# germ finitely determined

Does anyone know any result on finitely determined germs to help me prove that the germ $f(x,y)=x^3+ xy^3$ is $4$- determined? I tried using the definition of germs finitely determined, which is:$f: \mathbb{R}^n \rightarrow \mathbb{R}$ is $k$-determined if for any other germ $g: \mathbb{R}^n \rightarrow \mathbb{R}$ such that the $k$-jet of g is equal to $k$-jet of the $f$, then $f$ and $g$ are right equivalents, i. e., exist a difeomorfism $h$ such that $f=g\circ h$, but not getting success. I think there should be some results to help me prove it. Thanks!

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Your function germ $f$ has an $E_7$ singularity at zero. In fact it is the normal form of an $E_7.$ All simple singularities are finitely $\mathscr{R}$-determined. You are correct in thinking that your function germ is 4-$\mathscr{R}$-determined. Since the Milnor number $\mu(f) = 7$, there is a nice result that says that there exists a neighbourhood $U \subset \mathbb{C}\{x,y\}$ containing $f$ such that every $g \in U$ is 8-$\mathscr{R}$-determined.
A mini-versal deformation of $f(x,y)$ is given by: $$F((x,y),\lambda) = x^3 + xy^3 + \lambda_1y^4 + \lambda_2y^3 + \lambda_3y^2 + \lambda_4xy + \lambda_5x + \lambda_6y + \lambda_7$$
The diffeomorphism would be case specific and would depend on the function germ. I very much doubt, except for exceptional cases, that you'll be able to find an explicit diffeomorphism. Moreover, you don't need to find one! The result proves that $E_7$ is finitely determined, and so you can just "throw away" the tail safe in the knowledge that there exists a diffeomorphism. –  Fly by Night Sep 6 '12 at 18:55