Exploring Duality in $$S_{F,\Lambda}$$

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In summary, the author introduces a parent action and derives the equations of motion for the fields $F$ and $B$. By substituting the equation of motion for $F$ back into the action, the author obtains a dual theory with the coupling constants inverted. The author also explains the origin of this relation and shows how it leads to the final action of the dual theory.
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PhyAmateur
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(This question is only long because I tried to give all the details (necessary and not) so that you don't refer to the paper every now and then)

In this paper http://arxiv.org/pdf/hep-th/9705122.pdf

We have $$S_A = \frac{1}{4g^2} \int{d^4x F_{\mu\nu}(A)F^{\mu\nu}(A)}$$
where $$F_{\mu\nu}(A) = \partial_{[\mu A\nu]}$$. Its Bianchi Identity is $$\partial_\mu *F^{\mu\nu}$$ (Note that (*) represents Hodge Dual)

Great. Now the author went to parent action:
$$S_{F,\Lambda} = \int{d^4x(\frac{1}{4g^2} F_{\mu\nu} F^{\mu\nu} +a \Lambda_\mu \partial_\nu *F^{\nu\mu}} )$$

He first varied it w.r.t $$\Lambda_\mu$$ and then w.r.t $$F_{\mu\nu}$$.

1)He got in the first case, $$\partial_\mu *F^{\mu\nu} = 0$$ and thus he mentioned that our parent action reduces to $$S_A$$

2)He got in the second case, $$ \frac{1}{2g^2} F^{\mu\nu} = \frac{a}{2} \partial_\rho \Lambda_\sigma \epsilon^{\rho \sigma \mu \nu}= \frac{a}{2} *G^{\mu\nu} $$ and thus he mentioned that now *plugging this back into the action*:
$$S_{F,\Lambda} \rightarrow S_{\Lambda} = \frac{-g^2a^2}{4} \int{d^4x *G_{\mu\nu} *G^{\mu\nu}}$$
He then said knowing that $$*G_{\mu\nu} *G^{\mu\nu}=-2G_{\mu\nu} G^{\mu\nu} $$ We obtain perfectly $$ S_\Lambda = \frac{g^2}{4} \int{d^4x G_{\mu\nu}(\Lambda)G^{\mu\nu}(\Lambda)}$$And so this is duality with the coupling constants inversed. Perfect.

My questions:

A) When he said plugging this back into the action above (in italic). He plugged it in the first term of the parent action. What about the second term? Did he throw it away?

B) $$\frac{1}{2g^2} F^{\mu\nu} = \frac{a}{2} \partial_\rho \Lambda_\sigma \epsilon^{\rho \sigma \mu \nu}= \frac{a}{2} *G^{\mu\nu}$$ Where did this relation come from (The first and the second equality)?

Thank you for taking the time to read this.
 
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  • #2
PhyAmateur said:
My questions:

A) When he said plugging this back into the action above (in italic). He plugged it in the first term of the parent action. What about the second term? Did he throw it away?

B) $$\frac{1}{2g^2} F^{\mu\nu} = \frac{a}{2} \partial_\rho \Lambda_\sigma \epsilon^{\rho \sigma \mu \nu}= \frac{a}{2} *G^{\mu\nu}$$ Where did this relation come from (The first and the second equality)?

Thank you for taking the time to read this.


