If $G$ is a finite group and $\lbrace x_{g} \rbrace_{g\in G}$ are commuting formal variables, then one can form a matrix whose $(g,h)$ entry is $x_{gh^{-1}}$. The determinant of this matrix is a polynomial with integer coefficients and is called the group determinant. Considering the factorization of this determinant led Frobenius to discover seminal results in the representation theory of finite groups. Later on, it was shown that a group can be recovered from its group determinant.

Many people have asked me the following question over the years, and I haven't a good answer for it. One might think about looking at something like a generalized determinant over a polynomial ring in infinitely many variables indexed by the group.

Question: In the literature, does there exist a more or less direct attempt to generalize the Dedekind-Frobenius group determinant to the setting of infinite groups?

Moreover, if such a thing exists, can it determine the group from which it is constructed?

  • $\begingroup$ Jon, does the Fuglede-Kadison determinant have some of the properties you're after? $\endgroup$ – Yemon Choi Aug 27 '12 at 19:12
  • $\begingroup$ @Yemon: I hope so, actually. (This is why I've included the operator algebras tag.) Perhaps the composition of the thing I'm looking for, which might take values in a commutative operator algebra, with some state should give the KF determinant. I can't see clearly what should happen. $\endgroup$ – Jon Bannon Aug 27 '12 at 19:59
  • $\begingroup$ I meant to type "pure state" in the comment above... $\endgroup$ – Jon Bannon Aug 27 '12 at 20:00

Such a thing is defined in Steve Humphries' very interesting paper:

Humphries, Stephen P.(1-BYU) Cogrowth of groups and the Dedekind-Frobenius group determinant. Math. Proc. Cambridge Philos. Soc. 121 (1997), no. 2, 193–217

I am not sure it is as general as you might like, but...

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  • $\begingroup$ Thanks Igor! I'm happy that there is a positive answer to this question, even a less than fully general one. I'll have to interlibrary loan this ASAP. $\endgroup$ – Jon Bannon Aug 27 '12 at 20:31

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