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Does the Hodge dual (star) operator make the exterior algebra an involutive (*-) algebra?

https://en.m.wikipedia.org/wiki/Hodge_dual https://en.m.wikipedia.org/wiki/*-algebra

This would seem to be a different duality than that created by the involutive algebra created by taking transposes of the tensor algebra generated by the field underlying the vector space over which the exterior algebra is defined.

Another way to phrase this question, I think, is as follows: is the duality identifying vectors and pseudovectors via the Hodge star operator the same type of algebraic operation as the duality identifying (column) vectors with covectors (row vectors)?

This question is based on the discussions around another question of mine: https://math.stackexchange.com/a/1839461/327486

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Chill2Macht
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    I think you will note that $(x\wedge y)\neq (y)\wedge*(x)$ in general. This is seen for example by having $x$, $y$ be top degree forms. – s.harp Jun 25 '16 at 19:44
  • @s.harp I see what you are saying. Shouldn't the wedge product be the multiplication operation amd not the addition operation of the algebra though? I suppose it probably doesn't matter, but I am somewhat confused. – Chill2Macht Jun 25 '16 at 19:55
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    Yes, this is what is meant. For a $$ algebra you have $(xy)^=y^x^$, take here multiplication via $\wedge$ and $x^=(x)$. – s.harp Jun 26 '16 at 11:15
  • @s.harp Ah OK thank you now I actually see what you're saying. So even algebraically the two aren't analogous, this helps a lot, thank you! – Chill2Macht Jun 26 '16 at 15:33

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