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Extension of Riemannian Metric to Higher Forms

I have no problems with understanding the inner product of 1-forms on a Riemannian manifold. We have a metric tensor, it's inverse defines the metric in the cotangent space.

However I'm not very comfortable with the definition of inner product of $p$-froms with $p>1$. It is defined as

$$ \langle x^1 \land x^2 \land \ldots \land x^p, y^1 \land y^2 \land \ldots \land y^p \rangle = \det(\langle x^i y^j \rangle) $$

Is there a way to hide the determinant? Or at least to write it as a determinant of a sensible linear operator, not just some matrix.

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    Please take a look if [this](http://unapologetic.wordpress.com/2011/10/04/inner-products-on-differential-forms/) helps2012-08-02
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    To "hide" the determinant you may use the Hodge star as in [Geometry of Differential Forms](http://books.google.co.nz/books?id=5N33Of2RzjsC) on p. 151, see Proposition 4.7 (iv)2012-08-02
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    @YuriVyatkin I'm actually asking about inner product, because I'm learning how Hodge star is defined through it :-)2012-08-02
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    By the way, [this](http://math.stackexchange.com/questions/22550/) question is very relevant, and in fact answers yours.2012-08-21
  • 0
    Indeed, that what I was looking for.2012-08-21

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