# Question about the implicit function theorem. an example of a homogeneous form for which its implicit function satisfies certain conditions

Let $$F$$ be a homogeneous form with coefficients in $$\mathbb{R}$$. Suppose it defines a smooth projective variety, in other words at every point other than the origin at least one of the first partial derivatives is non-zero.

I am looking for $$F$$ which satisfies the following condition: There is a point in $$P \in \mathbb{R}^n$$ such that when we apply implicit function theorem and obtain a function $$G$$, all the second partial derivatives of $$G$$ is $$0$$ at $$P$$.

I have been trying to find such $$F$$ but I have not found one yet. If anybody has any leads I would appreciate it!

Clarification: Let $$P= (p_1, ..., p_n)$$. Since $$\nabla F(P) \neq 0$$ without loss of generality suppose $$\partial F/\partial x_n (P) \neq 0$$. Then the implicit function theorem tells us that there exists some function $$G$$ satisfying $$F(x_1, .., x_{n-1}, G(x_1, .., x_{n-1})) = 0$$ for $$(x_1, .., x_{n-1})$$ in some open neighbourhood of $$(p_1, .., p_{n-1})$$. I would like all the second partial derivatives of $$G$$ at $$(p_1, .., p_{n-1})$$ to be $$0$$.

• Could you explain what you mean by "apply the implicit function theorem"?
– abx
Nov 20, 2020 at 13:55
• I have added clarification. I hope the question makes more sense now. Thank you! Nov 20, 2020 at 18:17

Here is a simple example: Take $$F = w^3 +3 w u^2 -v^3$$ on $$\mathbb{R}^3$$ with coordinates $$(u,v,w)$$. At the point $$p=(u,v,w)=(1,0,0)$$, we have that $$F=0$$ can be solved for $$w$$ as a function of $$(u,v)$$. Meanwhile, via implicit differentiation, $$w_u = \frac{-2uw}{u^2+w^2}\quad\text{and}\quad w_v = \frac{v^2}{u^2+w^2}$$ while $$w_{uu} = \frac{-2w(w^2{-}u^2)(w^2+3u^2)}{(u^2+w^2)^3},\quad w_{uv} = \frac{2v^2(w^2{-}u^2)}{(u^2+w^2)^3},$$ and $$w_{vv} = \frac{2v(w^4{+}2w^2u^2{-}wv^3{+}u^4)}{(u^2+w^2)^3}.$$ Clearly, all of these partials vanish at $$p = (u,v,w) = (1,0,0)$$.
Remark: Of course, the point $$[p]=[1,0,0]$$ in $$\mathbb{RP}^2$$ is a flex of the smooth cubic curve defined by $$F=0$$. Meanwhile, there are smooth quartic projective curves that have no (real) flexes, so they would not have any points of the kind you are seeking.