5
$\begingroup$

On the one hand the determinant (technically the absolute value of the determinant) represents the "volume distortion" experienced by a region after being transformed.

On the other hand the scalar curvature represents the amount by which the volume of a small geodesic ball in a Riemannian manifold deviates from that of the standard ball in Euclidean space.

Is there a link between the two? Can the determinant of a matrix be interpreted as a scalar curvature? If I makes a difference, I am mostly interested in the context of general relativity.

$\endgroup$
4
  • 1
    $\begingroup$ Probably too vague to be a satisfying answer, but: the Ricci tensor is defined as a trace, and the derivative of the determinant function at the identity matrix (which should be thought of as representing the standard flat Euclidean metric) is the trace function. $\endgroup$
    – Ivo Terek
    Jun 27 at 3:18
  • $\begingroup$ Crossposted to physics.stackexchange.com/q/715747/2451 $\endgroup$
    – Qmechanic
    Jun 27 at 3:48
  • 1
    $\begingroup$ I do not understand the question. What is "Ricci scalar"? Do you mean the scalar curvature of a Riemannian metric? What is "the determinant"? The determinant is a function on the space of $n\times n$ matrices $M(n)$. Are you asking if there is a Riemannian metric on $M(n)$ whose scalar curvature equals the determinant function? $\endgroup$ Jun 27 at 13:54
  • $\begingroup$ I understand you are trying to relate the scalar curvature of a Riemannian metric in some smooth manifold to a determinant. But determinant of what matrix? It is unclear what determinant you are thinking about. $\endgroup$
    – Gold
    Jun 27 at 18:41

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.