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14 May2010 Wei-MingChen LagrangianFormulationsofSelf-dualGaugeTheoriesinDiverseDimensions

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Lagrangian Formulations of Self-dual Gauge Theories in Diverse Dimensions

Wei-Ming Chen

NTU

14thMay 2010

Based on arXiv:1001.3608 with Prof. P. M. Ho

(2)

Outline

1 Introduction 2 D= 2 + 4?

3 General Formulations in D= D+ D′′

4 Interaction with Charged Branes 5 Conclusion

(3)

1 Introduction 2 D= 2 + 4?

3 General Formulations in D= D+ D′′

4 Interaction with Charged Branes 5 Conclusion

(4)

Motivation

Recently, in the study of M theory, in large C-field

background , a new Lagrangain formulation of M5-brane was derived from the BLG model for multiple M2-branes.

P. M. Ho and Y. Mastuo,arXiv:0804.3629 P. M. Ho, Y. Imamura, et al,arXiv:0805.2898 P. M. Ho,arXiv:0912.0445

This formulation is a non-linear self-dual gauge field theory in 6 dimensions, where the 6 coordinates of the base space are divided into two sets of 3 coordinates{xa}3a=1

and{x˙a}3˙a=1

(5)

new M5-brane formulation:

self-dual gauge theory in D= 3®

{xa}

+ 3®

{x˙a}

old M5-brane formulation:

self-dual gauge theory in D= 1®

{xa}

+ 5®

{x˙a}

Does there exist a formulation corresponding to the decomposition D= 2 + 4? D = D+ D′′?

Non-linear self-dual theory is quite complicated, we will work in linearized self-dual theory for simplicity.

(6)

Hodge Duality and Self-duality Condition

Hodge-Star Operator (⋆A)µ1...µn−p = 1

p!ν1...νpµ1...µn−pAν1...νp Hodge Duality for Twice

⋆ ⋆ A = (−1)s+p(n−p)A

(7)

⋆A = A

Conditions for Self-duality

⋆A = A Ô⇒ n = 2p

⋆ ⋆ A = A Ô⇒ { p(n − p) & s = even ⇒ n = 4k p(n − p) & s = odd ⇒ n = 4k + 2

k = 0, 1, 2, . . . Definition for a Anti-self Dual Form

F − ˜F ≡ F, ⋆F = −F

(8)

The Difficulty to Have a Covariant Formulation

S∼ ∫ Fµ1µ2...Fµ1µ2...+ (constraints)

(constraints)= (F − ˜F)2≡ F2= Fµ1µ2...Fµ1µ2...= 0

×

Example: D= 1 + 5

FµνλFµνλ= 3F1ijF1ij+ FijkFijk = −FijkFijk+ FijkFijk = 0

⇒ not all constraints are independent

⇒ only add the independent ones

⇒ non-covariant formulations

Possible to have covariant formulations with the help of infinite or finite auxiliary fields, which are rather

complicated

We only consider non-covariant formulation here.

(9)

∼ ∫ F 1 2 + (constraints)

(constraints)= (F − ˜F)2≡ F2= Fµ1µ2...Fµ1µ2...= 0

×

Example: D= 1 + 5

FµνλFµνλ= 3F1ijF1ij+ FijkFijk = −FijkFijk+ FijkFijk = 0

⇒ not all constraints are independent

⇒ only add the independent ones

⇒ non-covariant formulations

Possible to have covariant formulations with the help of infinite or finite auxiliary fields, which are rather

complicated

We only consider non-covariant formulation here.

(10)

The Difficulty to Have a Covariant Formulation

S∼ ∫ Fµ1µ2...Fµ1µ2...+ (constraints)

(constraints)= (F − ˜F)2≡ F2= Fµ1µ2...Fµ1µ2...= 0

×

Example: D= 1 + 5

FµνλFµνλ= 3F1ijF1ij+ FijkFijk = −FijkFijk+ FijkFijk = 0

⇒ not all constraints are independent

⇒ only add the independent ones

⇒ non-covariant formulations

Possible to have covariant formulations with the help of infinite or finite auxiliary fields, which are rather

complicated

We only consider non-covariant formulation here.

