Infinitesimal Rotations Have a little question regarding infinitesimal rotations. In the Cohen Book, volumen 1, Complement B-VI, it says that the transform of a vector $\textbf{OM}$ under an infinitesimal rotation can be written, to first order in $d\alpha$ is
\begin{equation}
\Re_{\textbf{u}}(d\alpha)\textbf{OM}=\textbf{OM}+d\alpha \textbf{u}\times \textbf{OM}
\end{equation}
what I don't understand is the about the "first orden in $d\alpha$". Is a taylor expansion? Also I don't know how to deduce the that equation. Considering the differential that appears there, I think is this type of expansion, but I can't see why there's a vectorial product also.
 A: The rotation group of three dimensional space has three generators $T^a$ given by
$$ T^3 = \left(\begin{matrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{matrix}\right) \quad T^2 = \left(\begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0\end{matrix}\right) \quad T^1= \left(\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0\end{matrix}\right)$$
which yield rotations associated to a vector $\vec \phi \in \mathbb{R}^3$ by (summation over repeated indices implied in the following)
$$ R_{\vec \phi} = \mathrm{e}^{\phi^aT^a} = 1 + \phi^aT^a + \mathcal{O}(\phi^2)$$
and, by inspection, we see that they fulfill (for $e^i$ the $i$-th standard basis vector)
$$ (T^a v)^c = (T^a (v^b e^b))^c = (v^b T^a e^b)^c = v^b \epsilon^{abc} = (e^a \times \vec v)^c$$
since the cross product is given as
$$ (\vec v \times \vec w)^k = \epsilon^{ijk}v^i w^j$$
with $\epsilon$ the Levi-Civita symbol. Therefore,
$$ R_{\vec \phi} v = (1 + \phi^a T^a + \mathcal{O}(\phi^2))\vec v = \vec v + \vec\phi\times \vec v + \mathcal{O}(\phi^2)$$
and writing $\vec \phi = \phi \hat{\phi}$ with $\hat{\phi}$ being a unit vector and saying $\phi$ is infinitesimal, i.e. neglecting higher orders in $\phi$, yields your relation.
A: It is easy to verify that every proper rotation in space, which is the subgroup of the full group of rotations of $\mathbb R^3$ that is connected to the identity $I$, can be expressed in the form
$$R_u = e^{A_u},$$
where $A$ is a $3\times3$ skew-symmetric matrix. Indeed $(e^A)^Te^A = e^{-A}e^A = I$ and $\det(e^A) = e^{Tr(A)} = 1$ and therefore $e^A$ is a proper rotation. Since skew-symmetric $3\times3$ matrices are in one-to-one correspondence with vectors in $\mathbb R^3$ (by Hodge duality one has an isomorphism between $\mathbb R^3$ and $\bigwedge^2\mathbb R^3$), one can identify any skew-symmetric matrix $A$ with a vector $u_A$ representing the axis of rotation, with the magnitude of $u_A$ giving the idea of the extent of this rotation around $u_A$. For an infinitesimal rotation $\delta\alpha$ around the direction of the unit vector $u$ one then gets the rotation
$$e^{\delta\alpha A_u},$$
where $A_u$ is the skew-symmetric matrix corresponding to the vector $u$ through the isomorphism mentioned earlier. A series expansion around the identity then gives
$$e^{\delta\alpha A_u} = I + \delta\alpha A_u + o(\delta\alpha^2).$$
The way any $A_u$ acts on a vector $x$ of $\mathbb R^3$ can be seen to be
$$A_u x = u\times x,\qquad\forall x\in\mathbb R^3$$
and therefore the infinitesimal rotation of a vector $x\in\mathbb R^3$ has the approximation
$$e^{\delta\alpha A_u}x = x + \delta\alpha u\times x+o(\delta\alpha^2).$$
A: It is a Taylor series of the rotation matrix, but the idea is that we don't know the form of the matrix when we write the expansion. Instead we use the definition of rotations, as linear transformations which preserve the length of vectors, to determine the form of the expansion coefficients.
Its not hard to prove that every rotation matrix is orthogonal, i.e., $R^T=R^{-1}$.    
