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
Outline
1 Introduction 2 D= 2 + 4?
3 General Formulations in D= D′+ D′′
4 Interaction with Charged Branes 5 Conclusion
1 Introduction 2 D= 2 + 4?
3 General Formulations in D= D′+ D′′
4 Interaction with Charged Branes 5 Conclusion
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
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.
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
⋆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
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.
∼ ∫ 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.
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.
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!11 ˙a ˙b ˙c ˙d ˙eF˙c ˙d ˙e F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c=2!11 ˙a ˙b ˙c ˙d ˙eF1 ˙d ˙e
Action
S1+5= − 1
12 ∫ d6x[FµνλFµνλ− 3F1 ˙a ˙bF1 ˙a ˙b]
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!11 ˙a ˙b ˙c ˙d ˙eF˙c ˙d ˙e F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c=2!11 ˙a ˙b ˙c ˙d ˙eF1 ˙d ˙e
Action
S1+5= − 1
12 ∫ d6x[FµνλFµνλ− 3F1 ˙a ˙bF1 ˙a ˙b]
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!11 ˙a ˙b ˙c ˙d ˙eF˙c ˙d ˙e F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c=2!11 ˙a ˙b ˙c ˙d ˙eF1 ˙d ˙e
Action
S1+5= − 1
12 ∫ d6x[FµνλFµνλ− 3F1 ˙a ˙bF1 ˙a ˙b]
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!11 ˙a ˙b ˙c ˙d ˙eF˙c ˙d ˙e F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c=2!11 ˙a ˙b ˙c ˙d ˙eF1 ˙d ˙e
Action
S1+5= − 1
12 ∫ d6x[FµνλFµνλ− 3F1 ˙a ˙bF1 ˙a ˙b]
Equation of motion of A1 ˙a⇒ Additional Gauge Symmetry
∂˙bF˜1 ˙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 ˙e∂d˙Φ1 ˙e
⇒ F˙a ˙b ˙c= 0
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 ˙e∂d˙Φ1 ˙e If we define
F1 ˙d ˙e≡ ∂1Ad ˙e˙ − ∂d˙Φ1 ˙e+ ∂˙eΦ1 ˙d⇒ Fa ˙d ˙e= 0
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!1abc˙a ˙b ˙cFc ˙b ˙c Fa ˙a ˙b= ∂aA˙a ˙b− 2∂[ ˙a∣Aa∣ ˙b], F˜a ˙a ˙b=2!1abc˙a ˙b ˙cFbc ˙c F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c= −3!1abc˙a ˙b ˙cFabc
Action S3+3= − 1
12 ∫ d6x[FµνλFµνλ− FabcFabc− 3Fab ˙cFab ˙c]
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!1abc˙a ˙b ˙cFc ˙b ˙c Fa ˙a ˙b= ∂aA˙a ˙b− 2∂[ ˙a∣Aa∣ ˙b], F˜a ˙a ˙b=2!1abc˙a ˙b ˙cFbc ˙c F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c= −3!1abc˙a ˙b ˙cFabc
Action S3+3= − 1
12 ∫ d6x[FµνλFµνλ− FabcFabc− 3Fab ˙cFab ˙c]
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!1abc˙a ˙b ˙cFc ˙b ˙c Fa ˙a ˙b= ∂aA˙a ˙b− 2∂[ ˙a∣Aa∣ ˙b], F˜a ˙a ˙b=2!1abc˙a ˙b ˙cFbc ˙c F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c= −3!1abc˙a ˙b ˙cFabc
Action S3+3= − 1
12 ∫ d6x[FµνλFµνλ− FabcFabc− 3Fab ˙cFab ˙c]
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!1abc˙a ˙b ˙cFc ˙b ˙c Fa ˙a ˙b= ∂aA˙a ˙b− 2∂[ ˙a∣Aa∣ ˙b], F˜a ˙a ˙b=2!1abc˙a ˙b ˙cFbc ˙c F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c= −3!1abc˙a ˙b ˙cFabc
Action S3+3= − 1
