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Topic: How do we Evaluate This Form on S^1?
Replies: 11   Last Post: Jan 22, 2013 4:17 PM

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achille

Posts: 575
Registered: 2/10/09
Re: How do we Evaluate This Form on S^1?
Posted: Jan 22, 2013 8:04 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Tuesday, January 22, 2013 1:32:19 PM UTC+8, hbe...@gmail.com wrote:
> On Monday, January 21, 2013 8:18:16 PM UTC-8, bacle...@gmail.com wrote:
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> > On Monday, January 21, 2013 8:05:53 PM UTC-8, bacle...@gmail.com wrote:
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> > > On Monday, January 21, 2013 11:59:49 AM UTC-8, W. Dale Hall wrote:
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> > > > Shmuel (Seymour J.) Metz wrote:
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> > > > > In <x5qdnfD8tJNRFmHNnZ2dnUVZ5tidnZ2d@giganews.com>, on 01/20/2013
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> > > > > at 04:28 PM, "W. Dale Hall" <wdhall@alum.mit.edu> said:
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> > > > >> A tangent vector isn't *really* a real number
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> > > > > That much is true.
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> > > > >> [after all, tangent vectors to different points on S^1 aren't
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> > > > >> quite identifiable with one another since they belong to different
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> > > > >> tangent spaces],
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> > > > > That's a non sequitur and, in fact, is false. The standard affine
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> > > > > connection on S^1 is flat, so parallel transport establishes an
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> > > > > isomorphism between T_x and T_y, for x and y is S^1.
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> > > > Which statement is the "in fact, false" non-sequitur? My remark that
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> > > > spaces? That one can't "quite" identify tangents at one point to
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> > > > tangents at another?
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> > > > I maintain that my first statement is perfectly correct. Otherwise, why
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> > > > would one ever worry about the whole machinery of vector bundles? Even
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> > > > a trivial bundle has distinct fibres over distinct points, and I regard
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> > > > it as a pedagogical mistake to use the (what used to be called) "abuse
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> > > > of language" to ignore the distinction between tangent vectors at one
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> > > > point and tangent vectors at another point. A beginner needs to learn
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> > > > that the tangent bundle is something other than a vector space; the
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> > > > OP's question showed at least that level of confusion, and to dispel
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> > > > As for my second remark, please note my use of the weasel-word "quite".
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> > > > (from the original article):
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> > > > The tangent space to S^1 is R^1, so that a tangent vector is
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> > > > of the form c ; any non-zero real number can be a basis.
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> > > My mistake, sorry; I meant to say that the tangent space to any point p in S^1
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> > > is 1-dimensional, so that this tangent space is ( up to iso.) a copy of the reals. I'm aware of the bundle as the disjoint union of tangent spaces at all points, and of forms as sections of the bundle (1-forms as sections of the bundle; k forms as sections of powers of the bundle) , but I don't know how to use this fact to evaluate d\theta. I imagine that d\theta being a 1-form, there aren't really that many choices for its definition. I would appreciate it if you could explain how to evaluate d\theta from this perspective of sections and trivializations.
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> > > You have been very helpful already; I hope I'm not asking too much from you, please let me know if this is so.
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> > > Unfortunately, I know some disjointed bits, but I have trouble putting them together. I know the bundle TS^1 is trivial, since, e.g., (-sin(theta),cos(theta)) is a nowhere-zero tangent vector field. I just don't seem to put all this together. I would appreciate your suggestions for help in this respect.
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> > > > This statement betrays a fundamental misunderstanding. Note that there
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> > Sorry, hbert, and all, I confused this with one of my posts.No problem, bacle, you made some good points.
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> Still, Mr. Hall, I do know about the fact that the tangent spaces at points
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> I don't know about the triviality of the bundle that you and BacleH talk
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All the confusion will be cleared once you figure out what a tangent
vector really is.

A tangent vector at a point p on a curve (or any m-dim manifold M)
is an equivalent class of curves passing through that point p, tangent
to each other and moving at same speed. To make this precise, one need
to use a local coordinate system \phi over some neighbour of p.

Given two curves \gamma_1, \gamma_2 where \gamma_i is defined on
(t_i - \epsilon_i, t_i + \epsilon_i ) and \gamma_i(t_i) = p, they
are equivalent iff

d/dt( \phi(\gamma_1(t)) )|_{t_1} = d/dt( \phi(\gamma_2(t)) )|_{t_2}

On the surface, this equivalence depends on the choice of \phi. But a
manifold is essentially a topological space equipped with a collection
of compatible coordinate charts. Once two curves is equivalent under
one local coordinate chart around p, they are equivalent under other
coordinate charts containing p.

Associated with each equivalent class of curves X = { \gamma_i }, one
can define a linear map (in fact a derivative) upon C^1 real value
functions f defined over some neighbour of p:

D_{X} : f |-> d/dt ( f(\gamma_i(t) ) |_{t_i} where \gamma_i \in X.

It is easy to check:

1) D_{X} is well defined. ie. independent of choice of \gamma_i.
2) the correspondence X (equivalent class of curves) <-> D_{X} (derivative)
is a bijection.

As a result, a tangent vector X at p can also be identified as a derivative
D_{X} upon C^1 real valued functions defined around p.

Given this identification, any C^1 real values functions g around p naturally
induced a linear map on the set of tangent vectors X at p.

dg : X |-> D_{X}(g) = d/dt( g(\gamma_i(t)) )|_{t_i}

Back to your original question. What (d theta)[c] is depends on what
your c is. If your tangent vector c is the one corresponds to the
parametrization (a is constant):

\gamma : t -> (cos(at), sin(at)),

then d theta[c] = d/dt( theta(\gamma(t)) ) = d/dt(at) = a.

Finally, one more point to clear up.

If one has embedded a curve (or manifold M) in a bigger space R^N.
a curve \gamma in M is a curve in N. So any tangent vector X at p in M
corresponds to a unique tangent vector at p in R^N. Since T_p(R^N) is
naturally isomorphic to R^N. Only under such circumstances, one can
treat tangent vector X at p in M as some vector in R^N. Again, for the
parametrization above, one has:

c <-> d/dt( cos(at), sin(at) ) = ( -a sin(at), a cos(at) ).




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