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I have a doubt about the definition of the tangent space to a manifold $M$ using equivalence classes of smooth curves. I understood the definition using derivations, which in the end uses curves showing that each curve defines a derivation which is it's tangent vector.

But I hear there's a definition which says that a vector is an equivalence class of curves. I really don't understand what that's supposed to mean. I'm associating a vector with a curve? But the vector shouldn't be associated with the derivative of the curve? I know what's an equivalence class, but I really don't understand how this fits into this definition.

I know those questions may seem silly, but I think it's important to know how to construct the tangent space both as equivalence class of smooth cuvers and as derivations.

Thanks in advance for the aid.

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It will probably help if you write in detail the definition which you do not understand. –  Mariano Suárez-Alvarez Sep 22 '12 at 23:51
    
@MarianoSuárez-Alvarez, the definition is as follows: Two curves $\gamma_1$ and $\gamma_2$ at $p$ are tangent to each other if for a chart $(U, \phi)$ we have $(\phi \circ \gamma_1)'(0) = (\phi \circ \gamma_2)'(0)$. A tangent vector at $p$ is an equivalence class of mutually tangent curves at $p$. –  user1620696 Sep 23 '12 at 0:04
    
If you are already able to define $T_pM$, then you can recover the definition of tangent space at $p$ by declaring two curves through $p$ to be equivalent if their first derivatves at p (which are elements of $T_pM$) are the same. Then $T_pM$ is isomorphic to the set of equivalence classes. If you don't already know the definition of $T_pM$, you can still define it in terms of equivalence classes of curves, but you must be a little more careful. But the geometric idea is clear. –  treble Sep 23 '12 at 0:17
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@user1620696 take a look at math.stackexchange.com/questions/189385/… –  James S. Cook Sep 23 '12 at 2:19

2 Answers 2

up vote 2 down vote accepted

Let $M$ be a smooth Manifold. Let $p \in M$. Let $U$ be an open neighborhood of $p$. We denote by $C^{\infty}(U)$ the set of real valued smooth functions on $U$. Let $\Lambda_p = \bigcup C^{\infty}(U)$, where $U$ runs through all open neighborhoods of $p$. Let $f, g \in \Lambda_p$. Suppose $f \in C^{\infty}(U)$ and $g \in C^{\infty}(V)$. If there exists an open neighborhood $W$ of $p$ such that $W \subset U \cap V$ and $f|W = g|W$, we say $f$ and $g$ are equivalent. This is an equivalence relation on $\Lambda_p$. We denote by $\mathcal{O}_p$ the set of equivalence classes on $\Lambda_p$. Clearly $\mathcal{O}_p$ is an $\mathbb{R}$-algebra. Let $f \in C^{\infty}(U)$, where $U$ is an open neighborhodd of $p$. We denote by $[f]$ the equivalence class containing $f$.

A derivation of $\mathcal{O}_p$ is a linear map $D\colon \mathcal{O}_p \rightarrow \mathbb{R}$ such that

$D(fg) = D(f)g(p) + f(p)D(g)$ for $f, g \in \mathcal{O}_p$.

The set $T_p(M)$ of derivations of $\mathcal{O}_p$ is a vector space over $\mathbb{R}$ and is called the tangent space at $p$.

Let $\epsilon > 0$ be a positive real number. We denote by $\Gamma_p(\epsilon)$ the set of smooth curves $\gamma \colon (-\epsilon,\epsilon) \rightarrow M$ such that $\gamma(0) = p$. Let $\Gamma_p = \bigcup_{\epsilon>0} \Gamma_p(\epsilon)$. Let $(U, \phi)$ be a chart such that $p \in U$. Let $\gamma_1, \gamma_2 \in \Gamma_p$. Then $\gamma_1$ and $\gamma_2$ are called equivalent at $0$ if $(\phi\circ\gamma_1)'(0) = (\phi\circ\gamma_2)'(0)$. This definition does not depend on the choice of the chart $(U, \phi)$. This defines an equivalence relation on $\Gamma_p(M)$. Let $S_p(M)$ be the set of equivalence classes on $\Gamma_p(M)$. For $\gamma \in \Gamma_p(M)$, we denote by $[\gamma]$ the equivalence class containing $\gamma$.

We will define a map $\Phi\colon S_p(M) \rightarrow T_p(M)$. Let $c \in S_p(M)$. Choose $\gamma \in \Gamma_p(M)$ such that $c = [\gamma]$. Let $f \in C^{\infty}(U)$, where $U$ is an open neighborhood of $p$. We write $D_c([f]) = (f\circ\gamma)'(0)$ for $f \in C^{\infty}(U)$. Clearly $D_c$ is well defined and does not depend on the choice of $\gamma$. Clearly $D_c \in T_p(M)$. Hence we get a map $\Phi\colon S_p(M) \rightarrow T_p(M)$ such that $\Phi(c) = D_c$.

We claim that $\Phi$ is bijective. Let $c, e \in S_p(M)$. Suppose $D_c = D_e$. Suppose $c = [\gamma]$ and $e = [\lambda]$. Let $(U, \phi)$ be a chart such that $p \in U$. Let $\pi_i:\mathbb{R}^n \rightarrow \mathbb{R}$ be the $i$-th projection map: $\pi_i(x_1,\dots,x_n) = x_i$. We denote by $\phi^i$ by $\pi_i\circ\phi$. Since $D_c([\phi^i]) = D_e([\phi^i])$, $(\phi^i\circ\gamma)'(0) = (\phi^i\circ\lambda)'(0)$. Hence $(\phi\circ\gamma)'(0) = (\phi\circ\lambda)'(0)$. Hence $\gamma$ and $\lambda$ is equivalent. Thus $\Phi$ is injective.

Let $D \in T_p(M)$. Let $(U, \phi)$ be a chart such that $p \in U$. We assume that $\phi(p) = 0$. We define $\phi^i$ for $i = 1,\dots,n$ as above. Let $D([\phi^i]) = a_i$ for $i = 1,\dots,n$. There exists $\epsilon > 0$ such that $(a_1t,\dots,a_nt) \in \phi(U)$ for every $t \in (-\epsilon, \epsilon)$. Let $\gamma(t) = \phi^{-1}(a_1t,\dots,a_nt)$ for $t \in (-\epsilon, \epsilon)$. Then it's easy to see that $\Phi([\gamma]) = D$. Hence $\Phi$ is surjective and we are done.

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Because the equivalence class of smooth cuvers is determined by derivations,equivalence class and derivations are one to one correspondence. having derivations,we can give rise to equivalence class,conversely,having equivalence class,we know every element in the class have same derivations at p,that is, same tangent vector,so derivations and equivalence class are viewed as same object which can characterize tangent vector.

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