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Can an exterior algebra $$ k\langle x_{1},\dots,x_{n} \rangle/(x_{1}x_{2}-x_{2}x_{1},\dots,x_{1}^{2},\dots) $$ can be seen as a skew group algebra?

A skew group ring is defined for example in the introduction of this paper. I read this fact(?) somewhere but I cannot find a group action $G \rightarrow \mathrm{Aut}(k)$ that cooks up the exterior algebra.

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6  
Should it not be $x_ix_j+x_jx_i$? –  anon Aug 10 '12 at 8:16
    
I would be interested in knowing what $G$ you had in mind. –  rschwieb Aug 10 '12 at 12:02
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I'm guessing there are two usages of "skew" here. Probably the exterior algebra is "skew" in the sense that it's not quite commutative. –  Grumpy Parsnip Aug 10 '12 at 12:41
    
"skew group algebra" tends to mean "cross product of the group and a field" with a nontrivial action. –  Mariano Suárez-Alvarez Aug 10 '12 at 12:43
    
@JimConant The standard group ring construction causes coefficients to commute with the group basis elements. The paper referenced by the OP uses skew to mean "has a rule which modifies commuting coefficients with elements of $G$". This is conventional in noncommutative algebra. –  rschwieb Aug 13 '12 at 11:56

1 Answer 1

If there is such a thing as "Maschke's theorem for skew group rings" (I think one might be proven here), then the answer would often be "no".

Since the exterior algebra of a finite dimensional vector space is a local ring, there is no way it's going to be semisimple (meaning "semisimple Artinian") unless it's already a field (and of course it's not a field, since it has nilpotent elements).

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