Exercise 8.18

Question

extbf{Example 8.17.} Let $F: \mathbb{R} ightarrow \mathbb{R}^2$ be the smooth map $F(t)=(\cos t, \sin t)$. Then $d / d t \in \mathfrak{X}(\mathbb{R})$ is $F$-related to the vector field $Y \in \mathfrak{X}\left(\mathbb{R}^2ight)$ defined by
$$
Y=x \frac{\partial}{\partial y}-y \frac{\partial}{\partial x} .
$$

extbf{Exercise 8.18.} Prove the claim in the preceding example in two ways: directly from the definition, and by using Proposition 8.16.

Elu Thingol · Accepted Answer

\paragraph{Using the definition.} First, we calculate $dF\left(\frac{d}{dt}\right)$. 
    %By definition, it is a derivation $w \in T_{(x,y)}\mathbb{R}^2$ such that
    %$$\forall f \in C^{\infty}(\mathbb{R}^2):\ w(f) = \frac{d}{dt}(f \circ F)$$
    Notice that our manifold is identical to its coordinate representation, i.e., $f = \widehat{f}$ and $F= \widehat{F}$. Using the Chain Rule for Partial Derivatives (Corollary C.11), we get at any $T_{t_0}\mathbb{R}$:
    \begin{align*}
        \forall f \in C^{\infty}(\mathbb{R}^2):\ dF_{t_0}\left(\left.\frac{d}{dt}\right|_{t_0}\right)(f) &=  \left.\frac{d}{dt}\right|_{t_0}({f} \circ {F})\[5pt]
        &= \left.\dfrac{\partial {f}}{\partial x} \right|_{F(t_0)} \left.\dfrac{d {F}^1}{dt}\right|_{t_0}
        + \left.\dfrac{\partial {f}}{\partial y} \right|_{F(t_0)} \left.\dfrac{d {F}^2}{dt}\right|_{t_0}\[5pt]
        &= \left.\dfrac{\partial {f}}{\partial x} \right|_{F(t_0)} (-\sin(t_0))
        + \left.\dfrac{\partial {f}}{\partial y} \right|_{F(t_0)} \cos(t_0).
    \end{align*}
    In other words,
    \begin{equation} \label{eq1}
        dF\left(\frac{d}{dt}\right) = -\sin \dfrac{\partial}{\partial x} + \cos \dfrac{\partial}{\partial y}.
    \end{equation}
    Now we calculate $Y_{F}$. We have
    $$Y_{F_{t_0}} = F^1(t_0) \dfrac{\partial}{\partial y} + F^2(t_0) \dfrac{\partial}{\partial x} 
    =  \cos(t_0) \dfrac{\partial}{\partial y} - \sin(t_0) \dfrac{\partial}{\partial x}.$$
    In other words,
    \begin{equation} \label{eq2}
        Y_{F} =  \cos \dfrac{\partial}{\partial y} - \sin \dfrac{\partial}{\partial x}.
    \end{equation}
    We see from \ref{eq1} and \ref{eq2} that $dF(X) = Y_F$.

\paragraph{Using Proposition 8.16} This follows directly from the formula for $dF_{t_0}\left(\frac{d}{dt}\right)(f)$ we have derived in the previous part:
    $$\dfrac{d}{dt}(f \circ F) =  -\sin(t_0) \left.\dfrac{\partial {f}}{\partial x} \right|_{F(t_0)}
        + \cos(t_0) \left.\dfrac{\partial {f}}{\partial y} \right|_{F(t_0)} = (Yf) \circ F$$
    for any $f \in C^{\infty}(\mathbb{R}^2)$.

Exercise 8.18

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Exercise 8.18

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