# What is the trace in the Chern-Simons action?

Warning: This is a very stupid question regarding a basic misunderstanding that I have. I realize that the question is very elementary, but I guess asking stupid questions is better than remaining ignorant.

To be explicit, consider the $\mathrm{SU}(2)$ Chern-Simons action on some very nice $3$-manifold $M$, i.e. the number

$S(A) = \frac{k}{4\pi}\int_M \mathrm{tr} \left(A\wedge\mathrm{d}A + \frac{2}{3}A\wedge A\wedge A\right)$,

where $A$ is an $\mathfrak{su}(2)$-valued $1$-form on $M$.

What I simply cannot wrap my head around, and what is obviously a very silly basic question, is: What trace is this?! As I understand it, there is a certain abuse of notation in $\wedge$ on vector bundle-valued forms (namely, with $E$ the bundle, the wedge of two $E$-valued forms is an $E\otimes E$-valued one), but in the case of the Chern-Simons action this answer suggests that the $E\otimes E=\mathfrak{g}\otimes\mathfrak{g}\rightarrow\mathfrak{g}$ is supposed to be the Lie bracket on $\mathfrak{g}$. Anyway, that leads me to think that the trace is the trace on $\mathfrak{g}=\mathfrak{su}(2)$, which of course vanishes everywhere.

What am I misunderstanding here?

## 2 Answers

The trace is simply a (properly normalised) ad-invariant inner product on the Lie algebra; that is, a nondegenerate symmetric bilinear form $\langle-,-\rangle$ which obeys the "associativity" condition $$\langle [x,y],z \rangle = \langle x, [y,z] \rangle$$ for every $x,y,z$ in $\mathfrak{g}$.

Lie algebras admitting such inner products are said to be metric. The normalisation of the inner product is such that $k$ is an integer. This only makes sense for indecomposable metric Lie algebras; that is, those which are not isomorphic to the direct product of perpendicular proper ideals.

The notation "tr" stems from the fact that if $\rho: \mathfrak{g} \to \operatorname{End}(V)$ is a faithful finite-dimensional representation, then $$\langle x, y\rangle := c \operatorname{tr}\rho(x)\rho(y)$$ works for a suitable nonzero $c$. (For a simple Lie algebra, just take $\rho$ to be the adjoint representation.)

For the explicit case of $\mathfrak{g}$ the Lie algebra of SU(2) you can take $\rho$ to be the fundamental representation and $c= -\frac12$, I believe.

Edit

Notice that $\operatorname{tr}(A \wedge dA)$ is really $\langle A \stackrel{\wedge}{,} dA \rangle$, where $\langle -\stackrel{\wedge}{,}-\rangle$ means that we are both taking the wedge product of the forms and the inner product on the Lie algebra. Similarly, $$\operatorname{tr}(A \wedge A \wedge A) = \frac12 \langle [A\stackrel{\wedge}{,}A] \stackrel{\wedge}{,} A \rangle,$$ with a similar notational caveat about $[A\stackrel{\wedge}{,}A]$.

Further addition

In response to Anirbit's comment, I would say that there is, in general, no trace on vector-valued differential forms; although if the forms take values in endomorphisms, then of course there is: simply compose with the trace of endomorphisms to obtain a map $$\Omega^\bullet(M;\operatorname{End}(V)) \to \Omega^\bullet(M).$$

• I especially like the edit. You managed to put concretely into words something I've only been vaguely aware of in the past. – Willie Wong Aug 20 '10 at 10:57
• Ah, obviously! Thanks for enlightening me, José Figueroa-O'Farrill! – Anon Incog Aug 20 '10 at 12:22
• My pleasure, J. – José Figueroa-O'Farrill Aug 20 '10 at 16:00
• @Jose Given a vector valued k-form is there a definition of its trace? (if you could give some local coordinate expression also) Does the definition of a trace somehow need fixing of a n-dim representation of the Lie-Algebra? (where n is the dimension of the underlying manifold) Am I wrong in thinking thus for at least vector valued 1-form say $\omega$ that for a point $p$ on the underlying manifold we have a linear map $\omega _p: T_pM \rightarrow V$ (where $V$ supports a n-dim representation of the Lie-Algebra). Then $Tr(\omega)$ is the map $p \mapsto Tr(\omega _p)$ ? – Anirbit Aug 24 '10 at 12:01
• Further if you could elaborate on how this Tr works with taking deRham derivative and wedge products. Can you give a reference for this concept of taking trace of vector valued differential forms? – Anirbit Aug 24 '10 at 12:03

I will give the physicist's answer. I hope you are familiar with this notation.

$A=A_\mu dx^\mu$ and in this notation the action looks like

$S=\frac{k}{4\pi} \int_{\mathcal{M}} d^3x\ \epsilon^{\mu\nu\alpha}\ Tr \left( A_\mu \partial_\nu A_\alpha +\frac{2}{3} A_\mu A_\nu A_\alpha \right)$

where $\epsilon^{\mu\nu\alpha}$ is the Levi-Civita tensor in three dimensions.

$A_\mu=A_\mu^at^a$ where $A_\mu^a$ are real functions, $t^a$ are the matrix representatives of generators of Lie algebra $[t^a,t^b]=if^{abc}t^c$. Here $a,b,c=1,2,...,dim G$ where G is the gauge group and $f^{abc}$ are structure constants.

So, that trace in the action is the trace of products of these $t^a$ matrices.

For example,

for $G=SU(N)$ and in the fundamental representation $Tr(t^at^b)=\frac{1}{2}\delta^{ab}$

$(a,b=1,2,...N^2-1)$

• If a=b then the action would be zero, what does that mean in physics point of view?! – Gamma Jul 27 '14 at 16:51
• The action is not zero. Lets look at the trace of the first term $Tr(A_\mu \partial_\nu A_\alpha)= A^a_\mu \partial_\nu A^b_\alpha Tr(t^at^b)=\frac{1}{2}A^a_\mu \partial_\nu A^a_\alpha$ because $Tr(t^at^b)=\frac{1}{2}\delta^{ab}$. For the second term see: physics.stackexchange.com/questions/119953/… – TY1 Jul 28 '14 at 0:07
• I think you thought that way because of the wedge product. In this notation it is replaced by the Levi-Civita tensor and the volume element. Anti-symmetry is in $\epsilon$ here. – TY1 Jul 28 '14 at 0:13
• I mean $Tr(t^at^b)$ is not $Tr(t^a \wedge t^b)$. $Tr(t^at^b)$ is symmetric in $ab$ – TY1 Jul 28 '14 at 0:24