Exercise 3.17 (Change of coordinates)

Question

Let $(x,y)$ denote the standard coordinates on $\mathbb{R}^2$. Verify that $(	ilde{x},	ilde{y})$ are global smooth coordinates on $\mathbb{R}^2$, where
$$\widetilde{x}=x, \quad \widetilde{y} = y + x^3.$$
Let $p$ be the point $(1,0)\in\mathbb{R}^2$ (in the standard coordinates), and show that
$$\left.\dfrac{\partial}{\partial x}ight|_{p} 
eq \left.\dfrac{\partial}{\partial \widetilde{x}}ight|_{p},$$
even though the coordinate function $x$ and $\widetilde{x}$ are identically equal.

Elu Thingol · Accepted Answer

\paragraph{Part (a).} Before we start, let's quickly make sure that the coordinate map
$$\widetilde{\varphi}: \mathbb{R}^2 \to \mathbb{R}^2, \quad \begin{pmatrix} x\ y\end{pmatrix} \mapsto \begin{pmatrix}
        x\
        y+x^3
\end{pmatrix} \equiv \begin{pmatrix}\widetilde{x}\\widetilde{y}\end{pmatrix} $$
is indeed a smooth coordinate map. First, $\widetilde{\varphi}$ is smooth, since it is a composition of smooth functions. Second, $\widetilde{\varphi}$ is a bijection and it is easy to verify that its inverse $\widetilde{\varphi}^{-1}$ is smooth and is given by
\begin{equation*}
    \widetilde{\varphi}^{-1}: \mathbb{R}^2 \to \mathbb{R}^2, \quad \begin{pmatrix}\widetilde{x}\\widetilde{y}\end{pmatrix} \mapsto \begin{pmatrix}
        \widetilde{x}\
        \widetilde{y}-\widetilde{x}^3
\end{pmatrix} \equiv \begin{pmatrix} x\ y\end{pmatrix}.
\end{equation*}

\paragraph{Part (b).} Now take some $f \in C^{\infty}(U)$, for instance
$$f:\mathbb{R}^2 \to \mathbb{R}, \quad f\begin{pmatrix}
    x\y\end{pmatrix} := x+y.$$

On one hand, we have
$$\left.\frac{\partial f}{\partial x}\right|_p = 1.$$

On the other hand, we have
\begin{align*}
    \left.\frac{\partial f}{\partial \widetilde{x}}\right|_p 
    &= \left.\frac{\partial \left(f \circ \widetilde{\varphi}^{-1}\right)}{\partial x}\right|_{\widetilde{\varphi}(p)}\[7.5pt]
    &= \left.\frac{\partial \left(x \mapsto x + y - x^3\right)}{\partial x}\right|_{\widetilde{\varphi}(p)}\
    &= \left.1 -3(x)^2\right|_{\widetilde{\varphi}(p)}\
    &= 1 -3\widetilde{x}^2
\end{align*}

For $p=(1,0)$, the two values are not equal.

Exercise 3.17 (Change of coordinates)

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Exercise 3.17 (Change of coordinates)

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