Finite field trace property, what is the analog for characteristic 0

Question

Background:

Given an extension of finite fields $\mathbb{F}_{q^r} / \mathbb{F}_q$ where $q$ is a prime power, the field trace with respect to this extension is given by $$\text{tr}_{\mathbb{F}_{q^r} / \mathbb{F}_q}(\alpha) = \underset{i=0}{\overset{r-1}{\sum}} \alpha^{q^i}.$$ A property of the field trace here is that if we fix a basis $\mathcal{B} = \{ \zeta_0,\dots,\zeta_{r-1}\}$ for the extension, with corresponding dual basis $\{ \nu_0,\dots,\nu_{r-1}\}$ of $\mathcal{B}$, then for each $\alpha \in \mathbb{F}_{q^r}$, $$\alpha = \underset{i=0}{\overset{r-1}{\sum}} \text{tr}_{\mathbb{F}_{q^r} / \mathbb{F}_q}(\zeta_i \alpha) \nu_i.$$

Question:

I think the property above has to do with inner products more generally, and I was wondering what the result for more general inner product spaces is, particularly for fields of characteristic 0.

Answer 1

I believe I was thinking of inversion formulas; found something to cite in Tao's "An Epsilon of Room Part 1", I'll take the relevant parts of the exercise cited.

$\textbf{Exercise 1.4.19}$: Let $(e_\alpha)_{\alpha \in A}$ be an orthonormal system in $H$. Show that the following statements are equivalent:

• (i) The Hilbert space span of $(e_\alpha)_{\alpha \in A}$ is all of $H$.
• (iv) One has the $\textit{inversion formula}$ for all $x \in H$, $$x = \underset{\alpha \in A}{\sum} \langle x, e_\alpha \rangle e_\alpha.$$ where the coefficients $\langle x, e_\alpha \rangle$ are square summable.