Let $M$ be a manifold. By a round function we mean a function $M\to{\mathbb{R}}$ whose critical points form connected components, each of which is homeomorphic to the circle $S^1$

For example, let $M$ be the torus. Let $K=]0,2\pi[\times]0,2\pi[$ Then we know that a map $X\colon K\to{\mathbb{R}}^3$ given by $$X(\theta,\phi)=((2+\cos\theta)\cos\phi,(2+\cos\theta)\sin\phi,\sin\theta)$$ is a parametrization for almost all of $M$ Now, via the projection $\pi_3\colon{\mathbb{R}}^3\to{\mathbb{R}}$ we get the restriction $G=\pi_3|_M\colon M\to{\mathbb{R}}$ whose critical sets are determined by $$\nabla G(\theta,\phi)=\paren{{\partial G\over \partial\theta},{\partial G\over \partial\phi}}(\theta,\phi)=(0,0)$$ if and only if $\theta={\pi\over 2},\ {3\pi\over 2}$

These two values for $\theta$ give the critical set $$X\paren{{\pi\over 2},\phi}=(2\cos\phi,2\sin\phi,1)$$ $$X\paren{{3\pi\over 2},\phi}=(2\cos\phi,2\sin\phi,-1)$$ which represent two extremal circles over the torus $M$

Observe that the Hessian for this function is $d^2(G)= \left(\begin{array}{cc} -\sin\theta & 0 \\ 0 & 0 \end{array}\right) $ which clearly it reveals itself as of ${\rm rank}(d^2(G))=1$ at the tagged circles, making the critical point degenerate, that is, showing that the critical points are not isolated.