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Question is from do carmo Differential Geometry of curves and surfaces Chapter 2.3

Parametrized surfaces are often useful to describe sets $\Sigma$ which are regular surfaces except for a finite number of points and a finite number of lines. For instance, let $C$ be the trace of a regular parametrized curve $\alpha:(a,b)\rightarrow R^{3}$ which does not pass through the origin $O=(0,0,0)$. Let $\Sigma$ be the set generated by the displacement of a straight line $l$ passing through a moving point $p\in C$ and the fixed point $0$ (a cone with vertex $0$). Related figure

(a) Find a parametrized surface ${X}$ whose trace is $\Sigma$.

(b) Find the points where ${X}$ is not regular.

(c) What should be removed from $\Sigma$ so that the remaining set is a regular surface?

My work:

(a) ${X}:(0,1)\times (a,b) \rightarrow \Sigma$ by $(u,v)\rightarrow u\alpha(t)$.

(b) Singular points are the points where differential $dX_{p}$ is not one-to-one. So $\alpha(t)\times u\alpha'(t)=0$

(c) I have no clue...

user376629
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1 Answers1

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First of all, you should be consistent in your variables. Decide on $t$ or $v$. Next, the picture shows rays, not lines, passing through the origin and the points $\alpha(t)$. So you should have $u\ge 0$, I agree, but I don't know why you stopped at $u=1$.

To answer (b), you see two times this can happen: $u=0$ or values of $t$ for which $\alpha(t)\times\alpha'(t)=0$. (What does this mean geometrically?)

To answer (c), you just need to remove the points you identify in part (b), right?

Ted Shifrin
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  • Why you just need to remove the points in (b) to answer (c)? How does it guarantee that $X$ is a homeomorphism? – rowcol Apr 02 '21 at 12:57
  • @rowcol The other criteria are straightforward to check whenever $\alpha$ is one-to-one. – Ted Shifrin Apr 02 '21 at 17:32
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    Even if $\alpha$ is one-to-one, there could be points where $\alpha(t)=\lambda \alpha (s)$ for some $\lambda$, with $t\neq s.$ So, the surface $\Sigma$ may have self intersection at the points $\mu \alpha (t)$ ( as suggest the related figure of the OP) – rowcol Apr 02 '21 at 17:51
  • Yes, you're right, @rowcol. You need the mapping $\alpha/|\alpha|$ to be one-to-one. I normally require the directrix of a ruled surface to be normalized to have length $1$. – Ted Shifrin Apr 02 '21 at 17:54