This is from Zee's Group Theory in a Nutshell for Physicists, page 172. Although this question is connected to physics, I think it's more a question about mathematics.

Zee talks about $N$ dots connected by springs, in a $D$-dimensional Euclidean space. The dots are indexed by $a=1\dots N$ and have coordinates $x^i_a$. The equations of motion are

$$ \frac{d^2x^i_a}{dt^2} = -\sum_{b=1}^N\sum_{j=1}^D H^{ia,jb}x^j_b $$

where the couplings $H^{ia,jb}$ contain both the spring strengths and the masses of the dots.

The book uses this example with $N=3$ and $D=2$ to explain how representation theory can reduce the complexity of the problem.

The invariance group $S_3$ acts on the dots by exchanging them. So in terms of the $DN$ components $x^i_a$, we have a $DN=6$ dimensional representation $\rho$ of $S_3$. It's easy to find out that $\rho=1\oplus 1\oplus 2\oplus 2$, so $\rho$ is reducible and contains both the trivial and the 2-dimensional representation of $S_3$ two times.

Now Zee explains that these representations can be visualized like this:

  • One trivial irrep corresponds to rotations of the triangle.
  • The other trivial irrep corresponds to scalings of the triangle.
  • One 2-dimensional irrep corresponds to translations of the triangle.
  • The other 2-dimensional irrep is a bit less intuitive, it corresponds to shears of the triangle.

I can see that rotations and translations leave the triangle invariant, but I don't see the connection to $S_3$. The group $S_3$ acts by permutations of the dots, which are discrete operations. After each permuation, the dots' locations are the same as before, up to permutations. A translation however is continuous, and it does not leave the dots' locations invariant.

How can the permutation group of the dots "contain" the translations?

Edit: To make things clearer, here's a sketch from the book that describes the situation:

enter image description here

And these are visualizations of the four representations:

enter image description here

a) the translations (2D)

b) the rotation (1D)

c) the scaling (1D)

d) the shears (2D)

  • $\begingroup$ As I understand, each dot is connected to 6 "dots" giving triangular "cells". Is that right ? (it could have been a connection with 4 neighbours and square "cells"). What make you think that a translation should be considered as continuous ? Consider instead that you "jump" directly from the initial position to the final position. $\endgroup$ – Jean Marie Dec 9 '16 at 9:35
  • $\begingroup$ @JeanMarie No, it's just one "cell", i.e. three dots, and each dot is connected to the other two. All the four resulting representations are continuous, including the translation (moving the triangle aroung). See also the images I added in the edit. $\endgroup$ – Andy Miles Dec 9 '16 at 11:05

I'm not yet sure if that's really the answer to my question, but I'm starting to suspect that in this context, group theory is just a tool to simplify the spectral analysis of $H$ (the Hamiltonian).

The system is invariant under the action of the invariance group $S_3$, which means that $H$ does not change under the action of $S_3$. If we denote by $\rho(g)$ the representation of $S_3$ on the $DN$ dimensional vector $x^i_a$, then this invariance means that $H\rho(g)=\rho(g)H$ for all $g\in S_3$.

Now Schur's Lemma tells us that $H=\lambda\mathrm{Id}$ on all irreducible subspaces of $\rho$. Although we do not yet know the complete spectrum of $H$, at least we know that each vector of an irrep of $\rho$ is an eigenvector.

So the continuous symmetries that make up the irreps inside $\rho$ (for example the translation) are not "contained" in the action of $S_3$. The action of $S_3$ is merely a tool to uncover the spectrum of $H$.

Please correct me if that does not make sense.


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