Metric should be a non-negative real number. If ej=-1, will gij>=0?

I hope in the handbook "Riemannian manifolds"

you will be able to find everything you need.

Greetings, Tadeusz

That can not be written because if we take value of chroneker delta as 1 and ej as -1 then product becomes negative while the Riemannian metric is positive. In particular  If possible choose the values under certain conditions as the same sign of both terms then you may consider that as a Riemannian metric.

If you mean the following:

"Is there a chart in the neighborhood of any point of a psedo-Riemannian manifold, such that the matrix of the metric tensor is diagonal with plus-minus ones in the diagonal?",

then the answer is "NO". The main reason is not the signature of the metric, but the curvature tensor. If you can find such a chart, then all the Christoffel symbols vanish and the curvature tensor will be constant 0. On the other hand, if the curvature tensor vanishes, then the answer is "YES". In the latter case, normal coordinate systems, defined with the help of the exponential map will do.

It can be written in the form you mentioned only when the pseudo-Riemannian metric is flat one. Also, it cannot be a Riemannian metric unless all ej=1.

Thank so much sir 

Prof. Bang-Yen Chen gave the appropriate answer. Of course one can obtain such a form of $g$ at every single point $p$ (using an orthonormal basis of the tangent space at $p$), and we even can achieve  $g_{ij} = e_j\delta_{ij}$ and $\partial_k g_{ij} = 0$ at $p$ for all $i,j,k$ (Riemannian normal coordinates).

Best regards

Jost

• Thank you so much sir for giving valuable suggestion