Timeline for The determinant curvature
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Nov 30, 2022 at 13:27 | review | Close votes | |||
| Dec 5, 2022 at 3:02 | |||||
| Aug 30, 2018 at 16:32 | comment | added | Willie Wong | If you are using determinant of Ricci curvature, then this paper may be of interest. | |
| Aug 30, 2018 at 16:31 | comment | added | Willie Wong | Out of curiosity: why $D_g^2$? Do you want all manifold with vanishing $D_g$ to be a critical point? In any case, the computation of the variation $\delta \mathrm{Ric} / \delta g$ is well-known, and the computation of $\delta D_g / \delta \mathrm{Ric}$ follows from Jacobi's formula, so it seems in this case you can actually write down the Euler Lagrange equations for this action explicitly. (Though this will depend on the inverse of the endomorphism induced by the Ricci curvature. The formula for that is probably not very pretty.) | |
| Aug 30, 2018 at 16:17 | comment | added | David Hughes | One would like to put $R$ in a nice form, and in particular see how the geometry relates to its eigenvalues. It is symmetric so has a real basis in $\wedge^2 T_xM$. $A \in GL(T_xM)$ acts as $A\wedge A$, and so seeing if there is a nice basis becomes a complicated algebraic problem. | |
| Aug 30, 2018 at 16:10 | comment | added | David Hughes | Just an observation: if the manifold admits a product metric then there will always be a plane along which the curvature vanishes. This corresponds to an eigenvalue of 0 for $R$. So you probably only want such $M$ that do not admit product metrics , and you will also need to normalize volume (consider the constant curvature case). | |
| Aug 30, 2018 at 13:08 | comment | added | Antoine Balan | I must recognize a lake of concistency in my definitions. I propose you to take the Ricci curvature $Ricc(g)$ as a endomorphism and to take the determinant $D_g =det(Ricc(g))$, so that we may define $F(g)=\int_M D_g^2 w$, with $w$ a volum form. I hope that now it is all right. | |
| Aug 30, 2018 at 12:04 | comment | added | Willie Wong | @AntoineBalan: that much is basic. You still haven't told me how you compute the determinant. | |
| Aug 30, 2018 at 6:10 | comment | added | Antoine Balan | I consider the riemannian curvature as a tensor $R^i_{jkl}$ and not $R_{ijkl}$. $D_g$ is the determinant of the endomorphism when we fix the two low indice corresponding to the 2-form as $R(X,Y)=\nabla_X \nabla_Y-\nabla_Y \nabla_X-\nabla_{[X,Y]} \in End(TM) \otimes \Lambda^2 (TM)$. | |
| Aug 30, 2018 at 0:15 | comment | added | Willie Wong | 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. | |
| Aug 29, 2018 at 23:15 | comment | added | Antoine Balan | $D_g (2X,Y)=2D_g(X,Y)$. $D_g^n$ 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. | |
| Aug 29, 2018 at 23:01 | comment | added | Willie Wong | 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 $r$?) | |
| Aug 29, 2018 at 22:54 | comment | added | Antoine Balan | 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) | |
| Aug 29, 2018 at 18:26 | comment | added | Praphulla Koushik | 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... | |
| Aug 29, 2018 at 15:19 | history | asked | A.Balan | CC BY-SA 4.0 |