We can write infinitesimal rotations, which are very close to the identity, as,
$$ R= (\mathbb{I} + \epsilon \ \rho + O(\epsilon^2)) \qquad (\epsilon << 1).$$
The orthogonality of $R$ requires that,
$$ \mathbb{I} = R^T R= (\mathbb{I}+\epsilon \ \rho^T + O(\epsilon^2))(1 + \epsilon \ \rho +O(\epsilon^2)),$$
$$ \mathbb{I} =  \mathbb{I}+\epsilon \ ( \rho^T+\rho) + O(\epsilon^2),$$
if we require that coefficient of each power of $\epsilon$ vanish independently then we get,
$$ \boxed{ \rho^T + \rho =0}.$$
We now have that $\rho$ must be an antisymmetric matrix. In three dimensions there are only three linearly independent antisymmetric matrices,
$$ \rho_1 = \left( \begin{array}{ccc} \ 0 & 1 & 0 \\
\ -1 & 0& 0 \\
\ 0 & 0 & 0 \\
 \end{array}\right);
\ \rho_2 = 
\left( \begin{array}{ccc} \ 0 & 0 & 1 \\
\ 0 & 0& 0 \\
\ -1 & 0 & 0 \\
 \end{array}\right);
\ \rho_3 = 
\left( \begin{array}{ccc} \ 0 & 0 & 0 \\
\ 0 & 0& 1 \\
\ 0 & -1 & 0 \\
 \end{array}\right).
$$
It turns out that we can write the cross product of two vectors in terms of these matrices, 
$$ \vec{\omega} \times \vec{v} = \left( \begin{array} \ 0 & -\omega_3 & \omega_2 \\
\ \omega_3 & 0& -\omega_1 \\
\ -\omega_2 & \omega_1 & 0 \\
 \end{array}\right)
\left( \begin{array} \ v_1 \\ v_2 \\ v_3 \end{array} \right)$$
Since a rotation about the axis $u$ is supposed to leave any vector parallel to that axis unchanged it is natural to identify $R_u(\epsilon)$ with the corresponding "cross product matrix",
$$ R_u(\epsilon) = \mathbb{I} + \epsilon \ \vec{u} \times, $$
if we want a finite rotation about this axis we can use the formula,
$$ R_u(\theta) = \lim_{N\rightarrow \infty} (\mathbb{I} + \frac{\theta}{N} \ \vec{u} \times)^N = \exp(\theta \vec{u} \times ).$$
A: Yes, is a Taylor expansion! Take the rotational direction and define this direction as $z$ of some cartesian coordinate system (you have freedom to do that, you are a free physicist). Ignore the $z$ component of the vector, because don't change with the rotation.
The rotation take the form of:
$$
\vec{v}(\theta)=\left( \begin{array} \ v_x(\theta) \\ v_y(\theta)  \end{array} \right)=\left( \begin{array} \ cos(\theta) & -sin(\theta)\\
\ sin(\theta) & cos(\theta)\\
 \end{array}\right)
\left( \begin{array} \ v_x \\ v_y \end{array} \right)
$$
By taylor expansion in $\theta=0$ we have: 
$$
\vec{v}(\theta)=\vec{v}(0)+\theta\frac{d}{d\theta}\left( \begin{array} \ cos(\theta) & -sin(\theta) \\
\ sin(\theta) & cos(\theta) \\
 \end{array}\right)_{\theta=0}
\left( \begin{array} \ v_x \\ v_y \end{array} \right)+...=\vec{v}(0)+\theta\left( \begin{array} \ 0 & -1 \\
\ 1& 0  \\
 \end{array}\right)
\left( \begin{array} \ v_x \\ v_y \end{array} \right)+...
$$
Then, we have $\vec{v}(\theta)=\vec{v}(0)+(\vec{z}\times\vec{v}(0))\theta+...$, and, without loss of generality, for small rotations $\delta \theta$ in any direction in $3d$ space $\vec{u}$ we have:
$$
\vec{v}(\delta \theta)=\vec{v}+(\vec{u}\times\vec{v})\delta \theta.
$$
