Prove that every continuous mapping $f$ from $\mathbb S^2$ to $\mathbb R$ has three vector which are mutually perpendicular and have same value i.e $f(v_1)=f(v_2)=f(v_3)$

  • $\begingroup$ Using spherical coordinates, what about $f(v)=\cos \theta$? $\endgroup$ – Ross Millikan Sep 1 '11 at 20:12
  • $\begingroup$ Which results to you have to build on? (Also, standard question: is this homework?) $\endgroup$ – Henning Makholm Sep 1 '11 at 20:13
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    $\begingroup$ @Ross, stand a unit cube on one of its corners at the origin. The three neighbor corners have the same height. $\endgroup$ – Henning Makholm Sep 1 '11 at 20:14
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    $\begingroup$ @Mahan: This is proven in Kakutani, S., A proof that there exists a circumscribing cube around any bounded closed convex set in $\mathbb{R}^3$, The Annals of Mathematics, Second Series, Vol. 43, No. 4 (Oct., 1942), pp. 739-741. $\endgroup$ – LostInMath Sep 1 '11 at 21:27
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    $\begingroup$ @Mahan: A generalization to higher dimensional spheres is proven in Yamabe, H. & Yujobo, Z., On the Continuous Function Defined on a Sphere, Osaka Mathematical Journal, Vol. 2, No. 1 (March 1950), pp. 19-22. It's available at Project Euclid for free. $\endgroup$ – LostInMath Sep 1 '11 at 21:42

Algebraic topology never really clicked for me, so here is a homotopy-based proof instead:

Let $\alpha$ be $e^{\frac{2\pi}{3}i}$, a complex third root of unity. For each ordered triple $T=(a,b,c)$ of points in $\mathbb S^2$, define

$$g(T)=g(a,b,c) = f(a) + \alpha f(b) + \alpha^2 f(c)$$

Because $f$ is real-valued, $g(a,b,c)=0$ if and only if $f(a)=f(b)=f(c)$. Note also that that $\alpha g(a,b,c) = g(c,a,b)$.

Let $T_\theta=(a_\theta, b_\theta, c_\theta)$ be the standard orthogonal triple $(\mathbf e_x, \mathbf e_y, \mathbf e_z)$ rotated by $\theta$ around the line $x=y=z$. Note that $(a_{\theta+2\pi/3}, b_{\theta+2\pi/3}, c_{\theta+2\pi/3}) = (c_\theta,a_\theta,b_\theta)$, and therefore

$$g(T_{\theta+2\pi/3}) = \alpha g(T_\theta)$$

Now we seek the winding number of $g(T_\theta)$ around the origin as $\theta$ goes from $0$ to $2\pi$. The argument of $g(T_\theta)$ must increase by $2\pi(n+1/3)$ between $\theta=0$ to $\theta=2\pi/3$, for some integer $n$, and it must increase in the very same way from $2\pi/3$ to $4\pi/3$ and again from there to $2\pi$, so the total winding number is $3n+1$, which is nonzero no matter what $n$ is.

Now, it's just a matter of contracting the path in triple space to a point and seeing the corresponding path in $\mathbb C$ contract -- because it has nonzero winding number to begin with, it must pass through the origin somewhere along the way. But not so fast! The space of ordered orthogonal triples (even just right-handed ones) is not simply connected! Luckily it is well known -- because right-handed ordered triples is really the same as SO(3) -- that if we rotate through $4\pi$ instead of $2\pi$, the resulting curve is homotopic to null, and doing this results in a winding number of $6n+2$, which is just as nonzero.

  • $\begingroup$ Very... Very... nice !!! :wow: $\endgroup$ – Mahan Sep 1 '11 at 22:20
  • $\begingroup$ This is exactly Kakutani's original proof from the paper LostInMath mentioned. $\endgroup$ – Omar Antolín-Camarena Oct 7 '13 at 19:39

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