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Tangent vector space question

by rtharbaugh1
Tags: space, tangent, vector
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rtharbaugh1
#1
Sep13-06, 10:48 AM
P: 310
I see in my notes (I don't carry The Encyclopedia Britannica around with me) that George Mostow, in his artical on analytic topology, says "The set of all tangent vectors at m of a k-dimensional manifold constitutes a linear or vector space of which k is the dimension (k real)." Well ok, maybe it is more a paraphrase than a quote.

Shouldn't the dimension of the tangent vector space be k-1? I am imagining the tangent vector space at a point on a three-sphere as a 2-D disk originating at the point, rather as if I had tacked a CD onto my globe of the Earth.

Then on the real Earth, I am at a point, and my tangent space would be the space between me and the horizon? Say I am at sea far from any coast. Should I rather think of the tangent space as the 2d surface of the ocean, or as the 3d space in which the ocean waves occur?

Thanks,

R
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matt grime
#2
Sep13-06, 12:57 PM
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The surface of the earth is 2-d (locally). That it lives ina 3d space is neither here nor there.

I don't know about anyone else, but my definition of a k-dimensional manifold is that locally (i.e. the tangent space) a k-dimensional vector space. So of course it should not be k-1. Unless you think that the surface of the earth is 1 dimensional.
daveb
#3
Sep13-06, 01:03 PM
P: 926
a point on a 3-sphere can be thought of as a point on the unit 3-d sphere or the 2-d unit shell. The tangent space to the unit shell is the 2-d plane that is tangent at that point. But for the 3-d sphere, the tangent space is 3 dimensional.

HallsofIvy
#4
Sep13-06, 05:54 PM
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Tangent vector space question

Quote Quote by rtharbaugh1
I see in my notes (I don't carry The Encyclopedia Britannica around with me) that George Mostow, in his artical on analytic topology, says "The set of all tangent vectors at m of a k-dimensional manifold constitutes a linear or vector space of which k is the dimension (k real)." Well ok, maybe it is more a paraphrase than a quote.

Shouldn't the dimension of the tangent vector space be k-1? I am imagining the tangent vector space at a point on a three-sphere as a 2-D disk originating at the point, rather as if I had tacked a CD onto my globe of the Earth.

Then on the real Earth, I am at a point, and my tangent space would be the space between me and the horizon? Say I am at sea far from any coast. Should I rather think of the tangent space as the 2d surface of the ocean, or as the 3d space in which the ocean waves occur?

Thanks,

R
As others have pointed out, the 2-D disk you are seeing as the tangent space is the tangent space to the 2-dimensional surface of the sphere, not to the 3- dimensional sphere itself.
jbusc
#5
Sep14-06, 12:24 PM
P: 212
All a k-dimensional manifold is, is a space which locally "looks" like euclidean k-space. So in sufficiently "small" regions you would expect vectors to behave like they would in euclidean k-space, meaning the vectors "live" in a k-dimensional space. When you consider the whole manifold again, those k-dimensional spaces appear as the tangent spaces since they change as you move along the manifold.
mathwonk
#6
Sep14-06, 06:12 PM
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a k manifold is something locally homeomorphic (or diffeomorphic) to R^k, while a k vector space is something linearly isomorphic to R^k.

the tgangent space is the linear space that best approxiamtes the manifold. It makes sense it should have the same dimension.

a sphere in R^3 is locally diffeomorphic to the plane, via stereographic projection, hence a sphere is 2 dimensional.
Data
#7
Sep14-06, 08:39 PM
P: 998
Quote Quote by HallsofIvy
As others have pointed out, the 2-D disk you are seeing as the tangent space is the tangent space to the 2-dimensional surface of the sphere, not to the 3- dimensional sphere itself.
And to speak precisely, sphere always means just the surface. If you want the volume contained in it, that's a ball!
rtharbaugh1
#8
Sep15-06, 10:34 AM
P: 310
Thanks to all. I think I get it now. R.


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