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Tangent spaces at different points on a manifold

  1. Nov 27, 2015 #1
    Why are tangent spaces on a general manifold associated to single points on the manifold? I've heard that it has to do with not being able to subtract/ add one point from/to another on a manifold (ignoring the concept of a connection at the moment), but I'm not sure I fully understand this - is it simply because, even if two points lie in the same coordinate patch the coordinate map will not be Cartesian (i.e. the identity map) in general, and so subtracting/adding their coordinate values will not correspond to subtracting/adding one point from/to another on the manifold?

    Also, why is it that we can compare vectors at different points, and also add/subtract points in Euclidean space? For points, is it simple because the coordinate map is the identity map and so adding/subtracting the coordinates of points is equivalent to adding/subtracting points in Euclidean space. And for vectors, is it because Euclidean space is affine and so the tangent spaces at two different points in Euclidean space are naturally isomorphic (with the isomorphism given by parallel translation).
     
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  3. Nov 27, 2015 #2

    andrewkirk

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    In 2D Euclidean space, the tangent space at any point is simply the 2D Euclidean space itself. Strictly speaking, there is a set-theoretic difference between the two, but it can be safely ignored for most purposes. So we can in most cases add and subtract vectors in different tangent spaces because they are members of the same vector space.

    Now consider the tangent spaces to points on the surface of a sphere. The tangent space at a point is the vector space corresponding to the plane that touches the sphere only at that point. For any two points, no matter how close they are to one another, the planes will be angled with respect to one another, and so will only intersect in a single line (or not at all, if the points are antipodal). So the vectors in one are in a different vector space from those in the other, and cannot be added or subtracted.
     
  4. Nov 28, 2015 #3
    How would one argue this in a more general case though? I can visualise this for a sphere (by considering it to be embedded in 3D Euclidean space), but I'm struggling to see why it would be true in general - the tangent space at each point doesn't really "stick out" of the manifold at each point.

    Also, why can't neighbouring points on a manifold be added together? Is the argument I gave in my original post correct at all?
     
  5. Nov 28, 2015 #4

    FactChecker

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    An example of the problem on a sphere is also an example in the general case (which must include the sphere as one instance). I guess you are asking why the tangent space is always different at each point. It is not always different. There are many examples where two points might be thought of as having the same tangent space. But the general definition of "tangent space" must work all the time. So the sphere example shows that the definition must be "point-by-point".
     
  6. Nov 28, 2015 #5
    I think I'm struggling to convince myself in a technical way the whole idea that in Euclidean space there is only one tangent space (or at least the tangent spaces at each point are trivially connected to one another) and also points in Euclidean space can be added to one another (by adding their coordinates together), but that this can't be done in general (I understand the sphere example, but only by visualising it embedded in a higher dimensional Euclidean space, so I find it very hard to shake off the whole idea of Euclidean notions)?!

    Has it got anything to do coordinate charts, because in Euclidean space one can construct a global coordinate system - the Cartesian coordinate system - in which the coordinate map is the identity map, in which case it seems to make sense that points can be added together as they are essentially equivalent to their coordinate values in ##\mathbb{R}^{n}## (which can be added together). But in general this is not possible, there won't be any global coordinates and what's more the local coordinate maps will be curvilinear and so adding the coordinate values of two points will not correspond to adding the points together on the manifold.

    If tangent vectors at a point on the manifold are defined in terms of equivalence classes of curves passing through that point, then I can kind of see why the tangent space at each point will be different in general as they will contain a different set of equivalence classes, right?

    I thought I had an understanding of this stuff, but I had a deeper think about it all the other day and how I would explain the concept to someone else and now I feel very unsure! :-/
     
    Last edited: Nov 28, 2015
  7. Nov 28, 2015 #6

    andrewkirk

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    It's perfectly OK to add the coordinates of two points to get a third point and, is that third point is within the same coordinate patch, it's even a valid operation. But the operation is not in general useful because

    1. In many cases the result of adding the coordinates will not be a point in the coordinate patch, and hence will not correspond to a point on the manifold; and
    2. The operation will not be coordinate independent. So the point obtained by adding coordinates of two points in one coordinate system will not necessarily be the same as that obtained from the same two points if a different coordinate system is used.
    Don't Panic has the right idea in his second last para. The most widely used definition of a tangent space is as the collection of equivalence classes of curves through the host point. That collection will be completely different from the collection associated with a different point, because many curves through one will not even pass through the other.

    Bear in mind that one does not need to argue or prove that tangent spaces are distinct. From the way they are defined there is no reason to expect them to overlap. It is only in exceptional cases like Euclidean Space that once can consider them as overlapping and even there one has to abuse notation and misuse concepts in order to 'see' that overlap. eg we have to assume that the vector (1 0) at point (0,0) means the same as the vector (1 0) at point (1,1). That's a little like saying 'North of Mumbai' means the same as 'North of Ankara'. In fact that might be one way to explain the concept to a non-mathematician. Compare the tangent space at a point to the directions somebody could point. Then observe that if two people are pointing at a dog, their arms will not be parallel.
     
