Prove that $\mathbf{SO}_3(\mathbf{R})$ is simple by topological methods. I am trying to solve an exercise consisting in showing that the group $\mathbf{SO}_3(\mathbf{R})$ is simple by using topological methods. A hint say that I should use the trace map, but I have still no idea...
 A: I translate the nice proof that one can find here in english :
Let $E$ be a $3$-dimensional euclidan space, so that we want to show that $SO(E)$ is simple. Consider the application $$\begin{array}{rcl}
\theta : SO(E) & \rightarrow &  [0,\pi]\\
g & \mapsto & \textrm{arccos}(\frac{\textrm{Tr}(g) -1}{2}).
\end{array}$$ Write $g$'s matrix in a nice basis where it has the form $$\left(\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos(\theta) & -\sin(\theta) \\
0 & \sin(\theta) & \cos(\theta) \\
\end{array}\right)$$ for some $\theta$, which shows that the previous application is well-defined.
Let $H$ be an non-trivial connected normal subgroup of $SO(E)$. The map $\theta$ is obviously continuous and as $H$ is connected, so is $\theta(H)$. It is therefore an interval included in $[0,1]$. As $\theta$ maps the identity to $0$ you see that $\theta(H)$ contains $0$ and is of the form  $[0, \alpha[$ or $[0, \alpha]$, with $\alpha\in [0,\pi]$. As $H$ is non-trivial, $H$ contains an element different from the identity, element whose image by $\theta$ is non zero, so that $\alpha\in ]0,\pi]$. Now let $n\in\mathbb{N}^*$ such that $\frac{\pi}{n}\in ]0,\alpha[ \subset \theta(H)$ and $h\in H$ such that $\theta(h) = \frac{\pi}{n}$. Then  $h^n \in H$ and $\theta(h^n) = \pi$ so that $H$ contains an element of angle $\pi$. As such elements are all conjugated and as $H$ is normal, $H$ contains all such elements. As this elements generate $SO(E)$, the group $H$ is equal to $SO(E)$.
Now we suppose only $H$ to be a proper subgroup of $SO(E)$. Let $H^0$ be the connected component (in $H$) of the identity. It is a subgroup of $H$ as the multiplication $m : H\times H \rightarrow H$ is continuous and transforms $H^0 \times H^0$ in a connected set contained in $H$ and containing the identity, and therefore contained in $H^0$ (by definition of the connected component of the identity). An analogue argument applied to the morphisms $$\begin{array}{rcl}
\textrm{int}(g) : H & \rightarrow & H\\
h & \mapsto & g h g^{-1}
\end{array}$$ for $g\in SO(E)$ ensures that $H^0$ is a normal subgroup of $SO(E)$ and the previous case ensures that $H^0$ is trivial (would it be non-trivial, the previuous case would imply that $H^0$, and therefore $H$ is equal $SO(E)$, contradicting the hypothesis made on $H$). Now take $h\in H$. Consider the map $$\begin{array}{rcl}
\varphi_h : SO(E) & \rightarrow & H\\
g & \mapsto & g h g^{-1} h^{-1} .
\end{array}$$ It is well-defined as $H$ is normal, and it is continuous. As $SO(E)$ is connected, its image is connected, and as it contains the identity, it is contained in $H^0$ which is trivial. So $\varphi_h$ is constant equal to the identity, implying that $h$ is in $SO(E)$'s center. As the latter is trivial (each element of $SO(E)$'s center fixes all lines as it commutes with all the rotations) $h$ is equal to the identity. It follows that $H$ is trivial, proving the assertion.
