Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Basis for tangent space and cotangent space

  1. Jan 5, 2009 #1
    Hello, I'm trying (somewhat haphazardly) to teach myself about differential forms. A question I have which is confusing me at the moment is about the tangent and cotangent spaces.

    In https://www.physicsforums.com/showthread.php?t=2953" the basis for the tangent space was described in terms of the directional derivatives [tex]\partial/\partial x_{i}[/tex], with the basis for the cotangent space being the differentials [tex]dx_{i}[/tex]. This is consistent with other materials I've found (such as the articles on Wikipedia regarding the tangent space). However, in Saunders Mac Lane's Mathematics: Form and Function he calls the cotangent space the set of all directional derivatives, with the tangent space being the space of tangent vectors [tex]<dx/dt, dy/dt>[/tex] to the points of the parameterized curve given by x = g(t), y = h(t). (He began by describing the chain rule [tex]dz/dt = (\partial z/\partial x)(dx/dt) + (\partial z/\partial y)(dy/dt)[/tex] on terms of the inner product [tex]<\partial z/\partial x, \partial z/\partial y>\bullet<dx/dt, dy/dt>[/tex], with the first vector (grad(z)) being an element of the cotangent space, and the second an element of the tangent space.) (All this was in chapter VI.9, if anyone has the book.)

    Also, in David Bachman's book A Geometric Approach to Differential Forms, he describes the differentials dx, dy, etc. as coordinate functions of the tangent space. I'm wondering why these books would mix up these two. The problem is that both explanations make intuitive sense to me, (at least for the tangent space basis) so are they just two different formulations of the tangent and cotangent spaces, or is someone wrong? Or am I just misunderstanding something?

    If anyone can shed some light on this I'd be grateful!
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Jan 6, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Well, it's true that the differentials dx, dy, etc. are, by definition, coordinate functions of the tangent space: since [itex](\partial/\partial x_{i})_{i=1}^n[/itex] is a basis of the tangent space, any given tangent vector X can be written [itex]X=\sum a^i\partial/\partial x_{i}[/itex] for some constants a^i and dx^i is defined by [itex]dx^i(X)=a^i[/itex]. I.e., dx^i is the ith coordinate function of the tangent space.

    I guess your confusion stemmed from the fact that you though the sentence "the differentials dx, dy, etc. are coordinate functions of the tangent space" somehow meant that the dx^i are elements of the tangent space. But as you see, they are actually linear functions from the tangent space to the real numbers.
  4. Jan 6, 2009 #3


    User Avatar
    Science Advisor

    And thus in the dual to the tangent space, i.e. the cotangent space.
  5. Jan 6, 2009 #4
    Thanks, great explanation.
  6. May 1, 2011 #5
    Bumping this old thread because I had a question that closely relates to it. I think this is an easy thing to confuse for someone just starting to learn these concepts.

    My question is, what exactly is meant by

    I understand that these differentials dx,dy, etc. map a vector to real number, but it seems like that real number is the "component of the vector w.r.t the basis of the tangent space", not the "coordinate". Perhaps there is a special meaning of "coordinate function of a tangent space" that is not the same as "coordinate functions for a basis in a region of space within a manifold"?

    Also, just for my own clarification, is the sentence

    equivalent to the statement:
    "They are linear functions whose domain is elements of the tangent space and whose range is the real numbers"
  7. May 1, 2011 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Indeed, I meant to say "component functions" instead of "coordinate functions" in this spot!

  8. May 2, 2011 #7
    Thanks quasar, but if I might, allow me to try to make one final connection.

    Take an arbitrary one-form [itex] \omega [/itex] and tangent vector [itex] \mathbf{v} [/itex] at some point of interest

    [itex] \omega = f_1 \ dx^1 + f_2 \ dx^2 + f_3 \ dx^3 [/itex] and
    [itex] \mathbf{v} = a^1 \frac{\partial}{\partial x_1} + a^2 \frac{\partial}{\partial x_2} + a^3 \frac{\partial}{\partial x_3} [/itex] (hope my sub/superscripting are correct)

    The a^i s are the components of a tangent vector w.r.t some basis, and the f_i s are components of a covector with respect to the dual basis, and what exactly are the coordinate functions? Are they neither of the above, since coordinates/points in space are only pseudovectors, or are they still considered vectors or covectors?

