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  AAA in Chern-Simons

+ 4 like - 0 dislike
2378 views

I am confused with the wedging operations of Lie algebra valued differential forms. Especially, for instance, I have some problems with the Chern-Simons 3-form

AdA+23AAA,

where A is a Lie algebra valued 1-form. My question is "how is the last term AAA defined?"

As far as I know, a lot of sources (e.g. Wedge Product of Lie Algebra Valued One-Form, http://en.wikipedia.org/wiki/Lie_algebra-valued_differential_form) define the wedge of Lie algebra valued 1-forms as follows.

[ωη](X1,,Xp+q):=(coefficient)×σSp+qsgn(σ)[ω(Xσ(1),,Xσ(p)),η(Xσ(p+1),,Xσ(p+q))],

where ω and η are Lie algebra valued p-form and q-form, respectively. The coefficient differs by authors. Other people utilises local description (e.g. http://math.stackexchange.com/questions/315235/reference-for-lie-algebra-valued-differential-forms, and also in the Wikipedia)

[ωη]=[ωaTa,ηbTb]:=ωaηb[Ta,Tb],

where Tc(c=1,,dimg) are generators of the Lie algebra g, and the implicit sums understood.

These definitions, as the notations suggest, force you to take Lie bracket explicitly. Therefore it is obvious that wedged one [ωη] is Lie algebra valued (p+q)-form.

Then what about wedged ones without brackets, such as AA,AAA?

I can show that AA is equivalent to [AA] up to coefficient, using either matrix representation, considering g=gl(n), or universal enveloping algebra. The basic idea is

AA=(AaTa)(AbTb)=(AaAb)TaTb.

This time, by graded commutation relation, the multiplication of generators can be converted to commutators. This seems ok. Then what about AAA? I could not convert it to an expression only using commutators of generators...

So, what I did was calculating [A[AA]], which gave zero. I am totally confused at this stage. Could you point out some pieces that I possibly keep missing??

This post imported from StackExchange MathOverflow at 2015-06-06 21:05 (UTC), posted by SE-user N. Shimode
asked May 27, 2015 in Mathematics by N. Shimode (30 points) [ no revision ]
retagged Jun 6, 2015

3 Answers

+ 5 like - 0 dislike

Option (1) Use the definition (ωS)(ηS)=(ωη)(ST) of the wedge product for Lie algebra valued forms. Define Lie bracket and Killing form as bilinear maps [ST]=[S,T] and ST=S,T. Then the formula that you want is A[AA],

where the commutator and Killing form apply only to the Lie algebra factors, ignoring the differential form factors.

Option (2) Use the definition (ωS)(ηS)=(ωη)ST of the wedge product of forms valued in a particular matrix representation of a Lie algebra. Then the formula that you want is tr(AAA),

where again the trace applies only to the matrix factors ignoring the differential form factors.

The two formulas agree up to a constant factor, as long as your Lie algebra is simple.

This post imported from StackExchange MathOverflow at 2015-06-06 21:05 (UTC), posted by SE-user Igor Khavkine
answered May 27, 2015 by Igor Khavkine (420 points) [ no revision ]
+ 4 like - 0 dislike

For Lie algebras of matrices (which is what you really care in Chern-Simmons theory) think of A as a form with matrix coefficients

A=iAidxi,

where Ai are r×r matrices.With this convention, use the usual wedge product

(iAidxi)(jAjdxj)(kAkdxk)=i,j,kAiAjAkdxidxjdxk,

where you need to recall that the product of matrices is not commutative.

This post imported from StackExchange MathOverflow at 2015-06-06 21:05 (UTC), posted by SE-user Liviu Nicolaescu
answered May 27, 2015 by Liviu Nicolaescu (110 points) [ no revision ]
Liviu, just want to note that your formula gives a matrix-valued 3-form. One still needs to take the trace of the matrix coefficients to get an ordinary 3-form that one could use as a Lagrangian density.

This post imported from StackExchange MathOverflow at 2015-06-06 21:05 (UTC), posted by SE-user Igor Khavkine
So, do you think that AAA does not have to be Lie algebra valued 3-form? Perhaps this is the point that I cannot understand.

This post imported from StackExchange MathOverflow at 2015-06-06 21:05 (UTC), posted by SE-user N. Shimode
When the notation is ambiguous, you have to be more precise about the context. If you want a formula for the Lagrangian density of the Chern-Simon's theory, then it cannot be Lie algebra valued.

This post imported from StackExchange MathOverflow at 2015-06-06 21:05 (UTC), posted by SE-user Igor Khavkine
+ 1 like - 0 dislike

This is not a direct answer to my question, but I think it is worth noting. The reason why AAA without trace should not be a Lie algebra valued 3-form is as follows.

Suppose AAA be a well-defined Lie algebra valued 3-form in some sense. Then it must have such a local expression

CaμνρTadxμdxνdxρ.

To make this Chern-Simons Lagrangian density, you have to take trace of it, which gives zero unless the Lie algebra g in consideration is abelian.

On the other hand, if the algebra were abelian, the expression AAA must vanish (due to antisymmetry). This would make the theory useless. Thus the expression should never be a Lie algebra valued form.

P.S. I did not come up this story when I asked my question. But the discussion here uncovered my poor understanding on the subject and enlightened how I should proceed. Thank you everyone!

This post imported from StackExchange MathOverflow at 2015-06-06 21:05 (UTC), posted by SE-user N. Shimode
answered Jun 1, 2015 by N. Shimode (30 points) [ no revision ]

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