Thanks for watching the talk! After the seminar, I realized I slightly misspoke when asked about the Clifford group. The orthogonal group is not a subgroup of it. Instead, the orthogonal group is isomorphic to a quotient of the Clifford group. It’s worth noting though that the Clifford group always acts through the orthogonal group. Further, I believe someone asked what the geometric product of a trivector and a bivector means geometrically. This depends entirely on the algebra one works with. In the standard (Euclidean) three-dimensional Clifford algebra, multiplying with a trivector is equivalent to taking the bivector's dual (orthogonal complement) up to a scalar factor. A bivector B can be written as B = a^b, where a^b is the wedge product of two vectors. Then, multiplying with a trivector T, i.e., (a^b)T, gives the orthogonal complement, yielding a vector orthogonal to the plane spanned by the vectors a and b, which is (again, up to a scalar factor) precisely their cross product.
Thanks for watching the talk! After the seminar, I realized I slightly misspoke when asked about the Clifford group. The orthogonal group is not a subgroup of it. Instead, the orthogonal group is isomorphic to a quotient of the Clifford group. It’s worth noting though that the Clifford group always acts through the orthogonal group. Further, I believe someone asked what the geometric product of a trivector and a bivector means geometrically. This depends entirely on the algebra one works with. In the standard (Euclidean) three-dimensional Clifford algebra, multiplying with a trivector is equivalent to taking the bivector's dual (orthogonal complement) up to a scalar factor. A bivector B can be written as B = a^b, where a^b is the wedge product of two vectors. Then, multiplying with a trivector T, i.e., (a^b)T, gives the orthogonal complement, yielding a vector orthogonal to the plane spanned by the vectors a and b, which is (again, up to a scalar factor) precisely their cross product.