Transwedge products: Difference between revisions
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The ''transwedge antiproduct'' of order $$k$$ is the dual of the transwedge product of order $$k$$. | |||
[[Image:TranswedgeAntiproducts.svg|720px]] |
Revision as of 05:21, 17 May 2025
The transwedge product is a generalization of the exterior product and interior product that also includes a transitional sequence of liminal products between exterior and interior.
The transwedge product of order $$k$$ is written with a double upward wedge having either a subscript or an underscript indicating the order. It is defined as
- $$\displaystyle\mathbf a \mathbin{\underset{k}{\unicode{x2A53}}} \mathbf b = \sum_{\mathbf c \in \mathcal B_k}{(\mathbf{\underline c} \vee \mathbf a) \wedge (\mathbf b \vee \mathbf{\tilde c^{\unicode{x2605}}})}$$.
The set $$\mathcal B_k$$ is the set of all basis elements having grade $$k$$. When $$k = 0$$, this set is simply $$\{\mathbf 1\}$$, and the transwedge product reduces to the wedge (exterior) product. That is,
- $$\displaystyle\mathbf a \mathbin{\underset{0}{\unicode{x2A53}}} \mathbf b = \mathbf a \wedge \mathbf b$$.
When $$k = \operatorname{gr}(\mathbf a)$$, the transwedge product reduces to the right contraction $$\mathbf b \vee \mathbf{\tilde a^{\unicode{x2605}}}$$, which is an interior product. If $$k > \operatorname{gr}(\mathbf a)$$, then $$\mathbf a \mathbin{\unicode{x2A53}_k} \mathbf b = 0$$.
For operands $$\mathbf a$$ and $$\mathbf b$$ having grades $$g$$ and $$h$$, the transwedge product of order $$k$$ generates a result having grade $$g + h - 2k$$, assuming it’s nonzero.
The sum of all possible transwedge products yields the geometric product. That is,
- $$\displaystyle\mathbf a \mathbin{\unicode{x27D1}} \mathbf b = \sum_{k=0}^n{\mathbf a \mathbin{\underset{k}{\unicode{x2A53}}} \mathbf b}$$,
where $$n$$ is the dimension of the algebra.
The geometric product of each pair of the 16 basis elements in the 3D rigid algebra is given by exactly one of the transwedge products. These are shown in the following table, which color codes the transwedge products of order 0, 1, 2, and 3. The transwedge product of order 4 is always zero in this algebra due to the degenerate metric.
The transwedge antiproduct of order $$k$$ is the dual of the transwedge product of order $$k$$.