$A \wedge A \wedge A$ in Chern-Simons

Question

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

$$A \wedge dA + \frac{2}{3}A \wedge A \wedge A,$$

where $A$ is a Lie algebra valued 1-form. My question is "how is the last term $A \wedge A \wedge A$ 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.

$$[\omega \wedge \eta](X_1,\dots,X_{p+q}) := \text{(coefficient)}\times \sum_{\sigma \in S_{p+q}} \text{sgn}(\sigma) [\omega(X_{\sigma(1)},\dots,X_{\sigma(p)}),\eta(X_{\sigma(p+1)},\dots,X_{\sigma(p+q)})],$$

where $\omega$ and $\eta$ 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)

$$[\omega \wedge \eta] = [\omega^a \otimes T^a, \eta^b\otimes T^b] := \omega^a \wedge \eta^b \otimes [T^a,T^b],$$

where $T^c (c=1,\dots,\dim \mathfrak{g})$ are generators of the Lie algebra $\mathfrak{g}$, 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 $[\omega\wedge\eta]$ is Lie algebra valued $(p+q)$-form.

Then what about wedged ones without brackets, such as $A\wedge A, A\wedge A \wedge A$?

I can show that $A \wedge A$ is equivalent to $[A \wedge A]$ up to coefficient, using either matrix representation, considering $\mathfrak{g}=\mathfrak{gl}(n)$, or universal enveloping algebra. The basic idea is

$$A \wedge A = (A^a \otimes T^a) \wedge (A^b \otimes T^b) = (A^a \wedge A^b) T^a T^b.$$

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

So, what I did was calculating $[A \wedge [A \wedge A]]$, 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

Answer 1

Option (1) Use the definition $(\omega \otimes S) \wedge (\eta \otimes S) = (\omega \wedge \eta) \otimes (S\otimes T)$ of the wedge product for Lie algebra valued forms. Define Lie bracket and Killing form as bilinear maps $[S\otimes T] = [S,T]$ and $\langle S \otimes T \rangle = \langle S, T\rangle$. Then the formula that you want is $$\langle A \wedge [A \wedge A] \rangle,$$ where the commutator and Killing form apply only to the Lie algebra factors, ignoring the differential form factors.

Option (2) Use the definition $(\omega \otimes S) \wedge (\eta \otimes S) = (\omega \wedge \eta) \otimes ST$ of the wedge product of forms valued in a particular matrix representation of a Lie algebra. Then the formula that you want is $$\operatorname{tr} (A \wedge A \wedge A),$$ 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

Answer 2

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

$$ A=\sum_i A_i dx^i, $$

where $A_i$ are $r\times r$ matrices.With this convention, use the usual wedge product

$$\left(\sum_i A_i dx^i\right)\wedge \left(\sum_j A_j dx^j\right)\wedge \left(\sum_k A_k dx^k\right) = \sum_{i,j,k} A_iA_jA_k dx^i\wedge dx^j\wedge dx^k, $$

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

Comments

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

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So, do you think that $A\wedge A \wedge A$ 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

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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

Answer 3

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

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

$$C_{\mu\nu\rho}^a T^a \otimes dx_{\mu}\wedge dx_{\nu} \wedge dx_{\rho}.$$

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

On the other hand, if the algebra were abelian, the expression $A \wedge A \wedge A$ 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