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I learned that for a Manifold the tangent basis in a point are the partial derivative in the coordinates direction. I understand that there aren't problems in defining these object and that they form a vector space, but can you give me a geometrical intuition about their meaning? I suppose that if the Manifold is the $R^3$ space these concepts should agree with the classical description in which vector are arrows so there should be a connection.

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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Apr 20 at 17:06
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In $\mathbb R^3$ equipped with Cartesian coordinates, there is a one-to-one correspondence between vectors $\vec A = A_1 \hat e_1 + A_2 \hat e_2 + A_3 \hat e_3$ and directional derivative operators $\nabla_\vec A := \vec A \cdot \vec \nabla = A_1 \frac{\partial}{\partial x^1} + A_2 \frac{\partial}{\partial x^2} + A_3 \frac{\partial}{\partial x^3}$. In elementary contexts, we might interpret $\vec A$ as an arrow which lives in $\mathbb R^3$, but this loose idea fails in more general spaces like the 2-sphere $\mathrm S^2$.

We are presented with two options. First, we can embed $\mathrm S^2$ in $\mathbb R^3$ and define a tangent vector to $\mathrm S^2$ as a vector in $\mathbb R^3$ which lies in a plane tangent to $\mathrm S^2$ at some point. For example, the tangent plane to the north pole contains vectors of the form $\vec A = A_1 \hat e_1 + A_2 \hat e_2 + 0 \hat e_3$:

enter image description here

This may seem intuitive, but it comes at a cost. In this relatively simple example, the set of allowed tangent vectors (by which I mean, allowed components) depends on the point to which they are attached. Vectors attached to the points $(0,0,\pm 1)$ are of the form $(A_1,A_2,0)$, while vectors attached to the points $(\pm 1,0,0)$ are of the form $(0,A_2, A_3)$. More generally, vectors attached to a point $(a,b,c)$ must obey the constraint $A_1 a + A_2 b + A_3 c = 0$.

It's not hard to see that for a generic space, this could very quickly become extremely annoying. $\mathrm S^2$ is extremely simple, and can be embedded in an obvious way in $\mathbb R^3$ - contrast this with e.g. the 2-torus (which is also simple, but whose embedding is slightly less so). It's true that an arbitrary manifold can be embedded in some $\mathbb R^n$ (though figuring out which $n$ is not always trivial), but the prospect of embedding a 4D manifold into e.g. a 6 dimensional space and then having vectors with $6$ components subject to 2 algebraic constraints which depend on position is ... not appealing.

The alternative approach is to identify the vector $\vec A$ with the directional derivative operator $\nabla_\vec A$. Directional derivatives can be defined intrinsically on any differentiable manifold without requiring an embedding into a higher dimensional space. In an arbitrary coordinate chart $\{x^1,x^2,\ldots\}$ they can be expressed as $A_1 \frac{\partial}{\partial x^1} + A_2 \frac{\partial}{\partial x^2} + (\ldots)$. Changes of basis follow immediately from the chain rule of elementary calculus, as $\frac{\partial}{\partial y^i} = \frac{\partial x^j}{\partial y^i} \frac{\partial}{\partial x^j}$. So on and so forth.

So, in differential geometry one tends to take the latter approach. Tangent vectors to a manifold are defined as directional derivative operators (evaluated at the point to which they are attached); from this point of view, in a coordinate chart $\{x^i\}$ the partial derivative operators $\frac{\partial}{\partial x^i}$ constitute a natural and convenient basis for each tangent space.

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  • $\begingroup$ If there is a force $\vec F$ applied to a certain point I should write it as $F_1 \frac{\partial}{\partial x^1} + F_2 \frac{\partial}{\partial x^2} + (\ldots)$. But what is the physical meaning of $F_i$? is it the force in the tangent direction to the $q_i$ coordinate? $\endgroup$
    – SimoBartz
    Apr 20 at 15:41
  • $\begingroup$ @SimoBartz Yes. $\endgroup$
    – J. Murray
    Apr 20 at 16:52
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Consider a surface like a sphere. What is the tangent space at some point? It would be a vector space corresponding to different directions and speeds one can move at, at that point. In 3-space these would all correspond to vectors along the tangent plane (the local horizontal plane) to the point. Now, if you move to a different point on the sphere you will get a different bundle along that tangent plane: each vector space is separate from each other.

We need to define some basis vectors for these spaces, and this is where the tangent basis comes in. Normally we just take the differentials of the local coordinates and get basis vectors for every tangent space. As a bonus, since coordinates of nearby points tend to be similar, there is some relation between vectors in one tangent space and another one; setting up this connection is usually the next step.

(One can be more careful and distinguish between 1-forms and vectors; $\partial_x$ is a very different thing from $\mathbf{e}_x$ but in simple cases there is an obvious 1-to-1 map between them.)

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  • $\begingroup$ Sorry I think I wasn't clear, I would like to have an intuitive gemoetric idea of why the derivative operator should be considered as basis of the tangent vector $\endgroup$
    – SimoBartz
    Apr 20 at 14:10

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