Find Riemannian metric $g$ that induces given family of norms.

Question

I have a problem that I am stuck at and I would really appreciate your help. Let's say we have defined a metric on a subset (say, a region) $\Omega \subseteq \mathbb{C}$ as a continuous function $\rho: \Omega \rightarrow [0, \infty)$ that is twice continuously differentiable everywhere where $\rho > 0$. For simplicity let's suppose that this is the case for all $z \in \Omega$. Then we defined the curvature of this metric as $\kappa_{\Omega,\rho}(z) = \kappa(z) = \frac{-\Delta \log \rho(z)}{\rho(z)^2}$. Now, given such a metric $\rho$, we get a family of norms $|\cdot|_{\rho,z}$ on each of the tangent spaces $T_z\Omega = \mathbb{C}$ by defining $|\xi|_{\rho,z} = \rho(z) \cdot ||\xi||$, where $||\cdot||$ denotes the Euclidean norm. That's the setup. I am now supposed to find a Riemannian metric $g$ that induces the above family of norms and then calculate the Gaussian curvature of $g$ to compare it with the expression $\kappa$ above. Sadly, I lack an approach. I am supposed to apply what I have learnt in my Differential Geometry 1 lecture last semester, but there we only made sense of curvature for submanifolds of co-dimension 1. So the idea I had was to regard $\Omega$ as a subset of $\mathbb{R}^2$ and immerse it into $\mathbb{R}^3$ such that the immersion $f$ gives a $g$ that induces the given metric. But that did not work, as I couldn't find $f$, such that it made sense. Online I found some explanations using differential forms, something I am not supposed to use. I would be immensly grateful for some tips!

Answer 1

The Riemannian metric is in this particular case it is a just a rescaling of the Euclidean norm, since each basis vector is scaled by the same amount (for the tangent space at a particular point, the amount varies between points). If you want to be rigorous you can use the parallelogram rule for inner products to justify that this. This is an especially great Riemmanian metric to work with because it will be orthogonal in the usual basis.

As a side note, $\Omega \subseteq \mathbb{C}$ rather than $\Omega \subseteq \mathbb{R}^2$ is an unusual choice that (I presume) the course instructor made - you may notice that we don't actually use the complex structure anywhere. Calling $\rho$ a metric is likewise quite unusual and potentially confusing.

I can't think of how to embed $\Omega$ in $\mathbb{R}^3$, I'm afraid, so unless you can come up with one, I think you should try to deal with it using the intrinsic definition of curvature, if you have seen it. Have you been shown the metric connection $\nabla$, and how curvature can be defined using that? If so you could then use the Koszul formula to calculate what you want. This will be less difficult than usual because the metric is orthogonal.

Explicitly, my approach (which you might have to adapt based on what you are and aren't allowed to use) would be:

1. Parametrize $\Omega$ with $x,y$ coordinates by treating $\mathbb{C}$ as $\mathbb{R}^2$.

2. Find the the values of $\langle X,Y \rangle, \langle X,X\rangle$ and $\langle Y,Y\rangle$ at each point in $\Omega$ by following Didier's comment, where $X$ and $Y$ are 'usual basis': the unit vectors that uniformly point in the $x$ and $y$ directions. You should find that $\langle X,Y\rangle=0$ (i.e. they are an orthogonal basis in the usual inner product), and this will make every subsequent step much, much easier.

3. Use the Koszul formula to evaluate $\langle\nabla_U V, W\rangle$, where $U,V,W$ can be any selection from $X,Y$. (Thanks to the orthogonality this won't be too bad, also if you're confused by the Lie bracket $[X,Y]$ just ignore it, it is just $0$ for this particular choice of $X$,$Y$.)

4. Use step three to derive $\nabla_U V$ for $U,V$ in $X,Y$ (i.e. $\nabla_X X$, $\nabla_X Y$, etc.)

5. Use the results of part 4 in the intrinsic formula for gaussian curvature:

$$\kappa = \frac{\langle(\nabla_X \nabla_Y - \nabla_Y \nabla_X) X, Y\rangle}{\langle X,X\rangle \langle Y,Y\rangle - \langle X,Y\rangle^2}$$

P.S. If you haven't seen the connection $\nabla$, you might have seen the Christoffel symbols, $\Gamma_{ij}^k$, they are two ways of representing the same concept and I can re-write this answer in terms of them if you prefer them.

Good luck!

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p.p.s. re-write with coordinate-dependent notation

In the following also going to change notation for $X$, $Y$ to $\mathbf{e}_x$ and $\mathbf{e}_y$ respectively.

I'll also change $\langle X,Y \rangle$ to $g_{xy}$ since maybe this is also more familiar.

The equation that links the Christoffel symbols of the first kind and the metric connection is:

$$\nabla_{\mathbf{e}_i} \mathbf{e}_j = \sum_k \Gamma_{ij}^k \mathbf{e}_k$$

Essentially the Christoffel symbols are a coordinate dependent expression, whereas the metric connection contains the same information but in terms of vector fields.

So you should be able to verify for yourself the connection between the metric connection and the Christoffel symbols of the second kind: $\langle \nabla_{\mathbf{e}_i} \mathbf{e}_j, \mathbf{e}_k \rangle = \Gamma_{kij}$.

Thankfully with a orthogonal basis $g_{ij} = 0$ if $i \neq j$, the relationship between the first and second Christoffel symbols is much simpler than usual, and I'll use this to simplify some expressions in the following.

Making heavy use of the following page suggested by Didier

We can translate the steps I gave above as:

2. Find the the values of $g_{xy}, g_{xx}$ and $g_{yy}$ at each point in $\Omega$

3. (and 4) find $\Gamma_{ijk}$, and from there find $\Gamma_{ij}^k$ for $i,j,k$ ranging through $x,y$. The formulas you need are the first on the linked page, they're equivalent to Koszul's formula I was talking about before, but you don't need to know that.

4. Calculate $\langle (\nabla_X \nabla_Y - \nabla_Y \nabla_X) X, Y \rangle = g_{yy} R^{y}_{xyx}$, again using a formula from the page given. Then divide your result by $\langle X,X \rangle \langle Y,Y \rangle - \langle X,Y \rangle^2 = g_{xx} g_{yy} - g_{xy}^2$ to get the Gaussian curvature (see also the last formula on this page)

Learning the coordinate independent versions of formulas was very useful for me in terms of understanding these concepts, so I would say it's worth your time, even if you have to compute most things using coordinate systems eventually.