metric in three sphere and SU(2) consider $S^3$ $i.e$ 
\begin{align}
x_0^2 + x_1^2 + x_2^2 +x_3^2 =1
\end{align}
note that in $\mathbb{R}^4$ with metric or $\mathbb{S}^3$ we have
\begin{align}
ds^2 = l^2 (dx_0^2 + dx_1^2 + dx_2^2 + dx_3^2) = l^2 (\cos^2(\theta) d\varphi^2 + \sin^2(\theta) d^2 \chi + d\theta^2)
\end{align}
what i want to do is interpreted this $S^3$ as a group manifold $SU(2)$ 
\begin{align}
g=\begin{pmatrix} x_0 + ix_3 & ix_1 +x_2 \\
ix_1 - x_2 & x_0-ix_3 \end{pmatrix}
=\begin{pmatrix} \cos(\theta)e^{i\varphi} & \sin(\theta) e^{i\chi} \\
-\sin(\theta)e^{-i\chi} & \cos(\theta)e^{-i\varphi} \end{pmatrix} \in SU(2)
\end{align}
In this case the metric is written as 
\begin{align}
ds^2 = l^2 dx_a dx_a = \frac{l^2}{2} Tr(dg^{\dagger} dg) = -\frac{l^2}{2} Tr(g^{-1} dg)^2
\end{align}
here i have few questions 


*

*How we can obtain 
\begin{align}
dx_a dx_a = \frac{1}{2} Tr(dg^{\dagger} dg)
\end{align}

*How we can obtain
\begin{align}
Tr(dg^{\dagger} dg) = - Tr(g^{-1} dg)^2
\end{align}
If you don't mind please recommend me some relevant textbooks. 

I think i figure out the second question 
For $g \in SU(2)$
\begin{align}
&g^{\dagger} g=1, \qquad g^{\dagger} = g^{-1} \\
& dg^{-1} g + g dg^{-1} =0, \qquad \Rightarrow \qquad d(g^{-1}) = - g^{-1} dg g^{-1}
\end{align}
Thus i checked the second one
 A: If you simply want to verify that $dx_adx_a=\frac{1}{2}\text{Tr}(dg^{\dagger}dg)$ in this specific case, you can do the following: 
Think $dg$ as a matrix valued one form, i.e.in your parametrisation, $dg=A_adx^a$ except now the coefficients takes matrix values instead of real values. Then a direct computation can show 
$$dg=\begin{pmatrix} -\sin(\theta) e^{i\phi} & \cos(\theta)e^{i\chi} \\-\cos(\theta)e^{-i\chi} & -\sin\theta e^{-i\phi} \end{pmatrix} d\theta+\begin{pmatrix}i\cos(\theta)e^{i\phi}& 0\\ 0& -i\cos(\theta)e^{-i\phi}\end{pmatrix} d\phi+\begin{pmatrix}0 & i\sin\theta e^{i\chi}\\i\sin\theta e^{-i\chi}& 0\end{pmatrix}d\chi, $$ 
and $dg^{\dagger}=(A_a)^{\dagger}dx^a$.
Now $dg^{\dagger}dg$ is thus $(A_a)^{\dagger}A_b dx^adx^b$.You can now compute $Tr((A_a)^{\dagger}A_b)dx^adx^b$ to obtain$2(\cos^2(\theta) d\phi^2 +\sin^2(\theta) d^2 \chi + d\theta^2)$. One can directly see it produces the correct coefficients for the metric in the diagonal, and a short computation can show other ones vanish. 
