$g$ is a smooth function from $R^2 \rightarrow R$ satisfy $g(0,0)=0$,$\frac {\partial g}{\partial x}(0,0)\not = 0$,then there are neighbourhoods $M,N$ of $(0,0)$ and diffeomorphism t from $M$ to $N$ satisfy $g(t(x,y))=0$ iff $x=0$. Rank theorem maybe a way.
How to find such open sets $M$ and $N$ and map bewteen them?
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differential-geometry
manifolds
1 Answers
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Let's look for a diffeomorphism of the form $t(x,y) = (x+f(y),y)$ where $f$ is a smooth function - the idea here is to just shift each horizontal line by the correct amount to line up the zero set of $g$.
Now, how to construct $f$? Well, $-f(y)$ is just the amount we need to $x$-translate $g$ by so that $(0,y)$ is a zero; i.e. we need $g(f(y),y)=0$ for each $y$. This is now a problem to which we can apply the Implicit Function Theorem: the condition $\partial g/\partial x|_0 \ne 0$ is exactly the invertibility the IFT requires, so it yields open neighbourhoods $U,V$ of $0 \in \mathbb R$ and a smooth $f : U \to V$ such that $f(0) = 0$ and whenever $(x,y) \in V \times U$, $g(x,y) = 0$ if and only if $x = f(y)$.
Letting $t: V \times U \to \mathbb R^2$ be defined by $t(x,y) = (x+f(y),y)$, note that $Dt$ is invertible and $t$ is injective; so $t$ is a diffeomorphism on to its image $t(V\times U).$ By continuity we can choose neighbourhoods $M,N$ of the origin such that $t|_M : M \to N$ and $M,N \subset V \times U$. Finally, check that we have indeed constructed a solution to the problem: $g(t(x,y)) = g(x+f(y), y) = 0$ if and only if $x+f(y) = f(y)$; i.e. $x = 0$.