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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.


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

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  • $\begingroup$ Do you have any Lie theoretical background? You don't need it to write out the expressions, but it would help me give some hints without writing it out in full (time is tight this morning!) You're essentially working out what the Killing form becomes when you translate your tangent space away from the tangent to the identity to the tangent to another group member. $\endgroup$ – WetSavannaAnimal Jul 5 '16 at 1:29
  • $\begingroup$ For 1, compute $dg$, which is a 2 by 2 matrix of $1$-forms. Then compute dg†, which is also a 2 by 2 matrix of $1$-forms. After that, take the matrix product dg†dg, which is a 2 by 2 matrix of $(0,2)$ forms (by a $(0,2)$ form I mean a sum of tensor products of $1$-forms, i.e. a section of $T^* \otimes T^*$), and then take the trace. $\endgroup$ – Malkoun Jul 5 '16 at 3:18
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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.

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