If I have some equation of the form $F(x,y,z)=0$ which defines level set of the function $F$ and I want to know what is the dimension of the set $A:=\{(x,y,z):F(x,y,z)=0\}$ Is there any special criteria to use?
for example if The gradient of $F$ is not $0$ at some point $p$ then the level set is locally surface. but if $\nabla F(p)=0$ what can I say on the level set $A$?
thank you
I think you will get into the Dragon lair if you don't assume further things on $F$. To show how horrible things can be: if $C\subset \mathbb{R}$ is a closed set, then there is a smooth function $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $f^{-1}(0)=C$.
Basically you should not think of the dimension as a thing coming from the non-degeneracy of $F$ ($\nabla F\not=0$), but the dimension of the codomain of $F$ as giving the codimension of your level surface. More specifically:
Let $F:\mathbb{R}^n\rightarrow \mathbb R^k$ is a smooth function and $0\in \mathbb R^k$ is a regular value. This means that at each point $p$, such that $F(p)=0$ the derivative $dF$ is surjective. It then $F^{-1}(0)$ is a submanifold of $\mathbb{R}^n$ of dimension $n-k$. You should thing of the function $F$ giving $k$ constraints, reducing the dimension by $k$.
Sard's theorem says that almost all values of a function are regular. This implies that almost all level surfaces of a function are nice smooth manifolds of the expected dimension.