You are mistaken, the Hodge star does not belong to geometric algebra more than exterior algebra, that's the wrong way to look at it. Exterior algebra is just only a sub algebra of geometric algebra, they both have the same Hodge star. Saying the Hodge star belongs more in one than the other is a bit silly.
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?
