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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 $RP^2$.

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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
@TedShifrin Couldn't we have that $\frac{\partial F}{\partial x_1}(1,\alpha,\beta) = \frac{\partial F}{\partial x_2}(1,\alpha,\beta) = 0$, and just $\frac{\partial F}{\partial x_0}(1,\alpha,\beta) \neq 0$? I just asked a linked question because I'm trying to understand this argument. – justin Mar 11 at 17:04
@justin: One way to see the answer to your question is to recall Euler's Theorem on Homogeneous Functions: We have $\sum_{i=0}^2 x_i\frac{\partial F}{\partial x_i} = k F(x_0,x_1,x_2) = 0$. – Ted Shifrin Mar 11 at 17:32
up vote 1 down vote accepted

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)

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