  8. Nov 28, 2015 #7
    Is this because tangent spaces in Euclidean space are naturally isomorphic (one can parallel translate a vector defined at one point to another point, and the result is independent of the path taken between the two points), whereas on a more general manifold, the result is path dependent i.e. there is no natural isomorphism and hence two vectors with the same components at different points are not equivalent (as is the case in Euclidean space). Would this be correct at all?
     
  9. Nov 28, 2015 #8

    andrewkirk

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    @"Don't panic!" That sounds like a good way of thinking about it. The ability to unambiguously parallel translate a vector from one place to another could well form the basis of an attempt to make rigorous the loose notion that all Tangent Spaces in a Euclidean Space are in some sense the same. One would need to do some set-theoretic construction, I imagine involving equivalence classes of something or other, in order to formalise it, but it sounds like it would work.
     
  10. Nov 29, 2015 #9

    lavinia

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    if one adds two tangent vectors at different points in Euclidean space then the resulting vector is no longer a tangent vector.

    While a tangent vector at a point in Euclidean space can be visualized as a vector, it loses its meaning unless the point of tangency is remembered. As soon as it moves it is either no longer a tangent vector or it is a tangent vector at another point. For instance the velocity vector of a curve is often pictured as a vector but its point of tangency is just as important as its length and direction. The same vector translated to another point is no longer tangent to the curve.
     
    Last edited: Nov 29, 2015
  11. Nov 29, 2015 #10

    lavinia

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    In Euclidean space, one needs only a single coordinate chart. The coordinate functions, ##x_i## and the tangent vectors, ##∂/∂x_i## 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 ≅ M##x##R^n##,

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

    So parallel translation has nothing to do with it. It has to do with a global coordinate system for the manifold.

    In fact, there are flat Riemannian manifolds other than Euclidean space whose tangent spaces can be "unambiguously" identified via parallel translation. Examples are the cylinder and the flat torus.
     
    Last edited: Nov 29, 2015
  12. Nov 29, 2015 #11
    What would be the technical argument then for why one can compare vectors at different points in Euclidean space, but in general one cannot do this?

    By trivial, do you mean that the tangent bundle is globally isomorphic to ##\mathbb{R}^{n}##?

    So the is the reason one can't do this in general because it's not well defined?
     
  13. Nov 29, 2015 #12

    lavinia

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    See my last post. It is explained (I hope) there.

    The word "compare" needs a precise definition. If one means parallel translation then unambiguous comparison happens only on flat Riemannian manifolds with trivial holonomy group. In my limited experience "compare" is usually meant to refer to parallel translation. If one takes this as the definition of "compare'" then a theorem says that the manifold must be flat with zero holonomy group. The only examples are flat Euclidean space, flat tori in all dimensions, and Cartesian products of flat tori and Euclidean space. If one removes the restriction that the comparison be unambiguous then parallel translation works on any Riemannian manifold.

    If one means that the tangent bundle is trivial then tangent vectors can be compared by projection. (This would depend on the trivialization.)

    For Lie groups there is a natural group of diffeomorphisms with the property that if g.x = h.x for any x in G the g = h. In this case one can "compare" tangent spaces through the differentials of the element of G. This leads to the idea of left and right invariant vector fields.

    Also for Lie groups it makes sense to "add" two points if one takes add to mean multiplication in the Lie group. Euclidean space under vector addition is a Lie group but there are many others. The conclusion that follows from the existence of this multiplication is that the tangent bundle is trivial. Interestingly, the addition does not need to form a group. The classic example is octonian multiplication on then 7 sphere. The tangent bundle to the 7 sphere is also trivial.
     
    Last edited: Nov 29, 2015
  14. Nov 29, 2015 #13
    So is it purely the fact that it is not possible to construct a global system on a manifold in general that one cannot add tangent vectors at different points on a manifold and why one cannot construct vectors as directed line segments (as you can do in Euclidean space)?

    Also, is the reason why one cannot add/subtract points on a manifold in general because there is no consistent, well-defined way to do so, as one has to add/subtract their coordinate values to do so, and (as noted by andrewkirk in his earlier post) this is coordinate dependent and is also not guaranteed to correspond to the coordinates of another point in the given coordinate patch?!
     
  15. Nov 29, 2015 #14

    andrewkirk

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    That's why I said
    It loses some of its meaning (eg see post 6). But not all of it. Special Relativity in particular relies on the meaning that remains - eg the velocity composition formula. Strictly speaking, the velocities referred to in SR are elements of a quotient space of the tangent bundle. But that technicality can be ignored while working with inertial frames in flat space.
     
  16. Nov 29, 2015 #15

    andrewkirk

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    I don't think there's ever a single reason for a property not being there. The default assumption should always be that a property is not there, until we identify a reason for it to apply. Euclidean space is both a manifold and a vector space. The ability to meaningfully add its points comes from the vector space property. The sphere ##S^2## is a manifold but not a vector space. So we have no reason to expect to be able to add its points. Indeed, we don't even have a defined operation of addition.

