# Maximizing metrics for the first eigenvalue

Can someone explain me what does the following abbreviation denote:

$$\lambda_1(\Sigma,g)\mathrm{area}(\Sigma,g)?$$

The context I am learning about it are closed Riemannian surfaces $(\Sigma,g)$ and

where the spectrum of the Laplace operator acting on smooth functions, is purely discrete and can be written as

$$0=\lambda_0<\lambda_1(\Sigma,g)\leq\lambda_2(\Sigma,g)\leq\lambda_3(\Sigma,g)... \to \infty$$

## 1 Answer

My guess is that if we let $\Delta_g = \nabla_g\cdot\nabla_g$ and consider $\lambda_i,\phi$ to solve the Helmholtz equation $\Delta_g\phi + \lambda_i\phi=0$ for all $i$, then $\lambda_1(\Sigma,g)\in\mathbb{R}$ is simply the $i$th eigenvalue of the spectrum of the Laplace-Beltrami operator, for the Riemannian manifold $(\Sigma,g)$; i.e. it is a single fixed number associated to the manifold.

The "area" of a Riemannian manifold is defined analogously to that of a surface in Euclidean space (which is a type of Riemannian manifold). However, because these "surfaces" generalize to arbitrary dimensions, one generally calls it "volume". Recall that the metric tensor $g$ lets you define the length of vectors, as well as inner products between such vectors, on $\Sigma$ (or more accurately its tangent spaces at each point). Then we can define lengths of curves using $g$. We can also define a volume form (more here, here, or here), which in coordinates $x^j$ is given by: $$\omega_g = \sqrt{\det g\,}\; dx^1\wedge\ldots\wedge dx^n$$ for an $n$ dimensional manifold. Then the volume is: $$\text{Vol}(\Sigma) = \int_\Sigma \omega_g$$ but note that this only makes sense on a single chart where $x^j$ are defined!

However, let's assume you are concerned with 2D Riemannian manifolds that are 2D surfaces in 3D. Then, the "volume" of the Riemannian manifold is just the surface area.

As an example, consider $(\Sigma,g)=(S,I)$, where $\{(0,0)\leq (x,y) \leq (a,b)\}=S\subset\mathbb{R}^2$. Then, $\det g = 1$ and we get: $$\text{Vol}(S) = \int_S \omega_g =\int_0^a\int_0^b dx^1\wedge dx^2 = \int_0^a b\, dx^2 = ab = \text{area}(S)$$ as we would expect.

Note there are some interesting connections between $\lambda_1$ and $\text{area}(\Sigma)$. For instance, Kac (1966) showed: $$\sum_{i=1}^\infty \exp(-\lambda_i t) \approx \frac{1}{4\pi t}\left[ \text{area}(\Sigma) - \sqrt{4\pi t\,}\ell(\partial\Sigma) + O(t) \right]$$ where $\ell$ denotes Riemannian length. (See here for a bit more). So for small $t$, $\lambda_1$ is the dominant factor in determining $\text{area}(\Sigma)$ in some sense!