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Let $\Phi$ be a (crystallographic) root system and $\Phi^{+}$ a choice of positive roots, with $\Delta$ the corresponding choice of simple roots. So the root lattice of $\Phi$ is just $\mathbb{Z}\Delta$ ($=\mathbb{Z}\Phi^{+}=\mathbb{Z}\Phi$).

Then $\rho := \frac{1}{2}\sum_{\alpha \in \Phi^{+}}\alpha$ (the "half-sum of positive roots", a.k.a. the "Weyl vector") is a distinguished vector associated to $\Phi$ with great significance in representation theory; see e.g. the following Mathoverflow question: What is significant about the half-sum of positive roots?.

A priori, $2\rho \in \mathbb{Z}\Delta$, but in fact it may happen that $\rho \in \mathbb{Z}\Delta$.

Question: is there any representation-theoretic significance to whether we have $\rho \in \mathbb{Z}\Delta$?

With collaborators we have discovered some combinatorial phenomenon that is apparently related to whether $\rho \in \mathbb{Z}\Delta$, and we would like to relate this phenomenon more to representation theory.

It is not hard to work out exactly when $\rho \in \mathbb{Z}\Delta$:

  • Type $A_{n}$: Have $\rho \in \mathbb{Z}\Delta$ iff $n$ is even.
  • Type $B_n$: Always have $\rho \notin \mathbb{Z}\Delta$.
  • Type $C_n$: Have $\rho \in \mathbb{Z}\Delta$ iff $n \equiv 0,3 \mod 4$.
  • Type $D_n$: Have $\rho \in \mathbb{Z}\Delta$ iff $n \equiv 0,1 \mod 4$.
  • Type $E_6$: $\rho \in \mathbb{Z}\Delta$.
  • Type $E_7$: $\rho \notin \mathbb{Z}\Delta$.
  • Type $E_8$: $\rho \in \mathbb{Z}\Delta$.
  • Type $F_4$: $\rho \in \mathbb{Z}\Delta$.
  • Type $G_2$: $\rho \in \mathbb{Z}\Delta$.

But I see no particular rhyme or reason to these root systems.

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    $\begingroup$ One observation you likely know anyways: as this half-sum $\frac{1}{2}\sum_{\beta \in \Phi^+}\beta = \sum_{1 \leq i \leq rk(W)} \omega_i$ is the same as the sum of the fundamental weights $\{\omega_1,\ldots,\omega_{rk(W)}\}$, this is the same as saying that the diagonal of the weight lattice lies in the root lattice. Which is then again the same as saying that sum of the columns of the inverse Cartan matrix is an integer vector. $\endgroup$ Commented Nov 30, 2016 at 18:37
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    $\begingroup$ The later then immediately explains the difference between types $B$ and $C$, as one Cartan matrix is the transpose of the other. $\endgroup$ Commented Nov 30, 2016 at 18:40
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    $\begingroup$ So you could say that $\rho \in \mathbb{Z} \Delta$ if and only if the irreducible representation of highest weight $\rho = \sum_i \omega_i$ is a representation for the adjoint group. Which happens if and only if the kernel of this representation for the simply connected group is the center. But this observation is a bit trivial. $\endgroup$
    – spin
    Commented Nov 30, 2016 at 20:28
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    $\begingroup$ The root lattice is the subset of the weight lattice consisting of those weights which occur inside tensor powers of the adjoint representation. In particular $\rho$ appears as a weight in a tensor power of the adjoint representation iff you are in one of those types. Any chance your combinatorial phenomenon is some guise of iterated tensoring with the adjoint representation? $\endgroup$
    – Nate
    Commented Dec 1, 2016 at 23:12

2 Answers 2

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There is no single answer to this question. Here is one observation.

The element $z_G=e^{2\pi i\rho^\vee}$, where $\rho^\vee$ is one-half the sum of the positive co-roots, is a canonical (independent of the choice of positive co-roots) element of $G$, fixed by every automorphism of $G$. If $G$ is simply connected $z_G=1$ if and only if $\rho^\vee$ is in the co-root lattice. For whatever reason this $z_G$ tends to come up. For a slightly more precise list of your cases see Bourbaki, Lie Groups and Lie Algebras, Chapters 7-9, Chapter IX, Section 4, Exercise 13.

For example the Frobenius-Schur indicator of a self-dual finite dimensional representation $V_\lambda$ (telling whether the invariant form is orthogonal or symplectic) is $e^{2\pi i\langle\lambda,\rho^\vee\rangle}=\lambda(z_\rho)$. So $\rho^\vee$ is in the co-root lattice if and only if every such representation of the simply connected group is orthogonal. See [loc. cit. Chapter 9, Section 7] and Math Overflow: Is there a formula for the Frobenius-Schur indicator of a rep of a Lie group?

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  • $\begingroup$ Thanks very much for this answer! And especially thanks for the link to that MathOverflow question, in particular this answer with a similar list of root systems: mathoverflow.net/a/227881/25028 $\endgroup$ Commented Dec 4, 2016 at 16:16
  • $\begingroup$ Co-root lattice indeed. $\endgroup$ Commented Dec 4, 2016 at 19:40
  • $\begingroup$ I'm accepting this answer since it gives (in my opinion) a very elegant equivalent condition. $\endgroup$ Commented Dec 7, 2016 at 0:01
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The significance of $\rho$ (and the associated dot-action of the Weyl group or other Coxeter group) in representation theory is discussed from many angles in the earlier linked question.

When dealing with representations of simple Lie or algebraic groups, the annoying problem is that for some isogeny types (as indicated in the list of cases above) $\rho$ might fail to lie in the character group of a fixed maximal torus $T$ if this isn't the full "abstract" weight lattice of the root system (which Bourbaki denotes $P$, having the root lattice $Q$ as a subgroup of finite index). It's common in the literature to assume right away that the group under study is simply connected (meaning in any characteristic that the abstract weight lattice agrees with the character group $X(T)$), even though this isn't always necessary in a specific case to ensure that $\rho$ lies in $X(T)$.

Of course, there is usually just one Lie algebra occurring for each Lie type $A - G$ of simple Lie or algebraic groups (except in characteristic $p>0$ when the structure of the Lie algebra can vary more). The main motivation for looking at Lie algebra representations is usually the algebraic study of associated group representations. So the isogeny type of the group is always in the background: the adjoint group is not always the only one to consider.

Whether all of this can be understood in a combinatorial way is unclear to me.

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