# Does a glass of water sing because of the SO(2) symmetry?

This might very well be a crucially flawed reasoning. But I think has to have something true behind.

I was trying to explain basic ideas of representation of Lie Groups to an 11 years old girl who asked what I was studying. What I wanted to explain was the relation between special functions and symmetries.

The mathematical thing I wanted to explain was that if we have a group $G$ acting on a space $X$, and we look at the space of infinitely derivable functions on $X$, i.e. $\mathcal C^\infty(X)$, then there is a natural representation of $G$ on $C^\infty(X)$. So I thought about the simplest example I had in mind which was the case of of the circle $G = S^1$ which has ($SO(2)$ symmetry). Then the representation are given by harmonic analysis of Fourier series.

To explain this I said to consider a glass of water which has cylindrical symmetry ("if I rotate the glass you cannot say how much I did rotate the glass, so it has a symmetry") then the vibration and deformations of the edge of the glass are the functions on $S^1$ that can be classified in harmonics...

That's why - I concluded - glasses are used to sing with water... I said that but really I'm not sure at all if it's effectively the case. I mean that I'm note sure until what extension my reasoning was correct. Is the role of the water just the one to annihilate every representation but one (or some) which get excited and that's why the glass emits only one definite sound? Do you think the reasoning has a tragic flaw somewhere? To what extension is valid?

1. Well, we can certainly have singing/ringing from non-axisymmetric objects, cf. e.g. this, this Phys.SE posts and links therein. So $SO(2)$ symmetry is not necessary in that sense.
2. However, according to e.g. Ref. 1, it is apparently a good mathematical model to consider a wineglass as having topology $S^1\times I$, where the interval $I$ has 1 free and 1 fixed endpoint. The circle $S^1$ has $SO(2)$ symmetry and can be analyzed via Fourier series. The interval $I$ typically has no symmetry. The lowest mode has 2 nodes along $S^1$ and 0 nodes along $I$. There are also higher harmonics/overtones.