Interior product

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  • Thread starter Silviu
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  • #1
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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!
 

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  • #2
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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})##.
 
  • #3
Orodruin
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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.
 

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