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I was trying to calculate the fundamental group of $SO(3)$. In order to represent the group I reasoned the following way:

In order to build the $3\times3$ orthogonal matrix I need an orthonormal positive basis. So I take the first vector $v_1$ in $S^2$, that corresponds to the first column of the matrix or to the image of $e_1$. Then, I have to choose another vector in the sphere orthogonal to this one, in other words, I have to take a vector in the circle normal plane of $v_1$. The third vector is given by conserving orientation.

So I have chosen a vector in $S^2$ then a vector in $S^1$, freely. So I should have $SO(3)=S^2 \times S^1$. But, of course, this is not the case.

I checked wikipedia construction of $SO(3)$ ( http://en.wikipedia.org/wiki/Rotation_group_SO%283%29#Topology ) and it makes sense, but I don't find the flaw on the previous reasoning.

Thanks in advance.

Physor
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Vladmir Sicca
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    This is not a full answer, but A) $v_1$ and $-v_1$ give the same normal plane so at some point you should mod out by a $\mathbb{Z}/2\mathbb{Z}$-action.

    Now B) if you start thinking about how this action should work, you notice that you have to answer another question first: how do you identify the cirles in these normal planes? There are several ways of hanging a S^1 above each point of S^2 and the 'correct' one is not S^2 \times S^1 but the Hopf fibration of $S^3$ (see wikipedia). If you mod this out by the 'identify antipodes'-action of $\mathbb{Z}/2\mathbb{Z}$ you do indeed get $SO(3)$.

     – Vincent Jan 24 '15 at 03:33
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    The space from which you choose the vector in the circle orthogonal to $v_1$ depends upon your choice of $v_1$; it is homeomorphic to $S^1$, but there is no preferred homeomorphism. In order to define a homeomorphism $SO(3) \to S^2 \times S^1$, you would need a method of producing a specific element of the circle. –  Jan 24 '15 at 03:46
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    As it is you have argued that the map $SO(3) \to S^2$ which sends a matrix to $v_1$, the first column vector is a map where the fibre over any pint is $S^1$. That is, you have produced a fibre bundle $$SO(3) \to S^2$$ with fibre $S^1$. It is not obvious that this bundle is trivial (as indeed it is not). If it were, you would indeed have $SO(3) = S^2 \times S^1$. –  Jan 24 '15 at 03:46
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    Why is a Mobius strip not a cylinder? After all, you can build it by starting with a circle and attaching a line segment at each point, right? –  Jan 24 '15 at 03:57
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    An excellent question for Math.SE! –  Jan 24 '15 at 07:23
  • Thanks for the comments, they really helped me clarify the problem. And sorry about the off-topic, I'll try to avoid it in the future. – Vladmir Sicca Jan 24 '15 at 21:39

1 Answers1

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The beginning of your argument is correct. The first vector $v = v_1$ of a three-by-three orthogonal matrix is indeed a point of $S^2$, and every point of $S^2$ arises this way. It is also true that the set of vectors perpendicular to $v$, call it $P(v)$, is homeomorphic to $S^1$.

However, after that your reasoning breaks down. While $P(v)$ is homeomorphic to a circle, there is no canonical choice of a homeomorphism. There is not even a continuous way (depending on $v$) to make such a choice.

For, suppose there was such a continuous choice. Let $f_v \colon\, S^1 \to P(v)$ denote the resulting family of homeomorphisms. Fix $1 \in S^1 \subset \mathbb{C}$. Define $X \colon\, S^2 \to S^2$ by $X(v) = f_v(1)$. We now note that $P(v)$ is canonically identified with $T_v^1 S^2$, the tangent vectors to $S^2$, based at $v$, of unit length. Thus $X$ gives a non-vanishing unit tangent vector field on $S^2$. This contradicts the hairy ball theorem.

Said another way, the circle has too many self-homeomorphisms. For another, simpler example, we have the interval, with its two, essentially different self-homeomorphisms. The first is the identity map and the second is the reflection about the point $1/2$. Using these to make spaces gives us the annulus and the Möbius band.

Sam Nead
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    Thanks! The answer was good and presented something to be explored further. – Vladmir Sicca Jan 26 '15 at 18:29
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    Did you mean $X:S^2 \to TS^2$ ? – Physor Jun 11 '22 at 13:33
  • No - since $P(v) \subset S^2$ the range of the map $X$ is correct as written. However, $P(v)$ is canonically identified with $T^1_v S^2$. Thus $X$ "gives" (better, "induces") a non-vanishing unit tangent vector field on $S^2$. – Sam Nead Jun 11 '22 at 15:06