# Tangent spaces at different points on a manifold

#### WWGD

Gold Member
Sorry, I meant to put fibre, not section (was just about to change it, but you beat me).

Would it be correct to say that one could add a vector from a tangent space at one point to a vector from a tangent space at another point, but this will in general not correspond to tangent vector in either space - it will not be tangent to any curves passing through either point?
Also, from another point of view could one argue that adding the two vectors component wise relies on choosing a basis and is this clearly coordinate dependent, hence such an operation has no geometrical meaning - the resulting vector will not be in the tangent bundle over the manifold?!

Sorry to labour the point, I don't know why I'm finding it so hard to conceptualise.
I think it may be good to first create a definition for the general addition of tangent vectors at different points, which may help illustrate when/how/if this is possible. Without this definition it seems too speculative to be able to answer more definitive questions. EDIT: it seems like the first "natural" definition would be that of choosing a vector space isomorphism to identify any two spaces. But then there goes the naturality issue. This would bring you to the concept of a connection.

#### andrewkirk

Homework Helper
Gold Member

Working with a single, concrete example of a meaningful addition of vectors at different points on a flat manifold, it may be much easier to see why the meaning disappears when one moves to a curved manifold.

What would be an example of addition or subtraction of vectors at different points on a flat manifold that would be meaningful to you?

#### "Don't panic!"

What would be an example of addition or subtraction of vectors at different points on a flat manifold that would be meaningful to you?
I think the normal Euclidean space endowed with the Euclidean metric - this is what I was always familiar with before being introduced to differential geometry.
I can intuitively see why on a sphere we can't add/subtract vectors at different points since neighbouring tangent spaces are not parallel to one another. The problem is I can't escape from visualising this by embedding in Euclidean space and also can't come up with a satisfactory reason why it is not possible to do so in general. I know you said that we should work on the basis that such a property doesn't intrinsically exist, but unfortunately I've been exposed to a top down rather than a bottom up approach, being exposed to Euclidean vectors and flat Euclidean space first and this has rather got me stuck in the mud trying to break away from this blinkered viewpoint.
I very much appreciate your input and would be grateful for an example/some examples.

Perhaps part of the issue hindering my understanding is that two vector spaces $T_{p}M$ and $T_{q}M$ at two distinct points $p$ and $q$ are both isomorphic to $\mathbb{R}^{n}$ so why can't we just take a vector from each tangent space to the common vector space $\mathbb{R}^{n}$, then add/subtract them (or compare them in some other way) and then go back to their original tangent spaces?

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#### andrewkirk

Homework Helper
Gold Member
I think the normal Euclidean space endowed with the Euclidean metric - this is what I was always familiar with before being introduced to differential geometry.
Certainly it's Euclidean space we're talking about when we think in terms of adding or subtracting vectors at different points. What I think is needed though to better focus the discussion is an example of such adding or subtracting in Euclidean space that is physically meaningful. Different examples will have different analogues (or in some cases no analogues) in curved space.

#### lavinia

Gold Member
There is no meaning to adding two vectors from two different vector spaces. This is true even if the vector spaces are isomorphic. If $φ:V \rightarrow W$ is an isomorphism and $v$ and $w$ are vectors in each then one can add $φ(v)$ to $w$ since $φ(v)$ and $w$ are both in the same vector space,$W$. One cannot add $v$ to $w$.

- If $L$ is a vector space, then each vector in $L$ naturally corresponds to a tangent vector at each of its points. For $v$ in $L$, differentiating functions along the curve, $c(t) = p + tv$, determines a tangent vector at $p$. This defines an isomorphism between $L$ and $TL_{p}$. It is this isomorphism that one thinks of as translating $v$ so that it starts at $p$ rather than at the origin. This is not the same as parallel translation under the Euclidean metric since that defines an isomorphism between $TL_{0}$ and $TL_{p}$ not between $L$ and $TL_{p}$. For that, one would first need an isomorphism between $L$ and $TL_{0}$.

- The tangent bundle of a vector space is an example of a trivial bundle. A n-dimensional vector bundle over a space,$M$, is trivial if it is bundle isomorphic to the product,$M$x$R^n$. For trivial tangent bundles, a vector in $R^n$ corresponds to a tangent vector at every point of $M$. If $φ:M$ x $R^n \rightarrow TM$ is a bundle isomorphism and $v∈R^n$ then $φ(M,v)$ is the set of tangent vectors determined by $v$. For the vector space,$L$, one can choose $φ$ to be the map, $(p,v) \rightarrow (p,$ differentiate along the curve, $c(t) = p + tv)$. 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 $R^n$. But this does not work because each fiber is still a different vector space.

