I am reading a book about differential geometry that introduces an operator $i$, called the "interior product," that takes vectors and produces something that can act on 1-forms. Their rules are, $$ i\left(\frac{\partial}{\partial x_i}\right)\mathrm{dx_j} = \delta_{ij} $$ and, $$ i\left(\frac{\partial}{\partial x_i}\right)f = 0 $$ Now, the second rule is unique, and is definitely not what the vector $\partial/\partial x$ would do normally. Still, I don't understand why the first one should apply. Vectors are already dual to one-forms, so why do we need a map $i$ that takes vectors to something that can act on one-forms?

This is especially bothersome to me, because interior products usually allow you to measure vectors against other vectors in the same space. I am not sure how to interpret these laws, which apparently describe an "inner product" between spaces that already have a bilinear function to be dual under!


1 Answer 1


No, the point is that interior product maps $(k+1)$-forms to $k$-forms (for all $k\ge 0$). This has nothing to do with inner products. Indeed, it's the process of undoing the wedge product (so-called "adjoint operation"), which sends $k$-forms to $(k+1)$-forms.

It's based on contraction (evaluating a $1$-form on a vector), and that, indeed, is the rule they gave you for the case $k=0$.

More generally, if we have a vector $v$ and, for example, two $1$-forms $\omega$ and $\eta$, then $$\iota_v(\omega) = \omega(v) \quad\text{and}\quad \iota_v{\eta} = \eta(v),$$ and then $$\iota_v(\omega\wedge\eta) = \omega(v)\eta - \eta(v)\omega.$$ (I'll leave it to you to figure out why the negative sign is there.)

An important application of this notion is the following: If you have an oriented surface $S$ in $\Bbb R^3$, with outward pointing unit normal $\vec n$, then you get the area $2$-form on $S$ (written $dS$ or $d\sigma$ in calculus books) by taking $${"}dS{"} = \iota_{\vec n} (dx_1\wedge dx_2\wedge dx_3).$$ This generalizes to oriented hypersurfaces in any dimension.


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