Let us start from the first order action
[tex]S [ F , B ] = \int d^{ 4 } x \left( \frac{ 1 }{ 4 g^{ 2 } } F^{ 2 } + a B_{ \sigma } \partial_{ \rho } *F^{ \rho \sigma } \right) .[/tex]
Before varying with respect to [itex]F_{ \mu \nu }[/itex], we first integrate the second term by part to obtain:
[tex]S [ F , B ] = \int d^{ 4 } x \left( \frac{ 1 }{ 4 g^{ 2 } } F^{ 2 } - a ( \partial_{ \rho } B_{ \sigma } ) *F^{ \rho \sigma } \right) , \ \ \ \ (1)[/tex]
and then substitute for
[tex]*F^{ \rho \sigma } = \frac{ 1 }{ 2 } \epsilon^{ \rho \sigma \mu \nu } \ F_{ \mu \nu } .[/tex]
So, [itex]\delta_{ F } S [ F , B ] = 0[/itex], leads to
[tex]\int d^{ 4 } x \left( \frac{ 1 }{ 2 g^{ 2 } } F^{ \mu \nu } - \frac{ a }{ 2 } ( \partial_{ \rho } B_{ \sigma } ) \epsilon^{ \rho \sigma \mu \nu } \right) \delta F_{ \mu \nu } = 0 .[/tex]
From this it follows that
[tex]F^{ \mu \nu } = a g^{ 2 } ( \partial_{ \rho } B_{ \sigma } ) \ \epsilon^{ \rho \sigma \mu \nu } . \ \ \ \ \ (2)[/tex]
We can rewrite this as
[tex]F^{ \mu \nu } = a g^{ 2 } ( \frac{ 1 }{ 2 } ) \ \left( \partial_{ \rho } B_{ \sigma } - \partial_{ \sigma } B_{ \rho } \right) \epsilon^{ \rho \sigma \mu \nu } . \ \ \ (2a)[/tex]
Introducing the field tensor [itex]G_{ \rho \sigma } \equiv \partial_{ \rho } B_{ \sigma } - \partial_{ \sigma } B_{ \rho }[/itex] and its dual [itex]*G^{ \mu \nu } \equiv ( 1 / 2 ) G_{ \rho \sigma } \ \epsilon^{ \rho \sigma \mu \nu }[/itex] into Eq(2a), we find
[tex]F^{ \mu \nu } = \frac{ a \ g^{ 2 } }{ 2 } \ G_{ \rho \sigma } \ \epsilon^{ \rho \sigma \mu \nu } = a g^{ 2 } \ *G^{ \mu \nu } .[/tex]
For later use, let us lower the indices of the field F in the equation of motion Eq(2)
[tex]F_{ \mu \nu } = a \ g^{ 2 } \ ( \partial_{ \rho } B_{ \sigma } ) \ \epsilon^{ \rho \sigma \alpha \beta } \ \eta_{ \mu \alpha } \ \eta_{ \nu \beta } . \ \ \ \ (2b)[/tex]
Now, contracting Eq(2) with [itex]F_{ \mu \nu }[/itex] gives
[tex]F^{ 2 } = 2 \ a \ g^{ 2 } \ ( \partial_{ \rho } B_{ \sigma } ) \ ( \frac{ 1 }{ 2 } \epsilon^{ \rho \sigma \mu \nu } \ F_{ \mu \nu } ) = 2 \ a \ g^{ 2 } \ ( \partial_{ \rho } B_{ \sigma } ) \ \ *F^{ \rho \sigma } .[/tex]
Thus
[tex]\frac{ 1 }{ 4 \ g^{ 2 } } \ F^{ 2 } = \frac{ a }{ 2 } \ ( \partial_{ \rho } B_{ \sigma } ) \ \ *F^{ \rho \sigma } . \ \ \ \ (3)[/tex]
Now, inserting Eq(3) in Eq(1) gives
[tex]S [ F , B ] = - \frac{ a }{ 2 } \int d^{ 4 } x \ ( \partial_{ \rho } B_{ \sigma } ) \ \ *F^{ \rho \sigma } = - \frac{ a }{ 4 } \int d^{ 4 } x \left( \partial_{ \rho } B_{ \sigma } \ \epsilon^{ \rho \sigma \mu \nu } \right) \ F_{ \mu \nu } .[/tex]
Using the equation of motion Eq(2b), we find (the dual theory action)
[tex]S [ B ] = - \frac{ a^{ 2 } g^{ 2 } }{ 4 } \int d^{ 4 } x \left( \partial_{ \rho } B_{ \sigma } \ \epsilon^{ \rho \sigma \mu \nu } \right) \left( \partial_{ \lambda } B_{ \tau } \ \epsilon^{ \lambda \tau \alpha \beta } \right) \eta_{ \mu \alpha } \ \eta_{ \nu \beta } .[/tex]
In terms of the field tensor [itex]G_{ \mu \nu }(B)[/itex], the above action becomes
[tex]S [ B ] = - \frac{ a^{ 2 } g^{ 2 } }{ 4 } \int d^{ 4 } x \left( \frac{ 1 }{ 2 } G_{ \rho \sigma } \ \epsilon^{ \rho \sigma \mu \nu } \right) \left( \frac{ 1 }{ 2 } G_{ \lambda \tau } \ \epsilon^{ \lambda \tau \alpha \beta } \right) \ \eta_{ \mu \alpha } \ \eta_{ \nu \beta } .[/tex]
Or
[tex]S [ B ] = - ( \frac{ a \ g }{ 4 } )^{ 2 } \int d^{ 4 } x \left( \epsilon^{ \rho \sigma \mu \nu } \ \epsilon^{ \lambda \tau }{}_{ \mu \nu } \right) G_{ \rho \sigma } \ G_{ \lambda \tau } .[/tex]
Using the identity
[tex]\epsilon^{ \rho \sigma \mu \nu } \ \epsilon^{ \lambda \tau }{}_{ \mu \nu } = ( - 2 ) ( \eta^{ \rho \lambda } \ \eta^{ \sigma \tau } - \eta^{ \rho \tau } \ \eta^{ \sigma \lambda } ) ,[/tex]
we find
[tex]S [ B ] = - ( \frac{ a \ g }{ 4 } )^{ 2 } \ ( - 2 ) \int d^{ 4 } x \ 2 \ G_{ \rho \sigma } \ G^{ \rho \sigma } .[/tex]
Or
[tex]S [ B ] = \frac{ g^{ 2 } }{ 4 } \int d^{ 4 } x \ ( a G_{ \mu \nu } ) \ ( a G^{ \mu \nu } ) .[/tex]
Finally, by rescaling the field [itex]a G_{ \mu \nu } ( B ) \equiv \mathcal{ G }_{ \mu \nu } ( B )[/itex], we obtain the action of the dual theory
[tex]S [ B ] = \frac{ g^{ 2 } }{ 4 } \int d^{ 4 } x \ \mathcal{ G }^{ 2 } ( B ) .[/tex]
 
