> [!definition] > > Let $X$ be a $C^p$-[[Manifold|manifold]] and $J \subset \real$ be an open interval[^1] containing the origin. A [[Manifold Morphism|morphism]] $\gamma: J \to X$ is a **curve in** $X$, with $\gamma(0)$ being its *starting point*. # Velocities > [!definition] > > Let $\gamma: J \to X$ be a curve, $(J, \text{Id})$ be the identity [[Atlas|chart]]. Let $t_0 \in J$ and $(V, \varphi)$ be a chart at $\gamma(t_0)$, then the **velocity** of $\gamma$ at $t_0$ is > $ > \gamma'(t_0) = d\gamma_{t_0} \cdot \ol{1} \in T_{\gamma(t_0)}X > $ > its [[Differential|differential]] evaluated at $1 \in T_{(J, \text{Id}, t_0)}J$. > [!theorem] > > Let $X$ be a $C^p$-manifold, a point $p \in X$, and a [[Tangent Space|tangent vector]] $v \in T_pX$, then there exists a curve starting at $p$ whose velocity is $v$. > > *Proof*. Let $(U, \varphi) \in X$ be a chart such that $\varphi(p) = 0$[^2]. Suppose that $v = (U, \varphi, p, \hat{v})$, and let $\varepsilon > 0$ such that $\varepsilon \hat v \in \widehat{U}$. > $ > \hat\gamma: J_\varepsilon \to \widehat{U} \quad t \mapsto t\hat{v} > $ > then $\hat \gamma$ is a curve in $\widehat U$ starting at $\hat p$ with velocity $\hat v$. Taking $\gamma = \varphi^{-1} \circ \hat\gamma$ yields the desired curve. # Derivatives Along Curves > [!theorem] > > Let $X$ be a smooth n-manifold, $\gamma: J \to X$ be a smooth curve, and $f \in C^\infty(X)$ be a smooth function, then > $ > (f \circ \gamma)'(t) = df_{\gamma(t)}(\gamma'(t)) > $ > *Proof*. > $ > \begin{align*} > df_{\gamma(t_0)}(\gamma'(t_0)) &= \gamma'(t_0)f \\ > &= d\gamma_{t_0}\paren{\frac{d}{dt}\bigg|_{t_0}}f \\ > &= \frac{d}{dt}\bigg|_{t_0}(f \circ \gamma) \\ > &= (f \circ \gamma)'(t_0) > \end{align*} > $ # Curve Segments > [!definition] > > Let $J \subset \real$ be an open interval and $I \subset J$ be a closed interval. A **curve segment** in $X$ is a continuous mapping $\gamma: J \to X$. If $\gamma$ is $C^p$, then it is a $C^p$ curve segment. > > If there exists a finite partition $a = a_0 < \cdots, a_k = b$ of $[a, b]$, such that $\gamma|_{[a_{i - 1}, a_i]}$ is $C^p$ along $[a_{i - 1}, a_i]$, then $\gamma$ is a piecewise $C^p$ segment. > [!theorem] > > Let $X$ be a [[Connected|connected]] $C^p$-manifold modelled on $E$, then any two points of $X$ can be joined by a piecewise $C^p$-curve segment. > > *Proof*. Let $x \in X$ and define > $ > \mathcal{C} = \bracs{p \in X: \exists \text{piecewise } C^p \text{ curve segment in }X \text{ from } p \text{ to }x} > $ > Since [[Connected Component|connected components]] are both [[Open Set|open]] and [[Closed Set|closed]], and $X$ is connected, only $X$ itself is both open and closed. Therefore it is sufficient to show that $\mathcal C$ is both open and closed. > > Let $p \in \mathcal C$ and $(U, \varphi)$ be a chart at $p$, then there exists $r > 0$ such that $B(\hat p, r) \subset U$. Let $\hat q \in B(\hat p, r)$, and let > $ > \wh \gamma: \real \to E \quad t \mapsto (1 - t)\hat p + t\hat q > $ > As $B(\hat p, r)$ is open, there exists $\varepsilon > 0$ such that $\wh\gamma((-\varepsilon, 1 + \varepsilon)) \subset B(\hat p, r)$. So $\gamma = \varphi \circ \wh \gamma: [0, 1] \to X$ is a $C^p$-curve segment that connected $p$ and $q$. By adding this segment to a piecewise curve segment that connects $x$ to $p$, we get that $q \in \mathcal{C}$ as well. So $\varphi\braks{B(\hat p, r)} \in \cn^o(p)$ is contained in $\mathcal C$, so $\mathcal C$ is open. > > Let $p \in \partial \mathcal C$ and $(U, \varphi)$ be a chart at $p$, then there exists a coordinate ball/half-ball $\wh B$ that intersects $\varphi(\cc \cap U)$. Using the same way to construct a line between $\hat p$ and a point $\hat q \in \varphi(\cc \cap U)$, and adding this segment to the piecewise curve segment that connects $x$ and $q$ connects $x$ to $p$. Therefore $p \in \cc$ and $\cc$ is closed. [^1]: Viewed as a manifold with the standard structure. [^2]: Such a chart can be created by translating an existing chart containing $p$ as translation is a smooth isomorphism.