Let $(M^n,g)$ be a Riemannian manifold and $f:\Sigma^k\to M^n$ a minimal submanifold.
The Morse index of $\Sigma$ is the number of negative eigenvalues of the stability operator $L:\Gamma(N\Sigma)\to \Gamma(N\Sigma)$ acting on (smooth) sections of the normal bundle $N\Sigma$ of $\Sigma$ in $M$.
The Morse index of $\Sigma$ need not be finite in general, however, under certain conditions it is.
For example: the classical Morse index theorem says that if $\gamma:[0,1]\to (M^n,g)$ is a geodesic, then the Morse index of $\gamma$ is finite and equal to the number of points on $\gamma$ which are conjugate to $\gamma(0)$ (counted with multiplicity).
Another example: a complete oriented minimal surface in $(\mathbb{R}^n,\delta)$ with finite total curvature has finite morse index.
Question: Suppose $\Sigma^{n-1}\subset (M^n,g)$ is a closed (compact without boundary) minimal hypersurface. Is the Morse index of $\Sigma$ finite?
If this is not true in general, which extra assumptions are sufficient to ensure the Morse index is finite?