Frobenius formula for the determinant - MathOverflow

Frobenius formula for the determinant

2012-03-01T08:25:33Z

Is there a formula for the determinant of an induced representation, e.g. in the fashion of the Frobenius character formula.

I would hope for something:

$$ det \; Ind_H^G \rho(g) = (-1)^\alpha \prod\limits_{\gamma, \gamma_1 \in G/H \atop \gamma^{-1}g\gamma_1 \in H} det \rho(\gamma^{-1}g\gamma_1).$$

If you have a reference or a quick proof, that would be most helpful.

Answer by Geoff Robinson for Frobenius formula for the determinant

2012-03-01T09:49:53Z

Well, yes there is, but it's slightly complicatd by the fact that the permutation action of $G$ on the (say right) cosets of $H$ introduces a sign. Also, you have to worry about "non-diagonal" blocks. Hence, if we let $T$ be a complete set of representatives for the right cosets of $H$ in $G,$ then ${\rm det} {\rm Ind}_{H}^{G}(\rho)[g] = \prod_{s,t \in T: sgt^{-1} \in H} {\rm det}(\rho(sgt^{-1}))$ if $\rho(1)$ is even, but is ${\rm sign}_{H}(g)\prod_{s,t \in T :sgt^{-1} \in H} {\rm det}(\rho(sgt^{-1})$ if $\rho(1)$ is odd, where ${\rm sign}_{H}(g)$ denotes the sign of the permutation of the right cosets of $H$ in $G$ induced by right multiplication by $g.$ In the 1970's, T. Yoshida did some work on "character-theoretic transfer", which exploited the determinant of the induced representation, especially when $\rho$ was linear. In some situations, Mackey type formulae for the determinant of the induced representation can simplify calculations. Later edit: Perhaps a word about the use of Mackey type formulae. It can simplify things to look at the orbits of $\langle g \rangle$ on right cosets of $H$ in $G,$ in other words, to group contributions from the $(H, \langle g \rangle)$-double cosets. The factor ${\rm sign}_{H}(g)$ , can be accounted for as above, suppose that thre are $k$ of these double costs, and that $t_1,t_2, \ldots, t_k$ are representatives, where the double coset of $t_i$ consists of $m_i$ right cosets of $H.$ Then the formula for the determinant of ${\rm Ind}_{H}^{G}(\rho)[g]$ may be expressed as ${\rm sign}_{H}(g)^{\rho(1)} \prod_{i=1}^{k} {\rm det}(\rho( t_{i}g^{m_i}t_{i}^{-1})).$

Answer by F. Ladisch for Frobenius formula for the determinant

2012-03-01T17:47:29Z

The answer can be formulated in compact form using the usual transfer map $V_H^G\colon G\to H/H'$ or the (not-so-well-known) construction of tensor induction: Namely,

$$ \det( { \rm Ind}_H^G\; \rho ) = ( {\rm sign}_{[G:H]})^{\rho(1)} ((\det\rho)\circ V_H^G) =( { \rm sign }_{ [G:H] })^{\rho(1)} (\det \rho)^{\otimes G}, $$

where ${\rm sign}_{[G:H]}$ is the permutation sign character of $G$ on the cosets of $H$. An exposition of tensor induction and a proof of this formula is contained in Curtis and Reiner, Methods of Representation Theory I (see Proposition 13.15). It seems this formula is due to Gallagher (MR0185017). Of course, when evaluating at $g\in G$, this yields the last formula in Geoff Robinson's answer.