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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.

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  • $\begingroup$ Let $G$ be the adjoint group of type $\Phi$, and $H$ the subgroup of $G$ generated by the roots in $S$. Then your index is the size of the centre of $H$. (This is true over, say, an algebraically closed field of characteristic $0$. I don't know how to make sense of it over the integers, but I also don't know that it can't be made sense of in that setting.) $\endgroup$
    – LSpice
    Commented Sep 18, 2018 at 21:57
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    $\begingroup$ In particular, I think that Borel–de Siebenthal theory should enumerate the possibilities, and you can just check the indices in the various cases. $\endgroup$
    – LSpice
    Commented Sep 18, 2018 at 21:58

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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 $\mathsf E_8$).

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    $\begingroup$ I went a large portion of my life—including attending at least one talk on it—without really understanding what BdS theory was. When Stephen DeBacker showed it to me, the simultaneous power and simplicity of it seemed like a magic trick (and still kind of does). $\endgroup$
    – LSpice
    Commented Sep 20, 2018 at 20:11
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    $\begingroup$ Good motivation for me then! $\endgroup$
    – Sam Hopkins
    Commented Sep 20, 2018 at 20:12

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