# Functions constant on boundary and topology of underlying manifold

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?

• 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. – d_b Mar 22 '19 at 19:25