4
$\begingroup$

I am trying to understand the representation theory of the infinite dihedral group, which appears to be calculated in the paper

Berman, S. D.; Buzási, K. Representations of the infinite dihedral group. (Russian) Publ. Math. Debrecen 28 (1981), no. 1-2, 173–187.

The MathSciNet review seems pretty complete, but I am having trouble understanding it. In particular, the review author draws a distinction between non symmetrical polynomials in the group ring $F[x]$ over a field $F$ and symmetrical polynomials, defined as follows

"A polynomial $f(x)\in F[x]$ of degree $m$ is said to be symmetrical if $f(x)=x−1$ or $x^mf(x)=f(x)$[sic]".

This must be a typo! Does anyone know what the correct condition is here?

Improve this question
$\endgroup$
2

1 Answer 1

Reset to default
3
$\begingroup$

It is a guess. Possibly what is meant is that the polynomial is palindromic: algebraically this means whenever $\alpha$ is a root $\alpha^{-1}$ is also a root, which translates to $f(x) = x^m f(\frac1x)$.

The exceptional one $f(x)=x-1$ though not palindromic, has a unique root which is 1 which satisfies the algebraic version of the condition about roots. So this is also to be considered part of this family.

Improve this answer
$\endgroup$
2
  • $\begingroup$ That is a good and probably correct guess, although the root condition you mention is not exactly equivalent to being palindromic since antipalindromic polynomials also satisfy that condition. $\endgroup$
    – Jim Conant
    Commented Aug 23, 2016 at 3:13
  • $\begingroup$ By anti-palindromic perhaps you mean a definition similar to anti-symmetric matrices [coeff of $x^k$ and $x^{m+1-k}$ being of opposite sign] As roots determine a polynomial equation only upto a scalar this should be ok. This also explains the exceptional $x-1$ $\endgroup$
    – P Vanchinathan
    Commented Aug 23, 2016 at 3:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .