- #1

Markus Kahn

- 112

- 14

- Homework Statement
- Let

$$\mathbb{S}^{2}\equiv \{\vec{x}\in\mathbb{R}^3 \,|\, |\vec{x}|^2 = 1\}.$$

Let further ##\psi^\pm_k## be charts on ##\{\pm x_k>0\}##. Assume the transition functions to be smooth. For this part consider in particular ##\psi^+_1 : (x_1,x_2,x_3)\mapsto (u,v)##. Find the components of the two basis vectors

$$ X_u=\partial_u = a^\mu \frac{\partial}{\partial x^\mu}\quad X_v=\partial_v = b^\mu \frac{\partial}{\partial x^\mu}, $$

where ##\mu\in\{1,2,3\}##, with respect to the partial derivatives of ##\mathbb{R}^3## by calculating

$$ X_u(f|_{\mathbb{S}^2}) = \partial_u (f\circ (\psi^+_1)^{-1}),\quad X_v(f|_{\mathbb{S}^2}) = \partial_v (f\circ (\psi^+_1)^{-1}),$$

where ##f## is a differentiable function of ##\mathbb{R}^3##.

- Relevant Equations
- All in the exercise statement above.

I'm a bit confused about the notation used in the exercise statement, but if I'm not misunderstanding we have

$$\begin{align*}(\psi^+_1)^{-1}:\begin{array}{rcl}

\{\lambda^1,\lambda^2\in [a,b]\mid (\lambda^1)^2+(\lambda^2)^2<1\}&\longrightarrow& \{\pm x_1>0\}\subset \mathbb{S}^2\\

(\lambda^1,\lambda^2)&\longmapsto &(\sqrt{1-(\lambda^1)^2-(\lambda^2)^2},\lambda^1,\lambda^2) ,\end{array}

\end{align*}$$

where ##a,b\in \mathbb{R}##. In the same fashion we can then construct the other five charts if needed. Now let ##f## be a differentiable function on ##\mathbb{R}^3##. My first task is now to calculate ##X_u(f|_{\mathbb{S}^2})##. My first attempt was

$$\begin{align*}X_u(f|_{\mathbb{S}^2})

&= \partial_u (f\circ (\psi^+_1)^{-1})(\lambda) = \partial_u f((\psi^+_1)^{-1}(\lambda)) =a^\mu\frac{\partial}{\partial x^\mu} f((\psi^+_1)^{-1}(\lambda))\\

& = a^\mu \sum_{k=1}^3 \sum_{\alpha=1}^2\frac{\partial f}{\partial [(\psi^+_1)^{-1}]^k} \frac{\partial [(\psi^+_1)^{-1}]^k }{\partial \lambda ^\alpha} \frac{\partial \lambda^\alpha}{\partial x^\mu},\end{align*}$$

where ##\lambda## is an element of the domain of ##(\psi^+_1)^{-1}##. So I basically used the chain rule.. The thing is, I don't really know how this is supposed to be useful in finding the ##a^\mu##.

Am I on the right track? If so, how do I proceed?

$$\begin{align*}(\psi^+_1)^{-1}:\begin{array}{rcl}

\{\lambda^1,\lambda^2\in [a,b]\mid (\lambda^1)^2+(\lambda^2)^2<1\}&\longrightarrow& \{\pm x_1>0\}\subset \mathbb{S}^2\\

(\lambda^1,\lambda^2)&\longmapsto &(\sqrt{1-(\lambda^1)^2-(\lambda^2)^2},\lambda^1,\lambda^2) ,\end{array}

\end{align*}$$

where ##a,b\in \mathbb{R}##. In the same fashion we can then construct the other five charts if needed. Now let ##f## be a differentiable function on ##\mathbb{R}^3##. My first task is now to calculate ##X_u(f|_{\mathbb{S}^2})##. My first attempt was

$$\begin{align*}X_u(f|_{\mathbb{S}^2})

&= \partial_u (f\circ (\psi^+_1)^{-1})(\lambda) = \partial_u f((\psi^+_1)^{-1}(\lambda)) =a^\mu\frac{\partial}{\partial x^\mu} f((\psi^+_1)^{-1}(\lambda))\\

& = a^\mu \sum_{k=1}^3 \sum_{\alpha=1}^2\frac{\partial f}{\partial [(\psi^+_1)^{-1}]^k} \frac{\partial [(\psi^+_1)^{-1}]^k }{\partial \lambda ^\alpha} \frac{\partial \lambda^\alpha}{\partial x^\mu},\end{align*}$$

where ##\lambda## is an element of the domain of ##(\psi^+_1)^{-1}##. So I basically used the chain rule.. The thing is, I don't really know how this is supposed to be useful in finding the ##a^\mu##.

Am I on the right track? If so, how do I proceed?