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Consider the two-sphere $S^2 \subset \mathbb{R}^3$. By a dipole field on $S^2$, I mean a continuous function $f \colon S^2 \to S^2$ such that (1) $x$ is perpendicular to $f(x)$ for all $x \in S^2$ (this means that $f$ is a continuous tangent vector field on $S^2$), and (2) $f$ vanishes at exactly one point.

Question: Does there exist a dipole field on $S^2$ with the property that $f(x) \in \text{span} \left\{f(-x)\right\}$ for all $x \in S^2$, except for the point $x$ where $f(x) = 0$?

Doug
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    Well, if that condition holds the function $f$ will vanish at two points at least! – Mariano Suárez-Álvarez Jun 16 '14 at 06:57
  • Edit the question accordingly, please. – Mariano Suárez-Álvarez Jun 16 '14 at 06:58
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    (Notice that if $f$ is continuous, if $f(x)=\pm f(-x)$ for all $x$ except at most one, then it holds for all $x$, so your "fix" does not fix the question) – Mariano Suárez-Álvarez Jun 16 '14 at 06:59
  • Oh, you know, I guess I don't care about the size of the vectors either. Thanks for making that apparent. – Doug Jun 16 '14 at 07:01
  • There, I think that is the question I meant to ask. – Doug Jun 16 '14 at 07:03
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    A continuous tangent field $f$ to the sphere has to have a zero somewhere. Call it $x$. If you know that $f(-x)$ is in the span of $f(x)$, then of course $f(-x)$ is also zero, and you have two zeroes... – Mariano Suárez-Álvarez Jun 16 '14 at 07:05
  • You are right. I apologize, I am not thinking straight evidently. This picture is my motivation: http://en.wikipedia.org/wiki/Hairy_ball_theorem#mediaviewer/File:Hairy_ball_one_pole.jpg – Doug Jun 16 '14 at 07:06
  • And I think I finally have the question how I should have had it the first time... – Doug Jun 16 '14 at 07:09
  • (A note to the community: I had a comment between Mariano's first and second comments along the lines of "Indeed, that is true. Remove the condition for the point where $f$ vanishes." Out of a bad habit, I deleted the comment after I made the edit. I should stop doing that.) – Doug Jun 16 '14 at 07:14

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I think this should work: construct a vector field om $\mathbb R^2-${$(0,0)$} with exactly one zero, and use the stereographic projection to pull it back to $\mathbb S^2$.

Maybe a good point to make here, is that the fact that the stereographic projection is a diffeomorphism allows you to pushforward vector fields; this is not always possible.

user99680
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  • Or a tangent vector field on the projective plane $P^2$ with exactly one zero and pull it back along the covering map $S^2\to P^2$. – Mariano Suárez-Álvarez Jun 16 '14 at 07:15
  • O.K, good idea. I think I need to add the condition that the vector field should go to $0$ as we "go to $\infty$ , in the compactification; I think this is necessary to preserve some basic properties. – user99680 Jun 16 '14 at 07:17
  • Hi, user99680, why do you construct your vector field on $\mathbb{R}^2$ subtract the origin? – Doug Jun 16 '14 at 07:26
  • @MarianoSuárez-Alvarez: Can you guarantee that the vector field is well-defined under the projection? Don't we need it constant in the sense p(x)=p(-x)? – user99680 Jun 16 '14 at 07:26
  • @DanDouglas: Hi Dan, I am using the fact that the two spaces ; $\mathbb R^2$ and $S^2-N$ ; $N$ is the north pole, are diffeomorphic under the stereographic projection , so that we can pushforward vector fields. WE cannot pushforward under any map, but we can if we have a diffeomorphism. – user99680 Jun 16 '14 at 07:28
  • Do we not need that the vector field be defined on all of $\mathbb{R}^2$ so that when we use stereographic projection, we get a vector field on the whole sphere? – Doug Jun 16 '14 at 07:28
  • I believe you have it flipped: It is $S^2 - N$ and $\mathbb{R}^2$ that are diffeomorphic, right? – Doug Jun 16 '14 at 07:30
  • @Dan Douglas: Yes, sorry, it is defined on all of $\mathbb R^2$; please see my edit of the above comment. A moment of disxeslya :). – user99680 Jun 16 '14 at 07:30
  • What is meant by "push forward" in this context? That the inverse stereographic projection of the vector field on $\mathbb{R}^2$ is a vector field on $S^2 - N$? – Doug Jun 16 '14 at 07:32
  • And I do not see why the Span condition is satisfied, but perhaps I will have to just do the calculation in order to see that? – Doug Jun 16 '14 at 07:33
  • The pushforward is another name for the tangent map; (one of the )the versions of the derivative for manifolds ;it takes vector fields to vector fields. – user99680 Jun 16 '14 at 07:33
  • Do you mean you want f(x)=f(-x)? If that is the case, then, yes, you need to define it on $\mathbb R^2$ so that f(x)=f(-x). – user99680 Jun 16 '14 at 07:34
  • No, that is too specific. See the statement of the question above. – Doug Jun 16 '14 at 07:35
  • I'm not sure, let me think it through. – user99680 Jun 16 '14 at 07:37
  • @DanDouglas: I'm out for tonight; it is almost 4 a.m here, I will try to answer any followups after sleep. – user99680 Jun 16 '14 at 07:46
  • @user99680 to pull back you just use the fact that the projection is a local diffeo. – Mariano Suárez-Álvarez Jun 16 '14 at 07:58
  • @MarianoSuárez-Alvarez: How exactly do you go from a tangent vector field $f \colon P^2 \to P^2$ to a tangent vector field $g \colon S^2 \to S^2$? How do you "choose" what element of $S^2$ should correspond to an element of $P^2$ when there does not exist a lift from $P^2$ to $S^2$? I am referring to your comment above. More specifically, what exactly do you mean by "pull it back along the covering map"? – Doug Jun 16 '14 at 08:00
  • Exercise: If $p:M\to N$ is a covering map of manifolds and if $X$ is a tangent field on $N$ (that is, a section of $TN$) there there exists exactly one tangent field $Y$ on $M$ such that $dp(Y)=X$. Here «$Y$ is obtained by pulling back the tangent field $X$ to $M$». – Mariano Suárez-Álvarez Jun 16 '14 at 08:14
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    (Tangent fields on $P^2$ are certainly not maps $P^2\to P^2$, by the way!) – Mariano Suárez-Álvarez Jun 16 '14 at 08:15
  • @MarianoSuárez-Alvarez: Thank you again for your help on this question. I am rather outside of my realm of mathematical familiarity (I was led to this question through some research that I am doing). Would Guillemin and Pollack be an appropriate book for understanding the concepts at hand more deeply and clearly? – Doug Jun 16 '14 at 20:39
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    @DanDouglas. Sure. The book on smooth manifolds by Lee is currently my favorite. – Mariano Suárez-Álvarez Jun 17 '14 at 01:03
  • @MarianoSuárez-Alvarez: I thought you were referring to pulling back by general covering maps, not that specific one, sorry. I think in general you need a smooth covering map to be able to pullback, and not just a topological covering map, but maybe when we talk about a covering amp between manifolds we assume smoothness. – user99680 Jun 17 '14 at 02:07
  • Of course. If the map is not smooth then I don't think it even makes sense of talking about pull backs! My exercise above is for any smooth coverings. Since the one we have here is smooth, that's enough. – Mariano Suárez-Álvarez Jun 17 '14 at 02:28