string Theory a - n -

sheet 13

Exercise n :

(a) We will perform a slightly more

general calculation for

TCH = I : 2C b : t C I - n ) : c 2b :

and evaluate the final result for 1=2.

other values of I also play a role in

superstring theory .

As usual the contractions are

performed with the singular part ofthe OPE

, i - e.

- -

beta ) CC Zz ) = ¥z,

( = - cha ) bcznl )-

⇒ CCH b C Zz ) =I

let =I = o

4- Zz

TCH Tczz ) = : ¢Job t C X - n ) Cst ) ( 2=1

422 abt Ct - at cab ) Cta ) :

+ cross - contractions

- 2 -

i contraction :

- -: 2,2 ( Jc b) Ct

. ) ( Ic bl Ces ) t 72 ( Kel Cal C KHAI

- -

t 7, Ct - n ) ( salt Cat ( a Jblcult that ( Jcblfhl Calked

- -1- 74 - y ( CINCH C soll let HH - No Iet At Goblet

- -

+ H - at' to INCH l c IN CH th - al

' KINCH KINCH :

none of these contribute to give ¥244

2 contractions :

- -72 ( Ic bl Cta ) ( Jc blitz ) + A Ct - n ) Golf I H ( c Jbl C 2-4-_-1-7H - H ( c Jb ) CH ( Dcb ) C Zz ) t H - H

' ( a Jbl CH Cc Ibl feel- -

Iiit t Ciii ) : 2X 17 - at Iz.221¥72 ) - ÷z

.

= -

414€( 2-

n- Zz ) 4

lilt Civ ) : H '

t H - at' ) Iz

.

I ¥-1 Izz ( ¥z )= It 't Ca- it I - ÷⇒t ¥⇒ '

⇒ lilt.

. . Hiv I -

- ¥Ey , ( - 4th - at - 72 - Ct - i')

- 3 -

Comparison with42

That TCA ) =

izz ,t ¥⇒ . TH.lt#zJTGd

+ finite

⇒ c = - 2 ( 412-4,2 t X 't 72 - 2x t n )

= - 12 f

'+ r2 f - 2

= - 3 ( 2,2 - n )'

t n

1. e. for 1=2 one obtains s

Ghost=

- 26

(b) TCH btw ) = 2 : 2cal but : Ccw )

t I I - n ) : CCH INCH : ccw )

= Xz?cw - ca - n ) £ t finite

Use CCH = CCW ) t 2C Cw ) C z - w ) t . . .

⇒ TCH blwl = - I - DIII t Eff t finite

⇒ c has weight - &- n ) =- A

.7=22

- 4-TCH btw ) = I : JCCHBCH : btw )

+ I - n ) : CCH 2b Cz ) : bcwl

= - 7 : a ECHTCw ) biz ) :

- C 7 - n ) : c#bCw) 2b Cz I : + finite

= - 7 beet ( - ¥wy ) - A-a) 2bGt¥wc- finite

Now use bCzl= btw ) t Iller ) Cz - w/ t . . .

⇒ TCH beg = a MI +ahhh

E-us- ⇐

t finite

⇒ b has weight 7=21=2

=

'

- 5-

Exercise 2 :

(a) The coordinates I ' in which the metricis the

unitmetric can be found by solving

the differential equation

← III seaa Too

= geo

Given that points I O'

,84 and Crn

,r 't t2-alm.nl

are identified,

the derivatives III have to

be periodic.

Thisimpliesthat also fr E equivalent pointsform a

Lattice. This can be seen as

follows :

ja ( C on,

o 's + 2-alm.nl )

= Eaton .mx#EiimaII.ncoi;r4i-T!Iii2zEicrazama4

Italo;r4 t m rings.ca :r4tn%"dorsi:c no 's

periodicity

of II2ft

Now one has - 6 -

O = § I E- . d 5 where the integral

is over the following parallelogram :

q2^

( of,

6 't 2h )L

( 0,2T ) cBaVDA

,co ;r4

( 0,01- 7 6^

The integrals over the sides Aand C canceleach other due to periodicity .

Thus :

§do " Cr ;r 4 = ! do ' ' 3Eco

,oi4=E- E ' 1901

= we

Without loss of generality ,we can set 8%01=0

Similarly :

do " III! C r' '

,r4 = Elza,

o ) = u"

Thus Talon,

r 't = Tahrir 't t2-alm.nl )

= Ea ( r ;r7t mu"

-1 noa

- 7-

Actually ,all linearly independent vectors

na,

va can arise in this way .