(11)

R1+5= R × R5

Gauge potential components

A1 ˙a, A˙a ˙b ( ˙a, ˙b = ˙1, ⋯, ˙5)

Field Strength and Its Dual

F1 ˙a ˙b= ∂1A˙a ˙b− 2∂[ ˙a∣A1∣ ˙b], F˜1 ˙a ˙b=3!11 ˙a ˙b ˙c ˙d ˙eF˙c ˙d ˙e F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c=2!11 ˙a ˙b ˙c ˙d ˙eF1 ˙d ˙e

Action

S1+5= − 1

12 ∫ d6x[FµνλFµνλ− 3F1 ˙a ˙bF1 ˙a ˙b]

(12)

Known Examples: D = 1 + 5

Decomposition of 6D Minkowski space R1+5= R × R5

Gauge potential components

A1 ˙a, A˙a ˙b ( ˙a, ˙b = ˙1, ⋯, ˙5)

Field Strength and Its Dual

F1 ˙a ˙b= ∂1A˙a ˙b− 2∂[ ˙a∣A1∣ ˙b], F˜1 ˙a ˙b=3!11 ˙a ˙b ˙c ˙d ˙eF˙c ˙d ˙e F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c=2!11 ˙a ˙b ˙c ˙d ˙eF1 ˙d ˙e

Action

S1+5= − 1

12 ∫ d6x[FµνλFµνλ− 3F1 ˙a ˙bF1 ˙a ˙b]

(13)

R1+5= R × R5

Gauge potential components

A1 ˙a, A˙a ˙b ( ˙a, ˙b = ˙1, ⋯, ˙5)

Field Strength and Its Dual

F1 ˙a ˙b= ∂1A˙a ˙b− 2∂[ ˙a∣A1∣ ˙b], F˜1 ˙a ˙b=3!11 ˙a ˙b ˙c ˙d ˙eF˙c ˙d ˙e F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c=2!11 ˙a ˙b ˙c ˙d ˙eF1 ˙d ˙e

Action

S1+5= − 1

12 ∫ d6x[FµνλFµνλ− 3F1 ˙a ˙bF1 ˙a ˙b]

(14)

Known Examples: D = 1 + 5

Decomposition of 6D Minkowski space R1+5= R × R5

Gauge potential components

A1 ˙a, A˙a ˙b ( ˙a, ˙b = ˙1, ⋯, ˙5)

Field Strength and Its Dual

F1 ˙a ˙b= ∂1A˙a ˙b− 2∂[ ˙a∣A1∣ ˙b], F˜1 ˙a ˙b=3!11 ˙a ˙b ˙c ˙d ˙eF˙c ˙d ˙e F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c=2!11 ˙a ˙b ˙c ˙d ˙eF1 ˙d ˙e

Action

S1+5= − 1

12 ∫ d6x[FµνλFµνλ− 3F1 ˙a ˙bF1 ˙a ˙b]

(15)

Equation of motion of A1 ˙a⇒ Additional Gauge Symmetry

˙b1 ˙a ˙b ≡ 0 ⇒ A1 ˙a→ Φ1 ˙a Equation of motion for A˙a ˙b is

˙cF˙a ˙b ˙c= 0 ⇒ F˙a ˙b ˙c = 1 ˙a ˙b ˙c ˙d ˙ed˙Φ1 ˙e

⇒ F˙a ˙b ˙c= 0

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Known Examples: D = 1 + 5 (the other equivalent formulation)

An equivalent formulation is to start with a theory without A1 ˙a; the only fields of this alternative formulation are A˙a ˙b.

Fa ˙d ˙e replaceÐ→ f1 ˙d ˙e= ∂1Ad ˙e˙

E.O.M. of F˙a ˙b ˙c replaceÐ→ F˙a ˙b ˙c−1

2˙a ˙b ˙c ˙d ˙ef1 ˙d ˙e = 1 ˙a ˙b ˙c ˙d ˙ed˙Φ1 ˙e If we define

F1 ˙d ˙e≡ ∂1Ad ˙e˙ − ∂d˙Φ1 ˙e+ ∂˙eΦ1 ˙d⇒ Fa ˙d ˙e= 0

(17)

R1+5= R1+2× R3 Gauge potential components

Aab, Aa ˙a, A˙a ˙b (a, b = 1, 2, 3; ˙a, ˙b = ˙1, ˙2, ˙3)

