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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?

Thanks in advance.

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    $\begingroup$ 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. $\endgroup$ Mar 8, 2017 at 22:11
  • $\begingroup$ Thanks for the detailed explanation. Roughly speaking, both representations can be considered algebraically equivalent but not analytically equivalente. $\endgroup$
    – jaogye
    Mar 9, 2017 at 17:06
  • $\begingroup$ 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). $\endgroup$ Mar 9, 2017 at 17:38
  • $\begingroup$ Wow, many thanks for your comment. $\endgroup$
    – jaogye
    Mar 10, 2017 at 8:46

1 Answer 1

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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.

This is the case you have here. One representation allows you to use the Rank Theorem, the other one does not. But the second representation does not tell you, that the space isn't a manifold. Comare this with necessary and sufficient conditions for local extrema.

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  • $\begingroup$ Dear @Martins Bruveris thanks for your answer. $\endgroup$
    – jaogye
    Mar 9, 2017 at 16:59

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