# Breaking symmetries

Back when I was studying electromagnetism and Maxwell's equations, our teacher told us a quote. I can't recall it exactly, but the meaning was roughly the following:

Symmetry in a problem is useless if you don't have the means to exploit it.

(by the way, I would be delighted if a nice soul could provide the source for it.)

It makes sense in the context of electromagnetism: the effect of symmetries in the initial condition is not as simple as one might naively think. For instance, a planar symmetry for the charges yields a planar symmetry for the electric field, while a planar symmetry for the current yields an antisymmetry for the magnetic field. Hence, the effect of a given symmetry in the initial conditions depends on the properties of the equations.

I was later quite surprised when I learned of some much more striking examples. The first one which comes to mind is the following:

What is the shortest graph which connects the vertices of a square?

The first reflex of most people would be to look at graphs which have the same symmetry as the square ($D_4$). That's an error. The solutions exhibit some degree of symmetry ($D_2$), but less than the square!

However, I don't know any other nice examples for which the solution is less symmetric than the problem (except perhaps sphere packings, but that's less surprising). I think it would be nice to have a list as diverse as possible, both to hone my intuition and to provide counter-examples to my students. And, frankly, because this kind of phenomenon is quite fun.

• I didn't understand the "shortest graph" problem until I clicked the link, because I was interpreting the word "graph" as an abstract graph. Maybe you could say something like "shortest set of curves" or "shortest drawing". – Generic Human Jun 27 '12 at 0:28
• The link in the penultimate paragraph is dead ("inaccessible")... – The Chaz 2.0 Jan 15 '15 at 15:52
• @The Chaz 2.0: Thanks for the notice. I have replaced the link. – D. Thomine Jan 15 '15 at 21:11

## 3 Answers

I think there are two very distinct effects at play here: whether the rules of the problem are invariant under symmetry, and whether the solutions of the problem are invariant under symmetry.

The first case is embodied by the electromagnetism example. When this happens you really can't say anything, and the apparent symmetry really isn't one. Nothing to see here.

The second case is the "shortest road" problem. In the same spirit but simpler, you can look at $$P(x)=x^2+1$$ which is a real polynomial, in other words symmetric with respect to the $x$ axis symmetry (conjugation): $P(\bar x)=0$ iff $P(x)=0$.

Yet no solution is symmetric with respect to conjugation: the roots are $i$ and $-i$, which are both away from the $x$ axis. The symmetry was broken.

However, taken as a set, the set of roots is $\{i,\,-i\}$, which is globally invariant under conjugation. This is a general feature of symmetric problems. In fact the set of symmetries of a problem can be defined as the set of $\sigma$ from a given family that leave the set of solutions invariant. (Of course, you may object that under this definition a problem with no solutions is maximally symmetric, even though these hidden symmetries do not appear in the structure of the problem and are of no use to actually solving the problem.)

Excellent question! I will try to add my bit.

I think that the archetypal example of such behavior happens in game theory. It is not counterintuitive, and exactly for this reason may be used to explain the problem.

Consider the following symmetric payoff matrix:

$$\left[\begin{matrix}0&1\\1&0\end{matrix}\right],$$

the only pure Nash equilibria are pairs $\langle A,B \rangle$ and $\langle B,A \rangle$. This is a situation analogous to your example (there are two optimal solutions: vertical and horizontal one), and provides some intuition: the symmetric solutions carry some cost that makes them suboptimal. There are means to extend the set of strategies so that there is a symmetric one (choose $A$ with probability $\frac{1}{2}$, or in the case of a square, build both vertical and horizontal solution with edge weights $\frac{1}{2}$), but it is often unsatisfying. The problem is, how to break the tie? It might not be possible to do it in a deterministic way and the harder it is, the more the result seems to be surprising!

Cheers!

There are infinitely many examples of that phenomenon.
For the importance in physics I want to mention Laplace equation (invariant for rotations) and its solutions.
But the simplest I can imagine is the equation $x + y = 0$, $x, y\in \mathbb R$.
It is invariant for the transformation $$x\to y\\ y\to x$$ but each of its solutions (apart from $(0, 0)$ of course) is not invariant and it is sent to another one.
The symmetry of the problem is mirrored by the symmetry of the solution space not by that of the single solution.
In that sense, the square problem is not an exception. Single solutions don't have $D_4$ symmetry, space of solutions does (since the rotate solution is still a solution).