metric in three sphere and SU(2)

Question

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

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

2. 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.

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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

Answer 1

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.