Field Strength and Its Dual

Fabc= 3∂[aAbc], F˜abc= abc˙a ˙b ˙cF˙a ˙b ˙c abc˙a ˙b ˙c= abc ˙a ˙b ˙c

Fab ˙a= 2∂[aAb] ˙a+ ∂˙aAab, F˜ab ˙a= −2!1abc˙a ˙b ˙cFc ˙b ˙c Fa ˙a ˙b= ∂aA˙a ˙b− 2∂[ ˙a∣Aa∣ ˙b], F˜a ˙a ˙b=2!1abc˙a ˙b ˙cFbc ˙c F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c= −3!1abc˙a ˙b ˙cFabc

Action S3+3= − 1

12 ∫ d6x[FµνλFµνλ− FabcFabc− 3Fab ˙cFab ˙c]

(18)

Known Examples: D = 3 + 3

Decomposition of 6D Minkowski space R1+5= R1+2× R3 Gauge potential components

Aab, Aa ˙a, A˙a ˙b (a, b = 1, 2, 3; ˙a, ˙b = ˙1, ˙2, ˙3)

Field Strength and Its Dual

Fabc= 3∂[aAbc], F˜abc= abc˙a ˙b ˙cF˙a ˙b ˙c abc˙a ˙b ˙c= abc ˙a ˙b ˙c

Fab ˙a= 2∂[aAb] ˙a+ ∂˙aAab, F˜ab ˙a= −2!1abc˙a ˙b ˙cFc ˙b ˙c Fa ˙a ˙b= ∂aA˙a ˙b− 2∂[ ˙a∣Aa∣ ˙b], F˜a ˙a ˙b=2!1abc˙a ˙b ˙cFbc ˙c F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c= −3!1abc˙a ˙b ˙cFabc

Action S3+3= − 1

12 ∫ d6x[FµνλFµνλ− FabcFabc− 3Fab ˙cFab ˙c]

(19)

R1+5= R1+2× R3 Gauge potential components

Aab, Aa ˙a, A˙a ˙b (a, b = 1, 2, 3; ˙a, ˙b = ˙1, ˙2, ˙3)

Field Strength and Its Dual

Fabc= 3∂[aAbc], F˜abc= abc˙a ˙b ˙cF˙a ˙b ˙c abc˙a ˙b ˙c= abc ˙a ˙b ˙c

Fab ˙a= 2∂[aAb] ˙a+ ∂˙aAab, F˜ab ˙a= −2!1abc˙a ˙b ˙cFc ˙b ˙c Fa ˙a ˙b= ∂aA˙a ˙b− 2∂[ ˙a∣Aa∣ ˙b], F˜a ˙a ˙b=2!1abc˙a ˙b ˙cFbc ˙c F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c= −3!1abc˙a ˙b ˙cFabc

Action S3+3= − 1

12 ∫ d6x[FµνλFµνλ− FabcFabc− 3Fab ˙cFab ˙c]

(20)

Known Examples: D = 3 + 3

Decomposition of 6D Minkowski space R1+5= R1+2× R3 Gauge potential components

Aab, Aa ˙a, A˙a ˙b (a, b = 1, 2, 3; ˙a, ˙b = ˙1, ˙2, ˙3)

Field Strength and Its Dual

Fabc= 3∂[aAbc], F˜abc= abc˙a ˙b ˙cF˙a ˙b ˙c abc˙a ˙b ˙c= abc ˙a ˙b ˙c

Fab ˙a= 2∂[aAb] ˙a+ ∂˙aAab, F˜ab ˙a= −2!1abc˙a ˙b ˙cFc ˙b ˙c Fa ˙a ˙b= ∂aA˙a ˙b− 2∂[ ˙a∣Aa∣ ˙b], F˜a ˙a ˙b=2!1abc˙a ˙b ˙cFbc ˙c F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c= −3!1abc˙a ˙b ˙cFabc

Action S3+3= − 1

12 ∫ d6x[FµνλFµνλ− FabcFabc− 3Fab ˙cFab ˙c]

(21)

