# Supercommutator and covariant derivation

Let $$(E,\nabla)\to X$$ be a vector bundle endowed with a connection and $$f\in End(E)$$ a bundle endomorphism. We can express the covariant derivative of $$f$$ using a commutator $$\nabla f=[\nabla,f].$$

Let now $$E=E^+\oplus E^-\to X$$ be a super vector bundle endowed with a connection $$\nabla=\nabla^+ \oplus \nabla^-$$ preserving the splitting. We define a superconnection on $$E$$ ("Heat Kernels and Dirac operators" p.44)

In this definition the brackets in (2) represent a supercommutator defined as $$[a,b]=ab - (-1)^{|a||b|}ba.$$

My question is the following: I do not understand how this definition of the covariant derivative of an End-valued form extends the standard definition above.

For example: Take the superconnection to be the connection $$\nabla=\nabla^+\oplus \nabla^-$$ and take an odd endomorphism $$f=\begin{pmatrix} 0 & f_1\\ f_2 & 0 \end{pmatrix}$$ where $$f_1:V^-\to V^+$$ and $$f_2:V^+\to V^-$$. The classical commutator gives $$\nabla\circ f - f\circ \nabla = \begin{pmatrix} 0 & \nabla f_1\\ \nabla f_2 & 0 \end{pmatrix}$$ while the supercommutator $$[\nabla,f]=\nabla \circ f + f\circ \nabla$$ does not seem to give the good answer.