Definition of a metric tensor in Barrett O'Neil's book

Question

I am studying Barrett O Neils's book on semi riemannian geometry with applications to relativity. In chapter 3 he states the definition of a metric tensor as follows:

"A metric tensor g on a smooth manifold M is symmetric nondegenerate (0,2) tensor field on M of constant index."

In the next line he says:

"In other words, g is a (0,2) tensor field which smoothly assigns to each point p in M a scalar product g(p) on the tangent space Tp(M) and the index of g(p) is the same for all p."

My question is, how are these two statements equivalent? I mean how can g take as input a point of the manifold and give as output a scalar product?

Thanks

Answer 1

Since you are reading from O'Neill, you need to go back and revisit Chapter 2.

In the section titled "Tensor Fields" you find the description of a tensor field $A$ on a manifold $M$ being a tensor over the $\mathfrak{F}(M)$-module $\mathfrak{X}(M)$.

This corresponds to one interpretation of the metric, that is $g$ is a $(0,2)$ tensor field, so that it takes as input two vector fields and outputs a scalar field.

Two sections later (titled "Tensors at a point"), the main result is that "a tensor field $A\in \mathfrak{T}^r_s(M)$ has a value $A_p$ at each point $p$ of $M$, namely the function $A_p: (T_p M^*)^r \times (T_p M)^s \to \mathbf{R}$ defined ..."

This second interpretation is exactly the interpretation of inputing a point $p\in M$ and outputing a scalar product at $p$.

Answer 2

If you have two vector fields $X$ and $Y$, and a point $p$, then you can get the value of the scalar field $g(X,Y)$ at the point $p$. This is a function of three variables, call it $h(X,Y,p)$.

This way of phrasing it, which seems to be also how you are thinking about it, views $g$ as the map $(X,Y) \mapsto h(X,Y,\cdot)$.

The second point of view is that $g$ is the map $p \mapsto h(\cdot,\cdot,p)$.