Formular for interior product, example In Nakahara's Geometry,Topology and Physics, the interior product is defined like this :
$$i_X: \Omega^{r}(M) \rightarrow \Omega^{r-1}(M).$$
Where $ X \in X(M)$ and $\omega \in \Omega^{r}(M)$ 
$$i_X \omega = \frac{1}{r!} \sum\limits_{s=1}^r X^{\mu_{s}} \omega_{\mu_{1}\ldots\mu_{s}\ldots\mu_{r}}(-1)^{s-1}dx^{\mu_{1}} \wedge \cdots \wedge\widehat{  dx^{\mu_{s}}} \wedge \cdots \wedge dx^{\mu_{r}}.$$
I can't make sense out of this. May someone please give me a concrete example? How does $$ i_{e_{x}}(dx \wedge dy) = dy$$ look explicitly? 
 A: The formula given is horribly unenlightening because it does not seem to use the fundamental fact about differential forms that they are alternating and thus adds $r$ equal terms, it also does not provide the connection to index notation that it supposedly tries to.
Let us first understand the idea of the interior product. An $r$-form is something with $r$ slots for vectors that is linear in each slot and changes sign if you interchange any two slots. If you put a vector field $X$ into the first slot, these properties continue to hold for the $r-1$ slots that are left; this is $i_X \omega$. This is a coordinate-free definition, so let us see how we can get Nakahara's formula.
For computations, given any basis of one-forms $dx^\mu$, an $r$-form $\omega$ has the expansion $$\omega = \sum \omega_{\mu_1 \ldots \mu_r} dx^{\mu_1} \otimes \cdots \otimes dx^{\mu_r} \tag{1}$$
where the $\omega_{\mu_1 \ldots \mu_r}$ are completely antisymmetric. Clearly the interior product is just contraction on the first index.
Since $dx^{\mu_1} \otimes \cdots \otimes dx^{\mu_r}$ is related to $ dx^{\mu_1} \wedge \cdots \wedge dx^{\mu_r}$ by an antisymmetrization and a normalization, $$\omega = \frac{1}{r!} \sum \omega_{\mu_1 \ldots \mu_r} dx^{\mu_1} \wedge \cdots \wedge dx^{\mu_r}$$ where the $r!$ accounts for the normalization. Since forms are usually constructed with wedge products this is how you would find the components of a form. Now, using (1) and the interior product as contraction on the first index, \begin{align}i_X \omega & = \sum X^{\alpha}\omega_{\alpha \mu_1 \ldots \mu_{r-1}} dx^{\mu_1} \otimes \cdots \otimes dx^{\mu_{r-1}} \\
& = \frac{1}{(r-1)!} \sum X^\alpha \omega_{\alpha\mu_1 \ldots \mu_{r-1}} dx^{\mu_1} \wedge \cdots \wedge dx^{\mu_{r-1}}.
\end{align}
But since $\omega$ is alternating, we could contract over all of the $r$ indices and get the same thing, provided we keep track of the sign, yielding a sum with $r$ equal terms. Compensating for that we have $r$ equal terms, we get $$i_X \omega = \frac{1}{r(r-1)!} \sum_s X^{\mu_s} \omega_{\mu_{1} \ldots \mu_{s} \ldots \mu_{r}}(-1)^{s-1}dx^{\mu_{1}} \wedge \cdots \wedge \widehat{dx^{\mu_s}} \wedge \cdots \wedge dx^{\mu_{r}}.$$
The omission of the $s$:th factor in the wedge product is the contraction over the $s$:th index. The $(-1)^{s-1}$ comes from that moving the $s$:th factor to the front gives $s-1$ minus signs, one for each factor it has to move past.
Now for the example of $i_{e^x} (dx\wedge dy)$. We have $$dx\wedge dy = \frac{1}{2} \big( \omega_{12} dx\wedge dy + \omega_{21} dy \wedge dy \big) = \frac{2}{2} \omega_{12} dx\wedge dy$$ (using antisymmetry) so obviously $\omega_{12} = 1$. The components of $e^x$ are $(1, 0, \ldots, 0)$, so contracting on the first index we get $dy$.