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That was brilliant explanation. Thank you a lot!
 

1. What is the concept of duality in $$S_{F,\Lambda}$$?

The concept of duality in $$S_{F,\Lambda}$$ refers to the idea that there are two distinct ways to represent the same mathematical object. In this case, the mathematical object is a function $$F$$ defined on a set $$\Lambda$$. The duality arises because $$F$$ can be represented either as a function on $$\Lambda$$ or as a linear combination of basis functions, known as the Fourier series.

2. How is duality used in $$S_{F,\Lambda}$$?

In $$S_{F,\Lambda}$$, duality is used to simplify and generalize mathematical concepts. By representing a function in two different ways, we can make connections between seemingly different mathematical ideas and use one approach to prove results in the other. Duality also allows for more efficient and elegant solutions to problems by using the properties of one representation to solve problems in the other representation.

3. What are some examples of duality in $$S_{F,\Lambda}$$?

One example of duality in $$S_{F,\Lambda}$$ is the Fourier transform, which is a method for converting a function from its representation in time or space to its representation in the frequency domain. Another example is the duality between continuous and discrete functions, where a continuous function can be approximated by a discrete set of values and vice versa.

4. How does duality affect our understanding of functions in $$S_{F,\Lambda}$$?

Duality enhances our understanding of functions in $$S_{F,\Lambda}$$ by providing multiple perspectives and representations of the same mathematical object. This allows us to see connections between seemingly different concepts and provides a more comprehensive understanding of the function and its properties. Duality also allows for more efficient and elegant solutions to problems by utilizing the properties of one representation to solve problems in the other representation.

5. Can duality be applied to other areas of mathematics?

Yes, duality is a fundamental concept in mathematics and can be applied to various areas, such as algebra, geometry, and topology. Many mathematical structures and objects have dual representations, and understanding the duality between them can lead to new insights and connections within different branches of mathematics. Duality is also a crucial tool in mathematical physics, where it is used to bridge the gap between abstract mathematical concepts and real-world phenomena.

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