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)Heat Kernels and Dirac operators

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.