My differential forms book (Flanders/Dover) defines an inner product on wedge products for vectors that have a defined inner product, and uses that to define the hodge dual. That wedge inner product definition was a determinant of inner products.(adsbygoogle = window.adsbygoogle || []).push({});

I don't actually have that book on me right now, but believe it was like so:

[tex]

\langle (a_1 \wedge \cdots \wedge a_k), (b_1 \wedge \cdots \wedge b_k) \rangle

= \lvert \langle a_i, b_j \rangle \rvert

[/tex]

(edit to correct small typo above).

This is a slightly awkward definition since you have to have the wedge explicitly specified in terms of factors to calculate, and it only works for like grades, and requires that the two wedge product elements be simple (ie: both sides must be blades, not neccessarily sum of multivectors of same grade). In it's favour it doesn't assume an euclianian metric like GA.

Anyways, ... I was trying to reconcile this with the GA definition. For the GA generalized dot product (lowest grade selector of a multivector product), I believe the equivalent explicit expansion in terms of determinant would be:

[tex]

(a_1 \wedge \cdots \wedge a_k) \cdot (b_1 \wedge \cdots \wedge b_k) =

(-1)^{k(k-1)/2}\lvert a_i \cdot b_j \rvert

[/tex]

I got this by repeated expansion of

[tex]

(a_1 \wedge \cdots \wedge a_k) \cdot (b_1 \wedge \cdots \wedge b_k)

=

a_1 \cdot (a_2 \cdot ( \cdots ( a_k \cdot (b_1 \wedge \cdots \wedge b_k))))

[/tex]

To be honest I found that I had to make a ``compensating error'' to adjust the sign to what I expected it to be (ie: to agree with manually explicit expansion for k=1,2,3,4). I don't see what my error is offhand, but was wondering if anybody had seen a determinant expansion of the GA dot product of like grade blades like this in any text to compare with.

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# Reconciling differential forms' inner product of wedge with geometric algebra dot

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