Take the constant metric I which is non - degenerateiff ha

,va are linearly

independent )

ds2=

Kitkat) dado - +2

luiiecio'tdonde '

+

Clothier't

'tdo ' do -

with Ira,

r 't = Cris 't t 2-alm.nl

Using the transformation En =wantv r'

[ 2=U2rn + v252the metric becomes ds ? DEDE '

+ DE 'd 't '

with I E,

54 = IF,

EY t 2-nfmlun.ci/tnlviv7 )

les g-

- ( he.9%1 ; ( II . )T real

⇒ detg

= T2 - e . I -- O ⇒

gdegenerate

Consider the metric ds ? Idr 't c- do ' 12

Under a large coordinate transformation C 2.71this becomes

- 8-

As'= Id ( dr " ) t b Idr ' ' I t I calls ' ' l + e a Cdr ' ' I I

I d Cdr 'T + b Cdr ' Y te a @r'T t Ea Cdr " ) )

= Idea 12 Cdr" I 't lb t at 12 Cdr ' 42

+ ( Ibd t

LaceI t fuel ( ad that ) Cdr't Cdr' 4

= Ice- id I' I Cdr't 't 14dm y

2

+latter ) I c Etd ) t C at tht l at + d ) (doin )fdfgJ

I CT toll-

= I cetdl ' I Cdr

'T'

+ I t/(day '

+ It a I Idr 't Cdr'D )

= Ice - idk IIdr'T

'

t ( T'

te' ) Cdr'T Cdr ' 4

+ IT't'

Cdo '2) ' I= lo et all

' I do "

+ I' do " 12

Concerning the periodicity : The relation C2.61can be inverted :

I kit =L: -Etc :c)

- 9-

First show that da- values which

are identified , necessarily differ by2 a times some integer .

( FI) I ( I! ) ( or 't 2am ,or 't 2am )

= (a @

^- b r 't Zama - 2am b

- contd r'

- 2am a t Land )= I kitten ( II:c )

-

C- 22

In fact 8"

t 2am = G' ^ ( and a "t2-an=o' 4

for any m ,me 2 :

e.g .f

' '

-12 am = A G'

- br 't 2am

IZ IZI a ( f - 2-a.am/-blM-2-acmlt2-um

= of' ^

- 2am ( ad - be ) t 2am

= f"

Exercise 3 : - no -

(a)z

Pt ,=

- 2¥ ,. , Jds! It Er ( e

- ¥ CN 'Tit -21 ee-aiscnt.tt, )- I

2

I a

= - 24¥ Idk. f dg⇒ ,-4 ( e

- " " " ÷ i '

. "

ez-aie.cn' .

its )-

I C" thisis actually

E =I

; we just rename 'd E → Ed

' Tr

= - zY⇒.den I d÷÷ Ecg .

" -

ofit - .

)

->

eZai

⇐t its ) ( Nt - ale

- Zai I Tn - it ) C Tt - a )

= eEai

Ten ( N'

- n - Itta )

e-Zara ( Nt - n + It - n )

= eZai

Tal Nt - Tt )e

- 2 a- ⇐ ( N 'tTt - 2)

( ul d 'T = de DE - rn -

de '= 41¥ - III. ode )

= de * + ad -

act- be

( Cet d) 2

de=

-

ad - beGetd) 2

and DE'

=

dC CE -1 d) 2

⇒ dk '=

dICE toll "

c-'

= c-n' + it,

'=

akntitz.lt#cCTntiTzItd=(aCTntiT4tb)(cCEn-i-z1td)

-

ICE -1dL'

. latter ) C- Cte ) tate ( ctntd )=

. ..

-1 '

ICT t d 12

=. .

. + i

-

¥52-

bcT2tatadtz

lett dt2

Iz=

.. . t i -

ad - bc= : lcttdl'

= ? Tz'

= TZICTTd 12

- n 2-

Thus,

the measure DILI is SLC 2,21 - invariant.

For the integrand Tin'

ly tell- " 8

it is

Sufficient to check invariance under the

generators S and T of SLC 421 .

T : I → T'

= Ttr,

i. e.

a=b=n,

c=d=0

⇒ Tz'

= Tz

2 ( I' I = l

' ÷y ( I ) ⇒ lyte'll -48=1 ye -41-48

1. e . Ti"

ly Cell- 4 '

invariant under T.

S : I → T'

= - E,

i.e.

a=d=0,

b= - c= a

⇒ Tz

'=

IT 12

yI I 's = Fit yet ) ⇒ ( 2ft'll

- " 8

=÷⇒uGH5*

⇒ Kit- "

lyle 'll- " '

= ¥,a 171-4548

= Time 121-41-48