Here are my thoughts: Say I have two manifolds $M$ (one dimensional in my thoughts) and $\mathbb{R}$. Thinking in physical terms; $M$ I imagine as my space of states: of possible configurations of my physical system.

Now, how do I perform any sorts of calculations as a physicist? I need a function $\Psi :M \to \mathbb{R}$, in order to be able to somehow label my states. Of course my states now acquire any other structure I have endowed $\mathbb{R}$ with: partial order (to be able to tell which state preceded the other); continuity (to state that I understand what nearby states are, but that I can't quite distinguish them) etc.

Suppose that I find out, via this $\Psi$, that $\Psi(x)=\Psi(x+L)$. Now I could say: look, honestly, I wasn't sure of the nature of the manifold $M$, but having found that $\Psi(x)=\Psi(x+L)$: because $\Psi$ is single-valued, actually, my domain wasn't $M$ but rather $M/_{\sim}$, under the identification of $x$ with $x+L$, i.e., $S^1$.

In this case, the Fourier transformation $\Psi = \sum_k c_k e^{ikx}$ is merely an expansion of a function defined on a circle. And the circle is the manifold which is invariant under $x\mapsto x+L$.

The transformation $x\mapsto x+L$ is non-linear; while the transformation $$ i\dfrac{d}{dx} $$ is Hermitian and linear--though acting on a function space--the eigenfunctions of which are the basis $e^{ikx}$.

Is there a duality under the non-linear shift and the derivative operator?

The shift operator gives us topological information about the manifold: it tells us that the basis functions are defined on the circle, while the linear operator (naively?) doesn't seem to carry such topological information.

Can we acquire topological information about $M$ by closer inspection of the Hermitian operator?

  • $\begingroup$ I'm not sure what you mean by duality. The translation/shift operator is the exponential of the derivative. (We say the derivative is the 'generator' of translations.) Generally infinitesimal generators don't know about global/topological data, but their exponentials might. $\endgroup$ – d_b Mar 22 '19 at 19:25

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