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The center symmetry in QCD comes from the

$$a\ \mathcal{P}\mathrm{exp}\left(ig_s \int_C dx^\mu \ A_\mu(x)\right) a^{-1} = \mathcal{P}\mathrm{exp}\left(ig_s \int_C dx^\mu \ A_\mu(x)\right),$$ where $C$ is a loop and $a \in \{e^{2\pi i k/3}I| \ k =0,1,2\} \simeq \mathbb{Z}_3$ is an element of the $\mathrm{SU}(3)_C$ center group. This makes the vacuum expectation value of the Wilson loop

$$W[C] = \mathrm{Tr}\left[\mathcal{P}\mathrm{exp}\left(ig_s \int dx^\mu \ A_\mu(x)\right)\right],$$ $\langle W[C] \rangle$ should be also center-invariant.

Then, when the center symmetry is broken (deconfinement), this implies that $W[C]$ is not more an element of $\mathrm{SU}(3)$ because it doesn't commute with center elements. This does make $\mathrm{det}(W[C]) \neq 1$?

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    $\begingroup$ I don't understand this question. The center symmetry is not what you say, but for a loop along $[0,\beta]$ in the periodic time dimension, you can have a gauge transformation with $g(\beta) = ag(0)$, where $a$ is an element of the center. The Polyakov loop is not invariant under this transformation while the gauge field $A$ itself is, as one can show just by plugging in (and any good source on Polyakov loops should) - it has nothing to do with the loop "being an element of SU(3)" but with the loop being extended in $[0,\beta]$ instead of a value at a point. $\endgroup$
    – ACuriousMind ♦
    Commented Apr 17 at 17:04

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