1
$\begingroup$

In the book 'Supergravity' by Freedman and van Proeyen, in exercise (7.27) it is written

To calculate [the variation $\delta\omega_{\mu ab}$ of the torsion-free spin connection], consider the variation of the Cartan structure equation (7.81) without torsion: $d\delta e^a+\omega^a{}_b\wedge\delta e^b+\delta\omega^a{}_b\wedge e^b=0$. This $2$-form equation is equivalent to the component relation \begin{align} D_{[\mu}\delta e_{\nu]}^a+(\delta\omega_{[\mu}{}^{ab})e_{\nu]b}=0 \quad\text{with}\quad D_\mu\delta e_\nu^a\equiv \partial_\mu\delta e_\nu^a+\omega_\mu{}^{ab}\delta e_{b\nu}\,. \tag{7.95} \end{align} With the structure of (7.92) in mind, you should be able to derive \begin{align} e_\nu^a\, e_\rho^b\,\delta\omega_{\mu ab}=(D_{[\mu}\delta e_{\nu]}^a)e_{\rho a}-(D_{[\nu}\delta e_{\rho]}^a)e_{\mu a}+(D_{[\rho}\delta e_{\mu]}^a)e_{\nu a}\,. \tag{7.96} \end{align}

Eq. (7.92) is

\begin{align} \omega_{\mu[\nu\rho]}(e)=\tfrac12(\Omega_{[\mu\nu]\rho}-\Omega_{[\nu\rho]\mu}+\Omega_{[\rho\mu]\nu})=\omega_{\mu ab}(e)\,e_\nu^a\, e_\rho^b \tag{7.92} \end{align}

where, as per eq. (7.89)

\begin{align} \Omega_{[\mu\nu]\rho}=(\partial_\mu e_\nu^a-\partial_\nu e_\mu^a)\,e_{a\rho}\,. \tag{7.89} \end{align}

I multpliy eq. (7.95) with the Minkowski metric $\eta_{ac}$ to get \begin{align} &\partial_{[\mu}\delta(\eta_{ac}e_{\nu]}^a)+\omega_{[\mu|cb}\,\delta e_{\nu]}^b+(\delta\omega_{[\mu|cb})e_{\nu]}^b=0\quad (\because \partial_\mu\eta_{ac}=0\,\, \& \,\, \delta\eta_{ac}=0\, \text{as}\,\,\eta_{ac}\,\text{is constant}) \\ \Rightarrow\,\,& \partial_{[\mu}\delta e_{\nu] a}+\omega_{\mu|ab}\delta e_{\nu]}^b+(\delta\omega_{\mu|ab}) e_{\nu]}^b=0 \\ \Rightarrow\,\,& D_{[\mu}\delta e_{\nu] a}+\tfrac12[(\delta\omega_{\mu ab}) e_{\nu}^b-(\delta\omega_{\nu ab}) e_{\mu}^b]=0 \\ \Rightarrow\,\,& (\delta\omega_{\mu ab}) e_{\nu}^b=2D_{[\mu}\delta e_{\nu] a}+(\delta\omega_{\nu ab}) e_{\mu}^b \end{align} Relabelling $\nu$ as $\rho$ and then multiplying the above equation with $e_\nu^a$, we get, \begin{align} e_{\nu}^a\, e_{\rho}^b\,(\delta\omega_{\mu ab})=2\,(D_{[\mu}\delta e_{\rho] a})e_\nu^a+(\delta\omega_{\rho ab}) e_{\mu}^b e_\nu^a\,. \tag{c1} \end{align} We can see that this equation is different from eq. (7.96). The book says that to go from eq. (7.95) to eq. (7.96), eq. (7.92) has been used. Does it mean that Freedman and Proeyen derived the expression for $\delta\omega_{\rho ab}$ appearing on the right side of eq. (c1), by varying eq. (7.92)? If yes, then why not simply derive the variation of the spin connection directly by varying eq. (7.92) instead of starting from the variation of the Cartan structure equation, which is eq. (7.95)?

Can someone show a derivation of eq. (7.96)?

$\endgroup$

1 Answer 1

2
$\begingroup$

I was looking for the same answer and I stumbled upon your question. It's very late but I guess leaving a response is probably gonna be useful for someone in our situation in the future! I'll just answer your last question, by showing a simple derivation of eq.(7.96). Consider the eq.(7.95)

\begin{equation} D_{[\mu}\delta e^{a}_{\rho]}e_{\nu a}=\dfrac{1}{2}(\delta\omega_{\mu ab}e^{a}_{\nu}e^{b}_{\rho}-\delta\omega_{\rho ab}e^{a}_{\nu}e^{b}_{\mu}). \end{equation} Now, let's permute the indices to obtain the two expressions: \begin{align} D_{[\nu}\delta e^{a}_{\mu]}e_{\rho a}&=\dfrac{1}{2}(\delta\omega_{\nu ab}e^{a}_{\rho}e^{b}_{\mu}-\delta\omega_{\mu ab}e^{a}_{\rho}e^{b}_{\nu})\\ D_{[\rho}\delta e^{a}_{\nu]}e_{\mu a}&=\dfrac{1}{2}(\delta\omega_{\rho ab}e^{a}_{\mu}e^{b}_{\nu}-\delta\omega_{\nu ab}e^{a}_{\mu}e^{b}_{\rho}). \end{align} By summing the first two and subtracting the third equation, using $\omega_{\mu (ab)}=0$, you easily get your (7.96).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.