How do dual vectors and tangent bases relate in coordinate functions?

[nigelscott]

I am trying to figure how one arrives at the following:

dx^μ∂_ν = ∂x^μ/∂x^ν = δ^μ_ν

Where,

dx^μ is the gradient of the coordinate functions = basis of cotangent space

∂_ν = basis of tangent space

I know that dual vectors 'eat' vectors to produce scalars. Is this demonstrated by absorbing d into ∂ so that dx^μ ≡ ∂_νx^μ or is such an operation illegal?

 

[andrewkirk]

I think those equations only hold if the coordinate system is orthonormal. The more general result would be that those expressions are equal to
$$\mathbf g(\partial_\mu,\partial_\nu)=g_{\mu\nu}$$
As regards proving the result, it depends on where one starts, in particular, how one defines $\partial_\nu$ and dx^μ. Several different approaches are possible. The nature of the proof would depend on the nature of the definition. For some definitions, the result may even be part of the definition so that no proof is required.

 

[Orodruin]

By definition of the differential $df$, it holds that
$$
df(\dot\gamma) = \frac{df}{ds}
$$
where $\dot\gamma$ is the tangent vector of a curve $\gamma$ parametrised by $s$. In a coordinate system representation, this expression can be rewritten
$$
df(\dot x^\mu \partial_\mu) = \dot x^\mu df(\partial_\mu) = \frac{df}{ds} = \frac{\partial f}{\partial x^\mu} \frac{dx^\mu}{ds}
$$
using the linear property of a dual vector and the chain rule, respectively. This is satisfied for all tangent vectors if and only if
$$
df(\partial_\mu) = \frac{\partial f}{\partial x^\mu}.
$$
If you let $f$ be your coordinate function $x^\nu$, you now obtain
$$
dx^\nu(\partial_\nu) = \frac{\partial x^\nu}{\partial x^\mu} = \delta^\nu_\mu.
$$

I think those equations only hold if the coordinate system is orthonormal.

No, the result is completely general and an effect of how the differential is defined. I would not call $g(\partial_\mu,\partial_\nu) = g_{\mu\nu}$ a result as much as the definition of the components $g_{\mu\nu}$. In contrast, the result above is general and does not even require a metric to be well defined (the natural product is between tangent vector and dual vectors, the metric is necessary only to define a product between vectors of the same type).

I find that it sometimes helps to go back to the Euclidean case. In the Euclidean case you can define two bases $\vec E^\mu = \partial \vec x/\partial x^\mu$ and $\vec E_\mu = \nabla x^\mu$, respectively. These are the equivalents of $\partial_\mu$ and $dx^\mu$, respectively (heuristically, essentially just remove the position vector). Each of these sets of vectors will generally not be orthonormal among themselves (in curvilinear coordinates), however it will still hold that
$$
\vec E^\nu\cdot \vec E_\mu = \frac{\partial x^\nu}{\partial y^i} \vec e_i \cdot \vec e_j \frac{\partial y^j}{\partial x^\mu} = \frac{\partial x^\nu}{\partial y^i} \frac{\partial y^i}{\partial x^\mu} = \frac{\partial x^\nu}{\partial x^\mu} = \delta^\nu_\mu,
$$
where $\vec E^\nu$ and $\vec E_\mu$ have been written down in component form in a Cartesian coordinate system $y^i$.

 

[nigelscott]

Thank you. Now I understand.