Tangent spaces at different points on a manifold

[lavinia]

So is the point that even in Euclidean space the tangent spaces at each point are distinct from one another, but can be related to one another by a connection?

As explained in post #30, each vector in Euclidean space determines a tangent vector at each of its points. This defines an isomorphism from Euclidean space and the fiber at any point.

Is this the heart of it then for why we can easily relate tangent vectors at different points in Euclidean space, because the tangent bundle can be assigned a global coordinate chart and so the tangent vectors at each point can be described in terms of the same set of basis vectors?

I strongly suggest that you read through post #30.

 

["Don't panic!"]

- The tangent bundle of a vector space is an example of a trivial bundle. A n-dimensional vector bundle over a space,MM, is trivial if it is bundle isomorphic to the product,MMxRnR^n. For trivial tangent bundles, a vector in RnR^n corresponds to a tangent vector at every point of MM. If φ:Mφ:M x Rn→TMR^n \rightarrow TM is a bundle isomorphism and v∈Rnv∈R^n then φ(M,v)φ(M,v) is the set of tangent vectors determined by vv. For the vector space,LL, one can choose φφ to be the map, (p,v)→(p,(p,v) \rightarrow (p, differentiate along the curve, c(t)=p+tv)c(t) = p + tv). With trivial bundles, it is tempting to think that one can add vectors from different fibers since there is a given isomorphism of all fibers with RnR^n. But this does not work because each fiber is still a different vector space.

I think this is starting to make a bit more sense. So is the point that we can use the fact that Euclidean space is a vector space (under the identification with $\mathbb{R}^{n}$) to construct vectors as directed line segments in Euclidean space (by using the vector space structure of $\mathbb{R}^{n}$ to take the difference of two points, i.e. identify a tangent vector with a line from a point $p$ to a point $p+v$). Each of these vectors then correspond to a tangent vector at each point in Euclidean space (through the isomorphism that you put, $(p,v)→(p,$ differentiate along the curve, $c(t)=p+tv)$). Would this be correct at all?
Does one then relate tangent vectors residing in different tangent spaces, $T_{p}\mathbb{R}^{n}$ and $T_{q}\mathbb{R}^{n}$ respectively, by parallel translating a tangent vector in $\mathbb{R}^{n}$ from $T_{p}\mathbb{R}^{n}$ to $T_{q}\mathbb{R}^{n}$ (or vice-versa), since each of the tangent spaces in $\mathbb{R}^{n}$ are parallel to one another (the trivial tangent bundle ensures that a manifold is parallelizable, right?). (Apologies, I may well have misunderstood what you wrote in post #30).

 

[WWGD]

I think this is starting to make a bit more sense. So is the point that we can use the fact that Euclidean space is a vector space (under the identification with $\mathbb{R}^{n}$) to construct vectors as directed line segments in Euclidean space (by using the vector space structure of $\mathbb{R}^{n}$ to take the difference of two points, i.e. identify a tangent vector with a line from a point $p$ to a point $p+v$). Each of these vectors then correspond to a tangent vector at each point in Euclidean space (through the isomorphism that you put, $(p,v)→(p,$ differentiate along the curve, $c(t)=p+tv)$). Would this be correct at all?
Does one then relate tangent vectors residing in different tangent spaces, $T_{p}$ and $T_{q}$ respectively, by parallel translating a tangent vector in $\mathbb{R}^{n}$ from $T_{p}$ to $T_{q}$ (or vice-versa), since each of the tangent spaces in $\mathbb{R}^{n}$ are parallel to one another (the trivial tangent bundle ensures that manifold is parallelizable, right?). (Apologies, I may well have misunderstood what you wrote in post #30).

Please read all posts carefully before asking more questions, or this will go on forever. Parallelizability does not imply a vector space structure; you cannot add vectors on $S^1$ , which is parallelizable. And like lavinia said, parallel translation exists whenever you have a connection.

 

["Don't panic!"]

Please read all posts carefully before asking more questions, or this will go on forever. Parallelizability does not imply a vector space structure; you cannot add vectors on S1S^1 , which is parallelizable. And like lavinia said, parallel translation exists whenever you have a connection.

Apologies, I haven't worded my post particularly well in that respect. I was attempting to distinguish between parallel transport and parallel translation in $\mathbb{R}^{n}$, the former is usually more complicated as when one parallel transports along a curve from one tangent space to another the basis vectors change (introducing connection coefficients), whereas in the latter case (specifically referring to Euclidean space) the procedure is much more simple since one just translates along a straight line from one tangent space to another and the basis remains constant due to the trivial nature of the tangent bundle.

Ok, having read through the previous posts I'm going to try and attempt a summary.