Equation of motion of Aab ⇒ Additional Gauge symmetry

cabc+ ∂˙cab ˙c≡ 0 ⇒ Aab → Aab+ Φab

Equations of motion of Aa ˙aand A˙a ˙b

δAa ˙a∶ ∂˙bFa ˙a ˙b= 0 ⇒ Fa ˙a ˙b=12abc˙a ˙b ˙c˙cΦbc

⇒ Fa ˙a ˙b= 0

Aab→ Aab+ Φab Ä¦

δA˙a ˙b∶ ∂aFa ˙a ˙b+ ∂˙aF˙a ˙b ˙c = 0 ⇒ F˙a ˙b ˙c = ˙a ˙b ˙cf(xa)

⇒ F˙a ˙b ˙c = 0

Aab→ Aab+ abcfc Ä

(22)

Known Examples: D = 3 + 3

Equation of motion of Aab ⇒ Additional Gauge symmetry

cabc+ ∂˙cab ˙c≡ 0 ⇒ Aab → Aab+ Φab

Equations of motion of Aa ˙aand A˙a ˙b

δAa ˙a∶ ∂˙bFa ˙a ˙b= 0 ⇒ Fa ˙a ˙b=12abc˙a ˙b ˙c˙cΦbc

⇒ Fa ˙a ˙b= 0

Aab→ Aab+ Φab Ä ¦

δA˙a ˙b∶ ∂aFa ˙a ˙b+ ∂˙aF˙a ˙b ˙c = 0 ⇒ F˙a ˙b ˙c = ˙a ˙b ˙cf(xa)

⇒ F˙a ˙b ˙c = 0

Aab→ Aab+ abcfc Ä

(23)

Equation of motion of Aab ⇒ Additional Gauge symmetry

cabc+ ∂˙cab ˙c≡ 0 ⇒ Aab → Aab+ Φab

Equations of motion of Aa ˙aand A˙a ˙b

δAa ˙a∶ ∂˙bFa ˙a ˙b= 0 ⇒ Fa ˙a ˙b=12abc˙a ˙b ˙c˙cΦbc

⇒ Fa ˙a ˙b= 0

Aab→ Aab+ Φab Ä¦

δA˙a ˙b∶ ∂aFa ˙a ˙b+ ∂˙aF˙a ˙b ˙c = 0 ⇒ F˙a ˙b ˙c = ˙a ˙b ˙cf(xa)

⇒ F˙a ˙b ˙c = 0

Aab→ Aab+ abcfc Ä

(24)

Known Examples: D = 3 + 3

Equation of motion of Aab ⇒ Additional Gauge symmetry

cabc+ ∂˙cab ˙c≡ 0 ⇒ Aab → Aab+ Φab

Equations of motion of Aa ˙aand A˙a ˙b

δAa ˙a∶ ∂˙bFa ˙a ˙b= 0 ⇒ Fa ˙a ˙b=12abc˙a ˙b ˙c˙cΦbc

⇒ Fa ˙a ˙b= 0

Aab→ Aab+ Φab Ä¦

δA˙a ˙b∶ ∂aFa ˙a ˙b+ ∂˙aF˙a ˙b ˙c = 0 ⇒ F˙a ˙b ˙c = ˙a ˙b ˙cf(xa)

⇒ F˙a ˙b ˙c = 0

Aab→ Aab+ abcfc Ä

(25)

Equation of motion of Aab ⇒ Additional Gauge symmetry

cabc+ ∂˙cab ˙c≡ 0 ⇒ Aab → Aab+ Φab

Equations of motion of Aa ˙aand A˙a ˙b

δAa ˙a∶ ∂˙bFa ˙a ˙b= 0 ⇒ Fa ˙a ˙b=12abc˙a ˙b ˙c˙cΦbc

⇒ Fa ˙a ˙b= 0

Aab→ Aab+ Φab Ä¦

δA˙a ˙b∶ ∂aFa ˙a ˙b+ ∂˙aF˙a ˙b ˙c = 0 ⇒ F˙a ˙b ˙c = ˙a ˙b ˙cf(xa)

⇒ F˙a ˙b ˙c = 0

Aab→ Aab+ abcfc Ä

(26)

