5

If we have a map from a Riemannian manifold $(M,g)$ let's call it $\phi$ to reals, with $x^i$ the local coordinates on $M$. The curved Laplace equation (Laplace-Beltrami) looks like $$ g^{ij}\left(\phi_{ij}-\Gamma^k{}_{ij}\phi_{k}\right)=0 $$ Where $\Gamma^k_{ij}$ are Christoffel symbols of $g$ and I denote derivatives by subscripts. Looking at the structure we can notice that adding to $\Gamma^k_{ij}$ a tensor let's say $S^k_{ij}$ which has the property that the vector $g^{ij}S^k{}_{ij}=0$, this change doesn't affect the equation. This would correspond to adding a metric $h$ to $g$ such that $S$ are Christoffel symbols of $h$.

This would for example mean there are more Riemannian spaces $(M,g)$ for which the Laplace equation looks the same, also there are Riemannian spaces $(M,g)$ for which the Laplace equation looks exactly like a flat one.

First of all, I would like to ask if there is some mistake in my reasoning? Secondly, I would like to ask for some references where this is explored, I tried googling but I didn't quite know what to write and that what I tried didn't quite find any papers or something like that.

Maybe this could be used to find the simplest metric to a particular solution.

Thank you very much.

Stanislav Hronek
  • 117

1 Answers1

5

Yes, there's a mistake in your reasoning. If you change the metric, the factor $g^{ij}$ in your formula will change too, not just the Christoffel symbols, and that will change the Laplace operator.

In fact, the Laplace-Beltrami operator on a smooth manifold completely determines the metric. Here's an outline of how that works (borrowed and slightly expanded from my answer to this question):

Suppose $M$ is a smooth manifold and $\Delta: C^\infty(M)\to C^\infty(M)$ is a linear operator which we know to be the Laplace-Beltrami operator of some Riemannian metric $g$. The cometric $g^*$ (i.e., the metric on $1$-forms) is the principal symbol of $\Delta$, which can be computed as follows: Given $x\in X$ and $\xi\in T_x^*X$, choose a smooth function $f$ defined on a neighborhood of $x$ satisfying $f(x)=0$ and $df_x = \xi$. Then $$ g^*(\xi,\xi) = \tfrac 1 2 \Delta(f^2) (x). $$ From this, $g^*$ is determined by polarization: $$ g^*(\xi,\eta) = \tfrac 1 4 \big(g^*(\xi+\eta,\xi+\eta) - g^*(\xi-\eta,\xi-\eta)\big). $$

Once $g^*$ is known, the Riemannian metric $g$ is determined just by inverting the matrix of $g^*$ in any local coordinates.

Jack Lee
  • 50,850