Riemannian metric tensor and coordinate charts

Question

Is a metric tensor different on different coordinate charts due to the change in coordinates and hence the basis elements of the tangent spaces?

Answer 1

To add to the comment, I think there is an additional layer that should be considered, which is that coordinate systems are on the manifold $M$, whereas the metric operates on elements of the tangent spaces $T_pM$, for $p\in M$ to the manifold.

We can consider $g$ as a function $g_p(u,v):M\times T_pM\times T_p M\rightarrow \mathbb{R}$. Clearly, the components of $g$ change as $p$ changes.

But the question is for coordinate changes. Generally, we take $\partial_i x = \partial/\partial x^i$ to form the basis to the tangent space; i.e. the tangent vectors to the coordinate lines. Then, fixing $p$, we compute the components of $g$ as: $$ g_{ij} = g(\partial_i x,\partial_j x) $$ But when we change to coordinates $\xi^j$, the tangent basis vectors are order 1 covariant tensors, so $$ \partial_i \xi = \frac{\partial x^\kappa}{\partial \xi^i} \partial_\kappa x $$ Thus, if $g$ is the metric in $x$, then $ \tilde{g} $ is the metric in $\xi$ with components: $$ \tilde{g}_{ij} = g_{\alpha\beta}\frac{\partial x^\alpha}{\partial \xi^i}\frac{\partial x^\beta}{\partial \xi^j} $$ Thus, if we let our coordinates $x$ determine our tangent space basis, changing basis affects the components of the metric in this way.

Note that tangent vectors transform as contravariant tensors, but basis vectors transform as covariant tensors (see here).

However, another thing we can do is leave the coordinates alone, but change the frame on the tangent space.

For instance, suppose we have a set of tangent vectors $a_i$. We want to linearly transform to a new frame $b_j$. The original metric is defined by: $$ g_{ij} = g(a_i,a_j) $$ Let a matrix $M$ determine a change of basis, i.e: $ b = aM $ where we treat $a,b$ as a matrix of vectors $a_i$ or $b_i$. I prefer to write this as: $$ b_j = a_km^k_j $$ where $m^k_j$ is the element at the $k$th row and $j$th column. Given $b$, we can get $M$ by the relation above since $ b_i = \sum_j [M]_{ji}a_j = m^j_ia_j $, so the $i$th column of $M$ is the coordinates of $b_i$ in the basis $a$. (See here)

Then the new metric, using the frame $b = aM$ is given by: $$ g' = M^T g M $$

So this transform is also able to change the components of the metric.

Of course, if you measure the length of the same vector in two different coordinate systems using $g$ with different components (or in two different tangent space bases), you should get the same answer (as the coordinates of the vector will change to compensate for the changes to the $g_{ij}$.