# Adjoint action on Lie algebra su(2) ($A \in SU(2), X \in \mathfrak{su(2)} \Rightarrow AXA^{-1}\in \mathfrak{su(2)}$)

I am trying to understand ho $SU(2)/\{\pm I\} \cong SO(3)$ (see: how to show $SU(2)/\mathbb{Z}_2\cong SO(3)$) but i am not sure about the adjoint action. In especially, as I understand, the adjoint action operates on $\mathfrak{su(2)}$ therefore the following statement should hold: $A \in SU(2), X \in \mathfrak{su(2)} \Rightarrow AXA^{-1}\in \mathfrak{su(2)}$

We know that $SU(2) = \{ \left( \begin{matrix}x & y\\ -\bar y & \bar x \end{matrix}\right): |x|+|y| = 1\}$ furthermore $\mathfrak{su(2)} = span\left( \left( \begin{matrix}i & 0\\ 0 & -i \end{matrix} \right), \left( \begin{matrix}0 & -1\\ 1 & 0 \end{matrix} \right), \left( \begin{matrix}0 & i\\ i & 0 \end{matrix} \right) \right)$.

So i wanted to proof the statement on the basis of $\mathfrak{su(2)}$ (since $A\in SU(2) \Rightarrow A^{-1} = \bar{A}^t$): $AXA^{-1} = \left( \begin{matrix}a+ib & c+id\\ -c+id & a-ib \end{matrix} \right) \left( \begin{matrix}i & 0\\ 0 & -i \end{matrix} \right) \left( \begin{matrix}a-ib & -c-id\\ c-id & a+ib \end{matrix} \right) =$ $\left( \begin{matrix}((-a-i b) (c-i d)+(a-i b) (c+i d) & (-a-i b) (a+i b)+(-c-i d) (c+i d)\\ (a-i b)^2+(c-i d)^2 & (a-i b) (-c-i d)+(a+i b) (c-i d))\\ \end{matrix} \right)$

but i can't find a way to write this in the basis of $\mathfrak{su(2)}$.

Am i doing something completely wrong or was my conclusion about the adjoint action wrong?

• You can write this in the basis. Try expanding out these terms. (Sorry I can't show you the steps right now- I'm on my mobile.)
– Max
Commented Dec 28, 2015 at 1:38

An element $A$ of $\mathfrak{su}(2)$ must satisfy $A=-A^*$ and $tr(A)=0$, because, respectively, of the conditions $AA^*=I$ and $\det(A)=1$ (see Lee's Book proposition 5.38, pg 117, to see how to calculate the tangent space. hint: use $\Phi:SL(2,\mathbb{C})\to SL(2,\mathbb{C})$, given by $\Phi(X)=XX^*$, and so $SU(2)=\Phi^{-1}(I)$.
Now, if $A\in SU(2)$ and $X\in\mathfrak{su}(2)$ you have $(AXA^{-1})^*=(A^{-1})^*X^*A^*=-AXA^{-1}$, because $AA^*=I$ and $X=-X^*$. That is, $AXA^{-1}\in\mathfrak{su}(2)$.
With your calculations (you messed up the product $AXA^{-1}$, when you correct) you'll get \begin{aligned} AXA^{-1} &= \left( \begin{matrix}a+ib & c+id\\ -c+id & a-ib \end{matrix} \right) \left( \begin{matrix}i & 0\\ 0 & -i \end{matrix} \right) \left( \begin{matrix}a-ib & -c-id\\ c-id & a+ib \end{matrix} \right)\\ &=(a^2+b^2-c^2-d^2)\left( \begin{matrix}i & 0\\ 0 & -i \end{matrix} \right)+(-2ad-2bc)\left( \begin{matrix}0 & -1\\ 1 & 0 \end{matrix} \right)+(-2ac+2bd)\left( \begin{matrix}0 & i\\ i & 0 \end{matrix} \right) \end{aligned}