Is the general idea that in Euclidean space we can take advantage of the fact that it is a vector space as well as a manifold and so we can add/subtract points in Euclidean space from one another in a meaningful way, since the result will be another point in Euclidean space. We can also take advantage of this fact to construct vectors in $\mathbb{R}^{n}$ as directed line segments. Each of these vectors can be mapped to a tangent vector at each point in Euclidean space by an isomorphism $\mathbf{v}\mapsto\mathbf{v}_{p}$, and in doing so we construct a tangent space $T_{p}\mathbb{R}^{n}$ to each point $p$ in Euclidean space. Each of these tangent spaces is a distinct vector space, since the vectors in a particular tangent space $T_{p}\mathbb{R}^{n}$ at a point $p$ are distinguished from those in another tangent space $T_{q}\mathbb{R}^{n}$ at another point $q$ by the fact that they are "attached" to their respective points $p$ and $q$. Since Euclidean space is Riemannian we can define a connection that relates tangent spaces at different points. In this particularly case the connection corresponds to parallel translating vectors from one tangent space $T_{p}\mathbb{R}^{n}$ at a point $p$ to another $T_{q}\mathbb{R}^{n}$ at point $q$ along a straight line in Euclidean space connecting the points $p$ and $q$ (since such a line is unique does this make the operation natural in this case?).
In a more general setting, a manifold will not possesses a vector space structure and so we cannot add/subtract points on a manifold and end up with another point on the manifold (in general). We can also not apply the same procedure (as in Euclidean space) to introduce vectors onto the manifold. Instead, we must introduce the notion of tangent vectors to curves on the manifold - this can be done by identifying a tangent vector at a point as an equivalence class of curves passing through that point, all of which have the same tangent at that point. Since tangent vectors are defined at particular points, the set of tangent vectors at each point form distinct tangent spaces. Consequently, as tangent vectors at different points cannot, in general, be compared since they "live" in different vector spaces. However, if the manifold is Riemannian, then we can introduce a connection which allows us to relate tangent vectors in a tangent space at a particular point to those in another tangent space at a different point. This can be done via the notion of parallel transport in which we transport a vector along a curve passing through the base points of both tangent spaces and demand that the components of the tangent vector remain constant as it "moves" along this curve. In doing so we can map a tangent vector from one tangent space to another on the manifold such that we can compare tangent vectors at different points.

Would this summary be correct at all? (I have tried to read carefully through all the previous posts so hopefully I'm getting somewhere, I do apologise though if I'm still getting things wrong, it is not my intention to annoy anyone).

 

[andrewkirk]

I guess something like adding force vectors to gain a resultant force, or adding position vectors at different points ("top to tail") to gain another position (relative to the origin), or obtain a velocity vector as the derivative of a position vector...

This leads nicely to the point I was working towards, which is that even in Euclidean space, comparing or adding vectors at different points can have different meanings, or in some cases none, depending on what the vectors represent.

The first case is an example of where the addition is as invalid in Euclidean Space as it is in curved space. We can only add forces to get a net force where those forces apply at the same point. One might doubt that, thinking about two horses pulling a wagon. But when we think about it, we realize that the rigid structure of the wagon transmits the forces from the two horses so that, for each point particle of the wagon, the particle is subject to two forces, one from each horse. There's really no difference between the curved and flat space instances of this example. Both only allow meaningful addition of forces at a single point.

Position vectors can be generalised to curved spaces using the exponential map. Since the position vector of a point Q in Euclidean space must be relative to some base point P (usually the origin), we can create an analog of this in curved space whereby a position vector of Q is a vector $\vec v\in T_PM$ such that $\exp_P(v)=Q$. In some curved space cases the 'position vector' will not be unique. For instance on a sphere there are always at least two ways of getting from one point to another. If the manifold is Riemannian we can make the position vector unique by requiring the geodesic curve connecting the two to have the minimal length (ie the 'shortest route'). That works for all cases except antipodal points.

The 'generalised position vectors' thus defined will not usually form a vector space though. Nor should we expect them to. The vector space axioms are about having a linear structure and curved spaces by definition do not preserve linear structure.

Your last example about defining velocity vectors as derivatives of position vectors can be generalised using the same approach. There will be points where the derivative does not exist (eg Q antipodal to P) but again, that's what we'd expect in a curved space.

 

[lavinia]

Apologies, I haven't worded my post particularly well in that respect. I was attempting to distinguish between parallel transport and parallel translation in $\mathbb{R}^{n}$, the former is usually more complicated as when one parallel transports along a curve from one tangent space to another the basis vectors change (introducing connection coefficients), whereas in the latter case (specifically referring to Euclidean space) the procedure is much more simple since one just translates along a straight line from one tangent space to another and the basis remains constant due to the trivial nature of the tangent bundle.

