# Implicit Representation of Surfaces - Basic Quesion

I am reading on implicit representations of surfaces and cant quite come around the following example. Take $F : \mathbb{R^3} \rightarrow \mathbb{R}$, where $F(x,y,z)=x^2+y^2+z^2$. Now we want to verify under what conditions the equation:

$$F(x,y,z)=c$$

define a $C^1$ surface. Now:

$$dF = \begin{pmatrix}2x&2y&2z\end{pmatrix}$$

Rank$(dF)=2$ if $(x,y,z)\neq0$. The author follows up concluding that the 2-dimensional surface is well defined if $c>0$. I have a gap somewhere and can't follow the argument. Any hints os references? Thanks.

• an $1\times 3$ matrix can't have rank two – janmarqz Feb 2 '14 at 6:51

What it is happening is that $F^{-1}(c)$ is the set in ${\Bbb{R}}^3$ of triples $(x,y,z)$ such that $x^2+y^2+z^2=c$, meaning that they are at a sphere of radius $\sqrt{c}$, in case of $c>0$.
For $c=0$ there is only triple in $F^{-1}(0)$ which is $(0,0,0)$.
And for $c<0$ we have $F^{-1}(c)=\varnothing$, because $x^2+y^2+z^2\ge0$
Basically in your case the preimage theorem says that if $F:R^3\rightarrow R$ is a smooth function and if $c\in R$ is a point such that for all $x$ in the preimage $f^{-1}(c)$ the Jacobian $DF_x$ has maximal rank (such a point is called a regular value), then the locus $\{(x_1,x_2,x_3)\in R^3:F(x_1,x_2,x_3)=c$ is a manifold of dimension $\mathrm{dim}(R^3)-\mathrm{dim}(R)=2$.
• Almost there... Now suppose $F(x,y,z)=x^2+y^2-z^2$. I suppose that, for the same exercise above, any $c\neq0$ would define a manifold of dimension $2$, but I can't see why zero would be a problem. I mean, it seems as if $f^{-1}(0)$ would still provide rank $1$ for $DF(x,y,z)=(2x,2y,-2z)=(2x,2y,-2(\pm\sqrt{x^2+y^2}))$, given $(x,y,z)\neq\mathbf{0}_{\mathbb{R^3}}$. – user191919 Feb 2 '14 at 12:10
• Nope, for the preimage theorem to hold the jacobian should have maximal rank for ALL points in the preimage $F^{-1}(c)$, and for your function $0\in F^{-1}(0)$! – GFR Feb 2 '14 at 13:27