I'm curretly working on this reference, where the interior product by a 1-form, denoted by $\iota_\alpha$, is defined as: $$\iota_\alpha\theta(\alpha_1,\dots,\alpha_{k-1})=\theta(\alpha,\alpha_1,\dots,\alpha_{k-1}), \ \ \ \theta\in\mathfrak{X}^k(M), \ \ \alpha_i\in\Omega^1(M), $$ where $M$ differentiable manifold. This definition is very similar to that of the interior product by a vector field. However, I'm not entirely sure if I'm doing the math correctly.
First, given a point $x\in M$ and taking coordinates $\{x^1,\dots,x^n\}$, I think the following should be true: $$\theta(\alpha_1,\dots,\alpha_k)(x)=\theta|_x(\alpha_1|_x,\dots,\alpha_k|_x)=\\\sum_{i_1<\dots<i_k}\theta^{i_1\dots i_k}(x)\left.\frac{\partial}{\partial x^{i_1}}\right|_x\wedge\dots\wedge\left.\frac{\partial}{\partial x^{i_k}}\right|_x\left(\sum_i\alpha_{1,i}(x)dx^i|_x,\dots,\sum_i\alpha_{k,i}(x)dx^i|_x\right). $$ Is that correct? If so, it will be necessary to evaluate $\displaystyle \left.\frac{\partial}{\partial x^i}\right|_x\left(dx^j|_x\right) $, which I think would not be absurd because the vector space $T_xM$ is finite-dimensional and it would be correct to say that $\displaystyle \left.\frac{\partial}{\partial x^i}\right|_x\left(dx^j|_x\right) =dx^j|_x\left(\left.\frac{\partial}{\partial x^i}\right|_x\right)=\delta^j_i $. Is this also correct?
Finally, with $X\in\mathfrak{X}^1(M)$, $\alpha\in\Omega^1(M)$ and taking into account the usual definition of the interior product by a vector field, I think that $\iota_X\alpha=\iota_\alpha X$. I'm right?
Thanks for your help.