Ok, having read through the previous posts I'm going to try and attempt a summary.

Is the general idea that in Euclidean space we can take advantage of the fact that it is a vector space as well as a manifold and so we can add/subtract points in Euclidean space from one another in a meaningful way, since the result will be another point in Euclidean space. We can also take advantage of this fact to construct vectors in $\mathbb{R}^{n}$ as directed line segments. Each of these vectors can be mapped to a tangent vector at each point in Euclidean space by an isomorphism $\mathbf{v}\mapsto\mathbf{v}_{p}$, and in doing so we construct a tangent space $T_{p}\mathbb{R}^{n}$ to each point $p$ in Euclidean space. Each of these tangent spaces is a distinct vector space, since the vectors in a particular tangent space $T_{p}\mathbb{R}^{n}$ at a point $p$ are distinguished from those in another tangent space $T_{q}\mathbb{R}^{n}$ at another point $q$ by the fact that they are "attached" to their respective points $p$ and $q$. Since Euclidean space is Riemannian we can define a connection that relates tangent spaces at different points. In this particularly case the connection corresponds to parallel translating vectors from one tangent space $T_{p}\mathbb{R}^{n}$ at a point $p$ to another $T_{q}\mathbb{R}^{n}$ at point $q$ along a straight line in Euclidean space connecting the points $p$ and $q$ (since such a line is unique does this make the operation natural in this case?).
In a more general setting, a manifold will not possesses a vector space structure and so we cannot add/subtract points on a manifold and end up with another point on the manifold (in general). We can also not apply the same procedure (as in Euclidean space) to introduce vectors onto the manifold. Instead, we must introduce the notion of tangent vectors to curves on the manifold - this can be done by identifying a tangent vector at a point as an equivalence class of curves passing through that point, all of which have the same tangent at that point. Since tangent vectors are defined at particular points, the set of tangent vectors at each point form distinct tangent spaces. Consequently, as tangent vectors at different points cannot, in general, be compared since they "live" in different vector spaces. However, if the manifold is Riemannian, then we can introduce a connection which allows us to relate tangent vectors in a tangent space at a particular point to those in another tangent space at a different point. This can be done via the notion of parallel transport in which we transport a vector along a curve passing through the base points of both tangent spaces and demand that the components of the tangent vector remain constant as it "moves" along this curve. In doing so we can map a tangent vector from one tangent space to another on the manifold such that we can compare tangent vectors at different points.

Would this summary be correct at all? (I have tried to read carefully through all the previous posts so hopefully I'm getting somewhere, I do apologise though if I'm still getting things wrong, it is not my intention to annoy anyone).

This is right. One small point: the manifold does not need to be Riemannian manifold in order for parallel translation to be defined. All that one needs is a connection on the tangent bundle. On a Riemannian manifold there is a unique connection,called the Levi-Civita connection, that is compatible with the Riemannian metric and is torsion free. Under this connection, lengths of vectors and angles between them are preserved under parallel translation . But for a general connection, parallel translation is defined without reference to length or angle.

One way to see how this works for a Levi-Civita connection is to look at the covariant derivative.

Compatibility with the metric means that for any two vector fields along a curve,$c(t)$,

$∂/∂t<X,Y> = <∇_{c'(t)}X,Y> + <X,∇_{c'(t)}Y>$ If $X$ and $Y$ are parallel then these covariant derivatives are both zero. So $∂/∂t<X,Y> =0$ and $<X,Y>$ is constant.

 

[lavinia]

Position vectors can be generalised to curved spaces using the exponential map. Since the position vector of a point Q in Euclidean space must be relative to some base point P (usually the origin), we can create an analog of this in curved space whereby a position vector of Q is a vector $\vec v\in T_PM$ such that $\exp_P(v)=Q$. In some curved space cases the 'position vector' will not be unique. For instance on a sphere there are always at least two ways of getting from one point to another. If the manifold is Riemannian we can make the position vector unique by requiring the geodesic curve connecting the two to have the minimal length (ie the 'shortest route'). That works for all cases except antipodal points. The 'generalised position vectors' thus defined will not usually form a vector space though. Nor should we expect them to. The vector space axioms are about having a linear structure and curved spaces by definition do not preserve linear structure.

To elaborate a little bit:

- If the position vectors are unique, then the Riemannian manifold is homeomorphic to Euclidean space. The exponential map is a smooth map from $R^n$ onto the Riemannian manifold. If it is 1-1, then it has no singularities. Singular values are always conjugate points and these always are reached by infinitely many geodesics. So for every Riemannian manifold except possibly one that is homeomorphic to Euclidean space, there are points for which exponential map defines multiple generalized position vectors.

