×

### Let's log you in.

or

Don't have a StudySoup account? Create one here!

×

or

17

0

4

# Class Note for MATH 796 at KU 8

Marketplace > Kansas > Class Note for MATH 796 at KU 8

No professor available

These notes were just uploaded, and will be ready to view shortly.

Either way, we'll remind you when they're ready :)

Get a free preview of these Notes, just enter your email below.

×
Unlock Preview

COURSE
PROF.
No professor available
TYPE
Class Notes
PAGES
4
WORDS
KARMA
25 ?

## Popular in Department

This 4 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Kansas taught by a professor in Fall. Since its upload, it has received 17 views.

×

## Reviews for Class Note for MATH 796 at KU 8

×

×

### What is Karma?

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 02/06/15
Wednesday 49 Irreducibility Indecomposability and Maschke s Theorem Today G is a nite group and all representations are nitedimensional De nition 1 Let p V be a representation of G A vector subspace W C V is Ginvariant if pgW C W equivalently if W is a Gsubmodule of V V is irreducible or simple or colloquially an irrep if it has no proper Ginvariant subspace For instance any 1dimensional representation is clearly irreducible It would be nice if every Ginvariant subspace W had a Ginvariant complement ie another Ginvariant subspace WL such that W O WL 0 and W WL V However funny things can happen in positive characteristic Example 1 Let e1e2 be the standard basis for F2 Recall that the de ning representation of 32 12 21 is given by pdef12 Pdef21 and that Pdef951 52 Peru9X51 52 Pdef951 52 Psign951 52 Therefore as we saw last time the change of basis map 1 71 5 l 2 H1 1 is a Gequivariant isomorphism between pdef and pmv EB psign 7 unless F has characteristic 2 In that case W spane1 e2 is certainly Ginvariant but it has no Ginvariant complement D ohl row row De nition 2 The representation V is decomposable if there are Ginvariant subspaces W WL with W 0 WL 0 and W WL V Otherwise V is indecomposable Clearly every representation can be written as the direct sum of indecomposables Moreover irreducible implies indecomposable But the converse is not true in general as Example 1 illustrates Fortunately this kind of pathology does not happen in characteristic 0 Indeed something stronger is true Theorem 1 Maschke s Theorem LetG be a nite group and letlF be a eld whose characteristic does not divide Then every representation p G A GLV is completely reducible that is every Ginvariant subspace has an invariant complement Proof If p is an irreducible representation then there is nothing to prove Otherwise let W be a Ginvariant subspace and let 7r2VgtW be any projection ie a surjective linear transformation with nothing assumed about its behavior with respect to p For 1 E V de ne 1 1 quotCW E Z 97r9 1v gEG Then Cv E W because W is Ginvariant Moreover for h E G we have 1 WOW E Z 97r9 1hv gEG Ilthggtnltlthggt1hvgt gEG 1 1 Z g7rg 1v h7rGv gEG that is a is Gequivariant Now de ne WL ker we Certainly V E W EB WL as vector spaces and by Gequivariance if v E WL and g E G then Cgv g7rgv 0 ie 92 E WL That is WL is Ginvariant D Maschke s Theorem implies that a representation p is determined up to isomorphism by the multiplicity of each irreducible representation in p By the way implicit in the proof is the following useful fact Proposition 2 Any Gequivaritmt map has a Gequivaritmt kernel and Gequivaritmt image Characters De nition 3 Let p V be a representation of C over F lts character is the function Xp G A F given by XM tr 99 Example 2 Some simple facts and some characters we ve seen before 1 A onedimensional representation is its own character 2 For any representation p we have Xp1 dim p because p1 is the n X 71 identity matrix 3 The de ning representation pdef of 37 has character Xdef039 number of xed points of a 4 The regular representation preg has character G ifala Xam 0 otherwise Example 3 Consider the twodimensional representation