In a book on general relativity I find this sentence:

Transformation rules of one-forms

Let us consider an open set $\mathbf{U}$ of the manifold $\mathbf{M}$, and choose a coordinate system $\{x^i\}$. We have seen that this defines a coordinate basis for vectors, $\{\vec{e}_{(i)}\} \equiv \left\{\frac{\partial}{\partial x^i}\right\}$ and the dual coordinate basis for one-forms $\{\tilde{\omega}^{(i)}\}_{i=1,...,n}$. If we make a coordinate transformation ...

I am struggling to understand it (and to understand all the rest actually) since it appears that in general relativity we first assume coordinates exist and then derive properties of bases. But in usual geometry we go the other way around! What are we doing here? Can we just assign random tuples of numbers to points on a manifold (not too random of course, at least biijectively and continuously) and then derive the basis in the tangent and cotangent spaces from this arbitrary assignment? Am I right in understanding that we are actually doing this?

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    $\begingroup$ You are mixing up vector spaces and smooth manifolds. In a vector space we first define a basis and then we expand vectors in that basis. The coefficients of expansion work as coordinates. In a smooth manifold the story is different. In particular we now have a vector space at each point (the tangent spaces). We define coordinates on the manifold, which are just homeomorphisms mapping open sets to some $\mathbb{R}^n$, and these coordinates induce a definition of a basis on each tangent space (the coordinate frame). I suggest you pick up the GR textbooks by Carroll or Wald. $\endgroup$
    – Gold
    Oct 12, 2022 at 17:52
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    $\begingroup$ Ah, and there's also a great differential geometry book for physicists called "Modern Differential Geometry for Physicists" by Isham. $\endgroup$
    – Gold
    Oct 12, 2022 at 17:53
  • $\begingroup$ @Gold, aren't your comments more answers than comments? $\endgroup$ Oct 12, 2022 at 21:25
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    $\begingroup$ Seeing this excerpt it's no surprise that you are confused. Go back to where they define "vectors". There should be some mention of "tangent spaces". Then try to square what you've just read with the answers which explain to you how to go from local coords to tangent spaces to vectors. It may be confusing because at first it may seem like tangent spaces depend on the choice of coords, but the discussion should alleviate that worry giving sth like $x \to x' = x'(x)$ hence $\frac{\partial}{\partial x} \to \frac{\partial}{\partial x'} = \frac{\partial x}{\partial x'} \frac{\partial}{\partial x}$ $\endgroup$
    – tobi_s
    Oct 13, 2022 at 3:02
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    $\begingroup$ Note the phrase "coordinate basis". You can have perfectly good bases for the tangent spaces that are not derived from coordinates (that are not "coordinate bases"). $\endgroup$ Oct 13, 2022 at 10:04

3 Answers 3


You can do either one.

  • Any choice of coordinates $x^i$ on a neighborhood $U\subseteq M$ induces a basis $\frac{\partial}{\partial x^i}$ for the tangent space $T_pM$ at each point $p\in U$.

  • Given a basis $\hat e_i$ of a tangent space $T_pM$, we can define a coordinate system in some (possibly small) neighborhood of $p$ by extending the basis at $p$ to entire vector fields, and then "following" the integral curves of those fields.

The former approach is immediate - $x^i \mapsto \frac{\partial}{\partial x^i}$. The latter is more involved, primarily because the extension from the basis vectors $\hat e_i$ at $p$ to basis vector fields $X_i$ defined on all of $U$ is highly non-unique, so a non-canonical choice needs to be made.

For example, if the manifold in question is $\mathbb R^2$ and we consider the basis $$\hat e_1 = \pmatrix{1\\0} \qquad \hat e_2 = \pmatrix{0\\1}$$ at the point $p=(1,0)\in \mathbb R^2$, then we could equally well extend this basis to create the standard cartesian coordinate system, the polar coordinate system, or a hyperbolic coordinate system.

You may object to this by saying that the cartesian coordinate system is the most natural, but the only general sense in which this is true is that the corresponding basis vector fields are obtained by parallel transporting the vectors $\hat e_i$ to each point in space. This is the generalization of "copying and pasting" the vectors at each point of $\mathbb R^2$ which can be applied to manifolds which possess intrinsic curvature.

So in your familiar Euclidean geometry, it may seem just as easy to define vectors and then coordinates, but this is true only in the special case of a Euclidean manifold - which is precisely what we're trying to generalize by studying differential geometry. In the more general case, it is far simpler to define basis vectors from coordinates, so this is what we do in the beginning. Note, however, that things like Riemann normal coordinates are an important example of us starting from basis vectors to define a coordinate system.

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    $\begingroup$ A cautionary remark so that future readers don’t erroneously extrapolate the reasoning here: coordinates $(U,x)$ give rise to a local basis of vector fields $\left\{\frac{\partial}{\partial x^i}\right\}_{i=1}^n$ on $U$. The converse is false: if we are given $n$ arbitrary vector fields $\{X_i\}_{i=1}^n$ defined on an open set $U$, which at each point $p\in U$ form a basis for $T_pM$, then it is not necessarily true that (even after shrinking $U$ sufficiently small) there exist coordinates $(x^1,\dots, x^n)$ such that $X_i=\frac{\partial}{\partial x^i}$ on all of $U$. $\endgroup$
    – peek-a-boo
    Oct 15, 2022 at 0:41

it appears that in general relativity we first assume coordinates exist and then derive properties of bases. But in usual geometry we go the other way around!

The issue is actually slightly different than merely "which comes first". The issue is that coordinates are defined on the manifold while basis vectors are defined in the tangent space. So conceptually there is no first or second* between them.

However, before you are introduced to (pseudo) Riemannian geometry the distinction between the manifold and the tangent space is often neglected. You were dealing with a simple flat manifold where parallel transport is unique and the manifold is isomorphic to the tangent space. In that context you could blur the line considerably and get away with it.

*There is one obvious exception between the general "no first or second" statement. Obviously, coordinates must be defined first before you can define specifically the coordinate basis.


I'd start with coordinates to describe the space, both in geometry and relativity (that can be interpreted as geometry of time-space), so that you can write a point in space as a function of the coordinates $q^i$, $\mathbf{r}(q^i)$.

Now, the position in space is a function of the coordinates as independent variables, and it's quite natural to define the natural basis induced by the coordinates $\{ q^i \}_i$ as $\{ \mathbf{b}_i \}_i$ with

$ \mathbf{b}_i = \dfrac{\partial \mathbf{r}}{\partial q^i}$.

Vectors of the natural basis are not unit vectors or orthogonal vectors in general. From here, it's useful to define:

  • the metric tensor, whose covariant components are the dot product of the vectors of the basis,

    $g_{ij} = \mathbf{b}_i \cdot \mathbf{b}_j$,

  • the reciprocal basis $\{ \mathbf{b}^i \}_i$, s.t. $\mathbf{b}_i = g_{ij} \ \mathbf{b}^j$

  • the contravariant components of the metric tensor, $\mathbf{b}^i = g^{ij} \ \mathbf{b}_j$.

With these bases, you can write a vector field using contravariant or covariant components

$\mathbf{v} = v^i \mathbf{b}_i = v_i \mathbf{b}^i$,

where it's not hard to prove that

$v^i = \mathbf{v} \cdot \mathbf{b}^i \qquad v_i = \mathbf{v} \cdot \mathbf{b}_i$
$v^i = g^{ij} v_j \qquad v_i = g_{ij} v^j$
$\mathbf{b}^i = g^{ij} \mathbf{b}_j \qquad \mathbf{b}_i = g_{ij} \mathbf{b}^j$

Take a look here at interactive hand-written notes (link in light-blue):

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    $\begingroup$ Thank you @basics. But to define the basis vector I need to define the derivative; to define the derivative I need to define limits; to define limits I need to define distances on the manifold. How do I define distances on a manifold? Coordinates are clearly not enough, because if I set a rule to evaluate the distance with one system of coordinates nothing assures me that the distance will be the same in another system of coordinates. I should use geometrical objects, independent of coordinates. But these are vectors, which I defined with derivatives... What's the way out of this circularity? $\endgroup$ Oct 12, 2022 at 17:59
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    $\begingroup$ distance doesn't depend on the coordinates you choose, like vectors and tensors don't depend on the coordinates. You can define it using the metric tensor, that is a geometrical object, independent from the coordinates you use $\endgroup$
    – basics
    Oct 12, 2022 at 18:07
  • $\begingroup$ independent means that a vector/tensor can be written as a linear combination of a set of basis vectors (or tensors). If you change the basis, the components change as well, s.t. the vector/tensor doesn't depend of the choice of the basis you made $\endgroup$
    – basics
    Oct 12, 2022 at 18:15
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    $\begingroup$ You're right. But the independence of distance and shape of geometrical objects from the coordinates you use to describe is one of the preliminary assumption for geometry, something like an absolute/intrinsic nature of geometry (and the physics) $\endgroup$
    – basics
    Oct 12, 2022 at 18:32
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    $\begingroup$ @SalvatoreManfrediD Just as a note - you don’t need to define distances on the manifold (which would require a metric tensor) to define derivatives. All of the structure needed to define vector fields and coordinate bases is present if the manifold is merely smooth. $\endgroup$
    – J. Murray
    Oct 12, 2022 at 19:17

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