# Hairy Ball Theorem and $SO(3)$

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 $$S^2$$?

• 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? – William Mar 29 '20 at 14:33
• 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. – hb12ah Mar 30 '20 at 0:27

## 1 Answer

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. $$\square$$

• 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. – Jason DeVito Mar 29 '20 at 18:43
• 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 $\Psi$". – hb12ah Mar 30 '20 at 0:14
• @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 ..... – Jason DeVito Mar 30 '20 at 2:18
• $\psi:A\rightarrow B$ 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 ... – Jason DeVito Mar 30 '20 at 2:23
• 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. – Jason DeVito Mar 30 '20 at 2:27