The sphere in general relativity

[Markus Kahn]

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?

 

[Antarres]

What you are doing here actually, is that you have a sphere embeded in 3D space, and you want to find basis vectors tangent to the sphere if embedding map is given. Now there is no reason why there should be only one set of basis vectors $(u,v)$ that satisfies this requirement. So we can find one of the bases, and others are found simply by rotation in the tangent space.

Your $\psi^+_1$ map looks fine to me for mapping 3D half-sphere onto it's surface. Your chain rule is a little messy though, so let's analyze it a bit.
$(\psi^+_1)^{-1}$ is a map that is transforming coordinates from a coordinate patch on the sphere to the Euclidean space. Therefore, you can actually say $(\psi^+_1)^{-1}(\lambda) = x(\lambda)$. Coordinates on the patch here are actually defined via this $\lambda$ parametrization, that's clear. Now it's common in differential geometry to say that you take -points- on a manifold and map them into some Euclidean space, this map which is called the chart, gives us coordinates of points, so how can we say that our half-sphere has coordinates on itself? Well that's because this map $\psi^+_1$ is not a chart, it is a map that embedds the sphere in 3D space, so the new $\lambda$ parametrization that we get from this embedding by lowering from 3D space onto the sphere, is actually the coordinate system we get by charting a patch of the sphere(in this case it's half-sphere). Hope that doesn't sound confusing.
We can now correct your chain rule:
$$X_u(f\vert_{\mathbb{S}^2}) =\partial_u(f((\psi^+_1)^{-1}(\lambda)) = a^\mu\partial_\mu(f(x(\lambda)))= a^\mu\frac{\partial f}{\partial x^\mu}$$
On the other hand, we also have(bear in mind that in your notation $(u,v) \equiv (\lambda^1, \lambda^2)$:
$$X_u(f(x(\lambda))) = \partial_{\lambda^1}(f(x(\lambda)) = \frac{\partial f}{\partial x^\mu}\frac{\partial x^\mu}{\partial \lambda^1}$$
So the problem with your chain rule was that you made difference between $(\psi^+_1)^{-1}(\lambda)$ and $x$, but those are the same by definition of this map, so your last two fractions in chain rule cancel out.
Now by comparison of these actions, since they are coordinate independent, we see that we get usual vector transformation, except here Jacobian is not square matrix, since this is embedding, not a usual coordinate transformation.
We get:
$$a^\mu = \frac{\partial x^\mu}{\partial \lambda^1}$$
You find similar law for $b^\mu$. Notation can look a bit confusing here, but I hope it wasn't confusing, what I said. You basically got everything right, just got stuck in the notation a bit.

 

[Markus Kahn]

Thank you very much! I really appreciate it. As you said, the notation so far just seems like a mess.. I completely missed that we have $x(\lambda) \equiv (\psi^+_1)^{-1}(\lambda)$, which in hindsight should have been pretty clear (I mean, if I'm not misunderstanding this is the whole point of having a chart in the first place...).

The explicit coefficient then just follow by actually calculating $\partial_{\lambda^1} x^\mu(\lambda)$.

 

[Antarres]

That's right! Glad I was able to help.