I am reading about Yang-Mills theory in Section $10.5.4$ of Geometry, Topology and Physics by Nakahara. In Equation $10.108$ he gives two different forms for the Yang-Mills action, and I am having trouble going between them.
Consider $G=SU(2)$ Yang-Mills theory on a $4$-manifold $M$. The first form of its action is: $$ S_{1} = -\frac{1}{4}\int_{M}tr(\mathcal{F}_{\mu\nu}\mathcal{F}^{\mu\nu}) $$ where $\mathcal{F}_{\mu\nu}=F_{\mu\nu}^{\alpha}T_{\alpha}$ are the components of the local curvature two-form $\mathcal{F}=\frac{1}{2}\mathcal{F}_{\mu\nu}dx^{\mu}\wedge dx^{\nu}$, and $T^{\alpha}=\frac{\sigma^{\alpha}}{2i}$ are the Lie algebra generators of $SU(2)$. This trace is over the Lie algebra indices.
The second form Nakahara gives is: $$ S_{2}=\frac{1}{2}\int_{M}tr(\mathcal{F}\wedge\star\mathcal{F}) $$ where $\star$ denotes the Hodge star operator.
It should not be too difficult to go between these two forms of the action, but when I try I am off by sign and can't seem to find my error.
To begin with, I write $\mathcal{F}=\mathcal{F}^{\alpha}T_{\beta}$, so that $\mathcal{F}\wedge\star\mathcal{F}=(\mathcal{F}^{\alpha}\wedge\star\mathcal{F}^{\alpha})T_{\alpha}T_{\beta}$
then according to Section $7.9.3$ of Nakahara, for two-forms $\mathcal{F}^{\alpha}$ and $\mathcal{F}^{\beta}$ on a $4$-manifold $M$: $$ \mathcal{F}^{\alpha}\wedge\star\mathcal{F}^{\beta} = \frac{1}{2} \mathcal{F}^{\alpha}_{\mu\nu} (\mathcal{F}^{\beta})^{\mu\nu}\sqrt{|g|}dx^{1}\wedge\cdots\wedge dx^{4} $$ So that: $$ \frac{1}{2}tr(\mathcal{F}\wedge\star\mathcal{F}) = \frac{1}{4}tr(\mathcal{F}_{\mu\nu}\mathcal{F}^{\mu\nu})dV $$ Which is of course the correct result up to a sign.
I have probably made a really silly error here, and I would really appreciate some help spotting it.
EDIT:
I am specifically having trouble proving $(10.108)$ of Nakahara, which reads $$ S_{YM} = -\frac{1}{4}\int_{m}tr(\mathcal{F}_{\mu\nu}\mathcal{F}^{\mu\nu}) = \frac{1}{2}\int_{M}tr(\mathcal{F}\wedge\star\mathcal{F}) $$
There is a related statement $(10.110)$ about the Euclidean action (the action obtained after a Wick rotation), which reads
$$ S_{YM}^{E} = \frac{1}{4}\int_{M}tr(\mathcal{F}_{\mu\nu}\mathcal{F}^{\mu\nu}) = -\frac{1}{2}\int_{M}tr(\mathcal{F}\wedge\star\mathcal{F}) $$
I understand how to obtain the Euclidean version from the Minkowski version, but I am specifically having trouble with the middle equality which should reduce to showing that $$ tr(\mathcal{F}\wedge\star\mathcal{F}) = -\frac{1}{2}tr(\mathcal{F}_{\mu\nu}\mathcal{F}^{\mu\nu}) $$
in either case.