On a Riemannian manifold $M$, the matrix representation is diagonalisable, cause the tensor is symmetric. What is the physical meaning behind this? I mean, in Riemannian geometry, we always get a coordinate system defined by the charts(for simplicity I only use one) $\phi : M \rightarrow \mathbb{R}^k $ which we express here by $x_i := \pi_i \circ \phi,$ where $\pi_i $ is the projection on the $i-$th component in $\mathbb{R}^k.$ Now, by diagonalising the metric tensor, we can apparently make the transition to a basis of the tangent space $\partial_1,….,\partial_k$ at each point so that the metric tensor is the identity matrix. I am not sure how to interpret this?
Does this mean that the manifold looks locally euclidean? If so, does this mean that we can define locally a chart so that this chart would actually induce this locally euclidean coordinate system on the manifold? Probably, this chart would, if it exists, not be in general compatible with the atlas of the manifold $M$, is this correct?
If my attemt to interpret this is wrong, I would love to here an explanation of this fact.
Best Answer
To amplify slightly on the two comments by Demosthene and JHance: it means that the manifold is very locally euclidean, i.e., pointwise. At each point, you can find a basis of tangent vectors in which things are Euclidean, but this basis cannot generally be extended even to a small neighborhood. A good approximation to extending it to a small neighborhood is given by "exponential coordinates"; on a surface, each ray emanating from your chosen point, in a kind of polar coordinates, is actually geodesic. The "circles" of the polar coordinate system, however, cannot be made as nice as the rays, alas.