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Let $f,g:D\rightarrow \mathbb{S}^2$ where $D$ is the disk $\{(x,y)|x^2+y^2\le 1\}$, such that for $(x,y) \in \mathbb{S}^1$ these functions are given by: $f(x,y)=(x,y,0)$ and $g(x,y) = (-y,x,0)$ show that there is a point $(x,y) \in D$ such that $f(x,y)=\pm g(x,y)$.

For context, this is a question on the consequences of $\pi_1(\mathbb{S}^1)=\mathbb{Z}$, my attempts were into finding a function $h:D \rightarrow D$(by projecting some composition of $f$ and $g$) such that it having a fixed point by Brower would imply the result, but I really can't see how the other conditions come into play in that case.

Another shot I've tried was to morph the disk to the upper hemisphere and finding $F,G:\mathbb{S}^2 \rightarrow\mathbb{S}^2$ extending $f,g$ through oddness but I can't see an easy way from which this would imply the result either. The main problem arises from the co-domain being the sphere, most of the results I've seen use that as the domain and prove the results arriving at a contradiction with the fundamental group of the circle.

David Melo
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1 Answers1

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Suppose for sake of contradiction that $f(x,y) \neq \pm g(x,y)$ for all $(x,y)$, i.e., $f(x,y)$ and $g(x,y)$ are linearly independent. Then $(f,g)$ defines a map $F$ from $D^2$ to the Stiefel manifold $V_2(\mathbb{R}^3)$ of $2$-frames in $\mathbb{R}^3$.

Based on the boundary conditions, we see $F|_{\partial D^2}: S^1 \cong \partial D^2 \to V_2(\mathbb{R}^3)$ represents a nontrivial generator of $\pi_1(V_2(\mathbb{R}^3))$ as it is modelled by the standard inclusion $SO(2) \hookrightarrow SO(3)$. However, by construction $F|_{\partial D^2}$ extends to the disk $D^2$ providing a nullhomotopy. This gives the desired contradiction.

Answering a question in the comments:

As I alluded to above, the map $F|_{\partial D^2}: \partial D^2 \to V_2(\mathbb{R}^3) \simeq SO(3)$ is given in coordinates by $$(x,y) \mapsto \begin{pmatrix} x & - y & 0 \\ y & x & 0 \\ 0 & 0 & 1 \end{pmatrix},$$ i.e., it is the inclusion of the stabilizer of the $z$-axis in $SO(3)$. Since $SO(3)$ acts transitively on $S^2$, this inclusion fits into a fiber sequence $$S^1 \cong SO(2) \to SO(3) \to S^2.$$ Using the long exact sequence in homotopy, we see that the map $\pi_1(S^1) \to \pi_1(SO(3))$ is onto, and hence $F|_{\partial D^2}$ represents a generator of $\pi_1(SO(3))$.

I don't see how a torus is relevant. There may be a clever trick to visualize the loop inside $SO(3)$ for this problem, but I'll have to think more about it.

JHF
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    I haven't studied the classical groups through the eyes of algebraic topology yet, so why is the restriction of $F$ nontrivial on $V_2$? Is there a way to extend this argument constructing the same $F$ on the Torus instead? – David Melo Jan 29 '21 at 23:10
  • Is it possible ilustrate me how can I get the map $F$ on $D^2$? Is it the orthonormalization of $ \begin{pmatrix} f(x,y) \ g(x,y)\ ( f\times g)(x,y)\end{pmatrix}$? Many thanks! – Quiet_waters Feb 05 '21 at 18:35
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    @Quiet_waters Yes, precisely. – JHF Feb 05 '21 at 18:48
  • @JHF thank you! – Quiet_waters Feb 05 '21 at 20:15