Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
It is the adjoint of $f$ –  smanoos Jan 4 '12 at 0:05
    
Ah ok, do you know where I can find a definition for the adjoint of a smooth function ? That is, what is the meaning of the adjoint in this context ? –  harlekin Jan 4 '12 at 0:21

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer
    
Just being a little curious. What will be the definition of the commutator of a function and an operator as you have in your answer. –  smanoos Jan 4 '12 at 1:29
    
For $s$ a section of the bundle, $[f,D](s) = fDs - D(fs)$. –  Eric O. Korman Jan 4 '12 at 3:52

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 $g\in\mathfrak g$ $$\mathrm{ad}f(g)=\left[f,g\right]$$

share|improve this answer
    
Note that the Lie bracket is the same as the commutator –  smanoos Jan 4 '12 at 0:44
    
What is the Lie algebra structure you are using on $C^\infty(\mathbb R^n)$? Further, in the question $\text{ad } f$ is being applied to a differential operator, not another function. –  Eric O. Korman Jan 4 '12 at 0:46
    
@Eric. I agree with you, but it is still useful in explaining the basic concept of the adjoint map, though it does not mention differential operators. –  smanoos Jan 4 '12 at 0:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.