# Geometry and Topology Seminar: Negative eigenvalues of the conformal Laplacian

Speaker: Guillermo Henry, Universidad de Buenos Aires

Abstract: Let $(M,g)$ be a compact Riemannian manifold of dimension $n\geq 3$. The conformal Laplacian is the linear elliptic operator defined as follows $$L_g:=\frac{4(n-1)}{(n-2)}\Delta_g +s_g,$$ where $\Delta_g$ and $s_g$ denote the Laplace-Beltrami operator and the scalar curvature of $(M,g)$, respectively. The geometric meaning of conformal Laplacian is the following. If $h$ is a Riemannian metric belonging to the conformal class of $g$, that is $h=u^{\frac{4}{n-2}}g$ where $u$ is a smooth positive function, then the scalar curvature of $(M,h)$ is given by $s_h=L_g(u)u^{-\frac{n+2}{n-2}}.$ The spectrum of $L_g$ is an unbounded, non-decreasing sequence of eigenvalues. The sign of each eigenvalue is a conformal invariant. According to the resolution of the celebrated Yamabe Problem, within each conformal class there exists a metric with constant scalar curvature, and its sign coincides with the sign of the first eigenvalue of the conformal Laplacian. Consequently, there are obstructions to the existence of metrics with a non-negative first eigenvalue. However, it is well known that there are no obstructions to the existence of metrics for which the first eigenvalue is negative. Let $\Lambda_0(M)$ be the minimum of non-positive eigenvalues that the conformal Laplacian of a metric on $M$ can have. In this talk we will show that for any $k$ greater than or equal to $\Lambda_0 (M)$, there exists a Riemannian metric on $M$ such that its conformal Laplacian has exactly $k$ negative eigenvalues. Also, we will discuss upper bounds for the invariant $\Lambda_0 (M)$. This talk is based on a joint work with Jimmy Petean.

Host: Quo-Shin Chi

This will be a Zoom talk broadcasted live in Room 207 in Cupples I.