Look, you already need a bilinear form to get the Hodge star. My point is that with that same bilinear form you also get an entire Clifford algebra, and a much

Look, you already need a bilinear form to get the Hodge star. My point is that with that same bilinear form you also get an entire Clifford algebra, and a much more natural definition of the Hodge star. That's all I'm saying. Is that mistaken?