    Also, I still don't really get the expression [itex]dx^i(X)=a^i[/itex]. A differential maps a vector to a real number called its components. Is there any more to it? Sometimes I see a tangent vector written as (dx,dy,dz). Is this sloppy notation or is it acceptable/customary to write the differentials as the "components" of the tangent vector?
  9. May 2, 2011 #8


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The vectors [itex]\partial/\partial x^1,\ldots,\partial/\partial x^n[/itex] are not just any basis of TpM. They are the basis induced by some local coordinate system [itex]x^1,\ldots,x^n[/itex] defined in a neighborhood U of p in the manifold M. The map [itex]x^i:U\rightarrow \mathbb{R}[/itex] is called the ith coordinate function.

    Not really no. A coordinate system [itex]x^1,\ldots,x^n[/itex] around a point p in an n-manifold M induces a basis on TpM written [itex]\partial/\partial x^1,\ldots,\partial/\partial x^n[/itex] and this in turn induces a basis [itex]dx^1,\ldots,dx^n[/itex] on Tp*M obtained by taking the dual basis of [itex]\partial/\partial x^1,\ldots,\partial/\partial x^n[/itex]. And this just means that if a vector X is written [itex]X=a^1\partial/\partial x^1+\ldots+a^n\partial/\partial x^n[/itex], then, by definition, dx^i is the linear functional [itex]dx^i(X)=a^i[/itex].

    I've never seen this notation before. Do you have a source?
  10. May 3, 2011 #9
    Possibly http://www.lightandmatter.com/html_books/genrel/ch03/ch03.html#Section3.4 [Broken] is an example. The expression

    [tex]ds^2 = dx^i \; dx_i = dx^2 + dy^2[/tex]

    is said in section 3.4.1, The Euclidean metric, to be a statement of the Pythagorean theorem. If we have a tangent vector with components (x,y) in the natural basis for R2, it's squared length is x2 + y2. So it looks like the dxi here, or perhaps the dxi, denote the components of a tangent vector. (He writes, "It's not particularly important to keep track of which is which, since the relationship between them is symmetric, like the relationship between row and column vectors.) One guess I've had about this is that he's using the widespread "y := y(x)" convention (letting a single symbol do double service, sometimes denoting a function, sometimes its value at a given input), so that, for a tangent vector, s, he's saying let "dxi := dxi(s)". But that's just a guess. Another possibility is that he's using the regular notation but giving it an unconventional interpretation: letting a symbol that normally stands for a tangent or cotangent vector stand, instead, for a hyperreal number. But I don't know enough about either differential geometry using real numbers or about nonstandard analysis, or about what system he might be using to translate between the two interpretations, to understand what that would entail.
    Last edited by a moderator: May 5, 2017
  11. May 3, 2011 #10


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    the way I learned it, tangent vectors are derivations.

    A derivation at at point is linear on functions in a neighborhood of the point and also satisfies the Leibniz rule. One can show that in a coordinates system, every derivation is a combination of partial derivatives with respect to the coordinate functions.

    A directional derivative assigns a number to a direction, the directional derivative of a function in that direction. Directions are just tangent vectors so the directional derivatives with repect to a function define a linear map from directions to numbers i.e. they define a cotangent vector. These may be thought of as infinitesimal displacements of the function.
  12. May 3, 2011 #11
    In elementary calculus, the vector that, together with a point of the manifold, defines a directional derivative in this way is required to be a unit vector, otherwise the value of the directional derivative of a scalar field at a point could be any real number we choose it to be. But unit vectors alone don't form a vector space. When tangent vectors are characterised in this way (as derivations), should we think of the tangent space as comprising "multiples of directional derivative operators", rather than simply "directional derivative operators"?

    "these" = tangent vectors, cotangent vectors, the real numbers which are the value of the directional derivative at a point, the first two (tangent and cotangent vectors), or all three?
  13. May 3, 2011 #12


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    right - if you just take directional derivative for unit vectors then you do not get an element of the cotangent space. But generally there is no metric to measure length so if you take direction derivative to mean for any tangent vector then you get a cotangent vector from it.
  14. May 3, 2011 #13
    Could we express this as: you would get an element of the cotangent space (i.e. an element of the tangent space would be defined), just not every element?

    So when there is a norm defined on a tangent space, directional derivative operators form the proper subset of the tangent space whose elements are tangent vectors with unit length; otherwise every element of the tangent space, except for the zero vector, can be characterised as a directional derivative operator, hence every element of the corresponding cotangent space can be defined in terms of one each of these directional derivative operators?

