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I am currently reading Kiritsis's string theory book, and something bugs in the CFT (fourth) chapter. He derives the equation that should satisfy an infinitesimal conformal transformation $x^{\mu} \rightarrow x^{\mu} + \epsilon^{\mu}(x)$ which is $\partial_{\mu}\epsilon_{\nu}+\partial_{\nu}\epsilon_{\mu} = \frac{2}{d}(\partial . \epsilon)\delta_{\mu\nu}$ from $g_{\mu\nu}(x)\rightarrow g'^{\mu\nu}(x')=\Omega(x)g_{\mu\nu}(x)=\frac{\partial x^{\alpha}}{\partial x'^{\mu}}\frac{\partial x^{\beta}}{\partial x'^{\nu}}g_{\alpha\beta}(x)$. I have been trying to do that with no success, is there an additional hidden hypothesis I'm missing, or am I just bad with math?

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What have you tried? It's hard to know what you might be missing without seeing at least an outline of your work. –  David Z Jan 26 '12 at 19:27

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up vote 7 down vote accepted

The quantity $\partial_\mu \epsilon_\nu +\partial_\nu\epsilon_\mu$ is just the variation of the metric $g_{\mu\nu}$ under the (infinitesimal) diffeomorphism you wrote, $x^\mu\to x^\mu+\epsilon^\mu$. Your equation just says $$\partial_\mu \epsilon_\nu +\partial_\nu\epsilon_\mu = C \delta_{\mu\nu}$$ for some $C$ which means the condition that the variation of the metric under the diffeomorphism is proportional to the flat background metric (in conformal gauge) itself so it can be compensated by a Weyl scaling of the metric by some $\Omega(x)$. The equation above is indeed equivalent to yours because one may calculate $C$ (related to $\Omega-1$ which is infinitesimally small, just like $\epsilon$). Just trace my equation above over $\mu=\nu$ and you get $2\partial\cdot \epsilon$ on the left hand side and $Cd$ on the right hand side which implies $C=2(\partial\cdot\epsilon)/d$, just like your equation says.

The transformation rule for the metric, $\delta g_{\mu\nu} = \partial_\mu \epsilon_\nu +\partial_\nu\epsilon_\mu$, may be computed from your $g\to g'$ rule. Just Taylor-expand your formula $g'=()()g$ with respect to $\epsilon$ up to the linear terms in $\epsilon$. Use the Leibniz rule – which will produce two terms in the variation, one from the first $()$ and one from the second, and the fact that that $\partial x^{\prime\mu}/\partial x^\nu = \delta^\mu_\nu + \partial_\nu \epsilon^\mu $ which is just the derivative of $x'=x+\epsilon$ and which is how you get each term in the symmetrized sum.

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Ha ha, this will probably help me too somewhere along the way (if I`m able to mend some math weak spots in a finite amount of time such that I can return to this stuff) :-) –  Dilaton Jan 26 '12 at 19:59

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