Question

Question 4 (20 points) Let F: R R be any homogeneous polynomial function (with degree no less than one) with at least one positive value. Prove that the function f:Rn R, f(x) F(x) 1, defines on f-1(0) a structure of smooth manifold.

Answer 1

By regular level set theorem, in order to show that this function gives a smooth manifold structure on $f^{-1}(0)$,  it suffices to show that $0$ is a regular value for $f$. Since $f(x)=F(x)-1$, it suffices to show that $1$ is a regular value of $F$.

It is given that $F$ is a homogeneous polynomial of degree $n\geq 1$. Therefore, it is a homogeneous function satisfying

$F(\lambda x)=\lambda^n F(x)$.

By Euler's theorem, at any $x=(x_1,\cdots,x_n)\in\mathbb R^n$ such that $F(x)=1$ we have

$x_1{\frac{\partial F}{\partial x_1}}+\cdots+x_n{\frac{\partial F}{\partial x_n}}=nF(x)=n$

Thus, the gradient vector

$\nabla F(x)=\begin{pmatrix}{\frac{\partial F}{\partial x_1}},\cdots,{\frac{\partial F}{\partial x_n}}\end{pmatrix}$

is not the zero vector at a point $x=(x_1,\cdots,x_n)\in\mathbb R^n$ such that $F(x)=1$. In other words, $1$ is a regular value of $F$, as desired.