Chern-Simons for 2n-dimensional manifolds

Question

In the literature I can only find Chern-Simons terms for odd-dimensional manifolds. For example, for a $G$-bundle over a 3-dimensional manifold we have $A \wedge dA + A \wedge A \wedge A$ with $A$ being a $\mathfrak{g}$-valued 1-form. Why can't I write such forms for even-dimensional manifolds?

This post imported from StackExchange at 2014-03-22 17:28 (UCT), posted by SE-user Daniel

Comments

Cross-listed: physics.stackexchange.com/questions/51603/chern-simons-term

This post imported from StackExchange at 2014-03-22 17:28 (UCT), posted by SE-user Chris Gerig

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Cross-listed: mathoverflow.net/questions/119353/…

This post imported from StackExchange Physics at 2014-03-22 17:28 (UCT), posted by SE-user Chris Gerig

Answer 1

It has to do with the fact that the characteristic classes (over the reals) of a principal $G$-bundle have even degree. We can associate Chern-Simons-like theory to each characteristic class of degree $2k$ together with a $G$-bundle $P$ over a manifold of dimension $2k-1$.

To be a bit more technical a Chern-Simons-like form is asssociated to the following data

1. A homogeneous polynomial $\Phi$ of degree $k$ on the Lie algebra of $G$ invariant under the action of $G$ by conjugation.

2. A principal $G$-bundle $P\to M$ over $M$.

3. A pair of connections $\nabla^0, \nabla^1$ on $P\to M$.

The Chern-Weil theory produces two closed forms

$$ \Phi(\nabla^0),\Phi(\nabla^1)\in \Omega^{2k}(M) $$

and a form

$$ T\Phi(\nabla^1,\nabla^0)\in \Omega^{2k-1}(M), $$

such that

$$ d T\Phi(\nabla^1,\nabla^0)= \Phi(\nabla^1)-\Phi(\nabla^0). $$

(For details see Chapter 8 of these notes.)

The transgression form $T\Phi(\nabla^1,\nabla^0)$ is the one used in Chern-Simons theories. It depends on two connections, but usually $\nabla^0$ is some fixed connection.

This post imported from StackExchange at 2014-03-22 17:28 (UCT), posted by SE-user Liviu Nicolaescu

Comments

Of course you can consider Chern-Simons forms on any manifold. But integrating them over the manifold - to get a numerical invariant - only works if the dimension of the manifold equals the degree of the form.

This post imported from StackExchange at 2014-03-22 17:28 (UCT), posted by SE-user user15817

Answer 2

The starting points for the construction of the Chern-Simons terms are objects called Chern-Pontryagin densities. On a 2n dimensional manifold, these are of the form $$ {\mathcal{P}}^{2n} = \alpha\epsilon^{\mu_1\mu_2...\mu_{2n}}Tr F_{\mu_1\mu_2}...F_{\mu_{2n-1}\mu_{2n}} $$ where F is the curvature 2-form of some G-connection (G is the gauge group). These are gauge-invariant, closed, and their integral over the manifold M (compact, no boundary) is an integer which is a topological invariant. These sorts of invariants are examples of characteristic classes.

Now ${\mathcal{P}}^{2n}$ can be locally expressed as a differential $$ {\mathcal{P}}^{2n} = d({\mathcal{C}}^{2n-1})$$ of a 2n-1 form. $ \ {\mathcal{C}}^{2n-1}$ is the Chern Simons form. (It can be written in the familiar form in terms of the connection form A). It has the remarkable property that if I perform a G-gauge transformation, the action obtained by integrating $ \ {\mathcal{C}}^{2n-1}$ is gauge-invariant. At no point is a metric involved in this construction, so it's a topological theory.

Anyway, getting back to the question, the Chern Pontryagin density ${\mathcal{P}}^{2n}$ which we started with is only defined on a 2n (i.e even) dimensional manifold, so consquently the Chern-Simon's term is only defined on an odd dimensional one.

Edit: Example in which Chern Simon's term in 3d is produced:

The Chern-Pontryagin class is the integral of the C.P. density on a 4d manifold $$ {\mathcal{P}} \propto \int d^4xTr({^*F}^{\mu\nu}F_{\mu\nu})$$ $$ \propto \int d^4x \ \epsilon^{\mu\nu\rho\sigma}Tr( F_{\mu\nu}F_{\rho\sigma})$$ We can write the C.P. density as the divergence of a 4-current (the Chern Simons current) $$Tr({^*F}^{\mu\nu}F_{\mu\nu}) = \partial_{\mu}C^{\mu}$$ where $$C^{\mu} = \epsilon^{\mu\nu\rho\sigma} \ tr(A_{\nu}\partial_{\rho}A_{\sigma}+\frac{2}{3}f^{abc}A^a_{\nu}A^b_{\rho}A^b_{\sigma}) $$ If we now pick a local coordinate system (on the 4 manifold) such that $\frac{\partial}{\partial x^0}$ (say) is in the direction of the vector $C^{\mu}$, then we just look at the other components in the epsilon symbol (they're guaranteed to be 1,2,3), then we just freeze the $x^0$ dependence of the $A_{\nu}(x^{\mu})$. We've now got ourselves a 3 form on the three-manifold $x^0$ = constant.

This post imported from StackExchange Physics at 2014-03-22 17:28 (UCT), posted by SE-user twistor59

Comments

Thanks for the answer but there is something I don't understand. For example, the chern-simons form in 3 dimensions is a 3-form. Wouldn't the exterior derivative be zero since $dw=0$ for $w$ a p-form in a p-dim'l manifold?

This post imported from StackExchange Physics at 2014-03-22 17:28 (UCT), posted by SE-user Daniel

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The exterior derivative referred to is applied on the 2n dimensional manifold, not on the 2n-1 dimensional one onto which the CS form descends.

This post imported from StackExchange Physics at 2014-03-22 17:28 (UCT), posted by SE-user twistor59

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again, i appreciate the answer but i am a bit even more confused now. what's the relationship between these two different manifolds you refer to. also, i don't understand how an exterior derivative takes a form from one manifold into another form over another (higher dim'l) manifold.

This post imported from StackExchange Physics at 2014-03-22 17:28 (UCT), posted by SE-user Daniel

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OK I added some stuff which might help a bit.

This post imported from StackExchange Physics at 2014-03-22 17:28 (UCT), posted by SE-user twistor59

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great, that's what i thought you initially meant, but i was expecting the answer to be independent of some embedding.

This post imported from StackExchange Physics at 2014-03-22 17:28 (UCT), posted by SE-user Daniel

Answer 3

When you look at the definition of the Chern–Simons form

$d\omega_{2k-1}={\rm Tr} \left( F^{k} \right)$

one needs to understand that $F$ is a $2$-form. So the right hand side $F^k$ will always be a $2k$-form (with $k \in Z^+$ ). So the exterior derivative $d$ of the Chern–Simons form $\omega$ needs then to be a $2k-1$-form (since the exterior derivative of a $n$-form is a $n+1$-form) and since $k$ is a positive integer $\omega$ needs to be an odd form by definition.

This post imported from StackExchange Physics at 2014-03-22 17:28 (UCT), posted by SE-user ungerade