Rewriting the Yang-Mills action

Question

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.

Answer 1

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 $S_1 \neq S_2$.

Answer 2

I think it has to do with the fact that is a pseudo-riemannian manifold and the Hodge star picks up a sign sometimes (depending on dimension and also if it is the mostly plus convention or the mostly minus convention in the metric). See the section on dimension 4 in https://en.m.wikipedia.org/wiki/Hodge_star_operator