I am a bit puzzled on the following circuit. According to this Quantum Computing SE thread it holds that $$ e^{i(Z\otimes Z)t} = {\rm CNOT} (I\otimes e^{iZt}){\rm CNOT} \qquad (1) $$
As a result we have the following circuit (C1):
Furthermore, for $e^{iZ\otimes Z\otimes Z t}$ we obtain the following circuit (C2):
I was reading arxiv:2003.13599 and in Figure 3 we see three unitaries corresponding to $I\otimes Z \otimes I$, $I \otimes Z \otimes Z$ and $Z \otimes Z \otimes I$ seen below (C3):
Which is very confusing. Specifically, let's focus for example on the middle part of Diagram (a) corresponding to $I\otimes Z \otimes Z$. The $I$ term (in the first qubit) can be ignored. Then, if we ignore the first wire we essentially should have the term $Z\otimes Z$ from Eq. (1) but nevertheless we see two pairs of CNOT gates just like in (C2) that corresponds to $e^{iZ\otimes Z \otimes Z t}$.
Why is this the case? I suspect this is somehow related to the extra wire we ( C2 has three wires but for some reason C3 has four).