In special relativity the metric is expressed as
$ds^2=-c^2dt^2+dx^2+dy^2+dz^2$
I have questions.
0- is this the metric tensor or the space metric? I am confused here, i.e. what is the differenence between them?
1-is this flat metric?
2- does the speed light $c^2$ coming from some special relativity equations?
3- what would happen if I dropped the minus sign in front of the speed of light and make all sings of the metric positive?
4- Now is this the same metric used in general relativity? I guess no, but in any event, could the metric expressed in form like above relating the all coordinates?
Thanks.
$(ds)^2$ is an infinitesimal line element. In 2-D Cartesian coordinates an infinitesimal displacement is given by Pythagoras theorem such that $(ds)^2=dx^2 + dy^2$. So the 4-D spacetime you reference is a generalisation of this where the factor of $c^2$ is incorporated for dimensional purposes. The metric tensor $g_{\alpha \beta}$ and associated components are the individual relativistic potentials associated with a spacetime. So for the spacetime you mention above the metric tensor is given by
$$ g_{\alpha \beta} = \begin{bmatrix} -1 & 0 & 0 &0 \\ 0 & 1 & 0 &0 \\ 0 & 0 & 1 &0\\ 0 & 0 & 0 &1 \end{bmatrix}. $$
The metric represents the 4-D flat spacetime associated with special relativity. In GR the components of the metric tensor are more complicated however, as the radial component $r \rightarrow\infty$ you should recover the flat spacetime given above. So, yes it represents flat space.
You can drop the minus sign, however you then need to change all the spatial components to be negative, otherwise the line element doesn't make sense. Think about the displacement of a photon in the sense that I described above for an infinitesimal displacement (We always refer to photons due to the first postulate of relativity being that the speed of light is $c$ for all observers).