Tangent spaces at different points on a manifold

[fresh_42]

I am not sure of the origin. My point was that this is a question of topology not geometry. But originally curved surfaces may have been the first manifolds considered - not sure. It seems that Riemann first defined manifold in his Habilitation Thesis. The definition is not geometric per se but is topological. It defines the idea of a "multiply extended quantity" - i.e. a space which can be locally coordinatized.

Riemann would have been my guess, too. And I don't see it different than you with the topology and the charts. Nevertheless I can't imagine that the absence of addition or stretching was intended since vector spaces are - although trivial - manifolds, too. But you made me rather curious. Compact, zero curvature, not homeomorphic to a Euclidean space - do you have an example at hand? I'm sure it can be constructed, however, I'm curious whether there is a somehow "common" example.

 

[lavinia]

Riemann would have been my guess, too. And I don't see it different than you with the topology and the charts. Nevertheless I can't imagine that the absence of addition or stretching was intended since vector spaces are - although trivial - manifolds, too. But you made me rather curious. Compact, zero curvature, not homeomorphic to a Euclidean space - do you have an example at hand? I'm sure it can be constructed, however, I'm curious whether there is a somehow "common" example.

The flat torus and the flat Klein bottle are the two compact flat surfaces(without boundary). In each dimension there are finitely many flat manifolds. I think all of the 3 and 4 dimensional ones are known - not sure. A famous flat 3 manifold is the Hansche-Wendt manifold. It is orientable and has zero first Betti number. Its holonomy group is $Z_2$x$Z_2$. I can show you how to define it if you like.

The flat torus can be realized in $R^4$ as the image of the map, $φ:R^2 \rightarrow R^4$ defined by $φ(x,y) = ( cos(x),sin(x),cos(y),sin(y))$ Note that this defines a covering of the flat torus by $R^2$. In general all flat manifolds are covered by Euclidean space.

There are no compact flat manifolds without boundary in $R^3$. A theorem of Hilbert says that any compact surface without boundary in $R^3$ must have a point of positive Gauss curvature.

 

[fresh_42]

The flat torus and the flat Klein bottle are the two compact flat surfaces. In each dimension there are finitely many flat manifolds. I think all of the 3 and 4 dimensional ones are known - not sure. A famous flat 3 manifold is the Hansche-Wendt manifold. It is orientable and has zero first Betti number. Its holonomy group is $Z_2$x$Z_2$. I can show you how to define it if you like.

Thank you. I've found something here and here which looks quiet interesting. I've never thought about Klein's bottle other than what it is, and certainly not flat. I wouldn't have expected manifolds to be so manifold. Usually some Lie groups lurk around or space-time isn't far. (cp. 1st link)
What might be interesting to know to which extend mathematicians like Legendre, Liouville or others before Riemann considered manifolds in their work on differential equations, possibly without explicitly defining them. My history book (J. Dieudonné) says on this item Riemann (analytic manifolds) and Gauß (differential geometry, 1827) were the first, both considering the geometric component of them although Riemann's definition is basically the modern topological one. But I just had a quick glimpse in it.

 

["Don't panic!"]

The key - as has been pointed out above - is that the manifold is not homeomorphic to Euclidean space. Then it cannot be given the structure of a vector space.

This was my understanding (from the responses I've been given on here) of why we can generally add points together on a manifold.

 

[lavinia]

This was my understanding (from the responses I've been given on here) of why we can generally add points together on a manifold.

right

 

[lavinia]

Thank you. I've found something here and here which looks quiet interesting. I've never thought about Klein's bottle other than what it is, and certainly not flat. I wouldn't have expected manifolds to be so manifold. Usually some Lie groups lurk around or space-time isn't far. (cp. 1st link)
What might be interesting to know to which extend mathematicians like Legendre, Liouville or others before Riemann considered manifolds in their work on differential equations, possibly without explicitly defining them. My history book (J. Dieudonné) says on this item Riemann (analytic manifolds) and Gauß (differential geometry, 1827) were the first, both considering the geometric component of them although Riemann's definition is basically the modern topological one. But I just had a quick glimpse in it.

