The Euclidean Metric on $\mathbf R^3$ Induces an Index-Lowering Isomorphism $b:\mathfrak X(\mathbf R^3)\to \Omega^1(\mathbf R^3)$.

Question

In Lee's Introduction to Smooth Manifolds, Second Edition, the line just before Equation 14.25 reads

The Euclidean metric on $\mathbf R^3$ induces an index-lowering isomorphism $b:\mathfrak X(\mathbf R^3)\to \Omega^1(\mathbf R^3)$.

I am unable to understand the meaning of this statement.

A map from $\mathfrak X(\mathbf R^3)$ to $\Omega^1(\mathbf R^3) $which I can think of is the following:

$$P\frac{\partial}{\partial x} + Q\frac{\partial}{\partial y} +Q\frac{\partial}{\partial z} \mapsto Pdx+Qdy+Rdz$$

But I don't see what this map has to do with the "Euclidean metric" or "index-lowering".

Answer 1

In that sentence, I was referring to the tangent-cotangent isomorphism, defined on pages 341-343 of the book. It probably would have been a good idea to include a page reference.

Answer 2

Why it has to do with index lowering:

Because it maps the vector of components $p^i$ to the form of components $p_i$.

Why it has to do with the Euclidean metric:

One cannot just copy the components of a vector and write them into a differential form. They are different objects in different spaces. What really happens is that you get a linear map from vectors to forms (technically a tensor, called metric tensor), with components $g_{ij}$, so that: $$ p_i = \sum_j g_{ij}\,p^j. $$

In your case the $g_{ij}$ are the components of the identity matrix, which means that your metric is the Euclidean metric. This seems overkill, but it's simply because the Euclidean metric has a particularly simple expression (the identity matrix), and so you don't see it at work. But it is there, working!

Answer 3

Metric can be thought of as a mapping that takes two vector fields as arguments and produces scalar field. By fixing one vector field (call it $X$) you get a map from vector fields to scalars, that is a diferential form $gX$. This differential form will map vector field $Y$ to its scalar product with $X$. If you use Cartesian coordinates in Euclidean space its components will look like you written but only then. In general $X_i = g_{ij} X^j$ where Einstein notation is used.