I know that a ruled surface is a surface with parametrization $x(u,v)$ = $c(u)$ + $vf(u)$ where if $I$ $\subset$ $R$ is an interval, $c$: $I$ $\mapsto$ $R^3$ and $f$ : $I$ $\mapsto$ $R^3$ are smooth curves with $f$ $\neq$ $0$ on $I$. The curve $c$ is called the directrix and the $f(u)$ are called the rulings. If I'm not mistaken, is a ruled surface regular at a point $p$ = $x(u,v)$ provided that $f(u)$ $\bigwedge$ c'(u) = $vf(u)$ $\bigwedge$ f'(u) and $x_u$ $\bigwedge$ $x_v$ $\neq$ $0$? I actually think that only the second condition is really required, or are both of them required? Are the two conditions equivalent, and is there any other extra requirement? I'd really appreciate some input on this, thanks.
When a ruled surface can be regular
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differential-geometry
surfaces
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0The second statement is more or less the definition of regular, regardless of whether or not the surface is ruled. I expect you miss-copied the first statement from somewhere. – 2011-04-09
1 Answers
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Let $x(u,v)=c(u)+vf(u)$ where $u\in I$. Note that by definition, $x(u,v)$ is a regular surface if $x_u\wedge x_v\neq 0$. Simple computation gives x_u(u,v)=c'(u)+vf'(u) and $x_v(u,v)=f(u)$, which implies that x_u\wedge x_v=c'(u)\wedge f(u)+vf'(u)\wedge f(u). Therefore, $x(u,v)$ is a regular surface if and only if c'(u)\wedge f(u)+vf'(u)\wedge f(u)\neq 0.