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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?

Qmechanic
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phy_math
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    Related: http://physics.stackexchange.com/q/4359/2451 http://physics.stackexchange.com/q/12461/2451 http://physics.stackexchange.com/q/46324/2451 http://physics.stackexchange.com/q/71476/2451 and links therein. – Qmechanic Mar 13 '16 at 13:18
  • 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. – ACuriousMind Mar 13 '16 at 14:31
  • @ACuriousMind, I mean, consider GR as $diff(M)$ and YM as $SU(N)$, group theory. – phy_math Mar 14 '16 at 00:28
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    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. – ACuriousMind Mar 14 '16 at 00:32
  • @ACuriousMind, Thanks, i think my senior is wrong in some sense and you are right. I understand your point. – phy_math Mar 14 '16 at 00:34

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