I have a few questions concerning Kostant's work on principal three-dimensional subalgebras (TDS). Let $\mathfrak{g}$ be a finite-dimensional complex semisimple Lie algebra, and $\mathfrak{a}\subseteq\mathfrak{g}$ a principal TDS. Is it true that the centralizer $Z_{\mathfrak{g}}(\mathfrak{a})=\{\xi\in\mathfrak{g}:\text{$[\xi,\eta]= 0$ $\forall\eta\in\frak{a}$}\}$ is trivial (or equivalently, that the trivial one-dimensional representation of $\mathfrak{a}$ is not an irreducible constituent of the $\mathfrak{a}$-module $\mathfrak{g}$)?

Secondly, if we require that $\mathfrak{g}$ be simple with exponents $m_1,\dotsc,m_r$ ($r$ is the rank of $\mathfrak{g}$), then the $\mathfrak{a}$-representation $\mathfrak{g}$ decomposes as $$\mathfrak{g}=\bigoplus_{i=1}^{r}\mathfrak{g}_i,$$ where $\mathfrak{g}_i$ is an irreducible representation of dimension $2m_i+1$. To what extent does this also hold for $\mathfrak{g}$ semisimple?



It's helpful to point out the original source, in one of Kostant's influential early papers: "The principal three-dimensional subgroup and the Betti numbers of a complex simple Lie group", Amer. J. Math. 81 (1959), available online via JSTOR. (Note also the Bourbaki report by Koszul here and here.)

In his paper Kostant brings together a number of important themes, but most of them make sense initially just for a simple Lie group or its Lie algebra. For instance, much depends on working with a Coxeter element in the associated Weyl group (unique up to conjugacy there) and its order $h$, together with the exponents $m_i$ (and degrees $d_i = m_i+1$ of fundamental invariants) for the Weyl group. To work in a semisimple Lie algebra or its Weyl group is somewhat artificial here. For example, if you work with the principal TDS in a simple Lie summand of the Lie algebra, Kostant provides a nice picture of the adjoint action and the resulting submodules of dimensions $2m_i+1$. But other simple summands of the Lie algebra will commute with the given one, so the TDS acts on them trivially. What definition would you give of "Coxeter element" or "Coxeter number" $h$ if the Weyl group is not irreducible?

Small example: in type $G_2$ the Coxeter number is 6, the exponents are 1, 5, and the dimension of the Lie algebra is 14 = 3 + 11. The degrees are 2, 6, and their product is the order of the Weyl group, a dihedral group.

While Kostant deals in part with a semisimple Lie algebra, by section 6 of his paper he has to limit the treatment to the simple case. Here all the ideas come together in a beautiful way, illuminating the appearance of the exponents in the topology of the Lie group (via the Poincare polynomial).


Is it true that the centralizer $Z_{\frak{g}}(\frak{a})=\{\xi\in\frak{g}:[\xi,\eta]= $0$ \text{ }\forall\eta\in\frak{a}\}$ is trivial (or equivalently, that the trivial one-dimensional representation of $\frak{a}$ is not an irreducible constituent of the $\frak{a}$-module $\frak{g}$)?

The answer is yes. Let's follow Kostant's setup in section 5. Thus, fix a Cartan subalgebra $\mathfrak h$, a set of simple roots $\{\alpha_i\}_{i=1}^r$, and corresponding root vectors $e_{\alpha_i}$. Then we may assume that $\mathfrak a$ is generated by the $S$-triple $\{h, e,f\}$ as described in the proof of Lemma 5.2 with (say) $e = \sum_{i=1}^r e_{\alpha_i}$. Now, $Z_{\mathfrak g}(\mathfrak a)$ lies inside the centralizer of $h$, which is $\mathfrak h$ by the proof of Theorem 5.2. But if $x \in \mathfrak h$ centralizes $\mathfrak a$ then $$ 0 = [x,e] = \sum_{i=1}^r (x,\alpha_i) e_{\alpha_i} $$ whence $(x,\alpha_i)=0$ for all $i$ and so $x=0$.

  • $\begingroup$ @Faisal: Yes, I should have commented more directly on this point in my answer. Up to a point, one can study nilpotent elements and adjoint action of a TDS in an arbitrary semisimple Lie algebra. Still, this is what I tend to thunk of as an artificially general preliminary to Kostant's main results. Until you specialize to a simple Lie algebra, his main insights into Coxeter elements and related numerical invariants don't really come into play. Anyway, note the opening paragraph of his section 6. $\endgroup$ – Jim Humphreys Apr 8 '13 at 18:42
  • $\begingroup$ This has been an exceedingly helpful conversation. I suspect I will make more posts as I continue to read through Kostant's work. $\endgroup$ – Peter Crooks Apr 9 '13 at 13:58

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