Is Gauss curvature a Morse function? Given a Gauss map $\nu: M \rightarrow S^k$ of a orientable, compact manifold, we define the shape operator $S_p = -d \nu: T_p M \rightarrow T_{\nu(p)} S^k$ to be the negative differential. Define the Gauss curvature $K: M \rightarrow \mathbb{R}$ as $K(p) = det(S_p)$.  
My question is whether $K$ is a Morse function. We say a function on a manifold $f: M \rightarrow \mathbb{R}$ is Morse if every critical point is nondegenerate.
EDIT: Assume that $M$ is not a surface of constant Gauss curvature.  
My motivation for asking this question: I am an undergrad studying homological filtrations of a manifold by its Gauss curvature. I know a bit of Morse theory and was hoping to be able to pull out a result or two from Morse theory in my project. 
 A: Not a complete answer, but too long for a comment.
First a couple of remarks on set-up: If an oriented $n$-manifold $M$ is smoothly embedded in some Euclidean space $E^{N}$, then the Gauss map usually refers to the map sending a point $p$ of $M$ to the oriented vector subspace parallel to $T_{p}M \subset E^{N}$, viewed as an element of the Grassmannian of oriented $n$-planes in $E^{N}$; the Grassmannian has dimension $n(N - n)$, so the determinant is defined if and only if $n(N - n) = n$ (i.e., $N = n + 1$), if and only if $M$ is a hypersurface. In this event, the Gauss map can be viewed as taking values in the sphere $S^{n}$ by identifying the oriented hyperplane $T_{p}M \subset E^{n+1}$ with the unit normal compatible with a fixed orientation of $E^{n+1}$.
For simplicity, let's assume $M$ is an oriented surface smoothly embedded in $E^{3}$.
Results of Kazdan and Warner give conditions for a function on a surface to be the Gaussian curvature of some Riemannian metric. Their 1971 survey article in the Bulletin of the AMS is a readable introduction for the compact, orientable case, and their 1975 paper in the Annals of Mathematics contains sharpened results. (The second link may require JSTOR registration, but one statement of interest is on the first page. The full article can be found here.)
The closing section of their 1975 paper notes, "[Nirenberg's solution of the Weyl embedding problem, Communications in Pure and Applied Mathematics 6 (1953) 337-394, implies that] given any smooth strictly positive function $K$ on $S^{2}$, there is a diffeomorphism $\varphi$ of $S^{2}$, a compact, strictly convex surface $\Sigma$ in $E^{3}$, and a conformal diffeomorphism $\psi:\Sigma \to S^{2}$ such that $K \circ \varphi \circ \psi$ is the Gaussian curvature of $\Sigma$."
A: Here is an attempt, but one that is far from complete if one decides to pursue this problem even further.

As Dr. Andreas Blass has pointed out, the $ n $-spheres $ \Bbb{S}^{n} $, equipped with the standard differential structures and metrics, have constant positive Gaussian curvatures, so these curvatures cannot be Morse functions.
The OP was then edited so as to exclude oriented and compact Riemannian manifolds that have a constant Gaussian curvature. Even so, the assertion is not true, as we now show.
Consider the well-known smooth embedding $ \eta $ of the torus $ \Bbb{T}^{2} = \Bbb{S}^{1} \times \Bbb{S}^{1} $ into $ \Bbb{R}^{3} $ defined by
$$
\eta \stackrel{\text{df}}{=}
\left\{ \begin{matrix}
\Bbb{S}^{1} \times \Bbb{S}^{1} & \to & \Bbb{R}^{3} \\
(u,v) & \mapsto & ([c + a \cos(v)] \cos(u),[c + a \cos(v)] \sin(u),a \sin(v))
\end{matrix} \right\}.
$$
Via this embedding, $ \Bbb{T}^{2} $ inherits a metric from $ \Bbb{R}^{3} $ whose Gaussian curvature is given by
$$
\forall (u,v) \in \Bbb{S}^{1} \times \Bbb{S}^{1}: \quad
K(u,v) = \frac{\cos(v)}{a [c + a \cos(v)]}.
$$
This is certainly non-zero. Next, observe that
$$
\forall (u,v) \in \Bbb{S}^{1} \times \Bbb{S}^{1}: \quad
  {\nabla K}(u,v)
= \left( 0,- \frac{a c \sin(v)}{a^{2} [c + a \cos(v)]^{2}} \right).
$$
Hence, the set of critical points is $ \Bbb{S}^{1} \times \{ 0,\pi \} $. Each critical point is then degenerate because the Hessian of $ K $ can easily be shown to have three $ 0 $ entries (due to the lack of a dependence on $ u $) and is therefore singular.

Let $ M $ be an oriented and compact smooth manifold. The space $ \text{Met}(M) $ of Riemannian metrics on $ M $ can be equipped with the $ C^{\infty} $-topology, which turns it into a Fréchet manifold.
As Thomas as mentioned, the set of metrics whose Gaussian curvature is a Morse function may be generic (i.e., open and dense) in $ \text{Met}(M) $. Of course, this is pure speculation at the present moment.