Known Examples: D = 3 + 3

Equation of motion of Aab ⇒ Additional Gauge symmetry

cabc+ ∂˙cab ˙c≡ 0 ⇒ Aab → Aab+ Φab

Equations of motion of Aa ˙aand A˙a ˙b

δAa ˙a∶ ∂˙bFa ˙a ˙b= 0 ⇒ Fa ˙a ˙b=12abc˙a ˙b ˙c˙cΦbc

⇒ Fa ˙a ˙b= 0

Aab→ Aab+ Φab Ä ¦

δA˙a ˙b∶ ∂aFa ˙a ˙b+ ∂˙aF˙a ˙b ˙c = 0 ⇒ F˙a ˙b ˙c = ˙a ˙b ˙cf(xa)

⇒ F˙a ˙b ˙c = 0

Aab→ Aab+ abcfc Ä

(27)

1 Introduction 2 D= 2 + 4?

3 General Formulations in D= D+ D′′

4 Interaction with Charged Branes 5 Conclusion

(28)

D = 2 + 4?

Decomposition of 6D Minkowski space R1+5= R1+1× R4 Gauge potential components

Aab, Aa ˙a, A˙a ˙b (a, b = 1, 2; ˙a, ˙b = ˙1, ⋯, ˙4)

Field Strength and Its Dual

Fab ˙a= 2∂[aAb] ˙a+ ∂˙aAab, F˜ab ˙a=3!1ab˙a ˙b ˙c ˙dF˙b ˙c ˙d ab˙a ˙b ˙c ˙d= ab ˙a ˙b ˙c ˙d

Fa ˙a ˙b= ∂aA˙a ˙b− 2∂[ ˙a∣Aa∣ ˙b], F˜a ˙a ˙b=2!1ab˙a ˙b ˙c ˙dFb ˙c ˙d F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c=2!1ab˙a ˙b ˙c ˙dFab ˙d

Action S2+4= − 1

12 ∫ d6x[FµνλFµνλ− 3Fab ˙cFab ˙c− 3Fa ˙b ˙cFa ˙b ˙c]

(29)

R1+5= R1+1× R4 Gauge potential components

Aab, Aa ˙a, A˙a ˙b (a, b = 1, 2; ˙a, ˙b = ˙1, ⋯, ˙4)

Field Strength and Its Dual

Fab ˙a= 2∂[aAb] ˙a+ ∂˙aAab, F˜ab ˙a=3!1ab˙a ˙b ˙c ˙dF˙b ˙c ˙d ab˙a ˙b ˙c ˙d= ab ˙a ˙b ˙c ˙d

Fa ˙a ˙b= ∂aA˙a ˙b− 2∂[ ˙a∣Aa∣ ˙b], F˜a ˙a ˙b=2!1ab˙a ˙b ˙c ˙dFb ˙c ˙d F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c=2!1ab˙a ˙b ˙c ˙dFab ˙d

Action S2+4= − 1

12 ∫ d6x[FµνλFµνλ− 3Fab ˙cFab ˙c− 3Fa ˙b ˙cFa ˙b ˙c]

(30)

D = 2 + 4?

Decomposition of 6D Minkowski space R1+5= R1+1× R4 Gauge potential components

Aab, Aa ˙a, A˙a ˙b (a, b = 1, 2; ˙a, ˙b = ˙1, ⋯, ˙4)

Field Strength and Its Dual

Fab ˙a= 2∂[aAb] ˙a+ ∂˙aAab, F˜ab ˙a=3!1ab˙a ˙b ˙c ˙dF˙b ˙c ˙d ab˙a ˙b ˙c ˙d= ab ˙a ˙b ˙c ˙d

Fa ˙a ˙b= ∂aA˙a ˙b− 2∂[ ˙a∣Aa∣ ˙b], F˜a ˙a ˙b=2!1ab˙a ˙b ˙c ˙dFb ˙c ˙d F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c=2!1ab˙a ˙b ˙c ˙dFab ˙d

Action S2+4= − 1

12 ∫ d6x[FµνλFµνλ− 3Fab ˙cFab ˙c− 3Fa ˙b ˙cFa ˙b ˙c]

(31)

R1+5= R1+1× R4 Gauge potential components

Aab, Aa ˙a, A˙a ˙b (a, b = 1, 2; ˙a, ˙b = ˙1, ⋯, ˙4)

