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.
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.