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I am working on an exercise from Tapp "Differential Geometry of Curves and Surfaces" (Ex 4.56) about the deformation of the Helicoid to the Catenoid. We are given for $U = \{ (\theta, t) \in \mathbb{R}^2 \mid -\pi < \theta < \pi \}$, define a parameterisation $\sigma_s:U \to \mathbb{R}^3$ $$\sigma_s(\theta,t) = (\cos s) (c \cosh t \cos \theta, c \cosh t \sin \theta, ct) + (\sin s) (c \sinh t \cos\theta, c \sinh t \sin \theta, c \theta)$$ For $s = 0$ this surface is a catenoid and $s = \pi/2$ is a helicoid. To show isometry between these two surfaces I want to show that the first fundamental form is invariant to $s$. My approach is to take the partials: $$\sigma_{s,\theta} = \cos(s)(-c\cosh(t)\sin(\theta),c\cosh(t)\cos(\theta), 0) \\+ \sin(s)(-c\sinh(t)\sin(\theta),c\sinh(t)\cos(\theta), c)$$ $$\sigma_{s,\theta} = \cos(s)w_1 + \sin(s)w_2$$

$$\sigma_{s,t} = \cos(s)(c\sinh(t)\cos(\theta),c\sinh(t)\sin(\theta), c) \\+ \sin(s)(c\cosh(t)\cos(\theta),c\cosh(t)\sin(\theta), 0)$$ $$\sigma_{s,t} = \cos(s)w_3 + \sin(s)w_4$$

Now computing the coefficients we have: $$E = \langle \cos(s)w_1 + \sin(s)w_2, \cos(s)w_1 + \sin(s)w_2 \rangle $$ $$E = \cos^2(s)\langle w_1, w_1 \rangle +2\sin(s)\cos(s)\langle w_1, w_2\rangle + \sin^2(s) \langle w_2, w_2 \rangle $$

We have: $$\langle w_1, w_1 \rangle = c^2\cosh^2(t)\sin^2(\theta) +c^2\cosh^2(t)\cos^2(\theta) $$ $$ = c^2\cosh^2(t)(\sin^2(\theta) + \cos^2(\theta))$$ $$ = c^2\cosh^2(t)$$ and: $$ \langle w_1, w_2 \rangle = c^2\cosh(t)\sinh(t)\sin^2(\theta) + c^2\cosh(t)\sinh(t)\cos^2(\theta) $$ $$= c^2\cosh(t)\sinh(t)(\sin^2(\theta) + \cos^2(\theta))$$ $$= c^2\cosh(t)\sinh(t)$$ and: $$\langle w_2, w_2 \rangle = c^2\sinh^2(t)\sin^2(\theta) +c^2\sinh^2(t)\cos^2(\theta) + c^2 $$ $$ = c^2\sinh^2(t)(\sin^2(\theta) + \cos^2(\theta)) + c^2 $$ $$ = c^2\sinh^2(t) + c^2 $$ $$ = c^2(1 + \sinh^2(t)) $$ $$ = c^2\cosh^2(t)$$ Plugging in: $$E = \cos^2(s)c^2\cosh^2(t) +2\sin(s)\cos(s)c^2\cosh(t)\sinh(t) + \sin^2(s) c^2\cosh^2(t) $$ $$ = (\sin^2(s) +\cos^2(s))c^2\cosh^2(t) +2\sin(s)\cos(s)c^2\cosh(t)\sinh(t)$$ $$ = c^2\cosh^2(t) +2c^2\sin(s)\cos(s)\cosh(t)\sinh(t)$$ Now the first term looks correct in that it aligns with the first coefficient in the helicoid first fundamental form but the second term definitely depends on $s$ and it seems to me it should be zero. Where have I gone wrong? Is my approach correct? Is there a simpler way?

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The fundamental form of your equation is: $$ g = \frac{c^2}2 \begin{pmatrix} 1+\cosh 2t+\sin 2s\sinh 2t & \sin 2s \\ \sin 2s & 1+\cosh 2t+\sin 2s\sinh 2t \end{pmatrix}. $$

It is not invariant of $s$. However, since $g$ depends only on $\sin 2s$, it is the same when $s=0$ and $s=\pi/2$.

Your approach is correct in general. But you could just have shown that $g(w_1,w_3) = g(w_2,w_4)$

Requested edit. When $s=0$, $(\sigma_\theta, \sigma_t)=(w_1, w_3)$. When $s=\pi/2$, $(\sigma_\theta, \sigma_t)=(w_2, w_4)$.

We need to show that $g(w_1,w_3) = g(w_2,w_4)$ or $$ \begin{pmatrix} w_1^2 & w_1\cdot w_3 \\ w_1\cdot w_3 & w_3^2 \end{pmatrix} =\begin{pmatrix} w_2^2 & w_2\cdot w_4 \\ w_2\cdot w_4 & w_4^2 \end{pmatrix} $$

It can be simplified further if you consider three orthonormal vectors $e=(\cos\theta, \sin\theta,0)$, $n=e_\theta = (-\sin\theta, cos\theta, 0)$ and $z=(0,0,1)$. Then:

$$ w_1 = n\cosh t, \qquad w_3 = e\sinh t + z $$

It's obvious now that $w_1\cdot w_3=0$ and $w_1^2=w_3^2=\cosh^2t$.

For $g(w_2, w_4)$: $$ w_2 = n \sinh t + z,\qquad w_4 = e\cosh t, $$ it's also true.

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  • $\begingroup$ Can I get some more details on the suggested approach in the last line? $\endgroup$ – boddypen Oct 4 at 11:44
  • $\begingroup$ I have added details $\endgroup$ – Vasily Mitch Oct 4 at 12:51

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