In the pre-quantum Hilbert space, Poisson bracket algebra of pre-quantum operators is reducible because pre-quantum wave-functions depend on all phase space variables. To get an irreducible representation of the algebra one introduces a polarization by using a covariant derivative where the connection is the symplectic potential. This covariant derivative acting on the pre-quantum wave-function is called the polarization condition. This procedure forces the wave-function to depend on half of the phase space variables that commute with each other. Generally there are there options: choosing the coordinate Hilbert space where the quantum wave-function is $\psi(x_i)$, choosing the momentum Hilbert space where $\psi(p_i)$ or choosing the Segal-Bargmann space $\psi(z_i)$, where $z=x+iy$.

So basically the covariant derivative is zero in the polarized section because that is the condition that defines the polarized section.

I hope I understood your question right.

Refs: [1] V.P. Nair. Quantum Field Theory: A modern perspective. Springer, 2005. [2] B. C. Hall. Quantum Theory for Mathematicians. Springer, 2013.

In the pre-quantum Hilbert space, Poisson bracket algebra of pre-quantum operators is reducible because pre-quantum wave-functions depend on all phase space variables. To get an irreducible representation of the algebra one introduces a polarization by using a covariant derivative where the connection is the symplectic potential. This covariant derivative acting on the pre-quantum wave-function equals zero is called the polarization condition. This procedure forces the wave-function to depend on half of the phase space variables that commute with each other. So the quantum wave-function cannot be a simultaneous eigenstate of non-commuting variables. Generally there are three options: choosing the coordinate Hilbert space where the quantum wave-function is $\psi(x_i)$, choosing the momentum Hilbert space where $\psi(p_i)$ or choosing the Segal-Bargmann space $\psi(z_i)$(holomorphic polarization), where $z=x+iy$.

So basically the covariant derivative is zero in the polarized section because that is the condition that defines the polarized section.

I hope I understood your question right.

Refs:

[1] V.P. Nair. Quantum Field Theory: A modern perspective. Springer, 2005.

[2] B. C. Hall. Quantum Theory for Mathematicians. Springer, 2013.