Rewriting the Yang-Mills action

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.

The difference in sign has to do with Euclidean metric. To find instanton solutions one needs to adopt an Euclidean metric not Minkowski, which is where the extra minus. That is you must perform a Wick rotation to get from $$S_1$$ to $$S_2$$. Explicitly $$x^0 = i x^4$$ in terms of coordinates.
That is specifically the base Manifold is not technically the same. As you can see in the book itself compare Eq. 10.108 with 10.112 (2003 Edition), the action is clearly differentiated by the supscript $$E$$, standing for Euclidean. Hence $$S_1 \neq S_2$$.
• Thanks for your answer, this was my first guess as well but I still have some doubts. In particular, I have deduced my result by applying $7.178$ which doesn't seem to reference the signature of the metric at all. Commented Jul 3, 2021 at 2:12
• It seems you believe $S_1 = S_2$ still which is wrong, you have not supplied a proof of their equality, because there is none other than a Wick rotation. My mention of the hodge dual is misleading I will delete it. Commented Jul 3, 2021 at 13:55