# What is the $K_i$ in Fubini-Study tensor?

Following this paper and tutorial from Pennylane, I'm trying to re-calculate the Fubini-Study tensor:

$$g_{i j}^{(\ell)}=\left\langle\psi_{\ell-1}\left|K_{i} K_{j}\right| \psi_{\ell-1}\right\rangle-\left\langle\psi_{\ell-1}\left|K_{i}\right| \psi_{\ell-1}\right\rangle\left\langle\psi_{\ell-1}\left|K_{j}\right| \psi_{\ell-1}\right\rangle$$ where $$\left|\psi_{\ell-1}\right\rangle=V_{\ell-1}\left(\theta_{\ell-1}\right) W_{\ell-1} \cdots V_{0}\left(\theta_{0}\right) W_{0}\left|\psi_{0}\right\rangle$$

with: $$K_i$$ is the generator of the parametrized operation. In this case, when calculating $$g_{0,1}^{(0)}$$, I understand that $$K_0=K_1=-\frac{1}{2}Z$$ with $$Z$$ is the 2x2 Pauli matrix. But $$|\psi\rangle$$ is the state of 3 qubits, so how to perform $$g_{0,1}^{(0)}$$? Or $$K_0=-\frac{1}{2}Z\otimes I\otimes I$$ and $$K_1=-\frac{1}{2}I\otimes Z\otimes I$$?

It is the latter. One usually doesn't spell it out, but it should be understood that $$Z$$ acting on a single qubit means that it has to be padded by identities acting on the other qubits. Often one uses a shorthand like $$Z_1$$ for $$Z$$ acting on the first qubit to make this explicit.