It follows from Eq. 4 that the square of the duality operator is minus the identity,  The minus sign here is due to the minus sign in Eq. 4, which is in turn due to the Minkowski signature. Had we used Euclidean signature, i.e. formula_60, instead there would have been a positive sign. We define formula_61 to be self-dual if and only if  (with Euclidean signature the self-duality condition would have been formula_63). Say formula_61 is self-dual, write it as a real and imaginary part,  Write the self-dual condition in terms of formula_66 and formula_67,  Equating real parts we read off  and so  where formula_23 is the real part of formula_72. The proof of Eq. 2 in straightforward. We start by deriving an initial result. All the other important formula easily follow from it. From the definition of the Lie bracket and with the use of the basic identity Eq. 3 we have  That gives the formula  Now using Eq.5 in conjunction with formula_75 we obtain  So we have   Consider  where in the first step we have used the anti-symmetry of the Lie bracket to swap formula_79 and formula_80, in the second step we used formula_81 and in the last step we used the anti-symmetry of the Lie bracket again. So we have   Then  where we used Eq. 6 going from the first line to the second line. Similarly we have  by using Eq 7. Now as formula_85 is a projection it satisfies formula_86, as can easily be verified by direct computation:  Applying this in conjunction with Eq. 8 and Eq. 9 we obtain  From Eq. 10 and Eq. 9 we have  where we have used that any formula_80 can be written as a sum of its self-dual and anti-sef-dual parts, i.e. formula_91. This implies:  Altogether we have,  which is our main result, already stated above as Eq. We also have that any bracket splits as  into a part that depends only on self-dual Lorentzian tensors and is itself the self-dual part of formula_95 and a part that depends only on anti-self-dual Lorentzian tensors and is the anit-self-dual part of formula_96  The proof given here follows that given in lectures by Jorge Pullin  The Palatini action  where the Ricci tensor, formula_98, is thought of as constructed purely from the connection formula_99, not using the frame field.