A Proving that Levi-Civita tensor density is invariant

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It's a problem from the textbook Supergravity ( Freedman, Proeyen ). We are asked to prove that under any infinitesimal change in frame-fields, there is no change in the Levi-Civita tensor density i.e. the variation equals zero.
This is a problem from the textbook Supergravity ( by Daniel Z. Freedman and Antoine Van Proeyen ). I am trying to learn general relativity from this book. I am attempting to do the later part of the Exercise 7.14 ( on page 148 ). Basically it asks us to explicitly show that the Levi-Civita tensor density doesn't change under any variation of frame fields. I am supposed to use the formula: variation of determiant of matrix M = determinant * trace ( M_inverse * variation in M ). But I can not even think of how to begin with the problem. Any hint will be appreciated.
 
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For the frame field you have ##\delta e = e e^{\mu}_a \delta e^a_{\mu}## where ##e = \mathrm{det}(e^{\mu}_a)##, which you can use when you take the variation of ##\epsilon^{a_1 \dots} = e \epsilon^{b_1 \dots} ({e^{a_1}}_{b_1})(\dots)##
 
That ##\epsilon^{\mu\nu\rho\sigma}## is a actually a tensor is confirmed by the identity for the determinant of a ##4\times 4## matrix that

\begin{equation}

\epsilon'^{\mu'\nu'\rho'\sigma'}{\rm Det[L]}=L^{\mu'}_{\mu} L^{\nu'}_{\nu} L^{\rho'}_{\rho} L^{\sigma'}_{\sigma}\epsilon^{\mu\nu\rho\sigma}.

\end{equation}

This shows that ##\epsilon^{\mu\nu\rho\sigma}## is an idempotent pseudotensor of rank four.
 
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I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
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