Let the Chern-Simons lagrangian for a group $G$ be, $$L= k \epsilon^{\mu \nu \rho} Tr[A_\mu \partial _ \nu A_\rho + \frac{2}{3} A_\mu A_\nu A_\rho]$$

Then it is claimed that on "infinitesimal" variation of the gauge field ("connection") the lagrangian changes by,

$$\delta L = k \epsilon^{\mu \nu \rho} Tr[\delta A_\mu F_{\nu \rho}]$$

where the "curvature" $F_{\mu \nu}$ is given as $\partial _ \mu A_\nu - \partial _ \nu A_\mu +[A_\mu,A_\nu]$

Under gauge transformations on $A$ by $g \in G$ it changes to say $A'$ whose $\mu$ component is given as $g^{-1}A_\mu g+g^{-1}\partial _ \mu g$ (This makes sense once a representation of $G$ has been fixed after which $A$ and $g$ are both represented as matrices on the same vector space)

Say the Lagrangian under the above gauge transformations change to $L'$ and then one has the relation,

$$L'-L=-k \epsilon^{\mu \nu \rho}\partial_\mu Tr[\partial_\nu g g^{-1}A_\rho]-\frac{k}{3}\epsilon^{\mu \nu \rho} Tr[g^{-1}\partial_\mu gg^{-1}\partial_\nu gg^{-1}\partial_\rho g]$$

The second term of the above expression is what is proportional to the "winding number density" of the Chern-Simons lagrangian and thats what eventually gets quantized.

I would like to know the following things,

  • Is there a neat coordinate free way of proving the above two variation change equations? Doing this in the above coordinate way is turning out to be quite intractable!

  • Since the Lagrangian is just a complex number one can talk of the "real" and the "imaginary" part of it. But I get the feeling that at times a split of this kind is done at the level of the gauge field itself. Is this true and if yes the how is it defined? (Definitely there is lot of interest in doing analytic continuation of the "level" $k$)

  • The Euler-Lagrange equations of this action give us only the "flat" connections and in that sense it is a topological theory since only boundary conditions seem to matter. Still all the flat connection configurations are not equivalent but are labelled by homomorphisms from the first fundamental group of the 3-manifold on which the theory is defined to $G$. How to see this? What is the background theory from which this comes? And why is this called "holonomy"? (I am familiar with "holonomy" as in the context of taking a vector and parallel transporting it around a loop etc)


4 Answers 4


The secret to understanding the Chern-Simons functional over a 3-manifold is to realise that it's a 4-dimensional functional in disguise.

If $X$ is a closed, oriented 4-manifold, and $P \to X$ a principal $SU(n)$-bundle, one has the Chern-Weil formula for the second Chern number, $$ c_2(P)[X] = \frac{1}{8 \pi^2} \int_X {Tr F_A^2}, $$ where $F_A$ is the curvature of a connection $A$ in $P$. If now $Y$ is a closed, oriented 3-manifold, and $P_Y \to Y$ a principal $SU(n)$-bundle (necessarily trivial) then we can define the Chern-Simons functional $$ cs: Connections(P_Y) \to \mathbb{R}/\mathbb{Z} $$
for a connection $A$ by choosing any compact oriented 4-manifold $X$ bounding $Y$ (one always exists), any $SU(n)$-bundle $P_X \to X$ extending $P_Y \to Y$ (one again exists), and any connection $B$ in $P_X$ extending $A$ (ditto), and setting $$ cs(A) = \frac{1}{8 \pi^2} \int_X {Tr F_A^2}. $$ This is well-defined (mod integers) because given another choice $(X',P_{X'},B')$, we can glue together the two manifolds along $Y$ to get a closed 4-manifold (we reverse the orientation of one of them), and express the difference in values of $cs(A)$ as a Chern-Weil integral as above, which is an integer.

To obtain a formula for $cs(A)$, take $P_X$ to be a trivial bundle, and take $B$ to be a connection that is trivial except over the collar $Y\times [-1,0]$. Over that collar, the bundle is $[-1,0]\times P_Y$, and we can let $B$ the 4-dimensional connection $A + t(A-A_0)$, where $A_0$ is a reference connection trivialising $P_Y$ (i.e., flat with trivial holonomy). Then we can explicitly perform the integral defining $cs(A)$ over the $[0,1]$-factor (try it!) and get a formula for real-valued lift of $cs(A)$ (here we use $A_0$ to trivialise $P_Y)$, $$ CS(A) = \frac{1}{8 \pi^2} \int_Y{ Tr (A \wedge dA + \frac{2}{3} A\wedge A\wedge A)}, $$ which is the formula you gave in indices (up to a factor).

From this formula, it's straightforward to compute the first variation of $CS$, and to interpret it in terms of $F_A=dA+A\wedge A$.

