7
$\begingroup$

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. https://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??

$\endgroup$
10
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

For Lie algebras of matrices (which is what you really care about in Chern-Simons 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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Igor Khavkine May 27 '15 at 12:24
  • $\begingroup$ 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. $\endgroup$ – N. Shimode May 27 '15 at 12:27
  • 3
    $\begingroup$ 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. $\endgroup$ – Igor Khavkine May 27 '15 at 12:53
0
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.