Let $(M,g)$ be a Riemannian $n$-manifold. I have one question about the following quote from Geometric Relativity by Dan A. Lee.
It is sometimes convenient to fix a background metric $\bar g$ and compare the geometry to that of $\bar g$. Note that in a single local coordinate chart, one can always choose the background metric to be the Euclidean metric determined by the local coordinates, that is, one can choose $\bar{g}_{ij}=\delta_{ij}$.
In case of possible confusion, let us leave alone the physical meaning of the term background and think of $\bar g$ simply as some other Riemannian metric imposed on $M$ (aside from $g$).
My question is, what is meant by choosing the background metric to be the Euclidean metric determined by the local coordinates? Seemingly, Lee is suggesting that given any coordinate chart $(x_i)$, we can always express the background metric as
$$\bar g=\delta_{ij}dx^i\otimes dx^j.$$
Is it a theorem that can be proved? I used to prove a theorem which states that a Riemannian metric $h$ on $M$ is flat if and only if every point of $M$ is in the domain of a coordinate chart in which the Riemannian metric has the representation
$$h=\delta_{ij}dx^i\otimes dx^j.$$
Are they talking about the same thing? If so, why is the background metric necessarily flat? Isn't $\bar g$ arbitrarily chosen? Isn't there any tiny possibility that we have a Riemannian metric $\bar g$ whose coordinate representation is always
$$\bar g=\delta_{ij}dx^i\otimes dx^j?$$
Thank you.
Best Answer
I believe that they are saying that we can locally construct a positive definite symmetric bilinear form given by $$ \overline{g} = \delta_{ij} dx^i \otimes dx^j $$ Hence they are using the term "metric" loosely, as they are not implying that there exists a metric which takes this form in every chart (which would never be true). Rather, for a given chart, there is such a form defined only on this chart that looks just like the restriction of the euclidean metric to this chart.