12 ∫ d6x[FµνλFµνλ− FabcFabc− 3Fab ˙cFab ˙c]
Equation of motion of Aab ⇒ Additional Gauge symmetry
∂cF˜abc+ ∂˙cF˜ab ˙c≡ 0 ⇒ Aab → Aab+ Φab
Equations of motion of Aa ˙aand A˙a ˙b
δAa ˙a∶ ∂˙bFa ˙a ˙b= 0 ⇒ Fa ˙a ˙b=12abc˙a ˙b ˙c∂˙cΦbc
⇒ Fa ˙a ˙b= 0
Aab→ Aab+ Φab Ħ
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
Known Examples: D = 3 + 3
Equation of motion of Aab ⇒ Additional Gauge symmetry
∂cF˜abc+ ∂˙cF˜ab ˙c≡ 0 ⇒ Aab → Aab+ Φab
Equations of motion of Aa ˙aand A˙a ˙b
δAa ˙a∶ ∂˙bFa ˙a ˙b= 0 ⇒ Fa ˙a ˙b=12abc˙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
Equation of motion of Aab ⇒ Additional Gauge symmetry
∂cF˜abc+ ∂˙cF˜ab ˙c≡ 0 ⇒ Aab → Aab+ Φab
Equations of motion of Aa ˙aand A˙a ˙b
δAa ˙a∶ ∂˙bFa ˙a ˙b= 0 ⇒ Fa ˙a ˙b=12abc˙a ˙b ˙c∂˙cΦbc
⇒ Fa ˙a ˙b= 0
Aab→ Aab+ Φab Ħ
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
Known Examples: D = 3 + 3
Equation of motion of Aab ⇒ Additional Gauge symmetry
∂cF˜abc+ ∂˙cF˜ab ˙c≡ 0 ⇒ Aab → Aab+ Φab
Equations of motion of Aa ˙aand A˙a ˙b
δAa ˙a∶ ∂˙bFa ˙a ˙b= 0 ⇒ Fa ˙a ˙b=12abc˙a ˙b ˙c∂˙cΦbc
⇒ Fa ˙a ˙b= 0
Aab→ Aab+ Φab Ħ
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
Equation of motion of Aab ⇒ Additional Gauge symmetry
∂cF˜abc+ ∂˙cF˜ab ˙c≡ 0 ⇒ Aab → Aab+ Φab
Equations of motion of Aa ˙aand A˙a ˙b
δAa ˙a∶ ∂˙bFa ˙a ˙b= 0 ⇒ Fa ˙a ˙b=12abc˙a ˙b ˙c∂˙cΦbc
⇒ Fa ˙a ˙b= 0
Aab→ Aab+ Φab Ħ
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
Known Examples: D = 3 + 3
Equation of motion of Aab ⇒ Additional Gauge symmetry
∂cF˜abc+ ∂˙cF˜ab ˙c≡ 0 ⇒ Aab → Aab+ Φab
Equations of motion of Aa ˙aand A˙a ˙b
δAa ˙a∶ ∂˙bFa ˙a ˙b= 0 ⇒ Fa ˙a ˙b=12abc˙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
1 Introduction 2 D= 2 + 4?
3 General Formulations in D= D′+ D′′
4 Interaction with Charged Branes 5 Conclusion
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!1ab˙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!1ab˙a ˙b ˙c ˙dFb ˙c ˙d F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c=2!1ab˙a ˙b ˙c ˙dFab ˙d
Action S2+4= − 1
12 ∫ d6x[FµνλFµνλ− 3Fab ˙cFab ˙c− 3Fa ˙b ˙cFa ˙b ˙c]
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!1ab˙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!1ab˙a ˙b ˙c ˙dFb ˙c ˙d F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c=2!1ab˙a ˙b ˙c ˙dFab ˙d
Action S2+4= − 1
12 ∫ d6x[FµνλFµνλ− 3Fab ˙cFab ˙c− 3Fa ˙b ˙cFa ˙b ˙c]
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!1ab˙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!1ab˙a ˙b ˙c ˙dFb ˙c ˙d F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c=2!1ab˙a ˙b ˙c ˙dFab ˙d
Action S2+4= − 1
12 ∫ d6x[FµνλFµνλ− 3Fab ˙cFab ˙c− 3Fa ˙b ˙cFa ˙b ˙c]
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!1ab˙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!1ab˙a ˙b ˙c ˙dFb ˙c ˙d F˙a ˙b ˙c= 3∂[ ˙aA˙b ˙c], F˜˙a ˙b ˙c=2!1ab˙a ˙b ˙c ˙dFab ˙d
Action S2+4= − 1
12 ∫ d6x[FµνλFµνλ− 3Fab ˙cFab ˙c− 3Fa ˙b ˙cFa ˙b ˙c]
D = 2 + 4?
