# Does a continuous scalar field on a sphere have continuous loop of “isothermic antipodes”

For a continuous scalar field on a circle, there is a diameter of the circle such that the endpoints of the diameter have the same value. If you think of the scalar field as "temperature", then what this says is that there are points on opposite sides of the circle that are the same temperature: isothermic antipodes.

So, for a continuous scalar field on a sphere, the same is true: there are isothermic antipodes. (Just consider any great circle.)

Now, is there more you can say? Can you say for example that there is a closed loop on the surface of the sphere such that every point on the loop has the same value as the other endpoint of its diameter?

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I'd use the word "isothermal" instead. – J. M. Sep 20 '10 at 21:29

[edit: as predicted, there is a counterexample with no continuous loop. See addition below.]

I think that you may not quite get a loop, only a topological continuum (a compact connected subset) on the sphere whose complement contains multiple components. The continuum can be gotten by elaborating Rahul's answer to choose a suitable component of the $g=0$ locus.

The existence of topologically wild continua such as the "Warsaw circle" suggests that you can draw such a creature on the sphere or projective plane and then extend to a continuous function that would give a counterexample. Or you could take a field that has the equator as the locus of isothermal antipodes ($g=0$) and try to perform a (antisymmetric) bending construction that modifies parts of the equator, turning it into a wild curve that cannot be traced by a continuous loop.

[added: the extension construction would work as follows. Take two opposite points on the equator. Join them with a wild continuum in one hemisphere, and the antipode of that continuum in the opposite hemisphere. Define $f(x)$ to be the distance to the wild thing, in one hemisphere, and the negative of distance to the wild thing, in the opposite hemisphere. Hence $f(x) = -f(-x)$ on the whole sphere, and $f=0$ only on the wild construction that cannot be traversed continuously by a path.]

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I don't quite understand your last sentence. If this bending constructions is continuous, then the image of the equator will be homeomorphic to the equator (note that we're talking about a map from a compact space - the equator - to a Hausdorff space). Or maybe it's better to think about it in the other direction: try to precompose your standard map with a map taking something wild to the equator. The map will have to do some collapsing in order not to be a homeomorphism. I don't see yet how to do that continuously... – Dan Ramras Sep 21 '10 at 19:13
@Dan: I posted a more detailed version of extension construction, but the collapse and iterated-bending constructions would be interesting to work out. For the latter I had in mind something that does not extend continuously to the equator. In particular, it will not map the equatorial points 1-1 to the topologically wild circle. – T.. Sep 21 '10 at 21:18

For any continuous scalar field $f$ on the sphere, you can define a continuous scalar field $g$ as $g(p) = f(p) - f(\bar{p})$, where $\bar{p}$ is the antipodal point of $p$. Unless $g$ is zero everywhere, there is a point $q$ for which $g(p) > 0$ and $g(\bar{q}) = -g(q) < 0$. As $g$ is continuous, a closed contour separating $q$ from $\bar{q}$ exists on which $g$ is zero.

Edit: The last statement "seems obvious" to me, but I don't have enough knowledge of compact topological spaces or manifolds to make it rigorous. Perhaps a counterexample exists.

Edit 2: See T..'s answer for a sketch of a counterexample. Apparently I was thinking about the existence of a curve while the question asked for a loop?

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This is interesting. I don't immediately see how to justify the last statement. If the function g is differentiable and 0 is a regular value, then the inverse image of 0 is a 1 dimensional manifold and you get your conclusion. But that's a very special situation. – Dan Ramras Sep 20 '10 at 21:01
@Dan Ramras, if the inverse image of 0 is larger than a 1-dimensional manifold, then one can still find a suitable closed contour within it, yes? The proof would only fail if the inverse image of 0 did not "go all the way around" the sphere. This feels like it's either an easy consequence of some facts in topology, or a common mistake made by amateurs -- sadly, I don't have the expertise to know which, but I've edited my answer to reflect this. – Rahul Sep 20 '10 at 22:02
I feel like I ought to know the answer one way or the other, but I guess I don't... If I think of anything, I'll let you know. – Dan Ramras Sep 20 '10 at 22:15
re "curve" vs "loop" -- I think that the locus where g=0 will contain pieces that are loop-like in the sense that removing them leaves multiple components, but if g is only required to be continuous then these pieces might not contain a loop (the image of F(t) where F is continuous and the domain of t is a circle) that disconnects an open set from the rest of the sphere. If that is true then there could also be examples where the zero set of g does not contain nonconstant curves: continuous images of an interval into the zero locus are points. – T.. Sep 21 '10 at 14:23

As for what more you can say in higher dimensions, take a look at the Borsuk-Ulam Theorem.

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The unique pair of "isothermal" antipodes for $\rho = \sin[\theta](1+\cos[\phi])$ is {north pole, south pole} ($\theta$ is co-latitude and $\phi$ is longitude). – whuber Sep 20 '10 at 19:35
@whuber: Aren't antipodes isothermal on the great circle $\phi=\pm\pi/2$? I'm not familiar with the co-latitude $\theta$, but I assume it varies over $[0,\pi]$. – Rahul Sep 20 '10 at 20:11
@whuber: Argh, your notation confused me, I use the Greek letters in spherical coordinates the other way around! :D – J. M. Sep 20 '10 at 21:01
@Rahul: "co-latitude" is "complementary latitude", or ninety degrees minus the latitude. Remember that latitude measures from the x-y plane; co-latitude measures from the z-axis. So what you were supposing is correct. – J. M. Sep 20 '10 at 21:03
@J.M.: Sorry! Everybody uses a different convention: one for the mathematicians, one for the physicists, yet another for the geodesists. I chose Mathematica's notation here and hoped that naming the coordinates would help clarify things. – whuber Sep 20 '10 at 21:53