In a paper 1105.5073, the authors took a simply-laced root system $\Delta$ of type $G=A,D,E$, and then counted the number of unordered triples $(\alpha,\beta,\gamma)$ of roots which sum to zero: $\alpha+\beta+\gamma=0$.

They found that there are $rh(h-2)/3$ such triples, where $r$ is the rank of $G$ and $h$ is the Coxeter number of $G$.

Of course we can show this by studying the simply-laced root systems one-by-one (which is what the authors did.)

My question is if there is a nicer way to show this without resorting to a case-by-case analysis, using the general property of the Coxeter element, etc.

**update**

Let me add a bit of background info why string theorists care about this.

In string theory, there are things called D-branes. If $N$ D-branes are put on top of each other, we have $U(N)$ gauge fields on top of it, and there are of order $\sim N^2$ degrees of freedom on it. There are slightly more involved constructions which give us gauge fields with arbitrary simple group $G$. Then the number of degrees of freedom is $\dim G$.

In M-theory, there are things called M5-branes. If $N$ M5-branes are put on top of each other, we do not know what is on top of it. But there is an indirect way to calculate how many degrees of freedom there should be; and the conclusion is that there should be $\sim N^3$ degrees of freedom (hep-th/9808060). It is one of the big unsolved problem in string/M theory to understand what mathematical structure gives rise to this $N^3$ degrees of freedom.

We can call $N$ M5-branes as the $SU(N)$ version of the construction. There are slightly more involved constructions as in the case of D-branes, but it's believed that there are only simply-laced ones. These are the so-called "6d $\mathcal{N}=(2,0)$" theory which plays an important role in physical approach to geometric Langlands correspondence, etc. See Witten's review.

There are various stringy arguments why there are only simply-laced variants in 6d, but I like this one by Henningson the best. Anyway, the number of degrees of freedom for D and E cases was calculated in this one and this one; it turned out to be given by $h_G \dim G /3$. (As there is only simply-laced ones, there's no distinction between the dual Coxeter number and the Coxeter number.)

This product of the dimension and the (dual) Coxeter number was obtained in a very indirect way, and we'd like to know more. Bolognesi and Lee came up with an interesting numerological idea in the paper cited at the beginning of this question. There idea is to think of $h_G\dim G$ as

$h_G\dim G/3 = hr + h(h-2)r/3 = $ (number of roots) + (number of unordered triples of roots which sum to zero).

They propose that the first term corresponds to "strings" labeled by roots, and the second term as the "junctions of three strings". To consistently connect three strings at a point, the roots labeling them need to sum to zero. (They didn't write this way in their paper; they use more "physical" language. I "translated" it when I made the original question here.) This makes a little bit more precise the vague idea that three-string-junctions are important in M5-branes.

I found this fact on the simply-laced root systems quite intriguing, and therefore I wondered a way to prove it nicely.

unordered, though that is brought in later on. Anyway, it still seems necessary to study the Dynkin diagrams one by one. Is there a way to avoid the classification, while keeping the simply-laced assumption? (And then there's the question of treating other root systems.) $\endgroup$ – Jim Humphreys May 27 '11 at 13:12dimensionand the Coxeter number, divided by 3. Note too that the paper refers to the dual Coxeter number, which suggests a broader question for non-simply-laced types where this differs from the Coxeter number. Early papers by Kostant are probably the best source for the use of the Coxeter number in root system questions like this. $\endgroup$ – Jim Humphreys May 27 '11 at 15:313more comments