Chern-Simons for 2n-dimensional manifolds

2013-01-19T18:42:58Z

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?

Answer by Liviu Nicolaescu for Chern-Simons for 2n-dimensional manifolds

2013-01-19T19:59:46Z

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.

Chern-Simons for 2n-dimensional manifolds - MathOverflow

Chern-Simons for 2n-dimensional manifolds

Answer by Liviu Nicolaescu for Chern-Simons for 2n-dimensional manifolds