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 $0$.

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

1 Answer 1


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.


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