3
$\begingroup$

Let $T$ be a maximal torus of a compact Lie group $K$, and let $\Psi \subset {\mathfrak t}$ be the (finite) set of coroots for $(K,T)$, where $\mathfrak t$ is the Lie algebra of $T$.

Assume now that $K'$ is another compact Lie group that admits a homomorphic embedding $i:K\hookrightarrow K'$ verifying the condition that $i(Z_K)\subset Z_{K'}$, where $Z_K, Z_{K'}$ are the centers of $K$ and $K'$, respectively. Let $T'$ be a maximal torus of $K'$ such that $i(T)\subset T'$. This gives a vector space inclusion ${\mathfrak t} \subset {\mathfrak t}'$, into the Lie algebra of $T'$.

Question: Is it true that $$ \Psi = {\mathfrak t} \cap \Psi' \quad ?$$

where $\Psi' \subset {\mathfrak t}'$ are the coroots of $(K',T')$.

If this is not always true, are there some simple conditions under which it becomes true?

$\endgroup$

1 Answer 1

6
$\begingroup$

This can fail. Consider the natural embedding $\operatorname{Sp}_4(\mathbb C) \to \operatorname{SL}_4(\mathbb C)$, where $\operatorname{Sp}_4$ is taken with respect to the symplectic form $(x, y) \mapsto x_1 y_4 + x_2 y_3 - x_3 x_2 - x_4 y_1$. (This is a map of complex Lie groups, but it carries maximal compact subgroups to maximal compact subgroups, so you can work in that setting if you prefer.) If we denote the simple (with respect to the upper-triangular Borel) roots of the diagonal torus in $\operatorname{SL}_4(\mathbb C)$ by $\alpha_1, \alpha_2, \alpha_3$, in the obvious fashion, then the restrictions of $\alpha_1$ and $\alpha_3$ to the diagonal torus in $\operatorname{Sp}_4(\mathbb C)$ are the short simple (with respect to the upper-triangular Borel) root $a$, and the restriction of $\alpha_2$ is the long simple root $b$. Then $a^\vee$ equals $\alpha_1^\vee + \alpha_3^\vee$, which is not a coroot of $\operatorname{SL}_4(\mathbb C)$.

(As discussed in comments, the original version of the question did not include the condition on centres, and so admitted any proper, maximal-rank subgroup as a counterexample. This example does satisfy the centre condition.)

$\endgroup$
6
  • $\begingroup$ Thank you, @LSpice, I think your first answer is a good counterexample. I was actually thinking in the situation when the embedding includes the centers as well. I added that extra condition in the question. So, if you want, you can remove the last sentence, and I accept the answer. $\endgroup$ Commented Feb 19, 2023 at 18:29
  • 1
    $\begingroup$ General argument might be something like this: for the adjoint representation of a semisimple group, $\Psi'$ is of type A, so $\Psi$ cannot be a part of it unless it is itself of type A. Correct? $\endgroup$ Commented Feb 20, 2023 at 10:40
  • $\begingroup$ @მამუკაჯიბლაძე, re, of course, you want the group to be adjoint, not jut semisimple, so that its adjoint representation is an embedding. I'm not completely sure why a subset of type $\mathsf A$ that is an irreducible root system is of type $\mathsf A$. Restriction of $W'$-invariant positive-definite pairing is $W$-invariant, so lengths are unchanged, so subsystem is simply laced. But how do we rule out $\mathsf D$ or $\mathsf E$? $\endgroup$
    – LSpice
    Commented Feb 20, 2023 at 13:46
  • $\begingroup$ @მამუკაჯიბლაძე, re, oh, I see, the subset that's a root system is a closed subsystem (since it's the intersection of a subspace with a (co)root system), and every proper subsystem of $\mathsf A$ is contained in a disjoint union of two smaller $\mathsf A$'s by Borel–de Siebenthal-type stuff. (Is it possible to put a $\mathsf D$ or a $\mathsf E$ inside a $\mathsf A$ as a non-closed subsystem? If not, then is it obviously not?) $\endgroup$
    – LSpice
    Commented Feb 20, 2023 at 14:02
  • 1
    $\begingroup$ Maybe I am neglecting something, but a system of type D or E contains three pairwise orthogonal (co)roots together with a fourth one not orthogonal to any of these three. This cannot happen with A, right? $\endgroup$ Commented Feb 20, 2023 at 21:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .