There is an interesting everyday depiction of cohomology that nicely demonstrates how cohomology represents an obstructions to a global constructions. This is the fact that the Penrose Triangle represents a nontrivial cohomology class in the "group of distances from the observer" of the circle. Penrose has an article on this here: On The Cohomology of Impossible Figures.
I find this interesting because I find cohomology interesting and the fact there is such an accessible example illuminating it is pretty neat.
A Shepard tone is a sort of auditory analogue of the Penrose Triangle (or Escher's staircase). This is the noise whose pitch appears to be continuously rising, but isn't really.
For a short interval of time it appears as if the pitch is increasing, but at the end of the interval we are back where we were started. In this way a Shepard tone resembles the impossible figures referenced above. Because of this I imagine there could be a similar analysis in terms of cohomology of Shepard tones as with the Penrose triangle.
My question: Can any exact analysis in terms of cohomology of Shepard tones really be made and if so how?
Why do I want an answer to this question? Supposing the answer to this question is yes: If this has an elementary explanation then I would find it a welcome addition to the Penrose Triangle example and if it is not so elementary then it must have some interesting math to it.
This is still a lousy question as it is so I'll try to show my work on it.
- I have a naive mental model of the Shepard tone as a function $\mathbb{R} \to \mathbb{R}$ (time to pitch) which is constant and yet has a positive derivative everywhere.
- In contrast with Escher's stairs the problem is more of one going from "infinitesimal to local" than "local to global." This is the basic problem.
- A solution around this may have to do with how the tone is comprised of waves, and these aren't really infinitesimal data.
As an intermediary question, in case the main question doesn't have an answer/has an answer in thetive negative, I'll ask: How can we mathematically recognize the auditory illusion here? How do we get from the more accurate wave description to the paradoxical description in the first bullet point above?
The Shepard tone is a periodic signal, and one way to speak of a periodic signal is as a function $S^1\to\mathbb{R}$, where $S^1$ is the unit circle (which is conveniently thought of as an interval $[0,T]$ with the times $0$ and $T$ identified). The value of the function is the amplitude of the signal at that particular moment in modulo-$T$-time.
A conventional tone, produced by a conventional musical instrument, decomposes into sine waves. If $f$ is the frequency of the lowest-frequency component, called the fundamental frequency, then all of the other frequencies present in the tone are multiples of $f$, and these are called harmonics. The fundamental frequency usually defines the pitch of a tone (perceptually, however, it depends on the relative loudness of all the harmonics).
In contrast, the Shepard tone has at any given moment all the frequencies $2^nf$ for all $n\in\mathbb{Z}$. This means the fundamental frequency, if there were one, can always be divided by two. Like for distance in the Penrose triangle, let's say the fundamental frequency is ambiguous. The ambiguity group (using Penrose's terminology) is the set of all $2^n$ multiplicatively, but this is isomorphic to $\mathbb{Z}$ as an additive group.
Although the fundamental is ambiguous, through time we can measure how much the fundamental, if there were one, has changed. If the Shepard tone doubles in frequency after one period, then the tone corresponds to a generator of $H^1(S^1;\mathbb{Z})$.
(One might think that the relative amplitudes of the frequencies in the Shepard tone are part of the illusion, but I think it's actually just so the sum of all of the frequencies converges. The Penrose triangle doesn't have this concern because it obstructs the itself that's behind itself, so to speak.)