#### lavinia

Gold Member
BTW: Flat Riemannian manifolds generally are neither vector spaces nor Lie groups. The only vector space examples are Euclidean spaces, and the only Lie groups are flat tori, Euclidean spaces, and their Cartesian products. In these, parallel translation is independent of path. But in each dimension there are flat Riemannian manifolds for which parallel translation is not path independent. There are always closed paths for which parallel translation brings a vector back to a different vector. For instance, parallel translation of a vertical vector along the equator of the Mobius band brings it back to its negative. The same is true for the flat Klein bottle. A classic theorem says that any finite group can be the holonomy group of a flat Riemannian manifold.

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#### WWGD

Gold Member
To be more precise, I guess having a trivial bundle is necessary but not sufficient to be able to add tangent vectors based at different points. I guess Don't Panic is looking for an intuitively-believable obstruction to being able to do this operation.

#### "Don't panic!"

What I think is needed though to better focus the discussion is an example of such adding or subtracting in Euclidean space that is physically meaningful.
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...

BTW: Flat Riemannian manifolds generally are neither vector spaces nor Lie groups. The only vector spaces are Euclidean spaces, and the only Lie groups are flat tori, Euclidean spaces, and their Cartesian products. In these, parallel translation is independent of path. But in each dimension there are flat Riemannian manifolds for which parallel translation is not path independent.
So, is it that parallel translation is a well-defined operation in Euclidean space and that it is globally $\mathbb{R}^{n}$, that allows one to add vectors at different points? Is it simply that the mapping from a tangent space at one point to a tangent space at another point is given by such a parallel translation?

I guess Don't Panic is looking for an intuitively-believable obstruction to being able to do this operation.
Yes, that'd be great if possible?!

#### lavinia

Gold Member
So, is it that parallel translation is a well-defined operation in Euclidean space and that it is globally $\mathbb{R}^{n}$, that allows one to add vectors at different points? Is it simply that the mapping from a tangent space at one point to a tangent space at another point is given by such a parallel translation?
- Parallel translation is well defined on any manifold with a connection.
- You cannot add vectors from different tangent spaces because you can not add vectors in different vector spaces.

Post #30 explains everything in detail.

#### "Don't panic!"

- Parallel translation is well defined on any manifold with a connection.
- You cannot add vectors from different tangent spaces because you can not add vectors in different vector spaces.

Post #30 explains everything in detail.
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?

I think part of what is confusing me is that I've read statements in one or two books that say something like:

"It is often the case that we wish to compare tangent vectors at different points on a manifold, i.e. tangent vectors residing in different tangent spaces. In Euclidean space this is straightforward since the tangent space at a point in the Euclidean space is the Euclidean space itself [are they meaning here that Euclidean space can be identified with $\mathbb{R}^{n}$ by introducing a global coordinate chart (Cartesian coordinates), and that the tangent space at each point is isomorphic to $\mathbb{R}^{n}$?]. Moreover, the affine nature of Euclidean space allows the translation of vectors from one point to another in a unique fashion [I assume here they are referring to the statement you made earlier that translating a vector from one point to another is independent of the path in Euclidean space]"

This seems to be somewhat vague - it makes somewhat intuitive sense, but at the same time seems to gloss over a lot of important technical concepts.

- In Euclidean space, one needs only a single coordinate chart. The coordinate functions, xi and the tangent vectors, ∂/∂xi then form a global coordinate system for the tangent bundle of Euclidean space. On a general manifold, there is no global coordinate system so even if the tangent bundle is trivial,

TM≅MxRn,

the tangent bundle can not be coordinatized by a set of coordinate functions and their associated tangent vectors.
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?

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#### lavinia

Gold Member
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 RnTMR^n \rightarrow TM is a bundle isomorphism and vRnv∈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).

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#### WWGD

Gold Member
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 possess 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

Homework Helper
Gold Member
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 realise 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

Gold Member
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 possess 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.

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#### lavinia

Gold Member
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.

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#### "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

Gold Member
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

Mentor
2018 Award
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

Gold Member
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.

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#### lavinia

Gold Member
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

Mentor
2018 Award
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

Gold Member
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.

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