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?
- 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)
- If we cannot define such an operator $\star$, what is the obstacle?
Best Answer
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.)