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).

`$G_2$`

.) – Jim Humphreys Sep 16 '10 at 22:12