S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
It is currently Fri, 22 Aug 2014 06:53:17 UTC

All times are UTC [ DST ]




Post new topic Reply to topic  [ 2 posts ] 
Author Message
 Post subject: Induced Representations
PostPosted: Thu, 17 May 2012 01:03:06 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame

Joined: Wed, 10 Jan 2007 19:16:15 UTC
Posts: 663
Location: England
It's an easy example (better to start off with something simple) but I'm trying to find
V := \mathrm{Ind}_{S_{2}}^{S_{3}} 1, where 1 is the trivial representation on S_{2}.

So, I want to use Froebenius Reciprocity. Then
< \chi_{V}, \chi_{1} > _{S_{3}} = < \chi_{1}, \chi_{1}|_{S_{2}} > _{S_{2}} = 1; doing a similar calculation with the character of the sign representation gives 0, so therefore
V =  1 \oplus nU, where U is an irreducible 2-dimensional representation on S_{3} and n is the multiplicity.
Now, the dimension of V is equal to [S_{3}: S_{2}] dim(1); but the dimension of the trivial representation is 1, S_{2} \cong C_{2} and [S_{3}: S_{2}] = |S_{3}| / |C_{2}| = 6/2 = 3, so dimV = 3 and hence n=1, i.e. V = 1 \oplus U.
Of course, to show this explicitly I would have to calculate the restriction of U on S_{2}, show that it is equal to 1 \oplus sign and then go back to characters, but without doing that bit (im still working through the restriction of U) does it look like a semi-convincing argument?

_________________
"It's never crowded along the extra mile"

Graduated, and done with maths forever :P


Top
 Profile  
 
PostPosted: Thu, 17 May 2012 07:59:56 UTC 
Offline
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6814
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
peccavi_2006 wrote:
It's an easy example (better to start off with something simple) but I'm trying to find
V := \mathrm{Ind}_{S_{2}}^{S_{3}} 1, where 1 is the trivial representation on S_{2}.

So, I want to use Froebenius Reciprocity. Then
< \chi_{V}, \chi_{1} > _{S_{3}} = < \chi_{1}, \chi_{1}|_{S_{2}} > _{S_{2}} = 1; doing a similar calculation with the character of the sign representation gives 0, so therefore
V =  1 \oplus nU, where U is an irreducible 2-dimensional representation on S_{3} and n is the multiplicity.
Now, the dimension of V is equal to [S_{3}: S_{2}] dim(1); but the dimension of the trivial representation is 1, S_{2} \cong C_{2} and [S_{3}: S_{2}] = |S_{3}| / |C_{2}| = 6/2 = 3, so dimV = 3 and hence n=1, i.e. V = 1 \oplus U.
Of course, to show this explicitly I would have to calculate the restriction of U on S_{2}, show that it is equal to 1 \oplus sign and then go back to characters, but without doing that bit (im still working through the restriction of U) does it look like a semi-convincing argument?


Well, you can do it that way. In this case, it is much faster to just work out \chi_V directly from the formula \displaystyle\chi_V(g)=\sum_{x\in T}1_{S_2}(g^x) (where T is a right transversal of S_2<S_3) to show \chi_V=\mathop{\mathrm{fix}}, since there is only three conjugacy classes of S_3.

But for more complicated examples, yes, Frobenius reciprocity is a great tool to work out the decomposition.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 2 posts ] 

All times are UTC [ DST ]


Who is online

Users browsing this forum: No registered users


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum

Search for:
Jump to:  
Contact Us | S.O.S. Mathematics Homepage
Privacy Statement | Search the "old" CyberBoard

users online during the last hour
Powered by phpBB © 2001, 2005-2011 phpBB Group.
Copyright © 1999-2013 MathMedics, LLC. All rights reserved.
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA