TITLE: How big can the index inside the root lattice of the lattice generated by a subset of roots be?
QUESTION [6 upvotes]: Let $\Phi$ be an irreducible crystallographic root system in a Euclidean vector space $V$. Let $S\subseteq \Phi$ be some subset of roots for which $\mathrm{Span}_{\mathbb{R}}(S)=V$.
Question: How big can $[\mathrm{Span}_{\mathbb{Z}}(\Phi):\mathrm{Span}_{\mathbb{Z}}(S)]$ be?
If $\Phi$ is of Type A, then I believe we always have $[\mathrm{Span}_{\mathbb{Z}}(\Phi):\mathrm{Span}_{\mathbb{Z}}(S)]=1$ because of total unimodularity.
But for instance, if $\phi=D_4$ and $S=\{\alpha_1,\alpha_1+2\alpha_2+\alpha_3+\alpha_4,\alpha_3,\alpha_4\}$, where $\alpha_2$ corresponds to the trivalent node, then I get $[\mathrm{Span}_{\mathbb{Z}}(\Phi):\mathrm{Span}_{\mathbb{Z}}(S)]=2$. (With the standard realization of $D_4$ this is $S=\{(1,-1,0,0),(1,1,0,0),(0,0,1,-1),(0,0,1,1)\}$.)
In particular I'd like to know if $[\mathrm{Span}_{\mathbb{Z}}(\Phi):\mathrm{Span}_{\mathbb{Z}}(S)]$ is absolutely bounded or not.

REPLY [4 votes]: As suggested in my second comment, one approach to the answer is via Borel–de Siebenthal theory, which classifies the subgroups $H$ that can arise as in my first comment as those whose Dynkin diagrams arise by deleting a root, other than the lowest root, from the extended Dynkin diagram.  (Deleting the lowest root always gives back the original group.)  The desired index, which is the same as the order of the centre of the resulting group $H$, is the coefficient of the deleted vertex in the highest root, hence is absolutely bounded (by 6, which occurs only in case $\Phi$ is of type $\mathsf E_8$).