    On a quick guess, I think it's because we can rigorously construct a vector space from the tangent bundle. We can do that by picking a Cartesian coordinate system and then taking a quotient by identifying all vectors that have the same coordinate representation, and defining addition and scalar multiplication as the result of applying those operations in the chosen coordinate system. It can be shown that in Euclidean Space this operation is well-defined (ie does not depend on the choice of Cartesian Coordinate system), and that the addition and scalar multiplication operations satisfy the required properties of closure, commutativity, associativity etc.

    For a general manifold we cannot do that because the quotient may not be well-defined (there is no global coordinate system and the operation may not be consistent between coordinate systems) and/or the vector space axioms may not be satisfied.
     
    Last edited: Nov 29, 2015
  17. Nov 29, 2015 #16

    WWGD

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    Addition in ##\mathbb R^n ## uses the fact that there is a "natural" (meaning basis-independent) isomorphism between tangent spaces based at different points. EDIT: maybe you can see first what happens while you add vectors in ## \mathbb R^n ## and what this addition becomes when you have the images of these added vectors under chart maps. In Euclidean space (in the standard form of the structure), there is just one chart, the ID chart. The pullback by this id map is the vectors themselves. Try pulling back sum by the standard charts in , say, ## S^2 ##. Like Lavinia said, it has to see with the triviality of the tangent bundle, which is the gluing of tangent spaces at different points.

    EDIT: I think it would have to follow that the tangent bundle would have a natural vector space structure associated with its charts. Something like in the trivial case ## p \times \mathbb R^n ##
     
    Last edited: Nov 29, 2015
  18. Nov 30, 2015 #17
    Ah ok, so the reason we can add points together in ##\mathbb{R}^{n}## is due to its vector space structure. Is it also to do with the fact that we can construct a global coordinate system given by the identity map?

    In W. Boothby's "An Introduction to Differentiable Manifolds and Riemannian Geometry", he states that the reason we can consider vectors as directed line segments (and also compare vectors at different points) in ##\mathbb{R}^{n}## is because the tangent spaces at any two points in Euclidean space are naturally isomorphic - there is a unique way to construct an equivalence between them that is independent of the coordinate chart that we use (is this what is intuitively meant by a natural isomorphism?). Thus, if we define a set of basis vectors at one point in ##\mathbb{R}^{n}##, then this will automatically determine a basis for the tangent space to every point in ##\mathbb{R}^{n}## (by parallel translation).

    However, would it be correct to say that, in general, there is no natural isomorphism between tangent spaces at different points on a manifold (there will be many isomorphisms between them and one has to make an arbitrary choice) and so one cannot trivially compare two vectors at different points (since there the tangent space at each point will, in general, be distinct - there will be no way to construct a natural equivalence between tangent spaces at different points). We can also not construct vectors as directed line segments on a more general manifold due to the non-existence of a global coordinate chart (and the manifold lacking a vector space structure?!)

    Sorry to harp on about this all, just want to clear it up in my head a bit more - all the introductory books that I've read so far don't really explain in any detail why we can't add points together on a manifold and why tangent spaces at different points are generally distinct and can't be directly compared (without the introduction of a connection).
     
  19. Nov 30, 2015 #18

    fresh_42

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    I'm not sure whether my question although it arose from my reading here maybe posted here, too. If not, I apologize.
    My question is: Are there any conditions on when a linear map on let's say left-invariant vectorfields has a counterpart on the underlying (differentiable) manifold, i.e. can be "integrated"?
     
  20. Nov 30, 2015 #19

    WWGD

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    There is a formal definition in category theory of the term natural transformation. As an example, the double dual map on a vector space is natural, but the single dual is not.
     
  21. Nov 30, 2015 #20

    andrewkirk

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    It's really best to avoid trying to identify something as being 'the reason' for a property. We can have property P being 'a reason' for property Q, which just means that ##P\Rightarrow Q## (ie P is a sufficient condition for Q), but calling it 'the reason' is generally problematic. There could also be other, different preconditions in whose presence Q will necessarily occur.

    A manifold being a vector space is a sufficient condition for the existence of a well-defined and closed point addition operation. That follows directly from the vector space axioms. Prima facie, existence of a global coordinate system allows meaningful definition of an addition operation, but it is not immediately obvious that the operation would be closed. My guess is that it may well be a provable theorem that it is closed, but I haven't tried to prove it. In fact I would guess that the only manifolds without boundary for which a global coordinate system exists are those that are homeomorphic to a Euclidean space.

    I also note that being a vector space provides more structure than is needed for pointwise addition, since it also includes scalar multiplication. But again, it is conceivable that, for any manifold that admits pointwise addition, scalar multiplication necessarily follows and with it, the vector space property.

    This is all just conjecture though. I haven't tried to prove any of these things.
     
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