What is the difference between differential 1-form and duality pairing?

Question

I am trying to learn the basics of differential geometry. The duality pairing between a coordinate basis and its canonical basis is given by the Kronecker delta, which is nothing but a scalar value. Whereas, the differential 1-form in my notes is given as $\alpha = \alpha_i (x^1,...,x^n) dx^{i}$. In the above relation, am i right in assuming that $(x^1,...,x^n)$ is the vector of expansion coefficients of any tangent vector, if $\alpha_i$ are the expansion coefficients of the corresponding cotangent vector? If so, how does the relation for differential 1-form has a dimension of 1? Also how does one physically interpret duality pairing and differential 1-form?

Answer 1

To fix the language: For a finite dimensional real vector space $V$, its dual space is defined as $$V^* := \{ \alpha: V \to \mathbb R \mid \alpha \text{ is linear} \}.$$

A dual pairing is now a map $$<\cdot,\cdot>: V \times V^* \to \mathbb R: (v,\alpha) \mapsto \alpha(v). $$

If $\{ e_1, \dots, e_n\}$ is a basis of $V$, we call $\{ e^1, \dots, e^n\}$ the dual basis of $V^*$, if $$ e^i(e_j) = \delta_{ij}, \text{ for all } i,j.$$

Is $\alpha(x^1, \dots, x^n)$ the vector of expansion coefficients of any tangent vector?

No. I would assume, that these are coefficients which depend on the current coordinates.

A differential 1-form is a smooth map $\alpha:TM \to \mathbb R$ which is linear on each fiber, i.e. at each point $p$, the restriction $\alpha_p: T_p M \to \mathbb R$ is a linear map, or equivalently $\alpha_p \in (T_pM)^* =: T^*_pM$

A given chart implies a basis of the tangent space $\left\{ \frac{\partial}{\partial x^1}|_p, \dots, \frac{\partial}{\partial x^n}|_p\right\} \subset T_pM$ and a dual basis of the cotangent space $\left\{ dx^1_p, \dots, dx^n_p \right\} \subset T^*_p M$.

Now at each point $p$ we can find coefficients $\lambda_i(p)$, to write the covector $\alpha_p$ as a linear combination of the dual basis vectors, i.e. $\alpha_p = \sum_i \lambda_i(p) dx^i_p.$

But these coefficients depend on the current point $p$! Therefore, if we want to describe $\alpha$ not only at $p$ but in the domain of a chart $x$, we may write $$ \alpha = \sum_i \lambda_i(x^1, \dots, x^n) dx^i.$$

Relation to the dual pairing?

The dual pairing is already hidden in the construction of the dual basis!

For tangential and cotangential vectors, the relation between a basis and a dual basis reads $$ dx^j\left( \frac{\partial}{\partial x^i} \right) = \delta_{ij}.$$

In particular $$\alpha\left(\frac{\partial}{\partial x^j}\right) = \sum_i \lambda_i dx^i\left(\frac{\partial}{\partial x^j}\right) = \lambda_j.$$

Physical interpretation

That's a question on its own. There are many important 1-forms in physics, for example momentum and (often) force. In physics these are usually called contravariant vectors, since they transform differently than tangent vectors.