The gauge transformation formula says that if $g$ is an automorphism of $P_Y$ of degree $d $ then $ CS(gA)-CS(A) = d$. This has a conceptual explanation: linearly interpolate from $A$ to $gA$ to obtain a connection over $Y\times [0,1]$ whose Chern-Weil integral is $CS(gA)-CS(A)$. Use $g$ as a gluing recipe to obtain a bundle over $Y\times S^1$ whose second Chern number is $d$ (this defines $d$, if you will), and apply the Chern-Weil formula.

Reference: S.K. Donaldson, "Floer homology groups in Yang-Mills theory".

Very briefly on the second question: one can work with $SL(n,\mathbb{C})$-bundles instead. The Lie algebra is the complexfication of that of $SU(n)$, so (after the bundle is trivialised), connections have real and imaginary parts which are 1-forms valued in $su(n)$.

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    $\begingroup$ +1. A very beautiful answer! $\endgroup$ Commented Jul 15, 2010 at 0:12
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    $\begingroup$ Jose, thanks - I should make clear that this is the story told in the original paper of Chern and Simons! $\endgroup$
    – Tim Perutz
    Commented Jul 15, 2010 at 1:05
  • 1
    $\begingroup$ I'm embarrassed to say I never read the original paper. Perhaps I should! $\endgroup$ Commented Jul 15, 2010 at 1:42
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    $\begingroup$ Konrad, sorry for the slow reply to your comment (I'm travelling...). The simple connectedness of the group has the convenient effect that bundles over an oriented 3-manifold are trivial so one needn't worry. In general, one would want to choose $X$ so that the bundle does extend over it, but I don't honestly know whether this is ever problematic (it's straightforward for $SO(3)$, for instance). $\endgroup$
    – Tim Perutz
    Commented Jul 17, 2010 at 21:27
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    $\begingroup$ @ungerade I don't have a reference to hand, but it follows by obstruction theory from the fact that a simply connected group is 2-connected, hence its classifying space is 3-connected. For vanishing of $\pi_2(G)$, see mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups $\endgroup$
    – Tim Perutz
    Commented May 26, 2017 at 15:07

I'll address your first question, in some computational detail; although I cannot do better than the answers which have already been given in terms of what is conceptually going on.

Let $G$ be a 1-connected Lie group with Lie algebra $\mathfrak{g}$. We assume that the group admits a bi-invariant metric or, equivalently, that the Lie algebra admits an ad-invariant inner product, denoted $\operatorname{Tr}$ by the traditional abuse of notation in the Physics literature. Let $M$ be an orientable 3-manifold which, for simplicitly, I will take without boundary. Then every principal $G$-bundle on $M$ is trivial and choosing a trivialisation once and for all we can identify connections with $\mathfrak{g}$-valued one-forms on $M$. Let $A$ be one such connection.

In this language, the Chern-Simons action for $A$ can be written as $$I[A] = \frac{k}{4\pi} \int_M \operatorname{Tr}\left(A \wedge dA + \frac23 A \wedge A \wedge A\right).$$ There is of course some abuse of notation here. Both $A$ and $dA$ are $\mathfrak{g}$-valued, whence the first term in the action is the composition of their wedge product with the ad-invariant inner product on $\mathfrak{g}$. This results in a 3-form, which can be integrated on $M$. The second term is perhaps best understood if one rewrites $$\frac23 \operatorname{Tr} A \wedge A \wedge A = \frac13 \operatorname{Tr} [A , A] \wedge A,$$ where $[A,A]$ is the composition of wedging and the Lie bracket on $\mathfrak{g}$, hence it is again a $\mathfrak{g}$-valued 2-form, and the second term in the action is (up to the factor of one-third) again the composition of the wedge product of $[A,A]$ and $A$ with the ad-invariant inner product. (Notice that $[A,A]$ is not zero, because although $[-,-]$ is antisymmetric, so is $\wedge$ between two one-forms.)

Now let us vary $I$: $$\delta I = \frac{k}{4\pi} \int_M \operatorname{Tr} \left ( \delta A \wedge dA + A \wedge d \delta A + 2 \delta A \wedge A \wedge A \right).$$ Using the fact that $d$ is an odd derivation, we have that $$\operatorname{Tr}(A \wedge d \delta A) = - d \operatorname{Tr}(A \wedge \delta A) + \operatorname{Tr}(dA \wedge \delta A)$$ whence, $$\delta I = \frac{k}{2\pi} \int_M \operatorname{Tr} \left ( \delta A \wedge \left( dA + A \wedge A \right) \right) + \frac{k}{4\pi} \int_{\partial M} \operatorname{Tr} (\delta A \wedge A)$$ The boundary term drops out because we have assumed $\partial M = \emptyset$. In summary, the Euler-Lagrange equation is the flatness of the curvature $F = dA + A \wedge A$, which can also be written as $F = dA + \frac12 [A,A]$.

