The Transwedge Product

Summary

The Transwedge Product Eric Lengyel • May 23, 2025 Introductory texts on geometric algebra often begin by showing how the geometric product is a combination of the wedge product and the dot product, giving us the formula[1] \(\mathbf a \mathbin{\unicode{x27D1}} \mathbf b = \mathbf a \wedge \mathbf b + \mathbf a \cdot \mathbf b\).(1) However, the above formula holds only for vectors \(\mathbf a\) and \(\mathbf b\). When \(\mathbf a\) and \(\mathbf b\) are allowed to assume values of higher grade, the geometric product generally yields more terms, especially in higher dimensions. When the operands have grades \(g\) and \(h\), the geometric product can generate a result containing terms having grades \(m\) ranging from \(|g - h|\) to \(g + h\) such that the difference between each grade \(m\) and \(g + h\) is an even number. For example, the geometric product between two bivectors \(\mathbf A\) and \(\mathbf B\) in a 4D algebra generates components having grades 0, 2, and 4. We can decompose this product as \(\mathbf A \mathbin{\unicode{x27D1}} \mathbf B = \mathbf A \wedge \mathbf B + \mathbf A \times \mathbf B + \mathbf A \cdot \mathbf{\widetilde B}\),(2) where the \(\mathbf A \times \mathbf B\) term is called the commutator product, defined as \(\mathbf A \times \mathbf B = \mathbf A \mathbin{\unicode{x27D1}} \mathbf B - \mathbf B \mathbin{\unicode{x27D1}} \mathbf A\). The commutator product corresponds to the grade-2 part of the geometric product in this case, but it generally produces more terms when the operands have higher grades. Furthermore, because the commutator product is defined in terms of the geometric product, it can’t help us decompose the geometric product into independent operations. That would be circular. When one of the operands of the geometric product is a vector \(\mathbf a\), and the other operand is an arbitrary multivector \(\mathbf B\), we can generalize Equation (1) a little bit as \(\mathbf a \mathbin{\unicode{x27D1}} \mathbf B = \mathbf a...