Recent content by gorkiana

I agree with you, the metric is specific to the Riemannian manifold, as you said, it is so by definition. The change of metric, and this was an abuse of language from me, I meant to be the appearance of the metric in different coordinate systems. So, as you say this change is not considered to...

I do not agree with what you say, at least in what you say as a reply to what I said. If I have a certain Riemannian manifold I can always find an embedding for it in a higher dimensional space, by Nash's theorem, right? This embedding is just a "visualization" of the Riemannian manifold, the...

First of all thank you for all the replies A metric is intrinsically connected to a specific basis for used for the tangent spaces. In this way one can have as many metrics as basis for the tangent spaces. But all this metrics are connected to each other by a pull back no? I say this...

Hello I know that the metric is an intrinsic property of a manifold, hence it should be possible to compute it without any use of an embedding in a higher dimensional space. That is, one can easily compute the metric on a surface of a 2-sphere just by computing the inner products of the...

Hi! I was doing some more reading and I found something that might be useful to your problem. Take a look at chapter 4 of this book: Differential geometry for physicists Bo-Hu Hou and Bo-Yuan Hou -artur palha

Hello I am not an expert on the field, just starting. But I think that a good approach is to try to find the Hodge dual of your basis p-vector (p-form), one by one. That is: A \in \Omega^{p} The basis p-forms of \Omega^{p} are for example \left\{\sigma^{i}\right\} with...