SU(2) kinetic term as a trace Is there a easy way to rewrite the SU(2) kinetic term as a trace? As in 
$$\mathcal{L} = -\frac{1}{4}\vec{F}_{\mu\nu}\vec{F}^{\mu\nu}\\[1cm]
= -\frac{1}{2}\mathrm{tr}\Bigg[\bigg(\vec{F}_{\mu\nu}\cdot \frac{\vec{\tau}}{2}\bigg)^2\Bigg]. $$
Does this follow from the properties of the Pauli matrices? 
 A: The trace is just the inner product for the Lie algebra. The field strengths are Lie algebra valued, i.e., $\mathbf{F}_{\mu\nu}$ is an element of the Lie algebra, and can be written as a linear combination of generators: $\mathbf{F}_{\mu\nu} = \sum_a F^a_{\mu\nu} t^a$. One usually normalizes the generators such that $\left \langle t^a, t^b\right\rangle = \operatorname{tr} t^a t^b = c \delta^{ab}$, for some constant $c$, from which you get $\operatorname{tr} \mathbf{F}_{\mu\nu}\mathbf{F}^{\mu\nu} = F^a_{\mu\nu} F^{a\mu\nu}$.
The motivation for this is that a hypothetical gauged scalar of spinor field gets coupled to $N$ gauge fields $A^a$ via the covariant derivative, and each of these $N$ gauge fields gets it's own kinetic term of the form $F_{\mu\nu} F^{\mu\nu}$.
A: Here's what confused me (I was simply too lazy to put in work). What answers my question is the following simple algebra: Starting with the trace, we find* 
$$\mathrm{tr}\left\{\left(\vec{F}_{\mu\nu}\cdot\frac{\vec{\tau}}{2}\right)^2\right\}
= \mathrm{tr}\left\{\left({F}^1\cdot\frac{{\tau}^1}{2}+\cdots\right)\left({F}^1\cdot\frac{{\tau}^1}{2}+\cdots\right)\right\}
\\[1cm]= 
\mathrm{tr}\left\{\left(F^1 F^1\frac{{(\tau}^1)^2}{4}+\underbrace{F^1 F^2\frac{{(\tau}^1\tau^2)}{4}+F^2 F^1\frac{{(\tau}^2\tau^1)}{4}}_{0}+\underbrace{\cdots}_{_{0+F^2 F^2\frac{{(\tau}^2)^2}{4}}}+F^3 F^3\frac{{(\tau}^3)^2}{4}\right)\right\}
\\[1cm]=^\dagger
\mathrm{tr}\left\{\left(F^1 F^1\frac{{(\tau}^1)^2}{4}+F^2 F^2\frac{{(\tau}^2)^2}{4}+F^3 F^3\frac{{(\tau}^3)^2}{4}\right)\right\}
\\[1cm]=\frac{1}{4}\vec{F}_{\mu\nu}\cdot\vec{F}^{\mu\nu}\mathrm{tr}\{\mathbb{1}_{2\times2}\}
\\[1cm]=\frac{1}{2}\vec{F}_{\mu\nu}\cdot\vec{F}^{\mu\nu}.
$$
The relation 
$$\tau_a \tau_b = i\sum_c\varepsilon_{a b c}\,\tau_c + \delta_{a b}I
$$
was used in $\dagger.$

*Suppressing Lorentz indices not to clutter.
