Luke Burns
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 Feb18 comment Special case of the Hodge decomposition theorem Ah. Thanks! The general result is: for a vector $a$ and r-vector $B_r$, we have $a \cdot B_r = \frac{1}{2}(aB_r - (-1)^r B_r a)$. Hence, $a \cdot B_0 = \frac{1}{2}(aB_0 - (-1)^0B_0a) = \frac{1}{2}(aB_0 - B_0a) = 0$. Feb18 comment What is a covector and what is it used for? Then don't we have: $e^3\frac{e_1 \wedge e_2}{e_1 \wedge e_2 \wedge e_3} = (e_1 \wedge e_2)(e_3 \wedge e_2 \wedge e_1) = e_3$? Or is there some subtlety? Feb18 awarded Supporter Feb18 comment Special case of the Hodge decomposition theorem You wrote $∇∧ϕ$ for a scalar field $ϕ$. What does the curl of a scalar field mean? Feb24 comment Are Clifford algebras and differential forms equivalent frameworks for differential geometry? Can you provide references to the alternatives to Geometric Calculus that you're talking about? Oct28 comment What is a vector with a single non-zero component called? Right, I'm interested in the general term, when the component is not necessarily equal to 1. Oct28 awarded Student Oct28 asked What is a vector with a single non-zero component called? Jan24 asked Proof for the divisibility of natural numbers by at least one prime Jan24 awarded Autobiographer