Computing characteristic polynomial of unitary operation

I am trying to replicate a calculation from the linked paper but I am unsure if I understand their math. A locally invariant function defined as the characteristic polynomial is as follows: $$F_U(t) = det[\Re[M^{\dagger}UM] + t \cdot \Im[M^{\dagger}UM]]$$ Then it is said for $$U$$ with interaction coefficients $$(x,y,z)$$, we have $$F_U(t) = (t^2+1)(Ct^2 + Bt + A) -t^2$$ I'm having trouble understanding how $$(x,y,z)$$ are related to $$t$$, and because neither $$M$$ nor $$U$$ are defined as functions of $$t$$, then it seems like $$F_U(t)$$ is not a degree-4 polynomial.

In this paper, equation (9), they define $$U$$ with parameters $$(\alpha, \beta, \gamma_1, \gamma_2, \delta_1, \delta_2)$$ and find the corresponding coefficient $$C=\frac{1}{16}(\cos{2\alpha}-\cos{2\beta})^2$$. When I plug the same values in, since $$U$$ is not a function of $$t$$, then $$F_U$$ is not degree-4 and hence $$C=0$$. Any guidance would be appreciated, thank you.

• how are the coefficients $(x,y,z)$ related to $U$ exactly?
– glS
May 17, 2022 at 20:06
• I have no idea how all of these quantities are defined, but if $U$ is a $4\times 4$ matrix (two qubits as indicated by the title), then $F_U(t)$ is of course of degree 4. Recall $\mathrm{det}(t A) = t^d \mathrm{det}(A)$ if $A$ is $d \times d$. May 18, 2022 at 6:53
• For $U \in SU(4)$, $(x,y,z)$ are defined as $U$'s coordinates in the Weyl chamber. I did not know about this rule $\text{det}(tA)=t^d \text{det}(A)$ which appears to be what I was looking for. Thanks! May 18, 2022 at 15:56

Resolved in comments, missing step was $$\text{det}(tA) =t^d\text{det}(A)$$