Let $(M,g)$ be a riemannian manifold and $R(X,Y)$ the riemannian curvature as a two form with values in the endomorphisms of the tangent bundle. I define: $$ D_g(X,Y)=det(R)(X,Y) $$ with $det$ the determinant. $D_g$ is the derterminant curvature of $M$. What are the critical points of the functional $F(g)$ over $M$, where: $$F(g)=\int_M D_g^n$$ ($dim(M)=2n$)?

$\begingroup$ I do not understand the question... I only know what is a curvature of a connection... it would be clearer to you if you write little extra about this question and it might also help me to understand better,, see if you can add some more details... $\endgroup$ – Praphulla Koushik Aug 29 '18 at 18:26

1$\begingroup$ The curvature has values in the endomorphisms so that I can take the determinant of the endomorphisms, finding a two form that can be integrated, defining a functional $F(g)$ over the metrics. (we could also ask ourselves if $D_g$ is symplectic for example) $\endgroup$ – Antoine Balan Aug 29 '18 at 22:54

$\begingroup$ Is $D_g$ a two form? Isn't $D_g(2X,Y) = 2^{\dim M} D_g(X,Y)$ by your definition? And what is $D_g^n$ in your post? I think I am having trouble following your definition of $D_g$. Can you give an explicit example of how to compute $D_g$ from $R$, in the settings of very simple 2 or 4 dimensional Riemannian manifolds? (Say, the sphere of radius $r$?) $\endgroup$ – Willie Wong Aug 29 '18 at 23:01

1$\begingroup$ $D_g (2X,Y)=2D_g(X,Y)$. $D_g^n$ is a $2n$ form (the $2$ form put at power $n$). For a $2$sphere at radius $r$, $D_g$ is constant, proportional to the volum form as $R(X,Y)$ is a constant inversible endomorphism. $\endgroup$ – Antoine Balan Aug 29 '18 at 23:15

$\begingroup$ Let's try something a little more concrete. Fix a Riemannian manifold with dimension 2. We know that its Riemann curvature (with all indices lowered) can always be written as $\rho \cdot \omega \otimes \omega$, where $\rho$ is some scalar function and $\omega$ is the volume form. How do you compute from this $D_g$? What is the explicit formula of $D_g$ in terms of $\rho$ and $\omega$? // I ask because if $D_g(2X, Y) = 2 D_g(X,Y)$ then obviously I've interpreted your definition wrong. I am hoping a concrete example can help me clear things up. $\endgroup$ – Willie Wong Aug 30 '18 at 0:15