Help with variation of the 3-dimensional ##\sigma##-model action

[user1139]

Homework Statement

See below.

Relevant Equations

See below.

Consider the following action

$$S=\int\mathrm{d}^3x\sqrt{h}\left[R^{(3)}-\frac{1}{4}\mathrm{Tr}\left(\chi^{-1}\chi_{,i}\chi^{-1}\chi^{,i}\right)\right]$$

where $h$ is the determinant of the 3-dimensional metric tensor $h_{ij}$ and $R$ is the Ricci scalar.

I want to get the equations of motion

\begin{align*}
\left(\chi^{-1}\chi^{,i}\right)_{;i}&=0,\\
R_{ij}&=\frac{1}{4}\mathrm{Tr}\left(\chi^{-1}\chi_{,i}\chi^{-1}\chi_{,j}\right).
\end{align*}

However, how do I perform the variation on the trace?

 

[samalkhaiat]

Homework Statement:: See below.
Relevant Equations:: See below.

Consider the following action

$$S=\int\mathrm{d}^3x\sqrt{h}\left[R^{(3)}-\frac{1}{4}\mathrm{Tr}\left(\chi^{-1}\chi_{,i}\chi^{-1}\chi^{,i}\right)\right]$$

where $h$ is the determinant of the 3-dimensional metric tensor $h_{ij}$ and $R$ is the Ricci scalar.

I want to get the equations of motion

\begin{align*}
\left(\chi^{-1}\chi^{,i}\right)_{;i}&=0,\\
R_{ij}&=\frac{1}{4}\mathrm{Tr}\left(\chi^{-1}\chi_{,i}\chi^{-1}\chi_{,j}\right).
\end{align*}

However, how do I perform the variation on the trace?

If your concern is only about the variation of the trace, you better ignore the coupling to gravity. Also, the manipulations hold in any number of dimensions. The relevant identities are $\delta \mbox{Tr}(\cdot) = \mbox{Tr}(\delta \cdot)$, $\delta \partial = \partial \delta$, $\mbox{Tr}(AB) = \mbox{Tr}(BA)$, and the following important two $$\delta u^{-1} = - u^{-1} (\delta u) u^{-1}, \ \ \ \ (1)$$$$\partial u^{-1} = - u^{-1}(\partial u) u^{-1}. \ \ \ \ (2)$$ To make your life easy, define the vector $$A^{\mu} = u^{-1} \partial^{\mu} u.$$ So, the NLSM action becomes $$S = - \frac{1}{4} \int d^{n}x \ \mbox{Tr}(A^{\mu}A_{\mu}).$$ Thus $$\delta S = - \frac{1}{2} \int d^{n}x \ \mbox{Tr}(A^{\mu} \ \delta A_{\mu}) .$$ Now, use (1) to obtain $$\delta A_{\mu} = u^{-1} \ \partial_{\mu}\delta u - u^{-1} \ \delta u \ A_{\mu} .$$ Substitute in $\delta S$, you get $$\delta S = \frac{1}{2} \int \ \mbox{Tr}\left( u^{-1}\delta u \ A_{\mu}A^{\mu} - A^{\mu} u^{-1} \ \partial_{\mu}\delta u \right).$$ Integrate the 2nd term by part and ignore the surface term to obtain $$\delta S = \frac{1}{2} \int \ \mbox{Tr}\left( u^{-1}\delta u \ A^{\mu}A_{\mu} + \partial_{\mu}(A^{\mu}u^{-1}) \ \delta u \right).$$ Finally, expand the differentiation in the second term and use (2) to get $$\delta S = \frac{1}{2} \int \ \mbox{Tr} \left( u^{-1}\delta u \ A^{\mu}A_{\mu} + \partial_{\mu}A^{\mu} \ u^{-1}\delta u - A^{\mu}A_{\mu} \ u^{-1}\delta u \right).$$ The first and the third terms add up to zero (because $\mbox{Tr}(ab)= \mbox{Tr} (ba)$), and you end up with $$\delta S = \frac{1}{2} \int \ \mbox{Tr}\left( (\partial_{\mu}A^{\mu}) u^{-1} \ \delta u \right).$$ This gives you the equation of motion $$\partial_{\mu}A^{\mu} \equiv \partial_{\mu} \left(u^{-1} \ \partial^{\mu}u \right) = 0.$$