2
$\begingroup$

In using the shift operator, which is really just an expression of a Maclaurin (or Taylor) series expansion:

$$f(x+a)=\sum_{n=0}^{\infty}\frac{1}{n!}\left(a\frac{d}{dx}\right)^{n}f(x)=e^{a\frac{d}{dx}}f(x)$$

Or, as a generalization:

$$f(\overrightarrow{x}+\overrightarrow{a})=e^{\overrightarrow{a}\cdot\overrightarrow{\nabla}}f(\overrightarrow{x})$$

How would I generalize the shift operator to deal with a condition of periodicity such as:

$$f(\overrightarrow{x}+\overrightarrow{a})=f(\overrightarrow{x})$$

I think of some kind of condition such as:

$$e^{\overrightarrow{a}\cdot\overrightarrow{\nabla}}=1$$

However, it's an operator, so that really doesn't make sense here. Intuition says it'll have to be an imaginary exponential? Or am I looking at this wrong and this is actually a differential equation putting restrictions on our (probably function) $f(\overrightarrow{x})$.

Note: I'm working on a fairly general (pseudo) Riemannian manifold, so the derivative will be the covariant one in general. Hence, the shift operator will be in general non-commutative.

$\endgroup$
  • 1
    $\begingroup$ You can view this as eigenvalue problem. Periodic functions will be eigenfunctions of your operator, with eigenvalue of $1$. $\endgroup$ – Ruslan Dec 19 '16 at 9:22

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.