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Prove that an equation of the form $f(x,y,z)=c$ determines a regular surface if $f$, defined on some open subset $S$ of $\mathbb{R}^3$, is smooth and $\nabla f\neq 0$ everywhere in $S$.

I know these following definitions:

(1) a smooth surface $S$ is regular if for any surface patch $\sigma$ of $S$, we have $\vert\vert \sigma_u \times \sigma_v\vert\vert \not=0$

(2) a topological surface $S$ in $\Bbb R^3$ issmooth if all coordinate charts are smooth.

(3) $f:U\to \Bbb R^m$ is smooth if all partial derivatives of $f$ of any order exist on $U$.

Please help me solving the question. Thank you.

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Hint: use the implicit function theorem.


Added: the implicit function theorem has many (seemingly) different statements, but the gist of it is that the level sets of a regular (i.e., one which admits a full-rank differential) $C^1$ (or $C^\infty$) map are locally graphs of a $C^1$ (or $C^\infty$, respectively) function. This function is precisely the patch (which I assume is a different name for a local coordinate system) you require, and if I may presume, I urge you to verify that you are comfortable with this 'language'.

More precisely, in this context, the implicit function theorem affords us for each $(x_0,y_0,z_0)\in S$ such that $f(x_0,y_0,z_0)=c$ and (w.l.o.g.) $\frac{\partial f}{\partial x}(x_0,y_0,z_0)\neq 0$, a neighborhood $U$ of $(y_0,z_0)$ and $V$ of $x_0$ such that $\frac{\partial f}{\partial x}$ doesn't vanish in $V$, $V\times U\subset S$, and a $C^1$ function $g:U\to V$ such that for all $(x,y,z)\in V\times U$ one has $f(x,y,z)=c$ if and only if $x=g(y,z)$.

Moreover, we have $$\frac{\partial g}{\partial y}(y,z) = -\left(\frac{\partial f}{\partial x}(g(y,z),y,z)\right)^{-1}\frac{\partial f}{\partial y}(g(y,z),y,z),$$ and similarly for $z$, which helps us to easily verify that $g$ is also $C^\infty$, whenever $f$ is.

This is precisely what we need to note that $\varphi:U\to S$ defined $\varphi(y,z) = (g(y,z),y,z)$ is a patch for the surface $f=c$ in a neighborhood $V\times U$ of $(x_0,y_0,z_0)$, and that $$\varphi_y(y,z)\times\varphi_z(y,z) = \left(g_y(y,z),1,0\right)\times\left(g_z(y,z),0,1\right) =\\ = (1,-g_y(y,z),-g_z(y,z)) =\\ = \left(\frac{\partial f}{\partial x}(g(y,z),y,z)\right)^{-1}\nabla f(g(y,z),y,z)\neq \overline{0}$$

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  • $\begingroup$ How? Can you show this? Dear Jonathan Y. $\endgroup$ – user315 Nov 29 '13 at 16:54
  • $\begingroup$ @B11b: yes, I will, but I'd like to give you some time to try on your own before I do. $\endgroup$ – Jonathan Y. Nov 29 '13 at 17:02
  • $\begingroup$ Sorry. I can't :( $\endgroup$ – user315 Nov 29 '13 at 17:11
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    $\begingroup$ @B11b this should be clearer, but for what it's worth, I firmly believe that in this case reading past the first paragraph of my addendum before you've dedicated considerably more time to handling the question on your own will do you a disservice. $\endgroup$ – Jonathan Y. Nov 29 '13 at 21:08
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    $\begingroup$ Thank you so much Dear @JonathanY. :)) I understand well:) $\endgroup$ – user315 Dec 1 '13 at 21:25

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