Frobenius formula for the determinant - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T00:08:37Z http://mathoverflow.net/feeds/question/89938 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89938/frobenius-formula-for-the-determinant Frobenius formula for the determinant Marc Palm 2012-03-01T08:25:33Z 2012-03-01T19:03:03Z <p>Is there a formula for the determinant of an induced representation, e.g. in the fashion of the Frobenius character formula.</p> <p>I would hope for something:</p> <blockquote> <p>$$det \; Ind_H^G \rho(g) = (-1)^\alpha \prod\limits_{\gamma, \gamma_1 \in G/H \atop \gamma^{-1}g\gamma_1 \in H} det \rho(\gamma^{-1}g\gamma_1).$$ </p> </blockquote> <p>If you have a reference or a quick proof, that would be most helpful.</p> http://mathoverflow.net/questions/89938/frobenius-formula-for-the-determinant/89945#89945 Answer by Geoff Robinson for Frobenius formula for the determinant Geoff Robinson 2012-03-01T09:49:53Z 2012-03-01T12:50:40Z <p>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 <code>${\rm det} {\rm Ind}_{H}^{G}(\rho)[g] = \prod_{s,t \in T: sgt^{-1} \in H} {\rm det}(\rho(sgt^{-1}))$</code> if $\rho(1)$ is even, but is <code>${\rm sign}_{H}(g)\prod_{s,t \in T :sgt^{-1} \in H} {\rm det}(\rho(sgt^{-1})$</code> if $\rho(1)$ is odd, where <code>${\rm sign}_{H}(g)$</code> 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 <code>${\rm sign}_{H}(g)$</code>, can be accounted for as above, suppose that thre are $k$ of these double costs, and that <code>$t_1,t_2, \ldots, t_k$</code> are representatives, where the double coset of <code>$t_i$</code> consists of <code>$m_i$</code> right cosets of $H.$ Then the formula for the determinant of <code>${\rm Ind}_{H}^{G}(\rho)[g]$</code> may be expressed as <code>${\rm sign}_{H}(g)^{\rho(1)} \prod_{i=1}^{k} {\rm det}(\rho( t_{i}g^{m_i}t_{i}^{-1})).$</code> </p> http://mathoverflow.net/questions/89938/frobenius-formula-for-the-determinant/89978#89978 Answer by F. Ladisch for Frobenius formula for the determinant F. Ladisch 2012-03-01T17:47:29Z 2012-03-01T17:47:29Z <p>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 <em>tensor induction</em>: Namely, </p> <blockquote> <p><code>$$\det( { \rm Ind}_H^G\; \rho ) = ( {\rm sign}_{[G:H]})^{\rho(1)} ((\det\rho)\circ V_H^G) =( { \rm sign }_{ [G:H] })^{\rho(1)} (\det \rho)^{\otimes G},$$</code></p> </blockquote> <p>where <code>${\rm sign}_{[G:H]}$</code> 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, <em>Methods of Representation Theory I</em> (see Proposition 13.15). It seems this formula is due to <a href="http://dx.doi.org/10.1007/BF02993247" rel="nofollow">Gallagher</a> (<a href="http://www.ams.org/mathscinet-getitem?mr=185017" rel="nofollow">MR0185017</a>). Of course, when evaluating at $g\in G$, this yields the last formula in Geoff Robinson's answer.</p>