p of the dihedral group D7 r s l r 32 0 srs r l by rotations and re ections 1 0 cos 9 sin 9 MS l0 Ell MT l7 sint9 cos 9 Its character is Xprl 2cos23919 0 S 239 lt n Xpsrl 0 0 S j lt On the other hand if p is the ndimensional permutation representation on the vertices then its character IS 71 if g 1 0 if g is a nontrivial rotation Xpg 1 if n is odd and g is a re ection 0 if n is even and g is a re ection through two edges 2 if n is even and g is a re ection through two vertices One fixed point No fixed points Two fixed points Proposition 3 Characters are class functions that is they are constant on conjugacy classes of G More over if p g p then Xp Xp Proof Recall from linear algebra that trABA 1 trB in generals Therefore tr 019175 tr 9hp9ph 1 tr MMMQWWYI 179 For the second assertion let lt1 p A p be an isomorphism iiei lt1 179 p g lt1 for all g E G treating lt1 as a matrix in this notation Since lt1 is invertible we have therefore lt1 179 lt1 1 pg Now take tracesi D What we d really like is the converse of this second assertions In fact much much more is true From now on we consider only representations over C Theorem 4 Let G be any nite group X X en a is a represen a ion is e ermine up 0 isomorp ism y i s c arac er 1 p pth pr Tht39 tt39 39dt 39d t39 h39 b39t h t e c arac ers o irre uci e represen a ions orm a asis or e uec or space 0 a c ass 2 Th h t 39 d 39bl tt39 b 39 th t CZG ll l functions of G Moreover this basis is orthonormal with respect to the natural Hermitian inner product de ned by 1 7 ltf fla m E gEG The bar denotes complex conjugate s a consequence e num er 0 i eren irre uci e represen a ions 0 equas e num er 0 3A th b d39 t39 d39bl tt39 G lth b conjugacy classes e regu ar represen a ion re sa is es 4 Th l t t39 p g t39 5 ea dim p meg 7 EB p irreps so in particular 0 2 ohm irreps 17 Example 4 The group G 33 has three conjugacy classes determined by cycle shapes 01 1a 02 12 lt13 23 Cs 123 132 We ll notate a character X by the bracketed triple XC1 XC2 XC3li We know two irreducible 1dimensional characters of 33 namely the trivial character va 1 1 1 and the sign character Xsign 1 71 1 Note that ltva Xtriv 1 ltXsign Xsigngt 1 ltXtriv Xsigngt 0 Consider the de ning representation lts character is Xdef 3 1 0 and 1 3 7 ltXtriv Xdaf 6 Zlcjl XtrivCj XdefCj j1 7113311210 1 03H 3 1 7 ltXsign Xdaf 6 Zlcjl XtrivCj XdefCj l 61137311210 0 This tells us that pdef contains one copy of the trivial representation as a summand and no copies of the sign representation If we get rid of the trivial summand the remaining twodimensional representation p has character Xp Xdef 7 va 2 0 71 Since 12 2 30 0 2 1 1 7 7 ltxp abf it follows that p is irreducible So up to isomorphism 33 has two distinct onedimensional representations pmv psign and one twodimensional representation p1 Note also that Xcriv Xsign 2Xp 111l1r11l 2 0 1 6 0 0 Xreg New Characters from Old In order to investigate characters we need to know how standard vector space or in fact Gmodule functors such as EB and 8 a ect the corresponding characters Throughout let p V p V be representations of G with V V ll 11 Direct sum The vectors in V EB V can be regarded as column block vectors for v E V v E V Accordingly de ne p EB p V EB V by 7 90739 0 paw gtlthgt fl 0 W 1 It is clear that 2 Xp Bp Xp l Xp Next time Tensor product dual and Hom

×

×

### BOOM! Enjoy Your Free Notes!

×

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

Steve Martinelli UC Los Angeles

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

Amaris Trozzo George Washington University

#### "I made \$350 in just two days after posting my first study guide."

Bentley McCaw University of Florida

Forbes

#### "Their 'Elite Notetakers' are making over \$1,200/month in sales by creating high quality content that helps their classmates in a time of need."

Become an Elite Notetaker and start selling your notes online!
×

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com