Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to prove that in $RP^2$, given a homogeneous polynomial $F$ of degree $k$, the hypersurface $Z(F)$ of the zeros of $F$ is smooth if $\frac{\partial F}{\partial x_0}$, $\frac{\partial F}{\partial x_1}$ , $\frac{\partial F}{\partial x_2}$ are not all simultaneously $0$ on $Z(F)$.

What I tried to do is to show that $Z(F)$ is a submanifold of $RP^2$: given a point $p=[x_0 x_1 x_2] \in RP^2$ w.l.o.g. assume $x_0 \neq 0$ so take the neigboorhood $U_0$ of points with $x_0 \neq 0$. In $U_0$, $Z(F)$ is given by the zeros of $f(x,y)$ which is defined as $f(x,y):=F(1,x_1/x_0,x_2/x_0)$. My idea is to prove that the rank of $f$ is 1 on $Z(F)$, and then in the new coordinates of $U_0$ (assuming $\frac{\partial f}{\partial x} \neq 0 $), $(f,y)$ the set $Z(F)$ is given by $f=0$.

So all the problem is reduced to prove that either $\frac{\partial f}{\partial x} \neq 0 $ or $\frac{\partial f}{\partial y} \neq 0 $ in $Z(F)$.

In a point $(\alpha,\beta)=[1,\alpha,\beta]$ of $Z(F)$,

$\frac{\partial f}{\partial x} (\alpha,\beta)= \frac{\partial F}{\partial x_1} ([1,\alpha,\beta]) \frac{\partial x_1}{\partial x}(1,\alpha) = \frac{\partial F}{\partial x_1} ([1,\alpha,\beta]) $. Similarly,

$\frac{\partial f}{\partial y} (\alpha,\beta)= \frac{\partial F}{\partial x_2} ([1,\alpha,\beta]) $,

and since $\frac{\partial F}{\partial x_0}$, $\frac{\partial F}{\partial x_1}$ , $\frac{\partial F}{\partial x_2}$ are not all simultaneously $0$ on $Z(F)$, $f$ has rank 1 on $Z(F)$.

Is this argument correct? I don't understand the meaning of $\frac{\partial F}{\partial x_i} $ since $F$ is not a function on $RP^2$.

share|improve this question
1  
You need to remember that $F$ is a function on $\mathbb R^3$, not $\mathbb RP^2$. You can think of this calculation as finding the tangent plane to the cone over $Z(F)$ in $\mathbb R^3-\{0\}$. There the partial derivatives make perfect sense. Working in coordinates, as you are, write $f(x,y)=F(1,x,y)$ (so we're restricting $F$ to the plane $x_0=1$ in $\mathbb R^3$), and then $\frac{\partial f}{\partial x} (\alpha,\beta) = \frac{\partial F}{\partial x_1}(1,\alpha,\beta)$ (no brackets). –  Ted Shifrin May 6 '13 at 14:53
add comment

1 Answer

Generally, to show that $N$ is an embedded submanifold of $M$ it suffices to exhibit an open cover $U_j$ of $M$ such that $N\cap U_j$ is an embedded submanifold of $M\cap U_j$ for every $j$.

A natural choice of such a cover for $\mathbb RP^2$ is $U_j=\{[x_0:x_1:x_2] : x_j\ne 0\}$, $j=0,1,2$. On $U_0$ we have the diffeomorphism $\Phi:U_0\to \mathbb R^2$ given by $[x_0:x_1:x_2]\mapsto (x_1/x_0,x_2/x_0)$. It maps the zero set of $F$ to the zero set of the function $f(x,y)=F(1,x,y)$. Since the gradient of $f$ does not vanish (as you've shown), the zero set is a submanifold.


(CW posted to fill this box)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.