Question 3 chapter 4 in Brian Hall's Lie groups, Lie algebras and their representations. I am not sure how to solve the following exercise from Hall's textbook:


*Show that the adjoint representation and the standard representation are equivalent reprensentations of the Lie algebra $so(3)$. Show that the adjoint and standard represnetations of the group $SO(3)$ are equivalent.


Now the adjoint representation is $ad_X(Y)=[X,Y]$ and the standard representation is an orthogonal matrix with determinant =1.
But I don't see which isomorphism interwines between these two represnetations.
Any help?
 A: $SO(3)$ consists of matrices $A\in GL(3,\mathbb{R})$ with ${}^tA\cdot A = I$. The standard representation of $SO(3)$ is just the inclusion map $SO(3) \subset GL(3,\mathbb{R})$.
On the level of Lie algebras we have ${\mathfrak{gl}}(3;\mathbb{R}) = M_3(\mathbb{R})$, and ${\frak{so}}(3)$ consists of matrices $B\in M_3(\mathbb{R})$ with ${}^tB = -B$. Again the inclusion ${\frak{so}}(3)\subset {\mathfrak{gl}}(3;\mathbb{R})$ is the standard representation.
Now ${\frak{so}}(3)$ is three-dimensional, so once we identify it with $\mathbb{R}^3$ by picking an appropriate basis, the adjoint representation $X \mapsto \rm{ad}_X$, which is faithful, gives another injection ${\frak{so}}(3) \hookrightarrow M_3(\mathbb{R})=\mathfrak{gl}(3;\mathbb{R}).$
We want to show that if we pick the right isomorphism $\mathbb{R}^3 \simeq \mathfrak{so}(3)$, we get the same inclusion $\mathfrak{so}(3) \subset M_3(\mathbb{R})$. This is given by the hat map $u \mapsto \widehat{u}$, where
$$ u = \left(\begin{array}{c}x \\y \\ z\end{array}\right)\in \mathbb{R}^3,\ \ \ \widehat{u} = \left(\begin{array}{ccc} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0\end{array}\right)\in \mathfrak{so}(3).$$
Let $X, Y\in \mathfrak{so}(3)$. Then $\mathrm{ad}_X(Y) = XY - YX$ as matrices in $M_3(\mathbb{R})$. We have to check that if $Y = \widehat{u}$, then $\mathrm{ad}_X(Y) = \widehat{X\cdot u}$. This can be checked on a basis of $\mathfrak{so}(3)$, so let $\epsilon_1 = \widehat{e_1}$, $\epsilon_2 = \widehat{e_2}$, $\epsilon_3 = \widehat{e_3}$. Then
\begin{align*} \mathrm{ad}_{\epsilon_1}(Y) &= \epsilon_1 \widehat{u} - \widehat{u} \epsilon_1 \\&=  \left(\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0\end{array}\right) \left(\begin{array}{ccc} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0\end{array}\right) -\left(\begin{array}{ccc} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0\end{array}\right)\left(\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0\end{array}\right)\\
&=\left(\begin{array}{ccc} 0 & 0 & 0 \\ y & -x & 0 \\ z & 0 & -x\end{array}\right)-\left(\begin{array}{ccc} 0 & y & z \\ 0 & -x & 0 \\ 0 & 0 & -x\end{array}\right) = \left(\begin{array}{ccc} 0 & -y & -z \\ y & 0 & 0 \\ z & 0 & 0\end{array}\right)
\end{align*}
and
$$ \epsilon_1 \cdot u = \left(\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0\end{array}\right)\left(\begin{array}{c}x \\y \\z \end{array}\right) =
\left(\begin{array}{c}0 \\-z \\y \end{array}\right)$$
Indeed we have $\widehat{\epsilon_1 \cdot u} = \mathrm{ad}_{\epsilon_1}(\widehat{u})$. The other two basis vectors I will leave for you to enjoy working out. This shows the adjoint representation of $\mathfrak{so}(3)$ is equivalent to its standard representation. Since these come from the adjoint and standard representations of the group $SO(3)$, the latter are also equivalent. 
The connection with the cross product is not quite relevant for this problem, but I'll briefly mention it. $\mathbb{R}^3$ is isomorphic to $\mathfrak{so}(3)$ as we saw. The cross product gives a representation of $\mathbb{R}^3$ on itself. It so happens that under the hat map this is the same as the standard action of $\mathfrak{so}(3)$: we have $\epsilon_i\cdot u = e_i \times u$. By the above, it's also equivalent to the adjoint representation.
