Convergence of Riemannian metrics spectra

Question

Consider a one-parameter real analytic family of metrics $g_t$ on a compact manifold $M$ converging to a metric $g$ in $C^k$-norm, for some $k$. It is known that the Laplace spectrum of $g_t$ will converge to the Laplace spectrum of $g$. However, is it true that $\lambda_n(t)$ will converge to $\lambda_n$, where $0 \leq \lambda_1(t) \leq \lambda_2(t) \leq \cdots$ denotes the Laplace spectrum of $g_t$, and $0 \leq \lambda_1 \leq \lambda_2 \leq \cdots$ denotes the Laplace spectrum of $g$? Thanks!

Addendum: I just found the following paper. Regarding the theorem on page 1, I wish to ask more generally: the theorem says that if the operators $A(t)$ are $C^M$, where $M$ could be $\omega, \infty$ or Holder class, then the eigenvalues of $A(t)$ can be parametrized (under certain conditions) to be $C^M$. But does that mean the following: if the operators $A(t)$ are positive, self-adjoint and have discrete spectrum, and for each $t$, we have $0 \leq \lambda_1(t) \leq \lambda_2(t) \leq....$ as the spectrum of $A(t)$, then are $\lambda_i(t)$ parametrized to be $C^M$?

Edit: I am new to MO and MSE and unfortunately was not aware of the practices. I have edited the MSE question requesting people to post their answers here instead.

Further edit: As commented below, the existence of the limit metric is not that important. All I want to know is, if we rearrange the eigenvalues at each step according to increasing order, will the rearranged functions still be $C^M$?

Answer 1

The answer is no, for the reason Fan mentioned above. When two parameterized curves of eigenvalues cross, the resulting ordered eigenvalues will only be Lipschitz.

For a simple example, consider the square $[0,1] \times [0,b]$ with Dirichlet boundary conditions. The eigenvalues are $(m\pi)^2 + (n\pi/b)^2$ for $m,n \in \mathbb{N}$, so the first eigenvalue is $\pi^2(1 + 1/b^2)$, which depends smoothly on $b$. However, the second eigenvalue is $\pi^2(4+1/b^2)$ for $b\leq 1$ and $\pi^2(1+4/b^2)$ for $b>1$, which is not differentiable at $b=1$.

The result you cited will apply to any eigenvalue that is simple, in particular the first one. (This is trivial for a closed manifold, since the first eigenvalue is always 0, but for the Dirichlet problem on a compact manifold with boundary the conclusion is more interesting.)