    In general, in a vector space with no norm, are all nonzero scalings of vectors equal? That is, for every vector a, does the following hold for every scalar, s, not equal to zero: sa = a?
  15. May 4, 2011 #14


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    You should check the definition of directional derivative.

    The key point is that a cotangent vector at a point is a linear map from the tangent space into the base field. So it must be defined on all of the tangent vectors at a point.

    You do not need a norm to define the derivative of a function with respect to a vector. Just choose a smooth curve whose velocity at the point is equal to the vector. Then take the derivative of the function at the point along the curve. This derivative is the value of the cotangent vector on that tangent vector.

    In a coordinate system the directional derivatives of the coordinate functions form a basis for the cotangent space.
  16. May 4, 2011 #15
    Is there a particular aspect of the definition which I'm missing and which I should be looking out for? Is it an aspect which only appears when the definition is generalised beyond the case where every vector space in sight is identified with Rn, or does it also appear in that basic context?

    Okay, I see what you mean.

    How does this define the directional derivative at a point? Given an oriented curve to represent a direction, couldn't we still make the value of the directional derivative at that point any number at all by a suitable choice of parametrization of the curve?

    Did you mean tangent space?
  17. May 10, 2011 #16
    Hate to revive this thread when you guys seem to have come to a conclusion, but I had another question about the cotangent space.

    The only example I see of a covector is the gradient of a scalar function, with the partial derivatives of some smooth scalar function f, being the components of a covector in the basis (dx,dy,dz).

    It makes sense that the gradient is always introduced as an example because of what we've talked about on this thread... there is a mapping into the real numbers without a metric, simply by virtue of some field f's gradient and a vector in the tangent space at some point.

    However, I never see any other examples of covectors. Certainly, there are other meaningful one-forms that are not the gradient of a scalar field, aren't there?
  18. May 12, 2011 #17


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    yes by changing the parametrization of the curve you would change the derivative by a scalar factor. What I meant was that the derivative is the same for any curve that has that exact same vector as its tangent at the point.

  19. May 12, 2011 #18


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    yes there are all kinds of 1 forms that are not gradients. But ... at a point every 1 form is a linear combination of the differentials of the coordinate functions.

    the gradient is not a covector. The differential of a function is. The concept of gradient requires an inner product in which case the gradient is the dual of the differential of the function and is a vector not a covector. The gradient changes as the inner product changes.
  20. May 12, 2011 #19
    Woah wait a sec... I always used "one-form" and "covector" interchangeably. Were you humoring me with the first sentence? Is a gradient a one-form but not a covector, or are they in fact the same thing?

    Thanks for the clarification regarding gradient vs. differential being dual to each other.
  21. May 12, 2011 #20


    User Avatar
    Science Advisor

    You just contradicted yourself. The gradient is, in general, a one - form not a vector. You don't get a vector when you take the exterior derivative of a scalar function.
  22. May 13, 2011 #21
    Okay, good. It was the name directional derivative that was confusing me; I was assuming it meant that only the direction was significant. I also got some help on this issue from Hurkyl and HallsofIvy here:


    Let [itex]x : M \rightarrow \mathbb{R}^n[/itex] be a chart on a manifold [itex]M[/itex]. Let [itex]p[/itex] be a point in the domain of [itex]x[/itex]. Let [itex]f:M \rightarrow \mathbb{R}[/itex]. Let [itex]x^i[/itex] be the coordinate functions associated with [itex]x[/itex]. Then, for all [itex]i \in \left \{ 1,...,n \right \}[/itex], let [itex]n[/itex] tangent vectors

    [tex]\left ( p, \frac{\partial }{\partial x^i} \right )[/tex]

    be defined by

    [tex]\left ( p, \frac{\partial }{\partial x^i} \right )(f) := \partial_i f \circ x^{-1} \bigg|_{x(p)}.[/tex]

    These tangent vectors form a basis for [itex]T_pM[/itex], called the coordinate basis, and, at least in one way of speaking, might be referred to as "directional derivatives" associated with the chart, and hence its coordinate functions. That's what I was thinking of. But having typed this all out, I see that "directional derivatives of the coordinate functions" isn't exactly what they are. So what are "the directional derivatives of the coordinate functions"? These particular tangent vectors act on the coordinate functions thus:

    [tex]\left ( p, \frac{\partial }{\partial x^j} \right ) (x^i) = \delta^i_j = \left \langle \mathrm{d}x^i, \frac{\partial }{\partial x^j} \right \rangle \bigg|_p.[/itex]

    Is it these things, (p,dxi), that you're calling "directional derivatives of the coordinate functions", being a basis for the cotangent space at p?
  23. May 13, 2011 #22
    Here's my current understanding.

