# Why does Group Theory not come in here?

Here is a list of questions that I find quite similar, for the one and only reason that they all show much "symmetry". Which is at the same time my problem, because I don't have a very precise notion of that supposed "symmetry". Here comes:

Take the last question as an example. Let $a=b=c=d$ , then $a^2+b^2+c^2+d^2=4\,$ is of course fulfilled, because we know that $a,b,c,d\ge 0$ , if and only if $a=b=c=d=1$ .
Nobody would be surprised (?) if $\;f(a,b,c,d)=a/(b+3)+b/(c+3)+c/(d+3)+d/(a+3)$ assumes its one and only maximum $f(a,b,c,d)=1/4+1/4+1/4+1/4=1$ precisely for these equal values of $(a,b,c,d)$ .
Very much the same phenomenon is observed with the other of the above questions. The optimizing parameters turn out to be all equal, which often may be suspected beforehand: "due to symmetry".

It's frustrating that there seems not to exist a theorem somewhere that guarantees a maximum or a minimum of a function when all variables in a problem with such high symmetry are just equal.
I mean: sort of formalization of "by symmetry" that helps us to find such solutions immediately.
Now it is supposed that group theory is the discipline that should teach us a lot about symmetries.
So the question is: why doesn't group theory routinely come in here?

Update. Unfortunately - and I think it's against the spirit of MSE as well - it often happens that the better answers are actually given as a comment . Indeed, a key reference for this sort of problems is:

It's only with help of this reference that I could (try to) answer another such question .

Update. It's a continuing story:

And the list keeps growing ..

• I think "symmetry" related to group theory is very different from what you mean in your question, @Han : think of a $\;3-$dimensional cube. Can you see there are several "axis of symmetry" aroundwhich we can revolve or turn the cube? This gives us a group (of symmetries) of the cube. – DonAntonio Jan 4 '14 at 21:08
• Well find the minimum of $abc$ given $a+b+c=1$ and $a,b,c\geq 0$...I think it's just the issue that it don't work in general. Now there are indeed symmetric polynomial, which does get studied in abstract algebra, and group come in as permutation of variable. – Gina Jan 4 '14 at 21:13
• @Gina: (and Geoff) The question suggests "a maximum or a minimum of a function when all variables in a problem with such high symmetry are just equal" and in your example it would be a maximum of $abc$ – Henry Jan 4 '14 at 21:25
• – lhf Jan 4 '14 at 21:32
• @Henry: it's very to make a modification to handle both case. Let $a,b,c\geq 0$ and $a+b+c=4$. Find maximum and minimum of $abc(1-abc)$. – Gina Jan 4 '14 at 21:34

Group theory does come in to these situations - and is in particular very important in physics - but you need to understand how.

There are plenty of examples in the comments of situations where symmetric (under the exchange of variables) problems have asymmetric (under the exchange of variables) extrema.

Suppose that a group $G$ acts smoothly on some variables $\mathbf x$, and is a symmetry of some function $f(\mathbf x)$. This means that $f(g\mathbf x) = f(\mathbf x)$ for all $g\in G$. For simplicity, let's suppose that $G$ is a matrix group, so $g \mathbf x$ is just a matrix multiplying a vector. For example, if $$G=\left\{\pmatrix{1 & 0\\ 0 & 1},\pmatrix{0 & 1\\ 1 & 0}\right\}$$ then this corresponds to leaving the variables alone or swapping them.

Now consider the set of, say, global maxima $S$. Take any $\mathbf y \in S$. Then notice that for any $g\in G$, $g \mathbf y$ has exactly the same properties. In particular, because $$f(g(\mathbf y + \delta \mathbf y)) = f(g\mathbf y + g\delta \mathbf y)$$ you can easily check that $g \mathbf y \in S$ too. (Clearly, $f$ has the same value here, and also if all small changes used to make $f$ smaller, then they still do.)

The result is that $G$ also acts on $S$ - it takes global maxima to global maxima. But it doesn't necessarily act trivially, which is the criterion you want: you want every element of $S$ to be invariant under every element of $G$.

A prominent example used in physics is the Mexican hat potential (which we'll use upside down for no good reason). Consider $f=-(|\mathbf x|^2 - a^2)^2$. This has the symmetry group $G = O(2)$ of all orthogonal two by two matrices because this preserves the length of $\mathbf x$. Clearly the function is biggest at $|\mathbf x| = a$, which is a circle, our set $S$. This is indeed acted upon by $G$: the circle is mapped into itself. However, the rotations and reflections act non-trivially - points on the circle are not left untouched by the group.

The set of local minima, by contrast, is $\{\mathbf 0\}$, which is acted upon trivially by $G$.

This in fact leads on to things like representation theory, where we ask what different possibilities there are for what $G$ can do to various vector spaces.

You can make the above into a single variable example by going to one dimension, where $|\mathbf x| = x$ and $G$ is roughly the example matrix group given above (really it is either leave $x$ alone or consider $x\mapsto -x$). The same thing applies.

Physicists make a big deal of these (spontaneous) symmetry breaking situations because they are key to the Higgs effect by which elementary particles apparently acquire mass. The group theory which governs the vacuum (the set $S$) is very important, and worth saying more about, but this isn't probably isn't the place!

• Nice answer ! But I'll be truly convinced once you can solve one of those puzzles in the list, especially the last one (I mean: in a somewhat more elegant manner as has been done there) – Han de Bruijn Jan 5 '14 at 21:51
• @HandeBruijn The point is that (as I remark in the second paragraph) there are many examples where global maxima and minima are not obtained at symmetric points. As such, there is no general principle whereby one can guarantee that one need only check these. In the excellent article lhf linked to, it is observed that one generically does get a local maximum or minimum (except in degenerate cases) at certain symmetric points. But this tells you nothing about global behaviour. – Sharkos Jan 5 '14 at 22:45