How to prove that $SO(3) \neq S^2 \times S^1$ using Hairy Ball Theorem? In other words, if assuming $SO(3) = S^2 \times S^1$, how to construct a non-vanishing vector field on $S^2$?
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$\begingroup$ Can you provide any more context or details? For instance do you know for a fact that "HBT $\implies SO(3)\neq S^2 \times S^1$" and you just need to prove it, or is this a conjecture that you're trying to verify? $\endgroup$ – William Mar 29 '20 at 14:33
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1$\begingroup$ Hi @William, this is an exercise from Introduction to Mechanics and Symmetry by Jerrold E. Marsden and Tudor S. Ratiu, Exercise 1.2-4. That's why I'm pretty sure this is correct. $\endgroup$ – hb12ah Mar 30 '20 at 0:27
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I'm not exactly sure what you're looking for, but here's one approach.
Inside of the tangent bundle to $S^2$, $TS^2$, we can consider the vectors of unit length: $T^1S^2 =\{(p,v)\in TS^2: |v| = 1\}$. Then we have a map $\pi:T^1 S^2\rightarrow S^2$ given by $\pi(p,v) = p$, and this is a fiber bundle with fiber $S^1$.
Proposition 1: $T^1S^2\rightarrow S^2$ is bundle isomorphic to $\rho:SO(3)\rightarrow S^2$ with $\rho(A) = A_1$, the first column of $A$.
Proof: Given $(p,v)\in T^1 S^2$, note that $p\in S^2$ so $|p| = |v| = 1$. Further, $p\bot v$ since $v\in T_p S^2$. Thus, using the cross product on $\mathbb{R}^3$, the triple $\{p,v,p\times v\}$ is an oriented orthonormal basis of $\mathbb{R}^3$. So, we can create a map $\phi:T^1 S^2\rightarrow SO(3)$ by defining $\phi(p,v) = A$ where $A$ is the matrix whose columns are $A_1 = p, A_2 = v, A_3 = p\times v$. A simple calculation verifies that $\det (A) = 1$, so this really does land in $SO(3)$.
It's also easy to write down the inverse - $\phi^{-1}(A) = (A_1, A_2)$.
Finally, simply note that $\rho(\phi(p,v)) = p = \pi(p,v)$, so $\phi$ is a bundle map.$\square$
Note that under $\phi$, the projection map $T^1S^2\rightarrow S^2$ turns into $\rho:SO(3)\rightarrow S^2$ with $\rho(A) = A_1$.
Now, let $\psi:T^1 S^2\rightarrow S^1\times S^2$ be a diffeomorphism (or even just a homeomorphism). Transporting the bundle structure on $SO(3)$ to $\psi$, we see that the map $\alpha:\rho \circ \psi^{-1}:S^1\times S^2\rightarrow S^2$ is a fiber bundle map with fiber $S^1$. In particular, it is a fibration.
Proposition 2: The bundle $\alpha:S^1\times S^2\rightarrow S^2$ has a section. That is, there is a map $f:S^2\rightarrow S^1\times S^2$ for which $\alpha \circ f= Id_{S^2}$.
Proof: The last portion of the long exact sequence in homotopy groups associated to the bundle $S^1\rightarrow S^1\times S^2\xrightarrow{\alpha} S^2$ looks like $$0=\pi_2(S^1)\rightarrow \pi_2(S^1\times S^2)\xrightarrow{\alpha_\ast} \pi_2(S^2)\rightarrow \pi_1(S^1)\rightarrow \pi_1(S^1\times S^2)\rightarrow \pi_1(S^2)= 0$$ which is the same as $$0\rightarrow \mathbb{Z}\xrightarrow{\alpha_\ast} \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow 0.$$ The last self map of $\mathbb{Z}$ must be surjective by exactness. Since $\mathbb{Z}$ is Hopfian, this last self map is an isomorphism. Now exactness shows that $\alpha_\ast$ is an isomorphism as well.
Because $\alpha_\ast$ is an isomorphism, it maps generators to generators. So, let $g:S^2\rightarrow S^1\times S^2$ be a generator of $\pi_2(S^1\times S^2)$. Then the composition $S^2\xrightarrow{g} S^1\times S^2\xrightarrow{\alpha} S^2$, perhaps after replacing $\alpha$ with $a\circ \alpha$ with $a$ the antipodal map, has degree $1$, so is homotopic to the identity map.
