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Let $x$ be a vector of an $n$-dimensional subspace $\mathcal A_n$ with $P$ the projection operator onto this subspace. Then, let $A= P(A) = \langle A\rangle_r$ be an $r$-vector. Then, I know that $\dot \partial (\dot x \cdot A) = A\cdot \partial x$ if $r=1$, but why is this true for general $r$?

Rodrigo
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1 Answers1

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It's easiest to just expand the inner product: $$ 2\partial(x\cdot A) = \partial xA - \dot\partial\hat A\dot x,\quad 2(A\cdot\partial)x = A\partial x - \dot\partial\hat A\dot x. $$ The result follows because $\partial x$ is a scalar.

Nicholas Todoroff
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  • What does the hat on the $A$ mean? – Rodrigo Feb 27 '24 at 12:38
  • This proof seems to work if $A$ is an $r$-vector and I substitute $\hat A$ with $(-1)^rA$. – Rodrigo Feb 27 '24 at 14:14
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    @Rodrigo Correct. Perhaps Hestenes and Sobczyk don't define it. It is the grade involution which has the effect you describe on homogenous multivectors and is otherwise linear. It can be described as the unique algebra homomorphism that negates vectors, similar to how the reverse is the unique antihomomorphism that is the identity on vectors. It is an important tool for writing identities that are valid on arbitrary multivectors that are not necessarily homogeneous; my formulas above are valid for arbitrary $A$, not just an $r$-vector. – Nicholas Todoroff Feb 27 '24 at 15:26