Field Strength and Its Dual

Fab ˙a= 2∂[aAb] ˙a+ ∂˙aAab, F˜ab ˙a=3!1ab˙a ˙b ˙c ˙dF˙b ˙c ˙d ab˙a ˙b ˙c ˙d= ab ˙a ˙b ˙c ˙d

Fa ˙a ˙b= ∂aA˙a ˙b− 2∂[ ˙a∣Aa∣ ˙b], F˜a ˙a ˙b=2!1ab˙a ˙b ˙c ˙dFb ˙c ˙d F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c=2!1ab˙a ˙b ˙c ˙dFab ˙d

Action S2+4= − 1

12 ∫ d6x[FµνλFµνλ− 3Fab ˙cFab ˙c− 3Fa ˙b ˙cFa ˙b ˙c]

(32)

D = 2 + 4?

Equation of motion of Aab ⇒ Additional Gauge symmetry

˙aab ˙a≡ 0 ⇒ Aab → Aab+ Ωab

Equations of motion of Aa ˙a

˙bFa ˙a ˙b= 0 ⇒ Fa ˙a ˙b= ab˙a ˙b ˙c ˙d˙cΦb ˙d

⇒ Fa ˙a ˙b= ∂˙aΦa ˙b− ∂˙bΦa ˙a

⇒ ∂˙aΦa ˙b− ∂˙bΦa ˙a= ab˙a ˙b ˙c ˙d˙cΦb ˙d

⇒ ∂˙b˙bΦa ˙a− ∂˙a˙bΦa ˙b = 0 The differential equation of Φ is invariant under the transformation

Φa ˙a→ Φa ˙a+ ∂˙aΛa

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˙aab ˙a≡ 0 ⇒ Aab → Aab+ Ωab

Equations of motion of Aa ˙a

˙bFa ˙a ˙b= 0 ⇒ Fa ˙a ˙b= ab˙a ˙b ˙c ˙d˙cΦb ˙d

⇒ Fa ˙a ˙b= ∂˙aΦa ˙b− ∂˙bΦa ˙a

⇒ ∂˙aΦa ˙b− ∂˙bΦa ˙a= ab˙a ˙b ˙c ˙d˙cΦb ˙d

⇒ ∂˙b˙bΦa ˙a− ∂˙a˙bΦa ˙b = 0 The differential equation of Φ is invariant under the transformation

Φa ˙a→ Φa ˙a+ ∂˙aΛa

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D = 2 + 4?

Equation of motion of Aab ⇒ Additional Gauge symmetry

˙aab ˙a≡ 0 ⇒ Aab → Aab+ Ωab

Equations of motion of Aa ˙a

˙bFa ˙a ˙b= 0 ⇒ Fa ˙a ˙b= ab˙a ˙b ˙c ˙d˙cΦb ˙d

⇒ Fa ˙a ˙b= ∂˙aΦa ˙b− ∂˙bΦa ˙a

⇒ ∂˙aΦa ˙b− ∂˙bΦa ˙a= ab˙a ˙b ˙c ˙d˙cΦb ˙d

⇒ ∂˙b˙bΦa ˙a− ∂˙a˙bΦa ˙b = 0 The differential equation of Φ is invariant under the transformation

Φa ˙a→ Φa ˙a+ ∂˙aΛa

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˙aab ˙a≡ 0 ⇒ Aab → Aab+ Ωab

Equations of motion of Aa ˙a

˙bFa ˙a ˙b= 0 ⇒ Fa ˙a ˙b= ab˙a ˙b ˙c ˙d˙cΦb ˙d

⇒ Fa ˙a ˙b= ∂˙aΦa ˙b− ∂˙bΦa ˙a

⇒ ∂˙aΦa ˙b− ∂˙bΦa ˙a= ab˙a ˙b ˙c ˙d˙cΦb ˙d

⇒ ∂˙b˙bΦa ˙a− ∂˙a˙bΦa ˙b = 0 The differential equation of Φ is invariant under the transformation

Φa ˙a→ Φa ˙a+ ∂˙aΛa

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D = 2 + 4?