    Definition of dual space and covector. Let V be a finite dimensional vector space over, for example, the real numbers. V* denotes its dual space, another vector space whose vectors are scalar-valued (i.e. in this example real-valued) linear functions on V. Elements of V* are called dual vectors or covectors (with respect to V).

    The dual space of the dual space. There's a natural isomorphism between V and the dual space to V*. Because of this, (V*)* can be seen as essentially the same thing as V itself.

    Definition of tangent vector, cotangent vector, vector field. Let M be an n-dimensional differentiable manifold. There's a unique vector space associated with point, p, in M, called the tangent space at p, and denoted TpM. It's dual space, denoted Tp*M, is called the cotangent space at p. A tangent vector field on M is a function which associates each point, p, in M, with a tangent vector in TpM. Likewise, a cotangent vector field associates each point, p, with a cotangent vector in Tp*M.


    It's conventional, in this context, to speak of the tangent space at a point, p, as the vector space, and its vectors as "vectors", with no qualification; hence vectors of the cotangent space at p may be called simply covectors or dual vectors. Sometimes the name 1-form means cotangent vector, or a dual vector in general. Sometimes it means a cotangent vector field. Some authors distinguish, in definitions, between a 1-form (cotangent vector) and a differential 1-form (a cotangent vector field whose coefficient functions in any coordinate basis field are differentiable), but even these authors, in practice, often use the name 1-form for both concepts: cotangent vector and cotangent vector field. By contrast, other authors actually define a 1-form as a cotangent vector field. Since other sorts of "form" usually refer to scalar valued functions of vectors, I'm guessing it was originally intended to mean a cotangent vector. Similarly, sometimes, people speak loosely, and use the name vector and covector to mean tangent vector field and cotangent vector field respectively.

    The object you called the gradient of a scalar field in #16 is a particular cotangent vector field. There are certainly authors who use this naming convention, for example Sean Carroll, Chapter 1, p. 12 of his General Relativity Notes. Kip Thorne does too, in a series of online lectures about gravity waves; he uses the del (nabla) symbol before the letter denoting the scalar field, whereas Sean Carroll uses the lower-case Roman "d". Thorne's gradient written with del is a special case of what's called a connection or covariant derivative. Carroll's gradient written with d is a special case of what's called the exterior derivative. It happens that these two notions coincide when del or d operates on a scalar field; only when these operators are applied to other, higher-dimensional tensor fields are they no longer equivalent.

    Others, however, reserve the name gradient (of a scalar field) for the tangent vector field which is dual to your (Carrol's, Thorne's) gradient, in the sense of the following definition:

    [tex](\text{grad}(f)) \cdot v := (\mathrm{d}(f))(v).[/tex]

    This definition depends on the existence of a metric tensor field (or pseudo-metric tensor field, in the context of relativity); that's a field on M which associates each point with an inner product function (pseudo-inner product function) on the corresponding tangent space. This is the kind of gradient lavinia means, the kind that's "not a covector".
  24. May 13, 2011 #23
    Great, for future people reading this I think the reason it is so easy to be confused is simply notation (mathwonk would be glowing), but correct me if i'm wrong.

    For smooth f with the standard cartesian basis,

    [tex] \nabla f = \frac{\partial f}{\partial x^1} \mathbf{e_1} + \frac{\partial f}{\partial x^2} \mathbf{e_2} + \frac{\partial f}{\partial x^3} \mathbf{e_3} [/tex]


    [tex] df = \frac{\partial f}{\partial x^1} dx_1 + \frac{\partial f}{\partial x^2} dx_2 + \frac{\partial f}{\partial x^3} dx_3 [/tex]

    These are dual to each other, the first is a vector field and the second is a covector field, but if you're like me just learning you have spent your whole life cavalierly plopping partial derivative values in a matrix and never write down the basis as being that of tangent vectors or covectors.
  25. May 17, 2011 #24


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    sorry a gradient is not a covector. I was being careless.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook