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I have 2 questions regarding homotopy groups.

My first questions comes from the fact that I've read different definitions of homotopy groups. The book by Manton and Sutcliffe defines them as:

''The set of homotopy classes of based maps $\Psi: S^n \mapsto Y$'', where $S^n$ is a n-dimensional unit sphere. But most other sources that I've read about about this topic do not define $S^n$ necessary as a unit sphere, they don't define the radius of the sphere at all, therefore I'm not sure what is right or wrong.

My second question is for the situation where we consider a mapping from $\Psi: S^n \mapsto S^n$, i.e. $\pi_n(S^n)=\mathbb{Z} \; \forall \; n$. Now I've read in a lot of sources (for instance in Magnetic Monopoles by John Preskill) that say the winding number is a conserved quantity. I understand that the winding number is "preserved by continuous deformations''. However, I wonder if the winding number is also conserved if the radius of either n-spheres is changing with time. For instance, there exist a situation where $\pi_1(S^1)=2$, but there is surely also a situation where $\pi_1(S^1)=3$. I.e. they have different winding numbers, in my mind (I'm most likely wrong) this is because the radius of either spheres have changed. I don't, intuitively, understand how else this is possible.

Just in case my background is important: I'm physics student, so have no mathematical skills regarding topology. I'm not very good in mathematical proofs, but I would really like to intuitively understand this topic.

Thanks in advance!

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    $\begingroup$ Spheres of different radii are not only homotopy equivalent but even homeomorphic, so algebraic topology can't tell the difference between them. $\endgroup$ – Qiaochu Yuan May 15 '13 at 21:00
  • $\begingroup$ Thanks! That makes sense to me, however I'm then not sure why the book by Manton and Sutcliffe define it to be a n-dimensional unit sphere. And it seems that Wikipedia also uses that definition: en.wikipedia.org/wiki/Homotopy_groups_of_spheres $\endgroup$ – Hunter May 15 '13 at 23:16
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    $\begingroup$ If you want to define the homotopy groups of a space X as maps out of a sphere, you need to be specific about the domain of those maps: if you consider the collection of all maps out of all spaces S that are homeomorphic to an n-dimensional sphere, this isn't even a set! So the easiest thing to do is to choose one particular topological space and use that as the domain of your maps. One common choice is the unit sphere defined as the subset of $\mathbb{R}^n$ consisting of unit vectors. Another is the quotient of $[0,1]^n$ formed by collapsing its boundary. These give isomorphic groups. $\endgroup$ – Dan Ramras May 16 '13 at 0:41
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$\pi_1(S^1) \simeq \mathbb{Z}$ means that the fundamental group of the circle is isomorphic to the entire group of the integers, so it does not make sense to write $\pi_1(S^1) = 2$ or $\pi_1(S^1) = 3$.

Perhaps you mean that there could be two maps $f$ and $g: S^1 \to S^1$ such that the winding number of $f$ is $2$ and the winding number of $g$ is $3$. This can definitely happen. Think of the unit circle as being parametrized by radians from $0$ to $2\pi$ with $0 = 2\pi$. (In topology, we'd say that the circle is homeomorphic to the interval $[0,2\pi]$ with its ends identified.) The map $\theta \mapsto 2\theta$ has winding number $2$, while the map $\theta \mapsto 3\theta$ has winding number $3$. If you prefer to think of the unit circle as sitting in $\mathbb{R}^2$, think about the maps $\theta \mapsto e^{2i\theta}$ and $\theta \mapsto e^{3i\theta}$.

So radius doesn't play a role here.

One more thing: algebraic topologists generally don't think of $S^n$ as sitting inside $\mathbb{R}^{n+1}$ -- they prefer a perspective like "$[0,2\pi]$ with the ends identified." Because of that, they call tend to call winding number "degree." I tell you this only because it might be helpful in finding out more from mathematicians -- I defer to your judgment about the correct term to use in physics!

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  • $\begingroup$ This clarified a lot for me. I was under the impression that $\mathbb{Z}$ was the winding number, but this is wrong. Thanks for your help! $\endgroup$ – Hunter May 16 '13 at 0:44
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The radius of the sphere doesn't matter. Moreover, if you take a map from a particular $n$-sphere to $\mathbb R^{n+1}-\{0\}$, then you really see it as a winding number around the origin. As long as you never hit the origin by mistake as $t$ varies, this number will stay constant.

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  • $\begingroup$ Ok, thank you for your response. But if the radius does not affect the winding number, then what does have an influence on it? In other words, how can you have a situation where $\pi_1(S^1)=2$, but you can also have a different situation where $\pi_1(S^1)=3$, right? What is the reason they have different winding numbers? $\endgroup$ – Hunter May 15 '13 at 22:06
  • $\begingroup$ Hunter, your notation doesn't quite make sense. You can have different closed curves that wind different numbers of times around the origin in the plane. Your $2$ and $3$ are those winding numbers; they correspond to different curves (functions). I'll draw you pictures if you need me to. $\endgroup$ – Ted Shifrin May 16 '13 at 3:20
  • $\begingroup$ Hi Ted, I think I understand where my mistake is (thanks to Adam Saltz); I thought $\mathbb{Z}$ denoted the winding number, but I believe this is not correct. I also believe, please correct me if I'm wrong, that the statement $\pi_n(S^n)=\mathbb{Z}$ just means that a map from an n-sphere to another n-sphere means it will always covers it an integer amount of times, and therefore it is isomorphic to the group of integers, $\mathbb{Z}$? $\endgroup$ – Hunter May 16 '13 at 11:26
  • $\begingroup$ Right, and a bit more: If two maps wind the same number of times, each can be continuously deformed to the other. $\endgroup$ – Ted Shifrin May 16 '13 at 13:01
  • $\begingroup$ Yes, and if they have different winding numbers you will need to "break" them in order to deform them into each other so they are not homotopic to each other. I think I understand it (at least at a very basic level). Thanks for your help! $\endgroup$ – Hunter May 16 '13 at 15:56

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