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I am working on an exercise in Tapp's matrix groups for undergraduates. It is a proof that $SO(3)$ is path-connected. $SO(3)$ is the group

$$ SO(3) = \{A \in O(n)\mid \det A = 1 \}$$

where $O(n)$ is the group of orthogonal matrices.

My work so far:

An element in $SO(3)$ is a matrix with columns $(p|v|p\times v)$ where $p \bot v$ and $\|p\|=\|v\|=1$. Let $A=(p_a|v_a|p_a \times v_a)$ and $B=(p_b|v_b|p_b\times v_b)$ be two elements in $SO(3)$. The goal is to find a path.

Let $R$ be the rotation of $\mathbb R^3$ that takes $p_a$ to $p_b$. Let $\varphi_R$ be a parametrisation of $R$ such that $\varphi_R(0)$ is the identity map and $\varphi_R({1\over 2})$ has rotated $p_a$ to $p_b$.

Similarly, if $R'$ is the rotation that rotates $v_a$ to $v_b$ around the axis $p_b = R(p_a)$ and $\varphi_{R'}:[{1\over 2}, 1]\to \mathbb R^3$ its parametrisation then $\varphi_{R'}\circ \varphi_R$ is a path from $A$ to $B$.

My problem is: this proof is missing all the details but I don't know how to write it out. Could someone please show me how to write out this proof rigourously and in detail?

learner
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    I wouldn't say "Let $R$ be the rotation of...", as there are in fact infinitely many of these. Also, you can simplify the proof some by showing that any point can be connected by a path to a convenient reference point, say $I$. – Travis Willse Dec 26 '14 at 07:14
  • Also, your proof takes for granted that there is are rotations with the properties you describe: one takes $p_a$ to $p_b$, the other fixes $p_b$ but rotates $v_a$ to $v_b$. You should at least justify the existence of the second of these, but you might like to be more explicit about how to construct both of these rotations in the first place. – Travis Willse Dec 26 '14 at 07:16
  • There are several related threads on our site. See for example this or this. Not really duplicates, but the ideas recur. – Jyrki Lahtonen Dec 26 '14 at 09:07

2 Answers2

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I suggest using the fact that every matrix $X \in SO(3)$ is of the form $AX(\theta)A^{-1}$, where $A \in O(3)$ and $$X(\theta) = \begin{pmatrix} 1 & 0 & 0\\0 & \cos(\theta) & -\sin(\theta)\\0 &\sin(\theta) & \cos(\theta)\end{pmatrix}.$$ Then for every $X \in SO(3)$, the map $F_X : [0,1] \to SO(3)$ defined by the equation $F_X(t) = AX(t\theta)A^{-1}$ gives a path in $SO(3)$ from the identity to $X$.

kobe
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  • Thank you but the book has not mentioned this fact so far. – learner Dec 27 '14 at 00:25
  • Can you link to a source where this is explained in detail? – learner Dec 27 '14 at 00:29
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    @learner review Theorem 7.25, and use that to prove the above fact. – kobe Dec 27 '14 at 19:57
  • In the meantime your answer gave me an idea for a proof that is easier (with my background anyway): I can use that any $X\in SO(3)$ can be written as a product of the rotations $X(\theta_x), Y(\theta_y)$ and $Z(\theta_z)$ (the rotations around the three axes). – learner Dec 28 '14 at 00:28
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One possible approach is to use that (1) every element has $1$ as eigenvalue, (2) any element of $SO(3)$ that fixes a given vector $v\neq0$ also fixes the plane $v^\perp$ and is determined by its restriction to $v^\perp$, and (3) that restriction is an orientation-preserving orthogonal linear transformation of that plane, i.e., a rotation of the plane. Now given $M\in SO(3)$ find a vector $v\neq0$ fixed by $M$ using (1) and proceed to show that the set $\{\, A\in SO(3)\mid Av=v\,\}$ containing $M$ and $I$ is path connected, which will suffice. By (2) and$~$(3), restriction defines a homeomorphism of that set to the set of rotations of $v^\perp$, which is path connected, completing the proof. Proving each of (1), (2), (3) is easy.

Marc van Leeuwen
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