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.

enter image description here

  • $\begingroup$ Can you reference the text answer please? $\endgroup$ Dec 1, 2013 at 18:22
  • $\begingroup$ Why do you want? @HoseynHeydari $\endgroup$
    – 1190
    Dec 1, 2013 at 18:24
  • $\begingroup$ Read the answer then explain it and add some helpful things to answer instead of solving it.:) $\endgroup$ Dec 1, 2013 at 18:25
  • $\begingroup$ Okay of course!:) @HoseynHeydari $\endgroup$
    – 1190
    Dec 1, 2013 at 18:31
  • $\begingroup$ I posted it:) @HoseynHeydari $\endgroup$
    – 1190
    Dec 1, 2013 at 18:33

1 Answer 1


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 $\Phi:(-1/2,1/2)\times(\pi,2\pi)\to(-1/2,1/2)\times(-\pi,0)$. $$\frac{\partial -x}{\partial x}.\frac{\partial y-2\pi}{\partial y}-\frac{\partial y-2\pi}{\partial x}.\frac{\partial -x}{\partial y}=-1\times 1 - 0=-1$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.