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There is an exercise on page 114 of Humphreys' Linear Algebraic Groups (GTM 21)

Prove that $SO(3, \mathbb R)$ (= group of $3 \times 3$ real orthogonal matrices of determinate $1$) is a connected subgroup of $SL(3, \mathbb C)$ consisting of semisimple elements, but not commutative.

The semisimplicity of the elements and the noncommutativity of the group is clear. For the connectedness, I think this proposition might be applied:

Let $G$ be an algebraic group, $I$ an idex set, $f_i: X_i \rightarrow G$ $(i \in I)$ a family of morphisms from irreducible varieties $X_i$, such that $e \in Y_i = f_i(X_i)$ for each $i \in I$. Set $M = \cup_{i \in I} Y_i$. Then the intersection of all closed subgroup of $G$ containing $M$ is a connected subgroup of $G$.

But I don't know how to determine the $X_i$'s.

Thanks for help.

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Hint: the group $SO(3)$ acts transitively on the set of positively oriented orthonormal bases of $\Bbb R^3$. – Andrea Mori Feb 8 '12 at 15:47
@Andrea Mori: Thank you very much for the hint. I am sorry but I don't know how is this fact related to the connectedness of the group. May I ask for some further hint? Thanks again. – ShinyaSakai Feb 13 '12 at 16:32
@ShiniaSakai: well, let $g\in SO(3)$ and let $\cal B$ the orthonormal basis of $\Bbb R^3$ obtained applying $g$ to the standard one. If you manage to obtain the basis $\cal B$ from the standard one by moving the latter continuously so that at each "time" $t$ the basis stays orthonormal, this amounts to finding a path in $SO(3)$ joining the identity to $g$, i.e. showing that $SO(3)$ is pathwise connected. It shouldn't be too hard to visualize it – Andrea Mori Feb 13 '12 at 17:54
(continued) (move the first standard vector to the first vector in $\cal B$ moving along the sphere $S^2$ and drag the other two vectors along keeping them on the orthogonal plane and at a right angle. When the first vector is in place, rotate the other two into the correct position. This works because $SO(3)$ kees the orientation) – Andrea Mori Feb 13 '12 at 17:56
@Andrea Mori: Thank you very much for the geometric answer~ Thanks a lot for taking time to explain it to me. – ShinyaSakai Feb 13 '12 at 18:42

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