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Reading Do Carmo, differential forms, when the tangent space is defined there it says that the vectors $e_i$ form a canonical basis for $\mathbb{R}^n_0$, and then it considers the "translates" for $p$ denoted by $(e_i)_p$ form a basis for $\mathbb{R}^n_p$, I don't understand what is it meant for the translates. If someone can please explain this to me. Thank you!

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  • $\begingroup$ There is no (unique) notion of what "translates" could be. Who is the manifold, is it considered embedded somewhere? Is there an affine connection? What they are saying indicates that they have a way to send vectors from one tangent space to another. $\endgroup$
    – ameg
    Commented May 28 at 14:36
  • $\begingroup$ As mentioned by a lot of people under this post, it is incorrect to think of tangent vectors at a point as translations of some given vectors at another point. The correct way is to regard them as the "pushforward" of the basis vectors along some coordinate chart. You'll understand what this means if you continue reading. $\endgroup$
    – Sardines
    Commented May 29 at 3:29

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I don't like using the word "translate", because it means adding a fixed vector to every vector in $\mathbb{R}^n$. This is not a linear map.

What Do Carmo means is that there is a standard linear isomorphism between $\mathbb{R}^n_p$ and $\mathbb{R}^n_0$. I would explain this using the definition of a tangent vector viewed as the velocity vector of a curve as follows:

Given any $v \in \mathbb{R}_p^n$, there is a curve $$c: I \rightarrow \mathbb{R}^n$$ such that $c(0) = p$ and $c'(0)=v$. You can now use translation to define a new curve $$\tilde{c}: I \rightarrow \mathbb{R}^n,$$ where $$ \tilde{c}(t) = c(t) - p. $$ Then $\tilde{c}(0) = 0$. Let $$\tilde{v} = \tilde{c}'(0) \in \mathbb{R}^n_0$$ You can show that this defines a linear isomorphism \begin{align*} L_p: \mathbb{R}^n_p &\rightarrow \mathbb{R}^n_0\\ v &\mapsto \tilde{v}. \end{align*} The standard basis of $\mathbb{R}_p^n$ is given by $$ L_p(e_1), \dots, L_p(e_n), $$ where $e_1, \dots, e_n$ is the standard basis of $\mathbb{R}^n_0$.

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In a manifold we don't add vectors that are based at different points. Think of two vectors tangent to a sphere at different points, for example; how would we add them, and where would we locate the sum? We don't. This is different from introductory physics or Euclidean geometry, where it is often convenient to think of moving the vectors. So the $(e_i)_p$ look like the $e_i$ but are in a different vector space centered at $p$.

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Ultimately, the different canonical bases (of the manifold, the tangent space of the cotangent space, etc.) are related to each others via isomorphism, which themselves are linked to the canonical basis of $\Bbb{R}^n$ by diffeomorphism. However, this diffeomorphism is associated to the local chart of the atlas equipping the considered manifold; due to the said locality, the canonical basis can be identified to the one of $\Bbb{R}^n$ everywhere, without being identical at the same time, because each space $\Bbb{R}^n$ "originates" from another point $p$. But all these canonical bases are linked through isomorphism, what Do Carmo calls here 'translation' by a little abuse of language.

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Translates are simply translations of the basis vectors. Think of a plane in R^3 . You can move tangent to the plane starting anywhere in this subspace. Not just at the origin.

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