Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    

1 Answer 1

up vote 12 down vote accepted

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.

share|improve this answer
2  
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. –  ThiKu Jan 21 '13 at 7:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.