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Let $V_1, V_2 \rightarrow M $ be smooth vector bundles over a manifold $M$ and $s_1: M \rightarrow V_1$ a smooth section transverse to the zero set and $s_2: M \rightarrow V_2$ a continuous section such that $ s_2 : s_1^{-1}(0) \rightarrow V_2$ is a smooth section, transverse to the zero set. Let $p\in M$ be a point such that $$ s_1(p)=0, \qquad s_2(p) =0. $$ Does there always exist a solution $p(t_1, t_2)$ near $p$ such that $$ s_1(p(t_1,t_2)) =t_1 \qquad s_2(p(t_1,t_2)) = t_2 $$ if $t_1$ and $t_2$ are sufficiently small? To make sense of sufficiently small we can just chose some metric on $V_1$ and $V_2$.

Note that if I said that the section $p \rightarrow s_1(p) \oplus s_2(p)$ is smooth and transverse to the zero set, then the answer is obviously yes (by implicit function theorem). It seems that one should be able to imitate the proof of implicit function theorem (ie contraction mapping principal) to justify what I am asking. But I am not sure of this.

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up vote 2 down vote accepted

Note that by your assumption $s_2$ is only smooth along the submanifold $s^{-1}(0)$ of $M$ and might even jump in directions transverse to it. So it is easy to come up with a counterexample: $M=\mathbb R^2$, $s_1(x,y)=x$, $s_2(x,y) = y$ for $x\le 0$ and $=y+1$ for $x>0$.

But if you assume that $s_2$ is strictly differentiable on $M$ along $s_1^{-1}(0)$, then is true in a somewhat weaker sense. See

  • MR0817719 Reviewed Ver Eecke, Paul Applications du calcul différentiel. (French) [Applications of differential calculus] Mathématiques. [Mathematics] Presses Universitaires de France, Paris, 1985. 397 pp.
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Thank you, this seems to be a very clear counter example. Regarding your second comment, suppose I assume that $s_2:M \rightarrow V_2$ is smooth. Does that mean the answer to my question is yes? Or do I have to modify the question slightly? You said it's true in some weak sense. Usually that means that I have to have a different notion of what it means to have a solution. – Ritwik Jun 14 '13 at 13:15
@Ritwik: Weak means, that the implicitly given function is just strictly differentiable at each point of $s_1^{−1}(0)$. This means $lim_{x≠y→x_0}\frac{f(x)−f(y)}{x−y}$ exists (1-dim version). – Peter Michor Jun 14 '13 at 18:18

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