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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.

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5  
I am not sure this works, but I think the number of such triples is the number of $A_2$-subsystems in $\Delta$. If you show that each $\alpha\in\Delta$ lies in $h-2$ such subsystems, then the number is $h-2$ times the number of roots (which is $rh$) divided by $3$ (since each subsytem is counted three times). Fix $\alpha\in\Delta$. We can assume $\Delta>0$. For simply-laced systems, I think $\alpha$, $\beta\in\Delta^+$ span an $A_2$ if and only if $\langle\alpha,\beta\rangle\neq0$. We need to count this number and it should equal $2h-4$ (since we will be counting twice each subsystem). –  Claudio Gorodski May 26 '11 at 23:54
    
I meant we can assume $\alpha>0$ on line 4 above. –  Claudio Gorodski May 26 '11 at 23:55
    
Ah, that's a good observation. Let me think about it... –  Yuji Tachikawa May 27 '11 at 8:10
    
Claudio's suggested approach in the first sentence doesn't at first take account of the assumption that the triples are 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.) –  Jim Humphreys May 27 '11 at 13:12
    
@Yuji: To follow up my other comment, I don't believe the authors of the paper you cite are talking about unordered triples. And their counting formula arrives at the product of the dimension and 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. –  Jim Humphreys May 27 '11 at 15:31

3 Answers 3

up vote 13 down vote accepted

Assuming all roots have norm 2, this is essentially the same as showing that the number of roots having inner product 1 with a fixed root $\beta$ is 2h-4, which in turn follows from the property that $\sum_\alpha(\alpha,\beta)^2/(\alpha,\alpha)(\beta,\beta)=h$. This equality is one of many standard properties of h, given in Bourbaki ch V no 6.2 corollary to theorem 1.

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To compensate for my unfocused earlier comments it may be useful to supplement Richard's efficient answer based on Bourbaki's treatment of Coxeter elements in finite reflection groups. There is a short and largely self-contained 1999 paper by Weiqiang Wang which deals directly with (crystallographic) root systems and dual Coxeter numbers for all simple types. This was inspired partly by Wang's thesis and his joint work with Victor Kac on vertex operator algebras, but is elementary in nature. The link to the paper is here.

Briefly, for the given irreducible root system $\Delta$ one first fixes a positive system, with $\rho$ the half-sum of positive roots and $\theta$ the highest root having normalized inner product $(\theta, \theta) =2$. Following Kac, the dual Coxeter number is defined by $h^\vee = (\rho,\theta)+1$. (Standard arguments show that for simply-laced types this agrees with the usual Coxeter number $h$.) Now a positive root $\alpha$ is called special if $\theta-\alpha \in \Delta$. Then (Lemma 2) the number of special roots is $2(h^\vee -2)$. Each defines an ordered triple $(-\theta, \alpha, \theta-\alpha)$ summing to 0.

The lemma (combined with $h = h^\vee$ for simply-laced $\Delta$) provides another approach to the question here: All roots have length equal to that of $\theta$ and are conjugate under the Weyl group ($\Delta$ being irreducible), so each root plays the role of $\theta$ for some positive system. But the number of roots is $rh$ as noted already and thus the total number of sets $\{\alpha,\beta,\gamma\}$ summing to 0 is $2(h-2)rh/6 = rh(h-2)/3$ because $6$ ordered triples yield just one such set. .

While only the simply-laced root systems seem to have physical significance here, I wonder whether Wang's formulation in terms of $h^\vee$ has further combinatorial implications. (He used it to compute the dimension of the minimal nilpotent orbit in an associated simple Lie algebra: namely, $2h^\vee -2$.)

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Thank you very much! It's such an honor to have your response! –  Yuji Tachikawa May 29 '11 at 2:27

There is another approach by the strange formula of Freudenthal and de Vries, which states that $h^\vee d = 12 \rho^2$ where the Weyl vector $\rho$ is one half of the sum of positive roots. The long roots have the length square two. Thus $$\frac13 h^\vee d = \sum_{\alpha>0} \alpha^2+ \sum_{\alpha,\beta>0,\alpha\neq\beta}\alpha\cdot\beta $$ For simple-laced group, the first term of RHS is the number of roots $hr=d-r$, and the second term of RHS should be $rh(h-2)/3$. For the non-simple-laced case, there may be still some interesting interpretation of the above decomposition.

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