# Chern-Simons for 2n-dimensional manifolds

+ 6 like - 0 dislike
61 views

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

retagged Apr 19, 2014
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
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

+ 7 like - 0 dislike

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
answered Jan 19, 2013 by (110 points)
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
+ 6 like - 0 dislike

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
answered Jan 19, 2013 by (2,490 points)
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
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
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
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
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
+ 4 like - 0 dislike

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
answered Jan 19, 2013 by (125 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOver$\varnothing$lowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.