A question about the Angular Velocity Vector First time asking here, so please forgive me if I'm doing anything wrong. 
I'm having quite a hard time with this equation:
$$\mathbf{\omega} = \frac{1}{2} \left[
\left(\mathbf{i}\wedge\frac{d\mathbf{i}}{dt}\right)+
\left(\mathbf{j}\wedge\frac{d\mathbf{j}}{dt}\right)+
\left(\mathbf{k}\wedge\frac{d\mathbf{k}}{dt}\right)
\right].$$
 I don't understand where it comes from. I know it represents the angular velocity vector but I have no clue how we got here and neither my textbook nor the internet have proved helpful. 
Reference: Elementi di Meccanica Razionale (Maria Letizia Bertotti, Giovanni Modanese).
 A: Looks very much like the formula at this link, except that wedge product is used instead of vector product and unit vectors instead of directional derivatives 
EDIT(11/01/2019): So let me derive the formula. It should be noted that it was FUBAR by an editor, who omitted plus signs in the following original formula of the OP:
$$\mathbf{\omega}=\frac{1}{2}\left(\mathbf{i}\wedge\frac{d\mathbf{i}}{dt}+\mathbf{j}\wedge\frac{d\mathbf{j}}{dt}+\mathbf{k}\wedge\frac{d\mathbf{k}}{dt}\right).$$
Following the KISS principle, I will use vector products $\times$, rather than wedge products $\wedge$. I assume that vectors $\mathbf{i},\mathbf{j},\mathbf{k}$ are standard unit basis vectors in a rotating frame. The time derivatives of these unit vectors are:
$$\frac{d\mathbf{u}}{dt}=\mathbf{\omega}\times\mathbf{u},$$
where $\mathbf{\omega}$ is the vector of angular velocity of the rotating frame and $\mathbf{u}$ is one of the unit basis vectors $\mathbf{i},\mathbf{j},\mathbf{k}$ (I guess the above formula is what the OP calls "Poisson formula").
Let us vector-multiply the above formula by $\mathbf{u}$ from the left and use the formula for vector triple product:
$$\mathbf{u}\times\frac{d\mathbf{u}}{dt}=\mathbf{u}\times(\mathbf{\omega}\times\mathbf{u})=\mathbf{\omega}-(\mathbf{u}\cdot\mathbf{\omega})\mathbf{u}.$$
Applying this formula separately to $\mathbf{i},\mathbf{j},\mathbf{k}$ and adding the results, we obtain:
$$\mathbf{i}\times\frac{d\mathbf{i}}{dt}+\mathbf{j}\times\frac{d\mathbf{j}}{dt}+\mathbf{k}\times\frac{d\mathbf{k}}{dt}=\\= \mathbf{\omega}-(\mathbf{i}\cdot\mathbf{\omega})\mathbf{i}+\mathbf{\omega}-(\mathbf{j}\cdot\mathbf{\omega})\mathbf{j}+\mathbf{\omega}-(\mathbf{k}\cdot\mathbf{\omega})\mathbf{k}=2\mathbf{\omega}.$$ 
A: This is not common notation and it is unclear what it means. It looks like something from fringe physics, not mainstream physics. You should ignore it.
The notation is completely at odds with the common usage of $\mathbf i$, $\mathbf j$, and $\mathbf k$ to mean Cartesian unit vectors in the $x$, $y$, and $z$ directions. These unit vectors are constants so their time derivatives are zero, making the formula nonsensical.
The role of the parentheses and how they “combine” the three products is also completely obscure.
The formula appears to be an example of either ignorance or obscurantism, both of which are common attributes of fringe physics.
ADDENDUM: Thank you, Akhmeteli, for revealing the egregious mis-edit that made the formula look fringe-y. It was neither ignorant nor obscure, just mangled beyond recognition by an editor.