Also, since a vector space is homeomorphic to $R^n$, any manifold that is not homeomorphic to $R^n$ can not be a vector space and for any Riemannian metric on the manifold, there must be points with more than one position vector.

- A space does not need to be curved for there to be multiple generalized position vectors for a point. Compact flat Riemannian manifolds have zero curvature tensor but are not homeomorphic to $R^n$. The exponential map is non-singular (no conjugate points)and is a covering of the manifold by Euclidean space. Since the covering is infinite, every point has infinitely many position vectors.

BTW: Your statement, "If the manifold is Riemannian we can make the position vector unique by requiring the geodesic curve connecting the two to have the minimal length (ie the 'shortest route'). That works for all cases except antipodal points." is slightly inaccurate. Geodesics (half great circles) from the north to the south pole of the sphere do have shortest length and they are not unique What is true is that if these geodesics are continued beyond the south pole then they fail to minimize length. This is a general property of conjugate points. A geodesic minimizes length up to and including the first conjugate point but never beyond.

What is also true is that at each point,$p$, there is an open neighborhood of zero in $T_{p}M$ that is mapped diffeomorphically into $M$ under the exponential map. This follows from the Inverse Function Theorem since the differential of the exponential map at zero is the identity map. In such a neighborhood the generalized position vector is unique.

 

["Don't panic!"]

any manifold that is not homeomorphic to RnR^n can not be a vector space and for any Riemannian metric on the manifold, there must be points with more than one position vector.

Can this statement be viewed as an argument for why one cannot meaningfully add/subtract points from one another, in general, on a manifold?

Thanks for all your time (and patience) by the way.

 

[lavinia]

Can this statement be viewed as an argument for why one cannot meaningfully add/subtract points from one another, in general, on a manifold?

Thanks for all your time (and patience) by the way.

Not quite. It is an argument for why one cannot meaningfully add/subtract points from one another and multiply them by scalars in a general manifold. There are examples of manifolds for which addition and subtraction (but not scalar multiplication) can be defined. The addition law is usually not commutative. But most manifolds do not have an addition law either.

 

["Don't panic!"]

Not quite. It is an argument for why one cannot meaningfully add/subtract points from one another and multiply them by scalars in a general manifold. There are examples of manifolds for which addition and subtraction (but not scalar multiplication) can be defined. The addition law is usually not commutative.

Ah ok, thanks for the clarification on that.

But most manifolds do not have an addition law either.

Is this why when one first introduces the notion of a manifold it is a priori assumed that points cannot be added/subtracted from one another?

 

[fresh_42]

Is this why when one first introduces the notion of a manifold it is a priori assumed that points cannot be added/subtracted from one another?

I guess it's rather because of the need to examine curved structures for otherwise there will be vector spaces as natural model. That points cannot be added is due to curvature and not initially intended. E.g. $GL(ℂ)$ is a manifold but there's no way to add elements for you may leave the group. It's not intended, it's just a consequence.

 

[lavinia]

Is this why when one first introduces the notion of a manifold it is a priori assumed that points cannot be added/subtracted from one another?

I don't know but I doubt it. Manifolds occur naturally in many situations, as surfaces, as phase spaces, as physical models - e.g.space time - , as geometric solutions to physical constraints - e.g. soap bubbles - as domains of meromorphicic functions -e.g. Riemann surfaces - and so forth. The applications are endless. The key idea is that one can draw a local coordinate system but not necessarily a global one. Because of this, one wants to know how different coordinate regions compare. These comparisons are called coordinate transformations. In Physics a quantity whose measurement is the same no matter which coordinates (which observer) are used is of special interest. For instance in space-time, the proper time is measured to be the same in all frames of reference.

 

[lavinia]

I guess it's rather because of the need to examine curved structures for otherwise there will be vector spaces as natural model. That points cannot be added is due to curvature and not initially intended. E

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. There are examples of compact manifolds that are not curved i.e. their curvature tensor is identically zero. So curvature isn't the reason that points can't be added or multiplied by scalars.

Further one can put a metric of non-zero curvature on $R^n$ and it still can be given the structure of a vector space.

 

[fresh_42]

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. There are examples of compact manifolds that are not curved i.e. their curvature tensor is identically zero. So curvature isn't the reason that points can't be added or multiplied by scalars.

Further one can put a metric of non-zero curvature on $R^n$ and it still can be given the structure of a vector space.

You mean the origin of the term manifold was not driven by curved spaces? That was what I was answering to.

 

[lavinia]

You mean the origin of the term manifold was not driven by curved spaces? That was what I was answering to.

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.

 

[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.