I don't know the history but Complex Analysis could be thought of as the study of solutions of the two dimensional Laplace equation. Riemann surfaces arise as the domains of multi valued complex functions. Klein's book "Riemann's Theory of Algebraic Functions and Their Integrals"" shows the interplay between topology and solutions of the Laplace equation.

 

[WWGD]

This was my understanding (from the responses I've been given on here) of why we can generally add points together on a manifold.

I hate to beat this one to death, but, aren't the tangent vectors objects in the tangent bundle, so we would want to make the tangent bundle itself into a vector space, and not just the manifold itself? EDIT If the manifold is a vector space, then we can add _points_ on the manifold, but, if we wanted to add tangent vectors at different points, don't we need to address whether the tangent bundle is a vector space, i.e., homeomorphic to $\mathbb R^n$? Then we can use homological invariants on the triviality of the bundle.
EDIT 2: Fittingly, the tangent bundle of a vector space is itself a vector space. It would be nice to see the other way around. Is there some result whereby , given some topological space X, how to know which manifolds can have the topological space X as a vector bundle?

 

[fresh_42]

It would be nice to see the other way around. Is there some result whereby , given some topological space X, how to know which manifolds can have the topological space X as a vector bundle?

May this be rephrased in: "Given a local Euclidean topological space X (to consider an arbitrary topological space wouldn't make sense imao) how many isomorphism classes of vector bundels can be defined on X?"?

 

[lavinia]

I hate to beat this one to death, but, aren't the tangent vectors objects in the tangent bundle, so we would want to make the tangent bundle itself into a vector space, and not just the manifold itself? EDIT If the manifold is a vector space, then we can add _points_ on the manifold, but, if we wanted to add tangent vectors at different points, don't we need to address whether the tangent bundle is a vector space, i.e., homeomorphic to $\mathbb R^n$? Then we can use homological invariants on the triviality of the bundle.

The tangent bundle is never a vector space because fibers at different points can not be added. In the case of trivial bundles, adding vectors from different fibers forgets the bundle structure. It is no longer a bundle.

If the manifold is not homeomorphic to $R^n$ then it cannot be a vector space, even if its tangent bundle is trivial For instance, the tangent bundle of every orientable compact 3 manifold is trivial. But no compact space can be a vector space.

 

[lavinia]

May this be rephrased in: "Given a local Euclidean topological space X (to consider an arbitrary topological space wouldn't make sense imao) how many isomorphism classes of vector bundels can be defined on X?"?

That is a good question. A theorem of Thom says that for Gl(n) bundles there is a 1-1 correspondence between isomorphism classes of these bundles over a paracompact base space and homotopy classes of maps of the space into the Grassmann manifold of $n$-planes in $R^{n+m}$ for sufficiently large $m$.

 

[WWGD]

The tangent bundle is never a vector space because fibers at different points can not be added. In the case of trivial bundles, adding vectors from different fibers forgets the bundle structure. It is no longer a bundle.

If the manifold is not homeomorphic to $R^n$ then it cannot be a vector space, even if its tangent bundle is trivial For instance, the tangent bundle of every orientable compact 3 manifold is trivial. But no compact space can be a vector space.

I understand, but the OP wanted to know how to add tangent vectors at different points, not points themselves. Tangent vectors live in the tangent bundle. And Isn't the tangent bundle of $\mathbb R^n$ equal to $\mathbb R^{2n}$?

 

[lavinia]

And Isn't the tangent bundle of $\mathbb R^n$ equal to $\mathbb R^{2n}$?

It is homeomorphic to $R^{2n}$ but not isomorphic as a bundle. By itself $R^{2n}$ is not a bundle.

 

[WWGD]

It is homeomorphic to $R^{2n}$ but not isomorphic as a bundle. By itself $R^{2n}$ is not a bundle.

And that was the argument I was presenting for why one cannot add tangent vectors at different points: the tangent bundle is not a vector space.

 

[lavinia]

Maybe a new thread should be started that discusses the theory of vector bundles - with structure group the general linear group - for manifolds. this is a highly researched area with many difficult Theorems but we could

And that was the argument I was presenting for why one cannot add tangent vectors at different points: the tangent bundle is not a vector space.

yes.