Equation of motion of Aab ⇒ Additional Gauge symmetry
∂˙aF˜ab ˙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
∂˙aF˜ab ˙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
D = 2 + 4?
Equation of motion of Aab ⇒ Additional Gauge symmetry
∂˙aF˜ab ˙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
∂˙aF˜ab ˙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
D = 2 + 4?
Equation of motion of Aab ⇒ Additional Gauge symmetry
∂˙aF˜ab ˙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
Choose the Lorentz gauge
∂˙aΦa ˙a= 0 ⇒ ˙∂2Φa ˙a= 0, where ˙∂2≡ ∂˙a∂˙a Impose the boundary condition Φa ˙a∣x˙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
2ab˙a ˙b ˙c ˙d∂d˙Ωab
⇒ F˙a ˙b ˙c = 0
D = 2 + 4?
Choose the Lorentz gauge
∂˙aΦa ˙a= 0 ⇒ ˙∂2Φa ˙a= 0, where ˙∂2≡ ∂˙a∂˙a Impose the boundary condition Φa ˙a∣x˙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
2ab˙a ˙b ˙c ˙d∂d˙Ωab
⇒ F˙a ˙b ˙c = 0
1 Introduction 2 D= 2 + 4?
3 General Formulations in D= D′+ D′′
4 Interaction with Charged Branes 5 Conclusion
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
a1⋯aD′˙a1⋯ ˙aD′′ = a1⋯aD′˙a1⋯ ˙aD′′
{Aa1...aj˙a1... ˙a2p−j∣j = 0, . . . , D′− δD′,D/2,} Componets of Field Strength F and its dual ˜F
Fa1...aj˙a1... ˙aD/2−j
= j∂[a1Aa2...aj] ˙a1... ˙aD/2−j+ (D/2 − j)∂[ ˙aD/2−jA∣a1...aj∣ ˙a1... ˙aD/2−j−1], F˜a1...aj˙a1... ˙aD/2−j
= (−)(D2−j)(D′−j)a1...aD′˙a1... ˙aD−D′Faj+1...aD′˙aD/2−j+1... ˙aD−D′
(D′− j)!(D2 − D′+ j)!
for j= 0, 1, ⋯, D′
Action of the General Formulations
SD′+D′′ = −1
4 ∫ dDx
⎡⎢⎢⎢
⎢⎢⎢⎢
⎢⎢⎣
Fµ1µ2...µD/2Fµ1µ2...µD/2 (D/2)!
− D
′
∑
j=⌈D′2⌉
Fa1...aj˙a1... ˙aD/2−jFa1...aj˙a1... ˙aD/2−j j!(D2 − j)!
⎤⎥⎥⎥
⎥⎥⎥⎥
⎥⎥⎦
= −1
2 ∫ dDx
D′
∑
j=⌈D′2⌉
F˜a
1...aj˙a1... ˙aD/2−jFa1...aj˙a1... ˙aD/2−j j!(D2 − j)!2δj,D′/2
= −1
2 ∫ dDx
D′
∑
j=⌈D′2⌉
Fa1...aD′−j˙a1... ˙aD/2−D′+jFa1...aD′−j˙a1... ˙aD/2−D′+j (D′− j)!(D′′2−D′ + j)!2δj,D′/2
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.
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
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?
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]
+λb ˙b(xb∂˙b− x˙b∂b)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
1 Introduction 2 D= 2 + 4?
3 General Formulations in D= D′+ D′′
4 Interaction with Charged Branes 5 Conclusion
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.
Interaction term of the action
SDint′+D′′ = − ∫ dDx
D′
∑
j=⌈D′2⌉
Fa1...aD′−j˙a1... ˙aD/2−D′+jQa1...aD′−j˙a1... ˙aD/2−D′+j (D′− j)!(D′′2−D′ + j)!2δj,D′/2 ,
Q = σ − ˜σ σ: source
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.
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
.
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˜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)
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