Can the metric tensor be calculated

Question

Consider a smooth manifold. As I understand it, modern differential geometry does NOT regard the manifold as embedded in a higher dimensional space but rather as a space for itself. This means, that I can only parametrise points on the manifold by their coordinates with respect to some coordinate chart. (Correct?)

Question:

Suppose now, that I have no knowledge whatsoever about the inner geometry of the manifold, which means that I do NOT know the metric tensor. Is it then even possible to calculate the metric tensor, i.e. must the metric tensor be known or can it be calculated without using further assumptions?

Actually same question but alternative phrasing:

In classical differential geometry, the metric can be calculated using the basis vectors of the tangent space for some given parametrisation. The metric tensor's components are then the scalar products of these basis vectors.

However, in modern differential geometry, the basis vectors are differential operators. Can the metric be calculated analogously via a scalar product? If yes, how is the scalar product of differential operators, say $\partial_{x^1}$ and $\partial_{x^2}$, defined?

Answer 1

The question is not entirely clear to me, but according to the clarification given in the comments, the answer is no.

I guess the source of the confusion lies in inheriting structure. To give an example:

• The set $\mathbb{S}^2 = \{ (x,y,z) \in \mathbb{R}^3 : x^2+y^2+z^2 = 1 \}$ is just that - a set.
• Since we usually think of $\mathbb{R}^3$ as equipped with the standard metric, we may also think of $\mathbb{S}^2$ as a metric space (with the metric inherited from $\mathbb{R}^3$).
• Similarly, if $\mathbb{R}^3$ is viewed as a topological space, one can consider the induced topology on $\mathbb{S}^2$.
• The standard smooth structure on $\mathbb{R}^3$ induces a smooth structure on $\mathbb{S}^2$. In contrast to all previous cases, this is not automatic - one has to check first that $\mathbb{S}^2 \subseteq \mathbb{R}^3$ is indeed a smooth submanifold.
• Finally, $\mathbb{R}^3$ has a standard Riemannian metric, which induces a Riemannian metric on any smooth submanifold, in particular on $\mathbb{S}^2$.

Nowadays, we have separate notions of sets, metric spaces, topological spaces, smooth manifolds and Riemannian manifolds. But without this more abstract point of view, it's easy to mix all these together; after all, the underlying set $\mathbb{S}^2$ is the same in all the above examples.

One could also imagine similar (somewhat ill-defined) questions; for each of them the answer is again no:

• Given a set $A$, can I decide which functions $f \colon A \to \mathbb{R}$ are continuous and which are not?
• Given a topological space $A$, can I calculate the distance between two points $x,y \in A$?
• Given a metric space $A$ and a Lipschitz function $f \colon A \to \mathbb{R}$, can I define its differential? (actually, there are some weak workarounds here)

And please don't get me wrong - I'm not listing these to suggest that your question was silly, quite the opposite. I just think other examples might help.