Interior Product: Definition & Understanding

[Silviu]

Hello! The interior product is defined as $i_X:\Omega^r(M)\to \Omega^{r-1}(M)$, with X being a vector on the manifold and $\Omega^r(M)$ the vector space of r-form at a point p on the manifold. Now for $\omega \in \Omega^r(M)$ we have $i_X\omega(X_1, ... X_{r-1}) = \omega (X,X_1, ... X_{r-1})$. I am not sure I understand it. Based on this definition, $i_X$ acts on an r-1 form and it turns it into an r-form, by making it acts on X, too. But by the definition in the first line, $i_X$ should act the other way around. What am I reading wrong here? Thank you!

 

[fresh_42]

There is no contradiction. Say $\omega \in \Omega^r(M)$. Then $\omega$ applies to $r$ vector fields. Now we want to define $i_X(\omega) \in \Omega^{r-1}(M)$, i.e. by its action on $r-1$ vector fields $X_1,\ldots , X_{r-1}$. We do this by $(i_X(\omega))(X_1,\ldots ,X_{r-1}) := \omega(X,X_1,\ldots ,X_{r-1})$.

 

[Orodruin]

To put it slightly differently, $i_X\omega$ is an $r-1$-form so it takes $r-1$ arguments as you can see on the LHS of your expression. On the RHS of the expression you have only $\omega$, which is an $r#-form and therefore takes$r$arguments. Clearly, there are$r-1## $X_i$ on the LHS and the RHS therefore needs the extra argument $X$, which is the $X$ of the interior product.