# Chern-Simons form of the Chern character

1. Aug 10, 2015

### sgd37

So I'm working my way through Nakahara Geometry Topology and Physics. In exercise 11.4 of section 11.5.2 I'm tasked to show

$$\text{tr}[\epsilon^{ijkl}\mathcal{F}_{ij}\mathcal{F}_{kl}]=\partial_{i}[2\epsilon^{ijkl}\text{tr}(\mathcal{A}_j\partial_{k}\mathcal{A}_l+\frac{2}{3}\mathcal{A}_j\mathcal{A}_k\mathcal{A}_l)]$$

where the field strength is lie algebra valued
$$\mathcal{F}_{ij}=\partial_{i}\mathcal{A}_j-\partial_{j}\mathcal{A}_i+[\mathcal{A}_i,\mathcal{A}_j]$$

this should be a straightforward bit of tensor algebra but I'm getting a factor of 2 discrepancy in the first term on the RHS

$$2\epsilon^{ijkl}\partial_{i}\text{tr}(\mathcal{A}_j\partial_{k}\mathcal{A}_l)=2\epsilon^{ijkl}\text{tr}(\partial_{i}\mathcal{A}_j\partial_{k}\mathcal{A}_l)$$

whereas on the LHS

$$\epsilon^{ijkl}(\partial_{i}\mathcal{A}_j-\partial_{j}\mathcal{A}_i+[\mathcal{A}_i,\mathcal{A}_j])(\partial_{k}\mathcal{A}_l-\partial_{l}\mathcal{A}_k+[\mathcal{A}_k,\mathcal{A}_l])=\epsilon^{ijkl}4(\partial_{i}\mathcal{A}_j+\mathcal{A}_i\mathcal{A}_j)(\partial_{k}\mathcal{A}_l+\mathcal{A}_k\mathcal{A}_l)=4\partial_{i}\mathcal{A}_j\partial_{k}\mathcal{A}_l+\ldots$$

what am I missing it's driving me crazy

Thanks

2. Aug 16, 2015