What does ad$f$ mean, for $f$ a smooth function? I am currently reading Nicole Berline "Heat Kernels and Dirac Operators". On page 64 Differential Operators are introduced that are generalized from operators acting on scalar functions to vector bundles. 
Thereby it is mentioned that if $D$ is an i-th order operator and $f$ is a smooth function, then (ad $f)^i D$ is a zeroth order operator.
I am not sure I understand this comment, in case anybody can help that would be great!! Please let me know in case more details are necessary.
 A: A function determines a 0th order differential operator by pointwise multiplication.  Then $(\text{ad } f) (D) = [f,D]$ is the commutator of the operator corresponding to $f$ with the operator $D$.
The statement that $(\text{ad } f)^i D$ is 0th order if $D$ is of order $i$ follows from the Leibniz (product) rule.
A: This basically explains the adjoint operator though it does not address the case of the differential operator, but still useful. I posted this as a result of your response to my comment above. I thought perhaps you might need some basic understanding. So here you are, as you requested, the definition of the adjoint of a smooth function;
In general, suppose  $\mathfrak{g}$ be is a Lie algebra with $X\in\mathfrak{g}$. The adjoint action of $\mathfrak{g}$ on itself is the map $\mathrm{ad}(X):\mathfrak{g}\longrightarrow\mathfrak{g}$ defined by $\mathrm{ad}(X)\left(Y\right)=\left[X,Y\right]$,    $\forall Y\in\mathfrak{g}$, where $\left[.,.\right]$ is the Lie bracket. 
So in particular, if $\mathfrak g=C^{\infty}(\mathbb R^n)$, the vector space of smooth functions on $\mathbb R^n$, and  $f\in \mathfrak g$, then  we dfine $\mathrm{ad}f$ as the operator such that for every $g\in\mathfrak g$ $$\mathrm{ad}f(g)=\left[f,g\right]$$
