# Transition map for Möbius band in differential geometry.

Calculate the transition map $\phi$ between the two surface patches for the möbius band. These two surface patches are the following

$U=\{(t,\theta) \ | -1/2\lt t\lt 1/2,\ \ 0\lt \theta \lt 2\pi\}$

$\sigma(t,\theta)=((1-t\sin (\theta /2))\cos (\theta), (1-t\sin (\theta/2))sin (\theta), t\cos (\theta /2))$

$\tilde U= \{(t,\theta) \ | -1/2\lt t\lt 1/2,\ \ -\pi \lt \theta \lt \pi\}$

$\tilde{\sigma(t,\theta)}= \sigma (t,\theta)$

Show that it is defined on the union of two disjoint rectangles in $\Bbb R^2$ and that the determinant of jacobian matrix of $\phi$ is equal to $+1$ on one other rectangles and to $-1$ on the other hand.

Again there is its answer but this is not understandable and clear for me. Please explain and show me its answer. Thanks a lot.

• Can you reference the text answer please? Dec 1, 2013 at 18:22
• Why do you want? @HoseynHeydari
– 1190
Dec 1, 2013 at 18:24
There are two maps $$\sigma:(-1/2,1/2)\times(0,2\pi)\to\Bbb{R}^3,\tilde{\sigma}:(-1/2,1/2)\times(-\pi,\pi)\to\Bbb{R}^3.$$ Operations of $\sigma,\tilde{\sigma}$ over $(-1/2,1/2)\times(0,\pi)$ are the same(I can't explain this correctly but If there was problems perhaps some figurs can help.) and we have $$\sigma(x,y)=\tilde\sigma(x,y)$$ so transition function over here is identity map $$\Phi(x,y)=\tilde\sigma^{-1}\circ\sigma(x,y)=(x,y)$$ so jacobian will be $1$. Operations of $\sigma,\tilde{\sigma}$ over $(-1/2,1/2)\times(\pi,2\pi)$ and $(-1/2,1/2)\times(-\pi,0)$ are the same too and we have $\sigma(x,y)=\tilde\sigma(\tilde x,y-2\pi)$ so we get: $$\sigma(t,\theta)=((1-t\sin (\theta /2))\cos (\theta), (1-t\sin (\theta/2))sin (\theta), t\cos (\theta /2))=\tilde\sigma(\tilde t,\theta-2\pi)=((1+\tilde t\sin (\theta /2))\cos (\theta), (1+\tilde t\sin (\theta/2))sin (\theta), -\tilde t\cos (\theta /2))$$ so $t=\tilde t$ correct all things, and transition function will be $\Phi(x,y)=\tilde\sigma^{-1}\circ\sigma(x,y)=(-x,y-2\pi)$. Now we can calculate jacobian matrix of $\Phi:(-1/2,1/2)\times(\pi,2\pi)\to(-1/2,1/2)\times(-\pi,0)$. $$\frac{\partial -x}{\partial x}.\frac{\partial y-2\pi}{\partial y}-\frac{\partial y-2\pi}{\partial x}.\frac{\partial -x}{\partial y}=-1\times 1 - 0=-1$$