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This comes from Guillemin and Pollack's book Differential Topology. The book claims that one cannot parametrize a unit circle by a single map. I thought we could (by a single angle $\theta$).

I think one possible answer might be the fact that if we let $\theta$ in [0, 2$\pi$), when $\theta$ approaches 2$\pi$, $f(\theta)$ approaches $f(0)$. But is this a problem?

Also for $n$-spheres, why is there always a point that can't be covered by a single map (I know we can always cover it by two maps, one being the stereographic projection)?

Thanks a lot for your help!

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Remember that parametrizations are supposed to be homeomorphisms onto their images.

The angle map $\theta\rightarrow e^{i\theta}$ (to use complex coordinates), as a map from $[0,2\pi)$ onto the circle $S^1$ is continuous and bijective. The issue is that it is not a homeomorphism. You can see this explicitly by noting that the inverse map is not continuous. If you approach $(1,0)\in S^1$ from "above", the angle approaches $0$, but if you approach from "below", the angle approaches $2\pi$.

There is a more abstract way to see this: $S^1$ is compact, so it cannot be homeomorphic to $[0,2\pi)$, which is not.

This also shows that you cannot parametrize the $n$-sphere $S^n$ by a single chart. $\phi:S^n\rightarrow\mathbb{R}^n$. If you could, the image $\phi(S^n)$ would be compact and thus closed in $\mathbb{R}^n$. On the other hand, as $\phi$ is a homeomorphism and $S^n$ is open in itself, the image $\phi(S^n)$ would also be open in $\mathbb{R}^n$. By connectedness, a closed and open set in $\mathbb{R}^n$ is all of $\mathbb{R}^n$, so this would mean the sphere is homeomorphic to the full Euclidean space. But again, one is compact and the other is not.

This is a very common type of argument in topology.

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  • $\begingroup$ Whoops, didn't see Henry's answer go up as I was writing mine. $\endgroup$ – GCD May 10 '13 at 16:55
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Let's look more closely at Guillemin and Pollack's definition of parametrization:

Let $X$ be a subset of $\Bbb R^N$. A parametrization is a diffeomorphism $\phi: U \longrightarrow V$ from an open set $U \subset \Bbb R^k$ to an open subset $V \subset X$ (where we give $X$ the subspace topology inherited from $\Bbb R^N$).

So we notice two things. First, a parametrization must be one-to-one (since it is a diffeomorphism) and it must be defined on an open subset of $\Bbb R^k$.

In your proposed parametrization, you have $U = [0, 2\pi)$, which is not open in $\Bbb R$. We at least need something like $U = (0, 2\pi + \varepsilon)$ for some $\varepsilon > 0$. But once we extend $U$ this far, the parametrization by angle is no longer injective, since the angles $\tfrac{\varepsilon}{2}$ and $2\pi + \tfrac{\varepsilon}{2}$, which are both in $(0, 2\pi + \varepsilon)$, correspond to the same point on the circle.

Note that if we could use a single parametrization $\phi: (a, b) \longrightarrow S^1$, then $S^1$ would be diffeomorphic to $\Bbb R$, which you likely know is not true.

Similarly, if we could use a single parametrization $\phi: U \longrightarrow S^n$, where $U \subset \Bbb R^n$ is open, then $S^n$ would be diffeomorphic to the open subset $U$ of $\Bbb R^n$, which is impossible since $S^n$ is compact.

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  • $\begingroup$ Thanks a lot Henry! Both your answer and GCD's are great. However I particularly like his second paragraph. Plus the fact that he needs more reputation than you do :p I decided to award the best answer to him. But thanks a lot for your effort! Your answer is great as well! $\endgroup$ – Evariste May 11 '13 at 2:25

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