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Can anybody please help me in understanding this question?

User
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1 Answers1

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Hint: They want you to define a metric in the annulus so that the shortest path between 2 points with the same radius is not the length of the straight line between them, but the length of the shortest arc joining them, as it is if you're forced to walk along the wall of a cylinder. This can then be extended to a metric for points with different radii.

Felipe Jacob
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  • When the points are of same radii i.e. they lie on a circle (concentric with the boundary of annulus ) so the distance between those two points will be the lenth of smaller arc. But how to extend it? – User May 21 '16 at 14:36
  • One way to do it is to map back and forth to a rectangle, which has the euclidean norm. If you find a continuous invertible map $f$ from your annulus to the rectangle $[0,1) \times [0,1]$ then $d(x,y) = |f(x) - f(y)|$ where $| \cdot |$ is the euclidean norm in $\mathbb{R}^2$. – Felipe Jacob May 21 '16 at 14:41
  • Adding to my previous comment, some care must be taken for points where the shortest path would cross the boundary of the rectangle, so what you really need is the quotient space $[0,1] \times [0,1] , / \sim $ where $(0, y) \sim (1, y)$, with the metric derived from the euclidean norm. Then set the distance in your annulus to be the distance of the image points in this rectangle. – Felipe Jacob May 21 '16 at 14:50