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.