Let $G$ be a semisimple connected complex Lie group, compact real Lie group or linear algebraic group. Let $\chi$ be the character of a finite dimensional irreducible representation of $G$ (I am particularly interested in the adjoint representation, or perhaps the fundamental ones, but general statements are useful too); lest there be any ambiguity, $\chi$ is considered a regular function on $G$ (viz., the trace of the representation).

I say that $g\in G$ is $\chi$-critical (for lack of a better term) when the differential of $\chi$ vanishes at $g$.

We can, of course, make this definition for $T\subset G$ a maximal torus, or for $\mathfrak{t}$ or $\mathfrak{g}$ the Lie algebras associated to $T$ and $G$. They all amount to essentially the same thing.

Basically, any information regarding $\chi$-critical (especially adjoint-critical) elements interests me, including a more standard term, or any mention of them in the literature. More specifically:

  • Is a $\chi$-critical element $g\in G$ that is semisimple (or equivalently, a $\chi$-critical $g\in T$) necessarily of finite order? If not, are there any natural conditions implying that it is? If yes, can we give a bound on their order or can we say something about it?

  • Is a $\chi$-critical element necessarily principal? "Principal" here is in the sense of belonging to a Kostant principal $\mathit{SL}_2$ or $\mathit{PGL}_2$: see e.g. Reeder, Torsion automorphisms of simple Lie algebras, §2.5, for definitions.

But again, other kinds of remarks concerning $\chi$-critical elements are welcome.

  • $\begingroup$ Not that it narrows it down much, but I'm sure I've seen some variant of this somewhere in Steinberg … it's always a pleasant experience to look through his collected works, so no harm in doing so even if it's not there. :-) $\endgroup$
    – LSpice
    Apr 26, 2017 at 17:18
  • $\begingroup$ Presumably you mean "not principal" in your second question? For example, for a simple group every polynomial invariant on ${\mathfrak g}$ is of degree $>1$, so $0$ will be $\chi$-critical (focussing on the Lie algebra case, which I find easier to work with). $\endgroup$
    – Paul Levy
    Apr 26, 2017 at 17:52
  • 1
    $\begingroup$ Ok, let me reword that - $0$ is of course contained in any $\mathfrak{sl}_2$. The reason this jarred for me is that Kostant's criterion for regularity (read principality) of an element of the Lie algebra is that the differentials of a basic set of invariants are linearly independent. So vanishing of the differential would appear to me to be more associated with failure of principality. $\endgroup$
    – Paul Levy
    Apr 26, 2017 at 18:02
  • 1
    $\begingroup$ For relevant ideas in arbitrary characteristic, it's worth consulting the classic paper on regular elements by Steinberg (as indicated by L Spice): numdam.org/item/PMIHES_1965__25__49_0 Note especially how he generalizes Kostant's characteristic 0 work, which characterizes certain regular elements as those at which the differentials of all fundamental representations are independent. (In any case, you are asking too many open-ended questions here, which makes an "answer" elusive.) $\endgroup$ Apr 26, 2017 at 23:50
  • $\begingroup$ @JimHumphreys is right; that's what I was remembering. $\endgroup$
    – LSpice
    Apr 28, 2017 at 1:12

1 Answer 1


The question "does every semisimple $\chi$-critical $g\in G$ necessarily have finite order?" (with $\chi$ being the character of the adjoint representation) has a negative answer, as shown by the following fairly trivial counterexample:

  • Let $g \in \mathit{SO}_6$ be the rotation that is the orthogonal direct sum (i.e., the block diagonal matrix) of the following three plane rotations: one of angle $0$ (i.e., the $2\times 2$ identity matrix), one of angle $\pi$, and one of angle $\theta$ arbitrary. The adjoint character¹ of $g$ is $-1$ (the smallest possible), it is critical, but $\theta$ can be arbitrary, so $g$ can fail to be of finite order.

Now this $\chi$-critical point is not isolated, so one might hope to salvage the question by asking "does every semisimple isolated $\chi$-critical $g\in T$ necessarily have finite order?" or parhaps "is every $\chi$-critical value necessarily witnessed by a $g\in G$ of finite order?". But the answer is still negative, as shown by the slightly following more interesting example of a $\chi$-critical rotation:

  • Let $g \in \mathit{SO}_{10}$ be the rotation that is the orthogonal direct sum of the following five plane rotations: one has angle $0$ and four have the same angle $\theta$ such that $\cos\theta = -\frac{1}{3}$. Then $\chi(g) = \frac{7}{3}$, it is isolated critical (in a maximal torus $T$ containing it), yet $g$ is not of finite order; furthermore, all $\chi$-critical elements with this particular critical value are similar to the $g$ I described, so none is of finite order.

This all results from not-particularly-enlightening computations. I have very little intuition on what makes a $g$ such as above remarkable, or, in the end, what "$\chi$-critical" really tells us.

The same counterexamples can be used against my question about principality.

One question I still have is whether $\chi$-critical values are necessarily rational, but I think I had better think more about it before I open a new question.

  1. Recall that if $g\in\mathit{SO}_{2n}$ is orthogonal direct sum of rotations of angles $\theta_1,\ldots,\theta_n$, then its adjoint character $\chi(g)$ is: $n+2\sum\cos(\theta_i\pm\theta_j)$ where the sum is on all $n(n-1)$ combinations of an unordered pair $\{i,j\}$ with $i\neq j$ together with a sign $\pm$.
  • 1
    $\begingroup$ Why don't you rather edit the question? $\endgroup$
    – YCor
    May 5, 2017 at 10:56
  • $\begingroup$ @YCor Because now I'm not sure what question I want to ask, except for a very vague "what can we say about $\chi$-critical elements or values?", which as Jim Humphreys pointed out, is really too broad; and if I figure out something more interesting, it might be quite different. In the mean time, I thought I might as well answer the question I did ask, because I learned something in the process (and since someone marked this question as favorite, I guess they want to hear the answer). $\endgroup$
    – Gro-Tsen
    May 5, 2017 at 12:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.