The Hodge dual of a $p$-form is explicitly metric dependent, but the function HodgeDual[] doesn't seem to ask for the metric as an argument. How does that work? Does it mean that HodgeDual[] can only do the computation in Euclidean space?

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    $\begingroup$ Yes, currently HodgeDual assumes the metric is Euclidean. $\endgroup$ – jose Oct 10 '16 at 18:13

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