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I understand that geometric algebra equips itself with the contraction operators $\rfloor$ and $\lfloor$. While these are awesome when one wishes to project a subspace onto another, it is not an inner product structure anymore.

What I am looking for is some analogue of the inner product. An operator $\star: G \times G \rightarrow \mathbb R$ which is bilinear, symmetric, and non-degenerate.

Does such an operator exist over a geometric algebra?

  1. If it does, can we assign some nice geometric meaning to it? (I would like something along the lines of "measure of mutual containment" or some such)
  2. If we cannot define such an operator $\star$, what is the obstacle?
Siddharth Bhat
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  • Here is something said about products. – Dietrich Burde Apr 23 '19 at 14:39
  • @DietrichBurde Indeed, there are plethora of products there (inner, outer, geometric), but none of them have the signature I ask for. All of them can spit out values that are larger than scalars – Siddharth Bhat Apr 23 '19 at 14:41
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    Seeking to define a structure and then assigning meaning to it is putting the cart before the horse. If you've studied geometric/Clifford algebra for any amount of time, you'll know it leverages the theory of inner product spaces by "building in" the inner product into the algebra product. It is not clear how defining an inner product on the entire algebra is going to be useful. Maybe you should take a look at this and you might see something you're looking for. – rschwieb Apr 23 '19 at 15:33
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    I'm pretty sure any nondegenerate form on a finite dimensional space is going to give you a semisimple ring, which is automatically a von Neumann algebra, and it appears those all have inner products compatible with the algebra operations, so that is one thing. – rschwieb Apr 23 '19 at 15:36
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    It seems like the natural candidate for developing an inner product to detect "degree of containment" would be to work with the wedge product in the geometric algebra, since that already gives an indicator of containment. – rschwieb Apr 23 '19 at 15:42
  • What did you make of the product labeled "the scalar product" in the wiki? – rschwieb Apr 23 '19 at 15:44
  • @rschwieb : Unfortunately, the wedge product does not work, since I need an actual measure of containment, in the sense of a real number I can optimise on. I'm phrasing a nonlinear optimisation problem in terms of geometric algebra's constructs for which I need a value to maximise on. – Siddharth Bhat Apr 23 '19 at 15:52
  • @SiddharthBhat I know the wedge product doesn't directly work but I'm saying it's an obvious candidate to adapt (say, by composing with a linear functional) to produce something that does work. – rschwieb Apr 23 '19 at 16:07
  • @rschwieb ah , interesting, I missed that one :) however, the page covers no geometric meaning of that product. Can you tell me what it is? – Siddharth Bhat Apr 23 '19 at 16:07
  • @SiddharthBhat I have no idea. I'd just start searching, e.g. for something like this and other resources. – rschwieb Apr 23 '19 at 16:10
  • @rschwieb: thanks a ton! This was great :) – Siddharth Bhat Apr 23 '19 at 16:15
  • And, as always, it bears repeating since some people don't seem to be aware: not everything has geometric significance. – rschwieb Apr 23 '19 at 16:30
  • Indeed, but I would hope a building block of geometric algebra would have some geometric significance ;) – Siddharth Bhat Apr 23 '19 at 17:53
  • There's a paper titled The Inner Products of Geometric Algebra that goes over several variations on possible inner-product-like operations. – Derek Elkins left SE Apr 23 '19 at 20:16
  • @rschwieb I've been reading the GA notes you linked, and I managed to finally get a computational definition of the scalar product: As far as I can tell, all it appears to do is to take a "dot product" of two elements of the geometric algebra by multiplying their components together. This is disappointing, since it does nothing to deal with objects that are contained within one another. – Siddharth Bhat Apr 27 '19 at 19:50
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    Btw, if you are trying to optimize on something that is not purely numerical, but is of the form scalar $\cdot$ "thing" (for a fixed choice of thing, like basis vector for the highest non-vanishing exterior power), you can still optimize on the scalar coefficient. – paul garrett Aug 11 '19 at 21:03

1 Answers1

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Yes. The natural inner product is

$$A\star B=\langle A^\sim B\rangle_0=\langle A\,B^\sim\rangle_0=\langle B^\sim A\rangle_0=\langle B\,A^\sim\rangle_0$$

$$=\langle A^\sim\bullet B\rangle_0=\langle A^\sim\,\lrcorner\,B\rangle_0=\langle A^\sim\,\llcorner\, B\rangle_0,$$

the scalar part of the geometric product (or any of these four products), with one factor reversed. This is analogous to the Frobenius inner product of matrices, $\text{tr}(A^TB)$. It's the familiar sum of products of corresponding components, with respect to a standard orthonormal basis $\{1,e_1,e_2,e_1e_2,e_3,e_1e_3,e_2e_3,e_1e_2e_3,\cdots\}$. For example in 2D,

$$(\alpha+\alpha_1e_1+\alpha_2e_2+\alpha_{12}e_1e_2)\star(\beta+\beta_1e_1+\beta_2e_2+\beta_{12}e_1e_2)=\alpha\beta+\alpha_1\beta_1+\alpha_2\beta_2+\alpha_{12}\beta_{12}.$$

Terms of different grade are orthogonal. Terms of the same grade have inner product

$$(a_1\wedge a_2\wedge\cdots\wedge a_k)\star(b_1\wedge b_2\wedge\cdots\wedge b_k)=(a_k\wedge\cdots\wedge a_2\wedge a_1)\bullet(b_1\wedge b_2\wedge\cdots\wedge b_k)$$

$$=\det[a_i\cdot b_j].$$

(I'm assuming that the underlying vector space has an inner product; then $\star$ is also an inner product. But this $\star$ extends any symmetric bilinear form on the vector space to the whole algebra, regardless of positive-definiteness or non-degeneracy.)

mr_e_man
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