I know that there is a theorem that a non-empty level set of a regular value of a smooth function $f:M\rightarrow\mathbb{R}$ on a smooth manifold is a regular submanifold (or embedded submanifold) of codimension 1.


Now I wonder if there is also a condition in which the converse holds true. I mean suppose $c\in\mathbb{R}$ is a critical value of $f$ is there a condition under which you know the level set of $c$ is not a regular submanifold. 


If $g:M\rightarrow\mathbb{R}$ is defined by mapping all of $M$ to $0$ then it seems to me that $0$ is a critical value of $g$. Now $g^{-1}(0)=M$ so definitely a regular submanifold of $M$. So I know that it is at least not true without an extra condition. 


Also if the function is second degree polynomial from $\mathbb{R}$ to $\mathbb{R}$ then the level set of a critical value is just a point which would again be a regular submanifold. 


If there is no such general condition on the function $f$ or the critical value then how in general does one go about showing that a critical level set is or is not a regular submanifold?


Thanks in advance 



Can the level set of a critical value be a regular submanifold?
Suppose that $0$ is a regular value of $f$. Then $0$ is a critical value of $g=f^2$, yet the level set $g^{-1}(0)$ is a regular submanifold. In this case all the points on $g^{-1}(0)$ are critical points of $g$.


The general answer is difficult. You need to assume something about $f$. A natural assumption would be that the critical points of $f$ are isolated and of finite type. Near such points one can find local coordinates so that in these coordinate $f$ looks like a polynomial. (This is a generalization of the classical Morse lemma due to Tougeron.) In such cases you need to understand the zero sets of real polynomials which can be challenging. A nice place to consult for such issues is the book by Arnold, Gussein-Zade and Varchenko on singularities of differentiable mappings.


