The Yang-Mills action are usually given by $$S= \int\text{d}^{10}\sigma\,\text{Tr}\left(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\theta^{T}\gamma^{\mu}D_{\mu}\theta\right)$$
with the field strength defined as $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}-ig\left[A_{\mu},A_{\nu}\right]$
, $A_{\mu}$
being a U(N)
Hermitian gauge field in the adjoint representation, $\theta$
being a $16\times1$
Majorana-Weyl spinor of $SO(9)$
in the adjoint representation and $\mu=0,\dots,9$
. The covariant derivative is given by $D_{\mu}\theta=\partial_{t}\theta-ig\left[A_{\mu},\theta\right]$. We are using a metric with mostly positive signs.
We re-scale the fields by $A_{\mu}\to\frac{i}{g}A_{\mu}$
and let $g^{2}\to\lambda$
which gives us $$S=\int\text{d}^{10}\sigma\,\text{Tr}\left(\frac{1}{4\lambda}F_{\mu\nu}F^{\mu\nu}-\theta^{T}\gamma^{\mu}D_{\mu}\theta\right)$$
with the field strength defined as $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}+\left[A_{\mu},A_{\nu}\right]$
and the covariant derivative $D_{\mu}\theta=\partial_{t}\theta+\left[A_{\mu},\theta\right]$.
Now we perform a dimensional reduction from $9+1$
to $0+1$
, so that all the fields only depend on time, thous all spatial derivatives vanish i.e. $\partial_{a}(\text{Anything})=0$
. The $10$
-dimensional vector field decomposes into $9$
scalar fields $A_{a}$
which we rename $X^{a}$
and one gauge field $A_{0}$
which we rename $A$
. This gives (note that $\gamma^{t}=\mathbb{I}$ and that $\gamma^{a}=\gamma_{a}$. $$F_{0a}= \partial_{t}X^{a}+\left[A,X^{a}\right],\quad F_{ab}=+\left[X^{a},X^{b}\right]
\gamma^{t}D_{t}\theta= \partial_{t}\theta+\left[A,\theta\right],\quad\gamma^{a}D_{a}\theta=\gamma_{a}\left[X^{a},\theta\right]$$
The action for this theory is then
$$S=\int\text{d}t\,\text{Tr}\left(\frac{1}{2\lambda}\bigg\{-\left(D_{t}X^{a}\right)^{2}+\frac{1}{2}\left[X^{a},X^{b}\right]^{2}\bigg\}-\theta^{T}D_{t}\theta-\theta^{T}\gamma_{a}\left[X^{a},\theta\right]\right)$$
with the covariant derivative defined as $D_{t}X^{a}=\partial_{t}X^{a}+\left[A,X^{a}\right]$
and $D_{t}\theta=\partial_{t}\theta+\left[A,\theta\right]$
Now to the question. I need the potential energy $V=+\frac{1}{2}\left[X^{a},X^{b}\right]^{2}$ to be negative, not positive.
Taylor has a discussion on this in his paper "Lectures on D-branes, Gauge Theory and M(atrices)" (http://arxiv.org/abs/hep-th/9801182) on page 10, where he writes:
"Because the metric we are using has a mostly positive signature,
the kinetic terms have a single raised 0 index corresponding to a change of
sign, so the kinetic terms indeed have the correct sign. The commutator term $\left[X^{a},X^{b}\right]^{2}$ which acts as a potential term is actually negative definite. This follows from the fact that $\left[X^{a},X^{b}\right]^{\dagger}=\left[X^{b},X^{a}\right]=-\left[X^{a},X^{b}\right]$. Thus, as expected,
kinetic terms in the action are positive while potential terms are negative."
But I don't understand where the Hermitian conjugate $^\dagger$ comes from, to me this term is just: $$\left[X^{a},X^{b}\right]^{2}=\left[X^{a},X^{b}\right]\left[X_{a},X_{b}\right]$$
Note that Taylor uses a little different conventions when he re-scales, instead of $A_{\mu}\to\frac{i}{g}A_{\mu}$ he uses $A_{\mu}\to\frac{1}{g}A_{\mu}$ and $\theta \to\frac{1}{g}\theta$. But this should not cause any troubles I think.
