::::::i function = O c- zero - - - t en Tn C , et , t = O = eye - - PowerPoint PPT Presentation

SLIDE 1

Independence of

charades

III :

Independence

• f

Characters

(OPTIONAL)

SLIDE 2

From

Part I

• f

In depone of

auratus :

÷÷÷÷:÷:÷÷÷÷÷:÷:÷:÷:÷÷÷i

C , et , t

• - - t enTn

= O c-

zero

function Then

e ,

=

• - -

= eye = O .

NewSt

Proving this

result

SLIDE 3

Det (Character)

For

a

group

G

and

afield E

,

a

character

• f

G

• ver

E

is

a

hem

X : G -

s E #

.

EI

let

G - Glen ( IR) and

let

E

• IR

, then

dot

:

Glen ( IR) → IR#

is

a

character .

SLIDE 4

Defy (Independence of chants)

A set at Chanute's

{ Hi ,

. ., Xr ) of I

• ver

E

is

independent

if

the

• nly

values of a,

. ., erE E

satisfying

e , X, t

• - t erXr
• =D

are

e,

• - -
• er

= 0 ,

theorem

Cfndependence of Charters) c-Dedekind

Any

finite

collection

• f

distinct

characters of Gour E are

independent

.

why do

we care? If

Er . .

. .,rn3E Aut LE)

are given ,

then

rile#

is

a character

• f

E#

• ver

E

. By

indepuue of charters

.

They

are

" E

• independent .

' '

SLIDE 5

PI

( by

induction

• n The

number of

characters )

Basics

!

let

X

be

a

chanter

. If

ex

• O

,

can

we

say

e

• O ?

Nate

Ng) toe E

for all

GEG

. If

eto

, then

eXCg) to

.

Hence

ex

• O

implies

e

• O

.

①

Inductive

:

Assume

my

collection

• f

r

• I
• r fewer

distinct

characters

is

independent

.

We

want

{ Hi ,

. ., Hr )

are

independent

where

the

Xi

are

duetinct

from

each other)

SLIDE 6

So

suppose

we

have

• e. ,

. -yer EE

so

that

e , X, t

• - -

t en Xr

=D

.

This

means

• e. X. (g)

t

• ter Xrlg)
• O

tg EG

First

,

• bserve

that

if

ay

eino , then

by induction

we

know

all

ei

are

zero . So : assume all

The

coefficients

ei

are

nonzero

.

By scaling

through

by

er

. '

, we

can

assume

①

• e. X. (g)

t

• ter. .Xmlgltxrlg)
• O

tg EG

.

SLIDE 7

Since

X , # Hr , we

have

some

HEG

so

• X. Ch)tXrlh)

.

Because

HG

• { hg :gEG}

= G ,eq'

n I

becomes

①

• e. X. (hg)

t

• ter..Xr.lk/tXrlhg)--OV-gEG

.

⇒ e , X. lhtxilglt

• - ten. Xralhtxr. . (g) + Xrlhlhlrlgl

.

• O tg

Multiplying through

by

Xrlh)

• '

:

• e. X.MARCH
• ' High
• - ter . . xrdhlxrlhtkr.ly/tXrlg)=ofg

②

Subtract

①

and

②

:

SLIDE 8

①

• e. X , (g)

t

• ter . .Xmlgltxrlg)
• O

Fg EG

.

• ②
• e. A.MARCH
• ' High
• - ter . . xrdhlxrlhtkr.ly/tXrlg)=ofg
• ell
• x.lhtxrlh)

")Hlg) t

.

• -ter . , ( l - Xrnlh)Xrlh5

') Hr , Ig) =D

Hg

This

is

an

E

• combination of

six..

. ., Xm )

that equals

O

. By

inductive hypothesis

, all

coefficients

are

.

But

e , ( I - x.lhlxrlhl

") to .

→

a-

Dna