# Combinatorial identity involving the Coxeter numbers of root systems

The setup is:

$R$ = irreducible (reduced) root system;

$D$ = connected Dynkin diagram of $R$, with nodes numbered $1,2,...,r$;

$\hat D$ = extended Dynkin diagram, nodes numbered $0,1,2,...,r$;

$\alpha_k$ = $k^{th}$ simple root ($1\le k\le r$); $\alpha_0$ = -(highest root);

label $n_k$ ($1\le k\le n$) = multiplicity of root k in the highest root; $n_0=1$;

$n^\vee_k$ = dual label = multiplicity of co-root $\alpha_k^\vee$ in the highest short co-root;

$\quad\quad=n_k$ ($\alpha_k$ long) or $n_k/c$ ($\alpha_k$ short), where $c=2$ (types $B_n,C_n,F_4$) or $3$ (type $G_2$) (also $n_0^\vee=1$);

$h(D)$ = Coxeter number of D = $\sum_{k=0}^rn_k$ (for example $h(E_8)=30$);

Now, delete node $k$ from $\hat D$ where $n_k>1$. Write $\hat D-\{k\}$ as a union of connected Dynkin diagrams $D_1,\dots, D_s$. Let $h_1,\dots, h_s$ be the Coxeter numbers of $D_1,\dots, D_s$.

Lemma: If $\alpha_k$ is long:

$$\sum_{i=1}^{s}\frac1{h_i}\sum_{j\in D_i}n_j=n_k$$ If $\alpha_k$ is short: $$\sum_{i=1}^s\frac1{h_i}\sum_{j\in D_i}n_j^\vee=n_k^\vee$$

Note that the sum over $D_i$ involves the labels $n_j$ coming from $D$, which have no obvious relation to those coming from $D_i$. On the other hand $h_i$ is the Coxeter number for $D_i$, so this is a mix of data from $D$ and $\{D_i\}$.

The proof is case-by-case.

Question: is there a conceptual proof of the Lemma?

Examples: (Bourbaki numbering)

1) $D=E_8$, $k=2$, $n_2=3$, $\hat D-\{2\}$ = type $A_8$, $s=1$, $h(A_8)=9$: $$\frac19(2+4+6+5+4+3+2+1)=3=n_2$$ (As mentioned above, the numbers $2,4,6,5,4,3,2,1$ come from $E_8$, not $A_8$.)

2) $D=E_8$, $k=4$, $n_4=6$, $\hat D-\{6\}=A1+A2+A5$, $s=3$, $h(A_i)=i+1$: $$\frac12(3) + \frac13(2+4) + \frac16(5+4+3+2+1) = \frac32 + 2 + \frac{15}6 = 6 = n_4$$ Note that the terms in the sum are not all integers.

3) $D=E_8$, $k=8$, $n_8=2$, $\hat D-\{8\}=E_7+A_1$, $s=2$, $h(E_7)=18$, $h(A_1)=2$: $$\frac1{18}(2+3+4+6+5+4+3)+\frac12(1)=\frac{27}{18}+\frac12=2=n_8$$

4) Here is an example involving $n_k^\vee$: $D=C_8$, $k=3$, $\alpha_k$ is short, $n_k=2$, $n^\vee_k=1$ (in fact all $n^\vee_j=1$). $D-\{3\}=C_3+C_5$, $h(C_n)=2n$: $$\frac16(1+1+1) + \frac1{10}(1+1+1+1+1)=\frac12+\frac12=1=n_3^\vee$$

5) By popular demand, $G_2$:

a) $k=2$ (long root), $n_2=2$, $n_2^\vee=2$, $\hat D-\{2\}=A_1+A_1$. Note that $n_1=3,n_1^\vee=1$.

$$\frac12(1)+\frac12(1)=1=n_1^\vee$$

b) $k=1$ (short root), $n_1=3$, $n_1^\vee=1$, $\hat D-\{3\}=A_2$, $h(A_2)=3$. Note that $n_2=n_2^\vee=2, n_0=n_0^\vee=1$: $$\frac13(2+1)=1=n_1^\vee$$

Remark: Write $H$ for the subgroup of $G$ corresponding to $\hat D-\{k\}$. This identity plays a role in some computations related to embedding elliptic elements of the Weyl group of $H$ in those of $G$ (the Coxeter element is an example).

-
This is an intriguing observation, which may not have a convincing explanation outside representation theory or the combinatorial geometry of affine Weyl groups. It might for example come from Langlands duality in some context not so visible in classical Lie theory (The tag weyl-group and/or lie-algebra might be appropriate. Also, it makes the dualization process clearer if you include the case $G_2$.) – Jim Humphreys Sep 16 '10 at 22:12