Ytukyg

Let $g$ be a metric in $mathbb{R}^{3}$ defined as $partial_{x}, partial_{y}, partial_{z}$ are orthogonal everywhere, and $g(partial_{y},partial_{y})=1, g(partial_{z},partial_{z})=f(x), g(partial_{x},partial_{x})=f(x)$ for $f$ being a positive function. Compute $nabla_{partial_{i}}partial_{j}$ for all $i,j$ being $x,y$ or $z$ and $R^{nabla}$ (the Riemannian curvature tensor). Moreover, show that for $M={x=c}$, the restricted metric is isometric to $mathbb{R}^{2}$.

For the first question, this is just computing the Levi-Civita connection for the metric $g$. This can be computed with the Cristoffel symbols for the metric. However, I'm not sure on how to compute them as we are not given a metric of the form $u dx^{2}+v dy^{2}+t dz^{2}$. What would be a way of writing the metric in this form? Or would it just be easier to compute it using an index-free argument? I'm not sure on how to proceed since we haven't covered much of this in class.

For the second question, since $R^{nabla}$ is tensorial in all slots, it suffices to compute $R^{nabla}(partial_{i},partial_{j})partial_{k}$ for all $i,j,k$ being $x,y,z$. Is this correct? If so, would it just be $nabla_{partial_{i}}nabla_{partial_{j}}partial_{k}-nabla_{partial_{j}}nabla_{partial_{i}}partial_{k}$ since the Lie bracket $[partial_{i},partial_{j}]$ vanishes? For the final question, I think this would be equivalent to showing that $g(partial_{i},partial_{j})=langle partial_{i}, partial_{j} rangle_{Euc}$. How would I go about this? Thank you for the help.

Once you work in the global coordinates $(x,y,z)$ this is just a computation together with a thorough understanding of the definitions.

First, the usual proof of the existence of the Levi-Civita connection gives the coordinate formula for the Christoffel symbols
begin{align} label{a}tag{1}
Gamma_{ij}^k = frac12 g^{kl}(partial_i g_{jl} + partial_j g_{il} - partial_l g_{ij})
end{align}
where I have adopted the usual summation convention, identified $g$ with its matrix in these coordinates and written $g^{kl} = (g^{-1})_{kl}$.

The information you're given about $g$ says precisely that in the obvious coordinates on $mathbb{R}^3$, $g$ has matrix representation
$$g_{(x,y,z)} = begin{bmatrix}
f(x)&0&0& \0&1&0& \0&0&f(x)&
end{bmatrix}.$$
From here, finding $Gamma_{ij}^k$ is just a computation using (ref{a}). You should get the non-zero Christoffel symbols to be
begin{align}
Gamma_{ij}^k(x,y,z) = begin{cases} -frac{partial_xf(x)}{f(x)} quad text{ if } i = j in {1,3} text{ and } k = 1, \ \
frac{partial_xf(x)}{f(x)} quad text{ if } {i,j} = {1,3} text{ and } k = 3.
end{cases}
end{align}
Computing the curvature tensor is again just a computation in the global coordinates using the identity
$$R = R_{ijk}^{,,,,,l} dx^i otimes dx^j otimes dx^k otimes partial_l$$
where $R_{ijk}^{,,,,,l}$ satisfies $R_{ijk}^{,,,,,l} partial_l = R(partial_i,partial_j) partial_k$. As you noticed, this computation is made easier by the fact that the Lie brackets $[partial_i, partial_j]$ all vanish and hence is straightforward (if lengthy) now you know the Christoffel symbols.

We are left to see that $M = {x=c}$ is isometric to $mathbb{R}^2$. First, let's see what the induced metric on $M$ is. It is clear that $(c,y,z) mapsto (y,z)$ is a global coordinate chart on $M$ and that $operatorname{Id}:M to mathbb{R}^3$ is an immersion. Let $tilde{g} = text{Id}^* g$ be the induced metric on $M$. It is not too hard to check that $tilde{g}$ has matrix
$$tilde{g}_{(c,y,z)} = begin{bmatrix} 1&0\0&f(c) end{bmatrix}$$ with respect to our global coordinates on $M$.

Notice here that $g$ and $langle cdot, cdot rangle_{text{Euc}}$ act on different spaces so one can only write $g(partial_i, partial_j) = langle partial_i, partial_j rangle_{text{Euc}}$ up to some identification between the spaces - i.e. if we are thinking of $partial_k$ on the RHS as really meaning $dPhi (partial_k)$ for some diffeomorphism $Phi: M to mathbb{R}^2$. I assume that your idea was to take $Phi(c,y,z) = (y,z)$ but it is clear from the matrix form of $tilde{g}$ that this won't work.

Instead, it is natural to define the map $Phi: M to mathbb{R}^2$ by $Phi(c,y,z) = (y, sqrt{f(c)} z)$. It is immediate that $Phi$ is a diffeomorphism so we want to check that $tilde{g} = Phi^* langle cdot, cdot rangle_{text{Euc}}$. For this, it is enough to check that $Phi^* langle cdot, cdot rangle_{text{Euc}}$ has the right matrix representation.

This is another computation. We have, for example,
begin{align}
Phi^* langle cdot, cdot rangle_{text{Euc}}(partial_z |_p, partial_z |_p) = langle dPhi_p (partial_z |_p), dPhi_p (partial_z |_p) rangle_{text{Euc}}
end{align}
Now, it isn't too hard to see that for $F in C^infty(mathbb{R}^2)$ and $p = (c,y,z)$ we have
$$dPhi_p (partial_z |_p)(F) = partial_z |_p (F circ Phi) = sqrt{f(c)} partial_z|_{Phi(p)} F.$$
Therefore $dPhi_p (partial_z|_p) = sqrt{f(c)}partial_z|_{Phi_p}$ and so
$$Phi^* langle cdot, cdot rangle_{text{Euc}}(partial_z |_p, partial_z |_p) = f(c) = tilde{g}(partial_z |_p, partial_z |_p).$$
The other entries in the matrix follow similarly.