# Two representations of the same manifold.

Consider two differentiable functions $f,g:\mathbb{R}^n \rightarrow \mathbb{R}$ which define the level set $M(a,b) = \{x \in \mathbb{R}^n | f(x)=a, g(x)=b\}$.

If the gradients $\nabla f$ and $\nabla g$ are linear independent on every $x \in M(a,b)$ then by the Rank Theorem (Theorem 4.12 of Introduction to Smooth Manifolds, John Lee) $M(a,b)$ is a manifold.

However if we change the representation of $M(a,b)$ by $\{x \in \mathbb{R}^n | h(x)=0\}$ where $h(x)=(f(x)-a)^2 + (g(x)-b)^2$. And we compute

$\nabla h = 2\left( (f(x)-a) \nabla f + (g(x)-b) \nabla g \right)$,

and evaluate $\nabla h$ at any $x \in M(a,b)$ then we obtain $\nabla h = 0$. Hence by the Rank Theorem we conclude that $M(a,b)$ is not a manifold?.

My question is: Where is my error?

• You can see this phenomenon in a simpler setting. Just consider the submanifold $\{0\}\subset\Bbb R$ defined by $h(x)=x^2=0$, rather than by $f(x)=x=0$. As @MartinsBruveris said in his answer, the implicit function theorem (or the rank theorem) only give sufficient conditions, not necessary ones. Mar 8, 2017 at 22:11
• Well, in the sense of commutative algebra or algebraic geometry, they're not algebraically equivalent, either :) A geometric analogue might be comparing the intersection of $y=x$ with the $x$-axis in the $xy$-plane with the intersection of $y=x^2$ with the $x$-axis. :) You get the point $(0,0)$ in both cases, but with different "scheme structure" (transverse intersection versus tangential intersection). Mar 9, 2017 at 17:38
The Rank Theorem is a sufficient condition for something to be a manifold, not a necessary one. If you can represent your space $M$ as $M=F^{-1}(0)$ with $F : \mathbb R^n \to \mathbb R^k$ and $DF(x)$ has full rank at every point $x \in M$, then you know that $M$ is a manifold. If $F$ fails to have full rank, then it still could be a manifold, you just don't know yet.