Since $\alpha$ is a fibration, we may therefore apply the homotopy lifting property. This yields a homotopy $g_t$ of $g$ for which $g_0 = g$ and $g_1$ has the property that $\alpha\circ g_1 = Id_{S^2}$. Now simply take $f= g_1$.$\square$
Proposition 3: Assuming $SO(3)$ is homeomorphic to $S^1\times S^2$, there is a non-vanishing vector field on $S^2$.
Proof: Such a homeomorphism $\psi:SO(3)\rightarrow S^1\times S^2$ gives the map $\alpha = \rho \circ \psi^{-1}: S^1\times S^2\rightarrow S^2$ the structure of a fiber bundle over $S^2$ with fiber $S^1$. Applying proposition $2$, there is a map $f:S^2\rightarrow S^1\times S^2$ with $\alpha \circ f = Id_S^2$.
Then the map $\psi^{-1}\circ f: S^2\rightarrow SO(3)$ has the property that $\rho\circ(\psi^{-1}\circ f) = (\rho \circ \psi^{-1})\circ f = \alpha \circ f = Id_{S^2}$. That is, the bundle $SO(3)\rightarrow S^2$ has a section. Since this bundle is isomorphic to the bundle $T^1 S^2\rightarrow S^2$ (Proposition 1), this bundle must have a section. But a section is a unit length vector field on $S^2$, so is, in particular, a non-vanishing vector field. $\square$
answered Mar 29 '20 at 18:30
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2$\begingroup$ Incidentally, I accidentally originally proved the converse: if $SO(3)\neq S^1\times S^2$, then $HBT$ is true. This proof is significantly easier: assuming $HBT$ is false, one can use the non-vanishing vector field to show $T^1 S^2 \cong S^1\times S^2$ by simply writing down a map. This, together with Proposition 1 above shows that $SO(3) = S^1\times S^2$. The contrapositive of all this gives the desired result. $\endgroup$ – Jason DeVito Mar 29 '20 at 18:43
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$\begingroup$ Thanks, excellent solution! I have a few questions that I hope you could help clarify. 1) Regarding your comment, if we have a non-vanishing vector field on $S^2$, how do we construct a section of $T^1 S^2$. One thing I can think of is showing the norms of the vectors in the vector field have a nonzero lower bound since $S^2$ is compact. 2) After proving the bundle isomorphism between $T^1 S^2$ and $SO(3)$, you could treat them as if they are the same? I found it a bit confusing when you said "Transporting the bundle structure on 𝑆𝑂(3) to $\Psi$". $\endgroup$ – hb12ah Mar 30 '20 at 0:14
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1$\begingroup$ @hb12ah: For 1) you don't even need compactness. Give $M$ some background Riemannian metric. Then, if $V(p)$ is a non-vanishing vector field, then $\frac{V(p)}{|V(p)|}$ is a unit length vector field. (Different choices of Riemannian metrics lead $T^1 M$ being a different subset of $TM$, but all "copies" of $T^1 M$ are bundle isomorphic.) For 2) Yes, $T^1 S^2$ and $SO(3)$ are essentially the same. For the confusing part, I should have written "Transporting the bundle structure on $SO(3)$ by $\psi$." This is a general principle: if $A$ and $B$ are two objects and ..... $\endgroup$ – Jason DeVito Mar 30 '20 at 2:18
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$\begingroup$ $\psi:A\rightarrow B$ is an isomorphism, then you can use $\psi$ to transport any "special structure" $A$ has (as long as its defined by maps into or out of $A$) to $B$. In this case, $SO(3)\rightarrow S^2$ is a bundle, so you can use $\psi$ to make $S^1\times S^2$ a bundle over $S^2$ by using the projection map $\rho\circ \psi^{-1}$. Then $\psi$ isn't just a diffeo, but a bundle isomorphism. But there are many more examples. For instance, if $(R,+,\cdot)$ is a ring and $(G,+)$ is a group with $f:R\rightarrow G$ an isomorphism from the additive group of $R$ to $G$, you can define a ... $\endgroup$ – Jason DeVito Mar 30 '20 at 2:23
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2$\begingroup$ If $V(p)$ is smooth, then $\frac{V(p)}{|V(p)|}$ is smooth. This is simply because the real function $f(x,y) = \frac{x}{y}$ (for $x,y\in\mathbb{R}$) is smooth wherever it is defined. For, in local coordinates, $V(p)$ is really just an $n$-tuple of $C^\infty$ functions. $\endgroup$ – Jason DeVito Mar 30 '20 at 2:27
