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.
Best Answer
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.