Hi all,

I am going to give a talk in a seminar about the general theme 'sum of squares'. My interests lie in Differential Geometry, so I recalled that the curvature is a local invariant, an obstruction to the metric being locally a sum of squares of coordinate differentials.

Can some people suggest me some good books which clearly illustrate the relation between the curvature and the metric? Also, do you guys have other suggestions about 'sum of squares' in Geometry? Thanks a lot!

  • 1
    $\begingroup$ Pythagoras' theorem ;) Also, its many generalizations. Cauchy-Schwarz inequality. Also, Laplace operators and Casimirs wrt orthonormal bases (ok, it's not like these are two different things). $\endgroup$ Jan 22, 2011 at 14:35
  • 6
    $\begingroup$ I suggest focusing on dimension 2 and Gauss curvature. $\endgroup$
    – Deane Yang
    Jan 22, 2011 at 15:18
  • $\begingroup$ Gromov has an expository paper on curvature: springerlink.com/content/0l71567x5131842q $\endgroup$
    – Ian Agol
    Jan 22, 2011 at 19:28

1 Answer 1


This is maybe late for your seminar, but a classical textbook about Riemannian geometry, including relations between curvature and metric tensor, with any signature, is

Riemannian geometry by L.P.Eisenhart

Another very interesting book for you could be

Spaces of constant curvature by J. A. Wolf

Indeed, it seems that you are searching for the Riemannian manifolds whose metric element can be written as sum and/or difference of squares of coordinate differentials. This implies that the curvature is constant and equal to 0. As shown in Wolfs'book, this can be locally realized by several manifolds with different "global geometry". As example in dimension 2, $ds^2=dx^2+dy^2$ can be realized on the (Euclidean) plane, on the cylinder, on the torus, on the Moebius strip and on the Klein bottle, while $ds^2=dx^2-dy^2$ on the (Minkowski) plane, on the torus and on the Klein bottle.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.