Confusion regarding differential forms and tangent space (Spivak,Calc. on Manifolds)


by madshiver
Tags: differential forms, spivak, tangent space, vector fields
madshiver
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#1
Nov27-12, 10:05 AM
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I have been working through Spivak's fine book, but the part about differential forms and tangent spaces has left me confused.

In particular, Spivak defines the Tangent Space [itex]\mathbb R^n_p[/itex] of [itex]\mathbb R^n[/itex] at the point p as the set of tuples [itex](p,x),x\in\mathbb R^n[/itex]. Afterwards, Vector fields are defined as functions F on [itex]\mathbb R^n[/itex], such that [itex]F(p) \in \mathbb R^n_p \ \forall p \in \mathbb R^n[/itex].

My analysis professor had defined a vector field to simply be a function [itex] f: \mathbb R^n \to \mathbb R^n [/itex]. Now it appears to me that the definition according to Spivak is way more elegant in the sense that it maches the geometric intuition behind a vector field. But at the same time, as I see it, neither definition includes more "information" than the other.

And then the actually confusion comes around: A differential k-form [itex]\omega[/itex] is defined to be a function with [itex] \omega (p) \in \bigwedge^k(\mathbb R^n_p) [/itex].([itex]\bigwedge^k(\mathbb R^n_p) [/itex] in Spivak's book corresponds to [itex]\bigwedge^k(\mathbb R^n_p)^*[/itex] in other books). Now the question is the following: Would the definition [itex] \omega (p) \in \bigwedge^k(\mathbb R^n) [/itex] not suffice?

For example, I find the Notation [itex] dx^i(p)(v_p) = v^i , v_p=(p,v) [/itex] confusing. Why not just define it as [itex] dx^i(p)(v) = v^i [/itex] (so that in fact [itex] dx^i [/itex]would not depend on p and be a constant function)?

What am I missing here?
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lavinia
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#2
Nov27-12, 10:27 AM
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On any given tangent space, a differential form is a multi-linear function. But this multi-linear function is not constant. It varies from point to point. That is, it is a different multilinear map on different tangent planes. So for instance dx[itex]^{i}[/itex] is not a constant function but is a collection of linear maps parameterized by R[itex]^{n}[/itex].

Usually though the point at which the form is evaluated is understood and is not explicitly n=mentioned unless necessary for clarity.
madshiver
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#3
Nov27-12, 10:33 AM
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Thank you for your reply. I realize that on every point it is a different function. The condition [itex] \omega (p) \in \bigwedge^k(\mathbb R^n) [/itex] would express this fact, except that the new function acts on the actual vector space and not on the tangent space. But if my analogy to that of vector fields is correct, then these two formulations should be equivalent.

Or am I missing something again?

jgens
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Nov27-12, 10:59 AM
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Confusion regarding differential forms and tangent space (Spivak,Calc. on Manifolds)


Quote Quote by madshiver View Post
Or am I missing something again?
You are missing the big picture. Spivak sets things up in this way because it parallels how you will construct these objects on manifolds.
lavinia
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Nov27-12, 11:03 AM
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Quote Quote by madshiver View Post
Thank you for your reply. I realize that on every point it is a different function. The condition [itex] \omega (p) \in \bigwedge^k(\mathbb R^n) [/itex] would express this fact, except that the new function acts on the actual vector space and not on the tangent space. But if my analogy to that of vector fields is correct, then these two formulations should be equivalent.

Or am I missing something again?
Yes they are equivalent.

BTW: on a surface you can define the tangent space at each point to be the plane tangent to the surface at that point. These planes vary from point to point but one can still have a differential form defined on the the surface that is a multilinear function on each tangent plane.
There may not be a natural or meaningful way to compare these functions since the planes they act on are different.
madshiver
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Nov27-12, 04:23 PM
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Ok, this makes sense. I guess the description by Lavinia corresponds to the construction mentioned by jgens.

Thank you again ;).
mathwonk
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Nov28-12, 02:17 PM
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this concerns a point of confusion in vector calculus. namely, consider a tangent vector field to the circle. do you picture a family of tangent vectors, each one with its foot ata point of the circle and its direction eoprpendicular to the radius there? or do you picture a family of vectiors, all with their feet at the origin and each with its direction perpendicular to the corresponding radius?


the first version is geometrically correct and the second is not. but the first is more complicated, so preferring simplicity over correctness, essentially all books and professors except spivak do it wrong and simpler, i.e. the second way.

of course the two are equivalent since euclidean space has "trivial" tangent bundle, i.e. there is natural way to slide any vector with foot anywhere, over to have its foot at the origin. but the right way to do it would be spivak's way first, to get it right, and then point out how to simpklify it. almost no one does this.

this means e.g. that a vector in the plane is not given by a pair of numbers, rather it is given by a pair of points, i.e. 4 numbers. but then you can get the parallel vector with foot at the origin by subtracting the first and last numbers from each other.

that gives the coordinates of the vector, but it only represents the length and direction of the vector, not the location of its foot.

\this is extremely confusing in teaching calculus. even some professors have gotten it wrong in my experience. i.e. if you want to know how to approximate where a moving point will be in one unit of time, whose position now is p, and whose velocity vector is v, the answer should be p+v, a point.

moreover as pointed out above, this is not true on most manifolds, which are not "parallelizable".

However, one can still use this simplifying incorrect version, if one has an embedded manifold in euclidean space. e.g. the tangent bundle of the sphere is not parallelizable, but in its usual embedding in three space one can take advantage of the fact that the tangent bundle of three space is trivial.


this causes a problem in complex manifolds however which if compact are never embeddable in complex euclidean space. one case where the tangent bundle is trivial is for a group manifold, like an abelian variety, or any compact complex torus. Here the group operation again allows us to translate tangent vectors to the origin.
madshiver
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#8
Dec12-12, 09:11 AM
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My professor ended up defining the tangent space as an affine vector subspace, which actually kind of matches the geometric intuition and in a sense allows one to calculate the length, direction and "foot" of the vector.

Then again, we did not really prove anything with the tangent space, just some intuitive explanations about orientation (which we rigorously defined using the jacobian of the transition functions).

So, I guess Spivak's approach is still better ;). Spivak is Spivak after all..

I'd also like to use this chance to thank you Mathwonk for all the high quality content you have been posting for so many years. You are one of the main reasons I've been lurking around these forums for several years now; in fact during the period you stopped posting, I also stopped visiting PF.
mathwonk
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Dec12-12, 09:19 PM
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wow! thank you so much. that is very thoughtful indeed. fortunately, i was happy to notice as well that since i began, we have been blessed with many new young people (lavinia, micromass,....) whose answers are really fine, and have more than taken up any slack from the absence of us oldsters.


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