Recent content by shooride

Yeah, since the covariant derivative is a covariant :biggrin:. I think I was a bit confused when I asked this question :oops:. Anyway, do you know whether there is a particular name for the identity ##g^{jk}\Gamma^i{}_{jk}=\frac{-1}{\sqrt{g}}\partial_j(\sqrt{g}g^{ij})##?!

I guess I should name ##g^{jk}\Gamma^i{}_{jk}##, ##a^i##:biggrin:.

Hi friends, I'm learning Riemannian geometry. I'm in trouble with understanding the meaning of ##g^{jk}\Gamma^{i}{}_{jk}##. I know it is a contracting relation on the Christoffel symbols and one can show that ##g^{jk}\Gamma^i{}_{jk}=\frac{-1}{\sqrt{g}}\partial_j(\sqrt{g}g^{ij})## using the...

You are right! For this, I used normal coordinates, then wrote $$ d^2(x,x_0)=d^2(x'+(x-x'),x_0)=d^2(x',x_0) + (x-x')^\mu\partial_\mu d^2(x',x_0)\\+1/2(x-x')^\mu(x-x')^\nu\nabla_\mu\partial_\nu d^2(x',x_0)+\dots $$ I see your points.. but the thing is that I'm not sure the squared distance is a...

Can one define an isomorphism/diffeomorphism map ## f: R^n\to M^m## when ##n>m##? ##M## is a non-compact Riemannian manifold..

I'm interested in a Riemannian manifold in which ##x+x'## and ##\alpha.x## where ##\alpha## is a scalar from some field have a meaningful definition..so I thought that this kind of riemannian manifold should has an additional vector space structure..if I understand it correctly, one can locally...

Yeah.. I think things are starting to be a bit clearer .. however, I couldn't understand what does really a Riemannian vector space look like?!

I couldn't get your point on this quote..how can one define the distance on M?!

Can one define a vector space structure on a Riemannian manifold ##(M,g)##?! By this I mean, does it make a sense to write ##x+y## where ##x,y## are arbitrary points on ##M##?

I'd like to consider the taylor expansion of the square of the distance function (geodesic distance) ##d^2:M\times M\to R ## on a smooth Riemannian manifold ##(M,g)##. BTW, AFAIK, one can only define the Euclidean distance on a Euclidean space, right?!

I couldn't get your point. Could you please explain it more?! Where am I doing anything wrong?! I think that I miss some points about the distance and displacement..

Does it make a sense to define the Taylor expansion of the square of the distance function? If so, how can one compute its coefficients? I simply thought that the square of the distance function is a scalar function, so I think that one can write $$ d^2(x,x_0)=d^2(x'+(x-x'),x_0)=d^2(x',x_0) +...

Yeah... I think everything is starting to be a bit clearer now!

You are right! AFAIK, one can consider normal coordinates for points which are near to each others, right?! However, it's a little cryptic to me, again.. Under what conditions can one consider the geodesic distance equals to Euclidean distance? The only thing I can say is...

Hi, I read somewhere the geodesic distance between an arbitrary point ##x## and the base point ##x_0## in normal coordinates is just the Euclidean distance. Why?! That's the part I don't understand. I know that one can write g_{\mu \nu} = \delta_{\mu \nu} - \frac{1}{6} (R_{\mu \rho \nu \sigma}...