Under a gauge transformation $g:M \to G$, $$A \mapsto A^g = g A g^{-1} - dg g^{-1}.$$ Again there's some abuse of notation in that one is assuming that $\mathfrak{g}$ is a matrix Lie algebra, but this only notational and it's perfectly possible to write this in a way that makes sense in general; namely, $$A^g = \operatorname{Ad}_g A + g^*\theta$$ where $\theta$ is the right-invariant Maurer-Cartan form on $G$.

In order to vary the action is perhaps best to write it in an equivalent way; namely, $$I[A] = \frac{k}{4\pi} \int_M \operatorname{Tr}\left(A \wedge F - \frac13 A \wedge A \wedge A\right),$$ since the curvature transforms covariantly: $F^g = g F g^{-1}$.

The next calculation is a little long and I will not reproduce it here. It is not difficult: just plug in the expressions for $A^g$ and $F^g$ into the action $I[A^g]$ and expand using the fact that $\operatorname{Tr}$ is Ad-invariant. One must also use the Maurer-Cartan structure equation $d\theta = - \theta \wedge \theta$. The result of the calculation is $$I[A^g]-I[A]= \frac{k}{4\pi} \int_M \operatorname{Tr}\left( d(g^{-1} dg \wedge A) + \frac13 (dg g^{-1})^3 \right)$$ where the first term vanishes by Stokes's theorem since $M$ has no boundary, leaving $$I[A^g]-I[A]= \frac{k}{12\pi} \int_M \operatorname{Tr}\left( (dg g^{-1})^3 \right)$$

If $G$ is a compact simple group, then one can rescale the inner product $\operatorname{Tr}$ in such a way that $$\frac{1}{12\pi} \int_M \operatorname{Tr}\left( (dg g^{-1})^3 \right) \in \mathbb{Z}$$ whence the Chern-Simons path integral is gauge-invariant provided that $k \in \mathbb{Z}$ as well, in units where $\hbar =1$.

  • $\begingroup$ Jose: not every G-bundle over a 3-manifold is trivial. It is if G is simply connected $\endgroup$
    – Paul
    Commented Jul 15, 2010 at 3:20

on your 3rd point: The holonomy of a connection, defined as you say by parallel transport, defines a function from the loop space of the manifold (of any dimension) to $G$ (parallel transport of vectors should be generalized to frames for $= GL(n)$, and more generally if you work in a principal $G$ bundle the holonomy lies in $G$:this is either a defintion of a simle exercise depending on what "connection" means to you. The connection is flat if and only if the holonomy is trivial for all small enough loops. expressing an arbitrary loop as a product of small loops you see this is equivalent to the holonomy being homotopy invariant, hence defining a function (easily seen to be a homomorphism) on the fundamental group. Gauge equivalent flat connections yield conjugate holonomy representations, and in fact Flat/Gauge is identified with a subset of $Hom(\pi_1,G)/conj$.

For your first 2 points, given any connection A, all others are of the form A+a for an appropriate Lie(G) valued 1-form a. Then the equations $F(A+t a)= F(A) + t d_A a + t^2 [a,a]/2$, $g.(A+a)=A + gAg^{-1} + dg g^{-1}$ (up to signs!) and $cs(A)= Tr(F(A)\wedge A -1/3 A\wedge A \wedge A)$ lead to answers to your questions. I'm guessing from your notation that you are a physicist: all of this is explained in any mathematical treatment of the Chern-Simons function on connections.


There is a nice geometrical treatment of the Chern-Simons action functional for a general Lie group $G$ - not necessarily simply connected - that relates it to "holonomy". It also clarifies why Chern-Simons theories are classified by $H^4(BG,\mathbb{Z})$.

This point of view is described in a paper by Carey et. al. (http://arxiv.org/abs/math/0410013).

The group $H^4(BG,\mathbb{Z})$ classifies geometrical objects over $G$ called "multiplicative bundle gerbes". Given a multiplicative bundle gerbe $\mathcal{G}$ and a closed oriented 3-manifold $M$ with a principal $G$-bundle $P$ with connection over it, Carey et. al. describe a beautiful construction of a bundle 2-gerbe $CS(\mathcal{G},P)$ over $M$, the Chern-Simons 2-gerbe. They prove that this 2-gerbe carries a connection such that its holonomy around $M$, $$ Hol_{CS(\mathcal{G},P)}(M) \in S^1 $$ is the (exponentiated) Chern-Simons action. This works without any assumptions on the manifold $M$ or on the Lie group $G$.

For a little background, recall that connections on $n$-gerbes have holonomies around closed oriented $(n+1)$-manifolds. Ordinary bundles are 0-gerbes. Part of a connection on an $n$-gerbe is a curving: a $(n+1)$-form on some covering space. In case of the 2-gerbe $CS(\mathcal{G},P)$, the covering space is the total space of the given bundle $P$, and the curving 3-form is the Chern-Simons 3-form $CS_A$ on $P$.


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