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Let $G$ be a finite group and $\mathbb{C}G$ its group ring. Left multiplication by $\alpha\in \mathbb{C}G$ is a linear map $\alpha:\mathbb{C}G \to \mathbb{C}G$, and so $\alpha$ has a left determinant $\det(\alpha)$. If $\alpha\beta = 0$ for some non-zero $\beta$ (i.e. $\alpha$ is a left zero-divisor in $\mathbb{C}G$) then $\det(\alpha)=0$. The converse is also true.

For $G$ abelian $\det(\alpha)$ is zero just in case $\alpha$ lies on a coordinate hyperplane associated with the character idempotents. (This is an easy consequence of this PlanetMath page.)

What is known about the locus $\det(\alpha)= 0$ in $\mathbb{C}G$ when $G$ is finite but not abelian? Are there any similar statements?

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  • $\begingroup$ The group ring is a product of matrix algebras, each coming from an irreducible representation. This tells you what the structure of the locus is. $\endgroup$
    – Angelo
    Nov 16, 2011 at 23:10
  • $\begingroup$ ... and if the characteristic is $p>0$, while $G$ is a $p$-group, then $kG$ is a local ring and the non-invertible elements are just those of augmentation $0$. For a general group in positive characteristic, I don't really know. $\endgroup$
    – Pierre
    Nov 16, 2011 at 23:22
  • $\begingroup$ Formanek and Sibley (Proc AMS 112 (3)) and Mansfield (Proc AMS 116 (4)) are articles to the effect that the group determinant of a finite group determines the group. The first paper gives the complete factorization of the group determinant, along the lines in Angelo's comment. All useful information. My interest is more in the geometry of the zero locus of the group determinant than in the algebra of the equation defining it. $\endgroup$
    – Jonathan Fine
    Nov 17, 2011 at 21:11

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