# Deformation of Helicoid to Catenoid

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?

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.

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