First, let me state the form of Lagrangian for YM and GR \begin{align} L_{YM} = \alpha \textrm{tr}(F^2), \qquad L_{GR} = \beta R \end{align} I heard, YM is a gauge theory but GR isn't a really gauge theory. Due to the fact of its Cartan-Killing form of YM is positive, but GR is negative. Is it valid approach?

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  • $\begingroup$ Why did you state the Lagrangians if you did nothing with it? What do you mean by the "Cartan-Killing form of YM/GR"? The Killing form is a property of a group, not of a theory. $\endgroup$ – ACuriousMind Mar 13 '16 at 14:31
  • $\begingroup$ @ACuriousMind, I mean, consider GR as $diff(M)$ and YM as $SU(N)$, group theory. $\endgroup$ – phy_math Mar 14 '16 at 0:28
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    $\begingroup$ If you want the correct analogy to the global gauge group $\mathrm{SU}(N)$, you need to take $\mathrm{GL}(N)$, not $\mathrm{Diff}(M)$. And again, the Killing form not being positive is just a group theoretical fact - only compact group manifolds can have a definite Killing form. It's nothing to do with being a gauge theory or not. $\endgroup$ – ACuriousMind Mar 14 '16 at 0:32
  • $\begingroup$ @ACuriousMind, Thanks, i think my senior is wrong in some sense and you are right. I understand your point. $\endgroup$ – phy_math Mar 14 '16 at 0:34

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