Equation of motion of Aab ⇒ Additional Gauge symmetry

˙aab ˙a≡ 0 ⇒ Aab → Aab+ Ωab

Equations of motion of Aa ˙a

˙bFa ˙a ˙b= 0 ⇒ Fa ˙a ˙b= ab˙a ˙b ˙c ˙d˙cΦb ˙d

⇒ Fa ˙a ˙b= ∂˙aΦa ˙b− ∂˙bΦa ˙a

⇒ ∂˙aΦa ˙b− ∂˙bΦa ˙a= ab˙a ˙b ˙c ˙d˙cΦb ˙d

⇒ ∂˙b˙bΦa ˙a− ∂˙a˙bΦa ˙b = 0 The differential equation of Φ is invariant under the transformation

Φa ˙a→ Φa ˙a+ ∂˙aΛa

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Choose the Lorentz gauge

˙aΦa ˙a= 0 ⇒ ˙∂2Φa ˙a= 0, where ˙∂2≡ ∂˙a˙a Impose the boundary condition Φa ˙ax˙a→∞= 0

Φa ˙a= 0 ⇒ Fa ˙a ˙b= 0 Equation of motion A˙a ˙b

˙c(F + F)˙a ˙b ˙c+ ∂aFa ˙a ˙b= 0 ⇒ F˙a ˙b ˙c =1

2ab˙a ˙b ˙c ˙dd˙ab

⇒ F˙a ˙b ˙c = 0

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D = 2 + 4?

Choose the Lorentz gauge

˙aΦa ˙a= 0 ⇒ ˙∂2Φa ˙a= 0, where ˙∂2≡ ∂˙a˙a Impose the boundary condition Φa ˙ax˙a→∞= 0

Φa ˙a= 0 ⇒ Fa ˙a ˙b= 0 Equation of motion A˙a ˙b

˙c(F + F)˙a ˙b ˙c+ ∂aFa ˙a ˙b= 0 ⇒ F˙a ˙b ˙c =1

2ab˙a ˙b ˙c ˙dd˙ab

⇒ F˙a ˙b ˙c = 0

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1 Introduction 2 D= 2 + 4?

3 General Formulations in D= D+ D′′

4 Interaction with Charged Branes 5 Conclusion

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Setup of the General Formulations

Product space of Minkowski space MD=MD1×MD2′′

Spacetime coordinates{xµ∣µ = 1, ⋯, D}:

MD1 ∶ {xa∣a = 1, ⋯, D} and MD2′′ ∶ {x˙a∣ ˙a = 1, ⋯, D′′}. D≤ D′′⇐⇒ 1 ≤ D≤ D/2

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{Aa1...aj˙a1... ˙a2p−j∣j = 0, . . . , D− δD,D/2,} Componets of Field Strength F and its dual ˜F

(D− j)!(D2 − D+ j)!

for j= 0, 1, ⋯, D

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Action of the General Formulations

SD+D′′ = −1

4 ∫ dDx

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎣

Fµ1µ2...µD/2Fµ1µ2...µD/2 (D/2)!

D

j=⌈D′2

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎦

= −1

2 ∫ dDx

D

j=⌈D′2

a

= −1

2 ∫ dDx

D

j=⌈D′2

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The self-duality conditionFµ1⋯µD/2 = 0 can be decomposed into D+ 1 different types of equations

Fa1⋯aj˙a1⋯ ˙aD/2−j = 0 (j = 0, 1, ⋯, D).

Thus, in general , the equations of motion can not be all independent (except for the special case(D,D′′) = (1, 1)).

# of equations of motion that are trivial should be

(D− δD,D/2+ 1) − ⌈(D+ 1)/2⌉ ⇒ # of adt. gauge sym.

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Equations of Motion and Self-Duality Conditions

The way from equations of motion to self-duality conditions:

1 Solve the e.o.m. of Aa1...aD′−δ

D′ ,D/2˙a1... ˙a2p−D′−δ

D′ ,D/2 by using gauge symmetries (D=odd) or boundary condition (D=even)

2 Substitute obtained self-duality conditions into one another e.o.m. of gauge fields and use Bianchi identity to rewrite e.o.m

3 Solve the substituted e.o.m. by using gauge symmetries If D=even, for obtaining all self-duality conditions, MD1

has to be Lorentzian andMD2′′ has to be Euclidean

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The action is manifestly invariant under the

SO(D− 1, 1) × SO(D′′), or SO(D) × SO(D′′− 1, 1) subgroup of the full Lorentz group SO(D − 1, 1)

The Lorentz symmetry transformations which mix xawith x˙aare no longer manifest.

