about the Jacobian? Would one say it's in the tangent space of x? (I assume not because it's a matrix not a vector.) What space is the Jacobian in? —Ben FrantzDale 21:57, 4 May 2007 (UTC)

as you say, the Jacobian at a given point is a linear map between tangent spaces. see Pushforward (differential). Mct mht 02:11, 5 May 2007 (UTC)
That isn't quite what I meant. I was curious what space you'd say it is in. That is, if grad f is in the tangent space (and so the field grad f is a section of the tangent bundle), then the Jacobian should be in a somehow related space. I suppose it should be in the square of the tangent space, that is  . However that space is constructed, the connection for the tangent space should define a connection for this new space of rank-two symmetric tensors. The word I'm looking for may be associated bundle. —Ben FrantzDale 12:50, 6 May 2007 (UTC)
This may become clear if you are more specific about the nature of f. If f is a map from M to the real numbers R, then df is a map from TM to TR, namely the push forward. It is also a form on M (as opposed to a vector) which takes in a section v of TM by df(v(p))=v(f)(p) (here v is a directional derivative.) In turn (f)(p) is a real number which may be thought of as an element of the (one dimensional) tangent space to f(p) in R.
If f is a map from M to another manifold N then it is nice to work with local coordinates on N; you can write the k-th component of f (giving the k-th coordinate of the image of a point) as a function from M to R and consider its differential. The collection of all these forms (one for each value of k) is is a representation of df specified by the choice of local coordinates on N. If local coordinates are used on M as well then df has a (coordinate system dependent) matrix representation (acquired using the coordinate bases on both M and N) called the Jacobian of f.
I hope this clears a few things up, but I'm not sure I know the answer to your question. I feel similarly to you about the possibility of considering it an element of a bundle associated to T*M, but I'm not sure how to reconcile this with the dependance on local coordinates on N. While I'm curious, I don't think that the addition of a discussion of such an associated bundle would add much to this page, or the related pages.Dewa (talk) —Preceding comment was added at 22:37, 31 May 2008 (UTC)

discussion at Wikipedia talk:WikiProject Mathematics/related articles

This article is part of a series of closely related articles for which I would like to clarify the interrelations. Please contribute your ideas at Wikipedia talk:WikiProject Mathematics/related articles. --MarSch 14:08, 12 June 2005 (UTC)

Is this definition right?

The subsection "Definition as directions of curves" defines a map (dφ)x : TxMRn by

(dφ)x(γ'(0)) = (φ o γ)'(0).

Now, given the given type of (dφ)x, I'd expect its argument to be an element of TxM, which is is defined as the set of all tangent vectors, where a tangent vector is an equivalence class of curves. So the argument should be a class of curves [γ] and I'd be willing to accept the following definition:

(dφ)x([γ]) = (φ o γ)'(0),

whose required independence of the choice of representative γ follows immediately from the equivalence relation. I'm not sufficiently familiar with the topic to be fully confident that this is the correct improvement and apply the change myself.  --Lambiam 07:47, 4 September 2007 (UTC)

The article says a couple of lines before: "The equivalence class of the curve γ is written as γ'(0)". So γ'(0) is just a different notation (and the usual one) for [γ]. -- Jitse Niesen (talk) 07:57, 4 September 2007 (UTC)
Thanks, but what a weird convention, seeing how γ'(0) in the l.h.s. has a completely other meaning then than (φ o γ)'(0) in the r.h.s.  --Lambiam 09:26, 4 September 2007 (UTC)
I see what you mean: if the prime means what Nissen says then on the LHS the prime denotes an equivalence class of curves on M with equivalence determined by behavior at x while on the RHS it denotes an equivalence class of curves on Rn with equivalence (implicitly) determined by behavior at φ(x). But this does not give the desired construction; it does not map an equivalence class of curves to an n-touple. So Lambiam believes that the prime on the RHS is meant to be a derivative WRT t, which would mean this article made double notational use of the prime. I will change it.Dewa (talk) 23:01, 31 May 2008 (UTC)
I'm not sure the change is correct, as (φ o γ)(0) is not a function of a variable t. The following would be correct (I think):
(dφ)x(γ'(0)) = D(0), where D(t) = (d/dt)(φ o γ)(t).
I don't see how to say that in an essentially simpler way without overloading the prime in one formula.  --Lambiam 07:33, 1 June 2008 (UTC)