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If $$U = \{(u,v)\mid 0 < u,\; 0 < v < 2\pi\}, $$ and $$ x(u,v) = (u\cos v,u \sin v,u+v). $$

Then how can we show that $x$ is a simple surface?

Is it necessary true that for being simple implies that we need:

smoothness 1-1 and class C^k or is there anything I have overlooked?

For being simple, we need to show that it is smooth, 1-1, and of class $C^k$. How does the geometry help? I see that $x$ is a helix and to show 1-1, I'm thinking if we let $x(u,v) = x(a,b)$ then this implies $u = a$ and $v = b$, but this seems just wrong

Thanks

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2 Answers 2

up vote 2 down vote accepted

We can see that $x(u,v)$ is smooth, because $u, v, \cos v$, and $\sin v$ are all smooth functions.

To see that $x(u,v)$ is one-to-one, suppose $u,u,>0$ and $0<v,v'<2\pi$ such that $x(u,v)=x(u',v,)$. This implies by definition that $$(u\cos v,u \sin v,u+v)=(u'\cos v',u' \sin v',u'+v')$$ or equivalently, $$\tag{1} u\cos v=u'\cos v', u \sin v=u' \sin v', u+v=u'+v'.$$ By the first two equations in $(1)$, we have $$(u\cos v)^2+(u\sin v)^2=(u'\cos v')^2+(u'\sin v')^2$$ which implies that $u^2=u'^2$. Since $u,u,>0$, we have $u=u'$. Now by the third equation in $(1)$, we have $v=v'$. This shows that $x(u,v)$ is one-to-one.

To show that $x(u,v)$ is regular, note that its differential is given by $$dx=\left[ \begin{array}{cc} \cos v & -u\sin v \\ \sin v & u\cos v \\ 1 & 1 \\ \end{array} \right].$$ To show that its rank is 2, note that the minor given by the first and second rows is $$\left[ \begin{array}{cc} \cos v & -u\sin v \\ \sin v & u\cos v \\ \end{array} \right],$$ which is nonsingular, because it has determinant $u\cos^2v+u\sin^2v=u>0$.

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You forgot to show that the map $(u,v)\mapsto x(u,v)$ is regular at all $(u,v)$, i.e., that ${\rm rank}\ dx(u,v)=2$ everywhere. –  Christian Blatter Sep 27 '12 at 7:58
    
@ChristianBlatter: Thanks. I edit it. –  Paul Sep 27 '12 at 8:18

For showing $1$-$1$. Try to think of an inverse of $x(u,v)$.

HINT: $\cos(v)^2+\sin(v)^2=1$

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