Example. Quaternion spinor gauge fields [fdip-006X]

✍️sourceAGENTDRAFTED

Take \(G=Q_8=\{\pm 1,\pm i,\pm j,\pm k\}\) in the unit quaternions and its defining complex spinor representation \(\sigma :Q_8\to SU(2)\). Its character is \(\chi _\sigma (h)=2\operatorname {Re}h\). The 4 linear characters factor through \(Q_8/\{\pm 1\}\cong C_2\times C_2\). The spinor has character norm \((2^2+(-2)^2)/8=1\), so is irreducible. These 5 inequivalent irreducibles are complete because \(4\cdot 1^2+2^2=8\). This is the quaternion representation already used in the group-algebra examples.

For the spinor Wilson weight, put \(q=e^{-\beta }\). Then \[f_\beta (1)=1,\qquad f_\beta (-1)=q^2,\qquad f_\beta (\pm i)=f_\beta (\pm j)=f_\beta (\pm k)=q.\] Its irreducible character coefficients are \[\begin {aligned} a&=\frac {1+q^2+6q}{8} &&\text {(trivial)},\\ b&=\frac {(1-q)^2}{8} &&\text {(each other linear character)},\\ c&=\frac {1-q^2}{4} &&\text {(spinor)}. \end {aligned}\] For a nontrivial linear character, the sum over the 6 noncentral units is \(-2\); for the spinor it is zero, while the central values are \(2,-2\). Substitution into the character inner product proves all three formulas. The surface theorem gives \[Z_\Sigma (f_\beta )=a^P+3b^P+ 2^{2-2g}\left (\frac c2\right )^P.\] At \(\beta =0\), only \(a=1\) remains. At the identity, \(a+3b+2c=1\), checking the coefficient normalization.

A one-face torus has one vertex and two loop edges \(x,y\). The square below identifies equally labelled sides in their arrow directions and all 4 corners. A positive boundary traversal is \(x,y,x^{-1},y^{-1}\). Writing \(A=U_x\) and \(B=U_y\), its holonomy is \(H=B^{-1}A^{-1}BA\).

The normalized Wilson loop is \(W_\sigma =\operatorname {Re}H\). Of the \(64\) ordered pairs \((A,B)\), exactly \(40\) commute: the two central elements have centralizer size 8 and the 6 others size 4. Their commutator is \(1\); the other \(24\) commutators are \(-1\). Hence the direct finite sum gives \[Z_{\mathbb T^2}(f_\beta )=\frac {5+3q^2}{8},\qquad \langle W_\sigma \rangle =\frac {5-3q^2}{5+3q^2}.\] The partition formula also follows from \(a+3b+c/2\). This single plaquette is the whole torus, not a disk bounded by a simple contractible loop of the earlier square-torus theorem. Its boundary word can have a nontrivial holonomy because the connection is not constrained to be flat.