I have looked at the formula for this for several sources but the notation is very confusing. Looking at a few limited examples leads me to believe that the method may be the perscription:
Let a manifold M have local coordinates $(x^1,\dots,x^n)\varepsilon \mathbb{R}^n$
For vector field $v \varepsilon TM \text{ defined } v = v^a\frac{\partial}{\partial x^a}$ and differential form $\omega \varepsilon \Lambda^q$ (with a confusing notation for general formula)
the interior product $\iota_v(\omega) \varepsilon \Lambda^{(q-1)}$ is calculated by:
For each element $v^a\frac{\partial}{\partial x^a}$ of the vector field, go through the expression for $\omega$ and, for any term containing $dx^a$:
- Eliminate the $dx^a$ in the term
- Multiply the term by $v^a$
- Multiply the term by -1 to the power of the number of times you had to swap dx's to get $dx_a$ to the front.
- Add this transformed term to your $\iota_v(\omega)$ evaluation
This is what I've pieced together from a few examples but I've only seen a few so I'm not very confident in this. Could someone please explain how to properly do this in a algorithmic way? (An example would be amazing but no worries if not)
Basically, you are evaluating your $q$-differential form $\omega$ on $(v,z_1,\dots z_{q-1})$ where the $z_i$ are free, resulting in a $(q-1)$-differential form acting on $(z_1,z_2,\dots z_{q-1})$. Here is an example.
Suppose you have a 2-form in $\mathbb{R}^3$: $$ \omega=a(x)dx_1\wedge dx_2+b(x)dx_2\wedge dx_3+c(x)dx_1\wedge dx_3 $$ where $x=(x_1,x_2,x_3)$, and a vector field $v=(v_1(x),v_2(x),v_3(x))$.
Then, given a generic vector field $z(x)$ (I will omit the argument $x$ for simplicity) $$ i_v(\omega)(z)=\omega(v,z)=a(v_1z_2-v_2z_1)+b(v_2z_3-v_3z_2)+c(v_1z_3-v_3z_1) $$ and, rearranging, $$ i_v(\omega)(z)=(-av_2-cv_3)z_1+(av_1-bv_3)z_2+(bv_2+cv_1)z_3, $$
So, the resulting $1$-form is $$ i_v(\omega)=(-av_2-cv_3)dx_1+(av_1-bv_3)dx_2+(bv_2+cv_1)dx_3 $$