Tangent space as best approximation

Goldbeetle
Messages
210
Reaction score
1
Dear all,
in what sense the tangent space is the best approximation of a manifold?
The idea is clear to me when we think about a surface in Rn and its tangent plane at a point.
But what does this mean when we are referring to very general manifolds?
In what sense "approximation" and in what sense "best"?

Thanks.
Goldbeetle
 
Physics news on Phys.org
Hum. Well, any manifold can be embedded in R^N for N large enough, so the case where M is in R^n is the most general case in a sense.

But yeah, in the abstract setting I don't think it makes sense in any way. But maybe I'm wrong.
 
I believe you use the definition of tangent space in terms of derivations when the manifold is stand-alone.
 
The idea of best linear approximation requires the manifold to be embedded in another manifold.

The abstract tangent space will be mapped to the best linear approximation under any embedding into Euclidean space. So the tangent space may be thought of as an abstract space whose geometric realization is always the best linear approximation.
 
lavinia said:
The idea of best linear approximation requires the manifold to be embedded in another manifold.

The abstract tangent space will be mapped to the best linear approximation under any embedding into Euclidean space. So the tangent space may be thought of as an abstract space whose geometric realization is always the best linear approximation.

I think you need the manifold to be embedded in Euclidean space (which , as Quasar stated, is always possible by Whitney's embedding. ) . In order to have the local-linear approximation the tangent plane gives you, you need to have some properties that you cannot find in any manifold.

There is an equivalence between derivations and tangent vectors, in that each can be seen as being the other, i.e., every derivation can be seen as a tangent vector and viceversa.
 
Bacle2 said:
I think you need the manifold to be embedded in Euclidean space (which , as Quasar stated, is always possible by Whitney's embedding. ) . In order to have the local-linear approximation the tangent plane gives you, you need to have some properties that you cannot find in any manifold.

There is an equivalence between derivations and tangent vectors, in that each can be seen as being the other, i.e., every derivation can be seen as a tangent vector and viceversa.

I think linear approximation can be made local.
 
lavinia said:
I think linear approximation can be made local.

But, how do you define a plane in a generic ambient manifold? Don't you need the space

to have a vector space structure to talk about planes? There is a such a thing as

an abstract general plane over a field ( the set of combinations f1p+f2

y for fi in the field --this is how we work in abstract projective spaces

over a field) , but I don't see how to get the structure to have both planes and a norm ,

for , what do you mean when

you say that ||v-w||<e , where v is in the tangent plane and w is in the manifold?

n allows this because it is a normed space. How do you do it in a generic

ambient manifold X?
 
Bacle2 said:
But, how do you define a plane in a generic ambient manifold? Don't you need the space

to have a vector space structure to talk about planes? There is a such a thing as

an abstract general plane over a field ( the set of combinations f1p+f2

y for fi in the field --this is how we work in abstract projective spaces

over a field) , but I don't see how to get the structure to have both planes and a norm ,

for , what do you mean when

you say that ||v-w||<e , where v is in the tangent plane and w is in the manifold?

n allows this because it is a normed space. How do you do it in a generic

ambient manifold X?

You are right but my gut still tells me that this can be done. Perhaps using coordinate charts. Or like in Euclidean space, you have the idea of geodesics and geodesic planes in the ambient manifold. Locally these planes are close to linear.
 

Similar threads

Back
Top