Is the action S actually fully invariant under a Lorentz transformation rule for those transformations which mix xa with x˙a?

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Hidden Lorentz Symmetry

Modified Lorentz transformation rule δAa

1...ak˙a1... ˙a2p−k ≡ δ1Aa

1...ak˙a1... ˙a2p−k+ δ2Aa

1...ak˙a1... ˙a2p−k

where

δ1Aa1...ak˙a1... ˙a2p−k = kλ[a˙bkAa

1...ak−1] ˙b ˙a1... ˙a2p−k

−(2p − k)λb[ ˙a1A∣a

1...akb∣ ˙a2... ˙a2p−k−1]

b ˙b(xb˙b− x˙bb)Aa1...ak˙a1... ˙a2p−k

δ2Aa

1...ak˙a1... ˙a2p−k = δ⌊D/2⌋,kλb˙bx˙bFba1...ak˙a1... ˙a2p−k

k= 0, 1, ⋯, ⌊D/2⌋, λa ˙a: Lorentz trasf. parameter

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1 Introduction 2 D= 2 + 4?

3 General Formulations in D= D+ D′′

4 Interaction with Charged Branes 5 Conclusion

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A 2p-form gauge potential naturally couples to a

(2p − 1)-brane with a 2p dimensional worldvolume as the electric charge.

The interaction between the(2p − 1)-brane and the field should preserve all the gauge symmetry. Especially, the additional gauge symmetry are needed in order to have first order differential equations as equations of motion.

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Interaction term of the action

SDint+D′′ = − ∫ dDx

D

j=⌈D′2

Q = σ − ˜σ σ: source

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E.O.M. of Interaction Term

(F + Q)a1⋯ak˙a1⋯ ˙a2p−k˙b= 0

⇒ ∂˙b(F + σ)a1⋯ak˙a1⋯ ˙a2p−k˙b+ ∂b(˜F + ˜σ)ba1⋯ak˙a1⋯ ˙a2p−k

= ∂µσ˜µa1⋯ak˙a1⋯ ˙a2p−k

Although it appears that the effect of the source term is merely to replace F by F + σ in both equations of motion, one can not redefine F to absorb σ.

F and σ are both(2p + 1)-forms, F is a closed differential form while σ is not.

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D= D′′

µ˜σx1⋯xkx˙1⋯ ˙x2p−kµ= ∏2p+1`=k +1δ(x`) ∏2p+1`=2p−k +1δ( ˙x`) (0 ≤ k ≤ p)

D= odd

µ˜σx1⋯xkx˙1⋯ ˙x2p−kµ= ∏2q+1`=k +1δ(x`) ∏4p−2q+1`=2p−k +1δ( ˙x`) (0 ≤ k ≤ q)

D= even

µ˜σx1⋯xkx˙1⋯ ˙x2p−kµ= ∏2r +2`=k +1δ(x`) ∏4p−2r`=2p−k +1δ( ˙x`) (1 ≤ k ≤ r + 1)

Σ

x1...xk x.1...x2p-k

{x , x }l .l

.

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Examples of Source Configuration

D= D′′

µ˜σx1⋯xkx˙1⋯ ˙x2p−kµ= ∏2p+1`=k +1δ(x`) ∏2p+1`=2p−k +1δ( ˙x`) (0 ≤ k ≤ p)

D= odd

µ˜σx1xkx˙1⋯ ˙x2p−kµ= ∏2q+1`=k +1δ(x`) ∏4p−2q+1`=2p−k +1δ( ˙x`) (0 ≤ k ≤ q)

D= even

µ˜σx1⋯xkx˙1⋯ ˙x2p−kµ= ∏2r +2`=k +1δ(x`) ∏4p−2r`=2p−k +1δ( ˙x`) (1 ≤ k ≤ r + 1)

The x`’s and ˙x`’s appearing in the delta functions (there are 2p+ 2 of them) are the coordinates transverse to the brane.

k is a fixed number used to specify how the

2p-dimensional worldvolume of the source brane is divided intoMD1× MD2′′

We use the divergence of ˜σ to define σ, with all other components of the divergence of ˜σvanishing

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